Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

P K Nag Exercise problems - Solved

Thermodynamics
Contents
Chapter-1: Introduction
Chapter-2: Temperature
Chapter-3: Work and Heat Transfer
Chapter-4: First Law of Thermodynamics
Chapter-5: First Law Applied to Flow Process
Chapter-6: Second Law of Thermodynamics
Chapter-7: Entropy
Chapter-8: Availability & Irreversibility
Chapter-9: Properties of Pure Substances
Chapter-10: Properties of Gases and Gas Mixture
Chapter-11: Thermodynamic Relations
Chapter-12: Vapour Power Cycles
Chapter-13: Gas Power Cycles
Chapter-14: Refrigeration Cycles

Solved by

Er
...
However, it is still possible that there are a few errors (serious or
otherwise)
...
co
...


Chapter 1

Introduction
Some Important Notes

Microscopic thermodynamics or statistical thermodynamics
Macroscopic thermodynamics or classical thermodynamics
A quasi-static process is also called a reversible process

Intensive and Extensive Properties
Intensive property: Whose value is independent of the size or extent i
...
mass of the system
...
g
...

Extensive property: Whose value depends on the size or extent i
...
mass of the system (upper
case letters as the symbols)
...
g
...
If mass is increased, the value of
extensive property also increases
...
g
...

Specific property: It is a special case of an intensive property
...
(Lower case letters as symbols) e
...


Concept of Continuum
The concept of continuum is a kind of idealization of the continuous description of matter where
the properties of the matter are considered as continuous functions of space variables
...

Describing a fluid flow quantitatively makes it necessary to assume that flow variables
(pressure, velocity etc
...

Mathematical descriptions of flow on this basis have proved to be reliable and treatment of fluid
medium as a continuum has firmly become established
...
Considering
another extreme if ∀ is very small, random movement of atoms (or molecules) would change
their number at different times
...
This is called
continuum limit and is denoted by ∀C
...
It is the distance between the molecules which is characterized by mean free
path (λ)
...
If the mean free path is very small as compared with some
characteristic length in the flow domain (i
...
, the molecular density is very high) then the gas
can be treated as a continuous medium
...

A dimensionless parameter known as Knudsen number, Kn = λ / L, where λ is the mean free
path and L is the characteristic length
...

Usually when Kn> 0
...

In this, Kn is always less than 0
...

Other factor which checks the validity of continuum is the elapsed time between collisions
...

In continuum approach, fluid properties such as density, viscosity, thermal conductivity,
temperature, etc
...


The Scale of Pressure
Gauge Pressure

Absolute
Pressure

Vacuum Pressure

Local
atmospheric
Pressure

Absolute Pressure

Absolute Zero
(complete vacuum)

At sea-level, the international standard atmosphere has been chosen as Patm = 101
...
314 kJ/kmole − K
Characteristic gas constant, Rc =
For Air R =

Ru
M

8
...
287 kJ/kg- K
For water R =

8
...
461 kJ/kg -K
Units of heat and work is kJ
Units of pressure is kPa
1 atm = 101
...


Page 5 of 265

Introduction

By: S K Mondal

Chapter 1

Questions with Solution P
...
Nag
Q1
...
032 m3/s
...
50 m in diameter and 4
...
Find the density of the liquid and the mass flow rate of the liquid
handled by the pump
...
12
...
502
× 4
...
422 m3

mass
3000 kg
=
= 404
...
422
mass flow rate = Vloume flow rate × density

density =

= 0
...
203 kg
= 12
...
2

s

s

The acceleration of gravity is given as a function of elevation above sea
level by
−6
g = 980
...
086 × 10 H
Where g is in cm/s2 and H is in cm
...
89,716
...
315%)

Solution:

g´ = 980
...
086 × 10−6 × 10,000 × 100

= 977
...
77514 m 2
s2
s
90,000
Wsea = 90,000 N =
kgf
9
...
054 kgf
Wete = 9178
...
77514 N = 89716
...
765
% less =
× 100%
90,000
= 0
...
3

Solution:

Prove that the weight of a body at an elevation H above sea-level is given
by
2
mg ⎛ d ⎞
W =
g0 ⎜ d + 2 H ⎟
⎝
⎠

Where d is the diameter of the earth
...
3

2

2

from equation
...
4

m

If we place it in a surface of earth
then
Force of attraction =

∴

…

2

2

Pr oved
...
Taking the mean diameter of the earth
to be 12,680 km, and assuming the orbit to be circular, evaluate the value
of the gravitational acceleration at this height
...

Estimate the gravitational force acting on the satellite at the operational
altitude
...
8
...
65 N (Weight)

Q1
...
2 m H2O gauge
(d) 3
...
760 × 13600 × 9
...
16 Pa
101
...
7

1
...
2 × 1000 × 9
...
4 kPa
= 113
...
36 × 43
...
81 kPa
= 48
...
6

90 cm Hg gauge
= 0
...
81 × 10-3 + 101
...
4744 kPa

3
...
1 × 100 kPa = 310 kPa

A 30 m high vertical column of a fluid of density 1878 kg/m3 exists in a
place where g = 9
...
What is the pressure at the base of the column
...
544 kPa)
p = z ρg
= 30 × 1878 × 9
...
681 kPa
Assume that the pressure p and the specific volume v of the atmosphere
are related according to the equation pv1
...
3 × 105 , where p is in N/m2
abs and v is in m3/kg
...
81
m/s2
...
0132 bar at the earth’s surface? Consider the atmosphere as a fluid
column
...
64
...
4

pv

Zero line

1
×g
v

HO-h

h

g dh
dp

p = hρg

3

= 2
...
4 ⎛ 2300 ⎞
v=⎜
⎟ =⎜ p ⎟
⎝ p ⎠
⎝
⎠

dh

p

n

1
where n =
1
...
42 km
⎦
g (1 − n ) ⎣

Q1
...
Some steam condenses into water
...
Take the density of
mercury as 13
...
1 cmHg, and g as
9
...


Solution:

po + 0
...
9

Solution:

Hg

× g = 0
...
761 × 13
...
806 + 0
...
6 × 103 × 9
...
03 × 1000 × 9
...

= 167
...
66 mHg
...
3 kPa?
(Ans
...
3 kPa)
Absolute
= atm
...
3 – 0
...
6 × 103 × 9
...
24 kPa

Page 9 of 265

 

Page 10 of 265

Temperature

By: S K Mondal

Chapter 2

2
...
1

373K

F

Freezing Point

Relation:

212oF

32o F

80 o

x

K

2 73 K

30 cm

0o

10 cm

C−0
F − 32
K − 273
ρ −0
x − 10
=
=
=
=
100 − 0 212 − 32 373 − 273 80 − 0 30 − 10

Questions with Solution P
...
Nag

The limiting value of the ratio of the pressure of gas at the steam point and at
the triple point of water when the gas is kept at constant volume is found to be
1
...
What is the ideal gas temperature of the steam point?

(Ans
...
36605
pt

Solution:
∴

θ( v ) = 273
...
16 × 1
...
15°C

Q2
...
p
...
p
...
0
96
...
Plot the
ratio of Sb
...
:H2Ob
...
against the reading at the water boiling point, and
extrapolate the plot to zero pressure at the water boiling point
...
p
...
p
...
e
...
What is the boiling point of
sulphur on the gas scale, from your plot?
(Ans
...
p
...
0 100
200
300
Sulphur b
...
96
...
p

Ratio

= 1
...
93

Wb
...
926
Extrapolating

Solution :

Chapter 2

1
...
940

T1 = 100°C = 373K

0

T2 = ?

50

100

200

300

p1
= 1
...
3

T2 = 373 × 1
...
247 ohms at the steam point, and 28
...
Find the
constants A and B in the equation

R = R0 (1 + At + Bt2 )
And plot R against t in the range 0 to 660°C
...
595
11

y

0

x

R 0 = 11
...
247 = 11
...
( i )

or 3
...
887 = 11
...
( ii )

3
...


B = − 6 × 10 −7
A = 3
...
921 × 10 −3 t − 6 × 10 −7 t 2
or

(

Y = 11 1 + 3
...
595

Page 12 of 265

Temperature

By: S K Mondal
Q2
...
m
...
e of the
thermocouple is given by the equation

ε = at + bt2
Where a = 0
...
0 × 10-4 mV/deg2
(a)
(b)

Compute the e
...
f
...

Suppose the e
...
f
...

t* = a' ε + b'

Solution:
Q2
...
Find
the numerical values of a' and b' and draw a graph of ε against t*
...

(d) Compare the Celsius scale with the t* scale
...

The values of K are found to be 1
...
78 at the ice point and the
steam point, the temperatures of which are assigned the numbers 0 and
100 respectively
...
42 on the thermometer
...
21
...
83 + b
… (i)
100 = a x ln 6
...
6

⎛ 6
...
83 ⎠
a = 76
...
83
= − 46
...
35 ln k − 46
...
35 × ln 2
...
143
= 21
...
When operating at full load under steady
state conditions, the motor is switched off and the resistance of the
windings, immediately measured again, is found to be 93 ohms
...
00393 t]

Solution:

Where R0 is the resistance at 0°C
...

(Ans
...
41°C)
R25 = R0 [1 + 0
...
84 Ω
1 + 0
...
7

93 = 72
...
00393 × t}
t = 70
...
What is the
temperature reading on this new scale when the temperature is 150°C?
At what temperature both the Celsius and the new temperature scale
reading would be the same?
(Ans
...
)

Solution:

150 − 0
N − 100
=
100 − 0
400 − 100
or N = 550o N
let N= C for x o
C −0
N − 100
then
=
100 − 0
400 − 100
x
x − 100
=
or
300
100

or
or
or
or
Q2
...
The wire resistance was
found to be 10 ohm and 16 ohm at ice point and steam point respectively, and
30 ohm at sulphur boiling point of 444
...
Find the resistance of the wire at
500°C, if the resistance varies with temperature by the relation
...
31
...
6 + β × 444
...


Chapter 3

Work and Heat Transfer
Some Important Notes
-ive
W

+ive
W

+ive
Q

-ive
Q

Our aim is to give heat to the system and gain work output from it
...
K
...
1

(a)A pump forces 1 m3/min of water horizontally from an open well to a closed
tank where the pressure is 0
...
Compute the work the pump must do
upon the water in an hour just to force the water into the tank against the
pressure
...

(Ans
...
91
...
42
...
6 m/s)
Solution:
(a)
Flow rate 1m3/hr
...
101325 MPa
Pressure of outlet water = 0
...
9 − 0
...
31 kJ
(b)

1 m3
s
60

s

So that pressure will be 0
...
9 MPa

or

h=

0
...
743 m
1000 × 9
...
9 − 0
...
2 m / s
...
2

The piston of an oil engine, of area 0
...
00028 m3 of fresh air from the atmosphere
...
325 kPa, the difference being due to the flow resistance in the
induction pipe and the inlet valve
...

(Ans
...
375×10 m3
= 0
...
075m3
= 0
...
0003375 m3
as pressure is constant
= 80 kPa
So work done = pΔV
= 80 × 0
...
027 kJ = 27 J
Q3
...
12 m3 and contains gas at a
pressure of 1
...
The gas expands according to a process which is
represented by a straight line on a pressure-volume diagram
...
15 MPa
...
30 m
...
29
...
5 MPa
Final volume (V1) = 0
...
3m

Page 16 of 265

Work and Heat Transfer

By: S K Mondal

Chapter 3

= 0
...
15 MPa
As initial pressure too high so the volume is neglected
...
5 + 0
...
036 × 103 kJ
2
= 29
...
5 MPa
0
...


V
0
...
4

Solution:

A mass of 1
...
1
MPa to 0
...
The initial density of air is 1
...
Find the work done by the piston to compress the air
...
251
...
1 × 1
...
1
MJ
0
...
63 kJ

p1 V1
V

given p1 = 0
...
5
=
m3
ρ1
1
...
7 MPa

Q3
...
1 m3
to 0
...
03 m3
...

(Ans
...
83 kJ)
Given initial pressure ( p1 ) = 80kPa
Initial volume ( V1 ) = 0
...
4 MPa = 400 kPa
Final volume ( V2 ) = 0
...
6

Solution:

⎛ 400 ⎞
ln ⎜
⎟
⎝ 80 ⎠ = 1
...
3367 ≈ 1
...
20397
⎛ 0
...
03 ⎠
p V − p2 V2
Work done ( W ) = 1 1
n −1
80 × 0
...
03
=
= − 11
...
34 − 1

A single-cylinder, double-acting, reciprocating water pump has an
indicator diagram which is a rectangle 0
...
05 m high
...
The pump runs at 50 rpm
...
15 m and the piston stroke is 0
...

Find the rate in kW at which the piston does work on the water
...
43
...
75 × 10 −3 m2
Area of indicated diagram ( ad ) = 0
...
05 m
Spring constant (k) = 147 MPa/m

Page 18 of 265

Work and Heat Transfer

By: S K Mondal
Q3
...
15 m bore develops
an indicated power of 4 kW when running at 216 rpm
...
The length of the indicator
diagram is 0
...

(Ans
...
15 m
I
...
1 × Stoke (L)
Let Area of indicator diagram = ( ad )

∴

Mean effective pressure ( pm ) =

and

∴

or

pm LAN
[as 4 stroke engine]
120
a ×k L×A×N
I
...
= d
×
ld
120

I
...
=

ad =

or

ad
×k
ld

I
...
1L ⎥
⎣
⎦

I
...
1 L × 120 × 4
=
k × L × π × D2 × N
4 × 0
...
152 × 216
= 5
...
8

Solution:

A six-cylinder, 4-stroke gasoline engine is run at a speed of 2520 RPM
...
45 × 103 mm2 and its
length is 58
...
The spring constant is 20 × 106 N/m3
...
Determine the
indicated power, assuming that each cylinder contributes an equal
power
...
243
...
45 × 103
× 20 × 103 Pa
58
...
607 kPa

=

∴

mm2 N
mm × N ⎛ 1 ⎞
× 3 ⇒
=⎜
N / m2
mm m
1000 ⎟
m × m2
⎝
⎠

L = 0
...
142
=
4
4
N = 2520
n=6
A =

∴

I
...
=

pm LAN
×n
120

= 837
...
15 ×

[as four stroke]
π × 0
...
696 kW
Q3
...
25 m diameter is fitted with a light frictionless
piston
...
The volume on the other side of the piston is evacuated
...
The catch is released
and the piston travels along the cylinder until it comes to rest after a
stroke of 1
...
The piston is then held in its position of maximum travel
by a ratchet mechanism
...
Calculate the work done by
the compressed air on the piston
...
3
...
25m

1
...
2 m
∴ Work Done = 2560 × 1
...
m
= 3
...
2 m then 5000 = 120 + k × 1
...
2

∴

W=

∫ F dx
x

0

1
...
2

⎡
x2 ⎤
= ⎢120x + 4067 × ⎥
2 ⎦0
⎣
= 120 × 1
...
22
J
2

= 144 + 2928
...
24J = 3
...
l0

A steam turbine drives a ship’s propeller through an 8: 1 reduction gear
...
The turbine speed is 1450 rpm
...

(Ans
...
84 km N, (b) 14
...
765 MW)

Solution:

Power of the propeller = Power on turbine shaft

The net rate of working of the reduction gear
= (15 – 14
...
7647 MW
Q 3
...
The cylinder diameter is 0
...
During the
stirring process lasting 10 minutes, the piston slowly moves out a
distance of 0
...
The net work done by the
fluid during the process is 2 kJ
...
Determine the torque in the shaft and the power
output of the motor
...
0
...
92 W)

Page 21 of 265

Work and Heat Transfer

By: S K Mondal
Solution:

Chapter 3

Change of volume = A L

πd 2
×L
4
π × 0
...
485 m3
4
= 0
...
4m

M

0
...
325 × 0
...
1754 kJ
Net work done by the fluid = 2 kJ
∴ Net work done by the Motor = 4
...
1754 × 103
W
10 × 60
= 6
...
96 × 60
=
2π × 840
=

= 0
...
12

At the beginning of the compression stroke of a two-cylinder internal
combustion engine the air is at a pressure of 101
...
Compression
reduces the volume to 1/5 of its original volume, and the law of
compression is given by pv1
...
If the bore and stroke of each
cylinder is 0
...
25 m, respectively, determine the power
absorbed in kW by compression strokes when the engine speed is such
that each cylinder undergoes 500 compression strokes per minute
...
17
...
15 )
=
× 0
...
00442 m3

Initial volume ( V1 ) =

Initial p r essure ( p1 ) = 101
...

V1
= 0
...
2
= p2 V2

Final volume ( V2 ) =

p1 V11
...
2
= 699
...
2
V2

Work done / unit stroke − unit cylinder ( W )
⎛ 1
...
2 − 1 ⎠
⎛ 101
...
00442 − 700 × 0
...
2
1
...
2
kW
60
= 17
...
13

Determine the total work done by a gas system following an expansion
process as shown in Figure
...
2
...
4 – 0
...
3 = c
C

0
...
4

0
...
4 − 20
...
8
W
=
1
...
251MJ

⎡
⎢ Here
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣

pB = pB = 50 bar = 50 × 105 Pa
VB = 0
...
8m3
pC =

1
...
3
VC

=

50 × 105 × 0
...
3
0
...
3
= 20
...
251MJ
Q3
...
The macroscopic properties of the system obey the
following relationship:

a ⎞
⎛
⎜ p + 2 ⎟ (V − b) = mRT
V ⎠
⎝

Solution:

Where a, b, and R are constants
...

Calculate the work done by a system which contains 10 kg of this gas
expanding from 1 m3 to 10 m3 at a temperature of 293 K
...
7 × 10 Nm 4 , b = 1
...
278 kJ/kg-K
...
1742 kJ)
As it is constant temp-expansion then
a ⎞
⎛
⎜ p + 2 ⎟ ( V − b ) = constant ( mRT ) ( k ) as T = constant
V ⎠
⎝

Page 24 of 265

Work and Heat Transfer

By: S K Mondal

Chapter 3

⎛
⎛
a ⎞
a ⎞
⎜ p1 + 2 ⎟ ( V1 − b ) = ⎜ p2 + 2 ⎟ ( V2 − b ) = ( k )
V1 ⎠
V2 ⎠
⎝
⎝

∴
2

W = ∫ p dV

∴

1

a ⎞ constant ( k )
⎛
⎜p+ V ⎟ =
V−b
⎝
⎠

2

a ⎞
⎛ k
= ∫⎜
− 2 ⎟ dV
V−b V ⎠
1⎝

or

2

a⎤
⎡
= ⎢ k ln ( V − b ) + ⎥
V ⎦1
⎣

p=
−∫

k
a
− 2
V−b V
1
1
dv = + c
V2
V

⎛ V − b⎞
⎛ 1
1 ⎞
= k ln ⎜ 2
−
⎟ + a⎜
⎟
⎝ V1 − b ⎠
⎝ V2 V1 ⎠
⎡⎛
⎛ 1
V −b
a ⎞
1 ⎞⎤
= ⎢⎜ p1 + 2 ⎟ ( V1 − b ) ln 2
+ a⎜
−
⎟⎥
V1 ⎠
V1 − b
V2 V1 ⎠ ⎦
⎝
⎣⎝
a ⎞
⎛
⎜ p + 2 ⎟ ( V − b ) = constant ( mRT ) ( k ) as T = constant
V ⎠
⎝

Given m = 10 kg; T = 293 K; R = 0
...
K
∴
Constant k = 10 × 293 × 0
...
54 kJ
a = 15
...
07 × 10-2m3
⇒ V2 = 10m3, V1 = 1m3
∴

⎛ 10 − 1
...
54 ln ⎜
−2 ⎟
⎝ 10 1 ⎠
⎝ 1 − 1
...
44 − a × 0
...
44 − 157 × 0
...
14 kJ

Q3
...

(Ans
...
2 kJ)
Initial volume ( v1 ) = 6000 cm3

= 0
...
002 m3
If final pressure ( p2 )
p V 2 100 × ( 0
...
002 )
2

∴

Page 25 of 265

Work and Heat Transfer

By: S K Mondal
work done on the system =

Chapter 3
1
⎡p2 V2 − p1 V1 ⎤
⎦
n −1 ⎣

1
⎡900 × 0
...
006⎤ kJ
⎦
2 −1⎣
= 1
...
16
Solution:

The flow energy of 0
...
Find the pressure at this point
...
8709 kPa)
If pressure is p1
Area is A1
Velocity is V1
Volume flow rate (Q) = A1V1
∴
Power = force × velocity
= p1A1 × V1
p1
= p × (Q)
1

∴
or

Q3
...
124
60
18 × 60
p1 =
kPa
0
...
71 MPa

18 = p1 ×

V1

A1

A milk chilling unit can remove heat from the milk at the rate of 41
...
Heat leaks into the milk from the surroundings at an average rate
of 4
...
Find the time required for cooling a batch of 500 kg of
milk from 45°C to 5°C
...
187 kJ/kg K
...
2h 13 min)
Heat to be removed (H) = mst
= 500 × 4
...
740 MJ
Net rate of heat removal

= Hrej − Hleak
= ( 41
...
187 ) MJ / h

Q3
...
683 MJ / h
83
...
683
= 2 hr
...
20 sec
...
The specific
heat of fish above freezing point is 3
...
717 kJ/kg K
...
5 kJ/kg
...
186
...
6%)
Heat to be removed above freezing point
= 680 × 3
...
146 MJ
Page 26 of 265

Work and Heat Transfer

By: S K Mondal

Chapter 3

Heat to be removed latent heat
= 680 × 234
...
460 MJ
Heat to be removed below freezing point
= 680 × 1
...
676 MJ
∴
Total Heat = 186
...
460
× 100 = 85
...
2816

Page 27 of 265

 

Page 28 of 265

First Law of Thermodynamics

By: S K Mondal

4
...

•

•
•
•
•
•

δQ – δW = dE

An isolated system which does not interact with the surroundings Q = 0 and W = 0
...

The Zeroth Law deals with thermal equilibrium and provides a means for measuring
temperatures
...

The Second Law of thermodynamics provides with the guidelines on the conversion heat
energy of matter into work
...

The Third Law of thermodynamics defines the absolute zero of entropy
...


Summation of 3 Laws
• Firstly, there isn’t a meaningful temperature of the source from which we can get the full
conversion of heat to work
...

• Secondly, more interestingly, there isn’t enough work available to produce 0K
...
This is precisely the Third law
...

You can’t get something for nothing:
To get work output you must give some thermal energy
...

You can’t get every thing:
However much work you are willing to give 0 K can’t be reached
...


Page 30 of 265

First Law of Thermodynamics

By: S K Mondal

Chapter 4

Questions with Solution P
...
Nag
Q4
...
The
measured torque of the engine is 10000 mN and the water consumption
of the brake is 0
...
Calculate the
water temperature at exit, assuming that the whole of the engine power
is ultimately transformed into heat which is absorbed by the cooling
water
...
20
...
ω
⎛ 2π × 1000 ⎞
= 10000 × ⎜
⎟
60
⎝
⎠
= 1
...
0472MW
Let final temperature = t°C
∴ Heat absorb by cooling water / unit = m s Δt
= vρs Δt
= 0
...
2 × ( t − 20 )
∴

0
...
2 × ( t − 20 ) = 1
...
4986 ≈ 0
...
5°C

6

Q4
...
7 kJ, – 25
...
56 kJ and +
31
...
What is the net work for this cyclic process?

Solution :

∑ Q = (14
...
5 − 25
...
56 ) kJ

(Ans
...
34 kJ)

-25
...
44 kJ

From first law of thermodynamics
(for a cyclic process)

+14
...
44 kJ

Q4
...
56kJ

31
...
1 MPa
...
The insulation is
then removed and 105 kJ of heat flow to the surroundings as the fluid
goes to state 3
...

State
v (m3)
t (°C)
1
0
...
3
370
3
0
...
E2 = – 29
...
7 kJ)

Page 31 of 265

First Law of Thermodynamics

By: S K Mondal
Solution:

Chapter 4

From first law of thermodynamics
dQ = ΔE + pdV
∴

Q = ΔE + ∫ pdV
2

∴

Q1−2 = ( E2 − E1 ) + ∫ pdV
1

or
or

[as insulated Q2−3 = 0]

= ( E2 − E1 ) + 0
...
3 − 0
...
7 kJ
3

Q2−3 = ( E3 − E2 ) + ∫ pdV
2

or

−105 = ( E3 − E2 ) + 0
...
06 − 0
...
7 + 0
...
06 − 0
...
7 − 24

or
Q4
...
7 + 24

= − 110
...
Evaluate the magnitude and direction of the third heat
transfer
...
– 6 kJ)
From first law of thermodynamics
W = -15kJ
∑ dQ = ∑ dW
1

∴

Q1 + Q2 + Q3 = W1 + W2

or

75 − 40 + Q3 = − 15 + 44

Q1 = 75kJ

W2 = 44kJ

Q3 = − 6kJ
i
...
6kJ from the system
Q = -40kJ

Q4
...
During
a certain period the machine consumes 1 kWh of energy and the internal
energy of the system drops by 5000 kJ
...

(Ans
...
6 MJ)
Q = ΔE + W
Q2 −1 = ( E2 − E1 ) + W2 −1

−1000 × 3600
kJ
1000
= − 8
...
6

Solution:

Q4
...
5 kg of liquid having a constant specific heat of 2
...
Find
Δ E and W for the process
...
Δ E = 56
...
25 kJ)
Heat added to the system = 1
...
5 × 15kJ
= 56
...
25kJ
As it is insulated then dQ = 0
∴
ΔQ = ΔE + W
or
0 = 56
...
25 kJ

Solution:

The same liquid as in Problem 4
...

During the process 1
...
Find
Δ E and W for the process
...
Δ E = 54
...
25 kJ)
As temperature rise is same so internal energy is same
ΔE = 56
...
7 kJ to
the system
So
W = – 56
...
7 kJ
= –57
...
8

The properties of a certain fluid are related as follows:

u = 196 + 0
...
287 (t + 273)

Solution:

Where u is the specific internal energy (kJ/kg), t is in °C, p is pressure
(kN/m2), and v is specific volume (m3/kg)
...

(Ans
...
718, 1
...
718t + 0
...
718 ∂t + 0
...
005 ⎥
∂T ⎦ p
⎣
= 1
...
718t ) ⎤
=⎢
⎥
∂T
⎣
⎦v
∂t ⎤
⎡
= ⎢0 + 0
...
718 kJ / kg − K
Q4
...
If there is no heat transfer, find the net work
for the process
...
100
...
718 ( T2 − T1 )
= − 0
...
26 kJ / kg

∴

Total work (W) = 2 × (-50
...
52 kJ

Q 4
...
9 is done on the moving
piston, show that the equation representing the path of the expansion in
the pv-plane is given by pvl
...


Solution:

Let the process is pV n = constant
...
11

=

mR
( T1 − T2 )
n −1
2 × 0
...
39972
n = 1
...
4

[∴

pV = mRT]

⎡R = ( c p − c v )
⎤
⎢
⎥
⎢ = 1
...
718
⎥
⎢ = 0
...
52

A stationary system consisting of 2 kg of the fluid of Problem 4
...
2 = constant
...
1 MPa
...
Why is the work transfer not equal to ∫ pdV ?

(Ans
...
35, Δ E = – 217
...
1 ⎞
=⎜
⎟
⎝ 1 ⎠

∫ pdV

= 434
...
2 −1
1
...
2

T2 = T1 × ( 0
...
2
= 322
...
25°C

From first law of thermodynamics
dQ = ΔE + dW

∴
∴

0 = ∫ Cv dT + dW
dW = − ∫ Cv dT
2

= − 0
...
718 × ( 200 − 49
...
2356kJ
= − 216
...
5kJ
p V1 − p V
∫ pdV = 1 n − 12 2
mRT1 − mRT2
=
n −1
mR ( T1 − T2 )
=
n −1
2 × 0
...
25 )
=
(1
...
65kJ

As this is not quasi-static process so work is not ∫ pdV
...
12

A mixture of gases expands at constant pressure from 1 MPa, 0
...
06 m3 with 84 kJ positive heat transfer
...
Find DE for the gaseous mixture
...
54 kJ)
The same mixture expands through the same state path while a stirring
device does 21 kJ of work on the system
...

(Ans
...
06 − 0
...
13

Solution:

A mass of 8 kg gas expands within a flexible container so that the p–v
relationship is of the from pvl
...
The initial pressure is 1000
kPa and the initial volume is 1 m3
...
If specific
internal energy of the gas decreases by 40 kJ/kg, find the heat transfer in
magnitude and direction
...
+ 2615 kJ)
T2 ⎛ p2 ⎞
=⎜ ⎟
T1 ⎝ p1 ⎠
∴

n −1
n

⎛V ⎞
=⎜ 1⎟
⎝ V2 ⎠

p2 ⎛ V1 ⎞
=⎜
⎟
p1 ⎝ V2 ⎠

n −1

n

1

or

V2 ⎛ p1 ⎞ n
=⎜ ⎟
V1 ⎝ p2 ⎠

or

⎛ p ⎞n
V2 = V1 × ⎜ 1 ⎟
⎝ p2 ⎠

1

1

∴

∴

Q4
...
2
3
= 1×⎜
⎟ = 82
...
7
=
= 2932
...
2 − 1
ΔE = − 8 × 40 = − 320 kJ
Q = ΔE + W = − 320 + 2932
...
5kJ

A gas of mass 1
...
The initial and final
pressures are 1000 kPa and 200 kPa respectively and the corresponding
volumes are 0
...
20 m3
...
5 pv – 85 kJ/kg
Where p is the kPa and v is in m3/kg
...

(Ans
...
3 kJ)

Page 36 of 265

First Law of Thermodynamics

By: S K Mondal

Chapter 4

Solution:

1000 = a + b × 0
...
( i )

200 = a + b × 1
...
( ii )

( ii ) − ( i ) gives
−800 = b
∴ a = 1000 + 2 × 800 = 1160
∴

p = 1160 − 800V

∴ W=

v2

∫ pdV

v1

1
...
2

1
...
2

(

)

= 1160 × (1
...
2 ) − 400 1
...
22 kJ
= 1160 − 560kJ = 600kJ
0
...
5
1
...
5 × 200 ×
− 85 = 155kJ / kg
1
...
5 × 1000 ×

∴ ΔU = mΔu = 40 × 1
...
5pv − 85kJ / kg
⎛ 1160 − 800v ⎞
= 1
...
5
⎝
⎠
2
= 1160v − 800v − 85kJ / kg
∂u
= 1160 − 1600v
∂v
∂u
1160
= 0∴ v =
= 0
...
725 − 800 × ( 0
...

U max

Q4
...
5kJ / kg
= 1
...
25kJ

The heat capacity at constant pressure of a certain system is a function
of temperature only and may be expressed as
C p = 2
...
87
J/°C
t + 100

Where t is the temperature of the system in °C
...

(a) Find the magnitude of the heat interaction
...
(a) 238
...
79 J)
373

Solution:

Q=

∫ C dT

t = T − 273

p

273

∴ t + 100 = T − 173
373

=

41
...
093 + T − 173 ⎟ dT
⎝
⎠

273

373

= ⎡2
...
87 ln T − 173 ⎤ 273
⎣
⎦
⎛ 200 ⎞
= 2
...
87 ln ⎜
⎟
⎝ 100 ⎠
= 209
...
87 ln 2
= 238
...
32 − 101
...
0024 − 0
...
32 − 40
...
79J
Q4
...
1 m3/kg
...
If the heat rejected by the engine in a
cycle is 1000 kJ per kg of working fluid, what would be its thermal
efficiency?
(Ans
...
19 kJ/kg, (b) 0
...
54 cm2
4
p
1 cm2 ≡ 300kPa × 0
...
54 × 30kJ / kg
= 2356
...
2
× 100%
Therefore, η =
2356
...
204%

Page 38 of 265

V

First Law of Thermodynamics

By: S K Mondal
Q4
...
5 m3 and U1 = 512 kJ
...
Neglecting KE and PE changes,
determine the heat interactions Q12 and Q31
...
74 kJ, 22 kJ)
Q1−2 = ΔE + ∫ pdV
v2

Q1−2 = ( u2 − u1 ) + p1 V1 ∫

v1

dV
V

⎛p ⎞
= ( 690 − 512 ) + 100 × 1
...
972
= 74
...
As W31 is +ive (positive) so
expansion is done
...
18

A gas undergoes a thermodynamic cycle consisting of the following
processes:
(i) Process 1–2: Constant pressure p = 1
...
028 m3, W12 = 10
...
4 kJ
...

(a)
(b)
(c)
(d)

Sketch the cycle on a p–V diagram
Calculate the net work for the cycle in kJ
Calculate the heat transfer for process 1–2
Show that ∑ Q = ∑ W
...
(b) – 8
...
9 kJ)
= 10
...
4 bar

⎛V ⎞
= p2 V2 ln ⎜ 3 ⎟
⎝ V2 ⎠
⎛V ⎞
= p2 V2 ln ⎜ 1 ⎟
⎝ V2 ⎠

u3

1
u1

2 u2
W12= 10
...
028m
V

⎛ 0
...
4 × 100 × 0
...
103 ⎠
⎡as
⎤
W 12 = p ( V2 − V1 )
⎢
⎥
10
...
4 × 100 ( V2 − 0
...
783kJ
⎢
⎢
⎥
V2 = 0
...
283 kJ

( c ) Q12

ans
...
4 + 10
...
9kJ

(d) Q23 = U3 − U 2 + W23
= 0 − 18
...
783 kJ
Q31 = U 2 − U3 + 0 = − 26
...
9kJ − 18
...
4
= − 8
...


Page 40 of 265

First Law Applied to Flow Process

By: S K Mondal

Chapter 5

5
...
F
...
E
...
F
...
E
...
K
...
1

A blower handles 1 kg/s of air at 20°C and consumes a power of 15 kW
...

Find the exit air temperature, assuming adiabatic conditions
...
005 kJ/kg-K
...
28
...
F
...
E
...

Here w1 = w2 = 1 kg / s ; Z1 = Z2 ;
dt
1002
1502
∴
+ 0 = h2 +
− 15
h1 +
2000
2000
⎛
1002 1502 ⎞
∴
−
h2 − h1 = ⎜15 +
⎟
2000 2000 ⎠
⎝
or
or

Q5
...
75
t2 = 20 +

8
...
7°C
1
...
2 MPa, temperature 188°C, enthalpy 2785
kJ/kg, velocity 33
...
The steam leaves the turbine
at the following state: Pressure 20 kPa, enthalpy 2512 kJ/kg, velocity 100
m/s, and elevation 0 m
...
29
kJ/s
...
42 kg/s, what is the
power output of the turbine in kW?
(Ans
...
51 kW)

Page 42 of 265

First Law Applied to Flow Process

By: S K Mondal
Solution:

Chapter 5

w1 = w2 = 0
...
2 MPa
t1 = 188°C
h1 = 2785 kJ/kg
V1 = 33
...
29 kJ/s
dW
=?
dt

3m
2

By S
...
E
...


2
p2 = 20 kPa
h2 = 2512 kJ/kg
V2 = 100 m/s
Z2 = 0

⎛
⎛
V2
g Z1 ⎞ dQ
V2
g Z2 ⎞ dW
w1 ⎜ h1 + 1 +
= w2 ⎜ h 2 + 2 +
⎟+
⎟+
2000 1000 ⎠ dt
2000 1000 ⎠ dt
⎝
⎝
⎧
⎧
⎫ dW
33
...
81 × 3 ⎫
1002
or
0
...
29 = 0
...
3

Solution:

1169
...
14 +

dW
dt

dW
= 112
...
At the inlet to a certain nozzle, the enthalpy of the fluid passing
is 3000 kJ/kg and the velocity is 60 m/s
...
The nozzle is horizontal and there is negligible
heat loss from it
...

(b) If the inlet area is 0
...
187
m3/kg, find the mass flow rate
...
498 m3/kg, find the exit
area of the nozzle
...
(a) 692
...
08 kg/s (c) 0
...
e
...
F
...
E
...
498 m /kg

Data for b
A1 = 0
...
187 m3/kg

Page 43 of 265

First Law Applied to Flow Process

By: S K Mondal

Chapter 5

or

dQ
dW
= 0 and
=0
dm
dm
V2
V2
h1 + 1 = h2 + 2
2000
2000
2
2
V2 − V1
= ( h1 − h2 )
2000
V22 = V12 + 2000 ( h1 − h2 )

or

V2 =

Here Z1 = Z2 and
∴
or

602 + 2000 ( 3000 − 2762 )m / s

=

( b)

V12 + 2000 ( h1 − h 2 )

= 692
...
1 × 60
kg / s = 32
...
187
Mass flow rate is same so
=

(c )

A 2 × 692
...
498
A 2 = 8
...
0855613 =
or

Q5
...
Under steady flow
conditions, the oil enters at 90°C and leaves at 30°C, while the water
enters at 25°C and leaves at 70°C
...
68 t + 10
...
78 kg/s of oil?
(Ans
...
47 kg/s)
wo (h oi + 0 + 0) + wH2 O (h H2 Oi + 0 + 0) + 0 wo (h o,o + 0 + 0) + wH2 O (h H2 Oo + 0 + 0) + 0

Oil

1
90°C

Water

∴

25°C
1
wo (h oi − h o,o ) = wH2 0 (h H2 Oo − h H2 Oi )

2
30°C
70°C
2

hoi = 1
...
5 × 10–4 × 902 kJ/kg = 159
...
68 × 30 + 10
...
395 kJ/kg
∴

Q5
...
78 × 108
...
187 (70 − 25)
= 1
...
6 kg/s

WH2o

=

A thermoelectric generator consists of a series of semiconductor
elements (Figure) heated on one side and cooled on the other
...
In a
Page 44 of 265

First Law Applied to Flow Process

By: S K Mondal

Chapter 5

particular experiment the current was measured to be 0
...
8 volt above that at
(2) Energy transfer as heat to the hot side of the generator was taking
place at a rate of 5
...
Determine the rate of energy transfer as
heat from the cold side and the energy conversion efficiency
...
Q2 = 5
...
073)
•

•

•

Q1 = E + Q2

Solution:

•

or
or

Q5
...
5 = 0
...
8 + Q2
Q2 = 5
...
5 − 5
...
273%
5
...
33 m3/s at 0
...
During the expansion, there is a heat transfer of 0
...
Calculate the turbine exhaust temperature if
changes in kinetic and potential energy are negligible
...
157°C)

Solution:

t = 93°C

C
...


V1 = 2
...
276 M Pa ; t = 930°C
1

1
dW
= 1860 kW
dt
2

2

dQ
= – 0
...
7

Chapter 5

m1R T1

p1V1
276 kPa × 2
...
091 kg/s
=
RT1
0
...
60
1
...
60
= 156
...
11 MPa, 20°C
which it delivers at 1
...
The power absorbed by the
compressor is 4
...
Determine the heat transfer in
(a) The compressor
(b) The cooler
State your assumptions
...
– 0
...
76 kJ/s)

Solution:
(a)

∴
∴

dQ
dW
= w1 h 2 +
dt
dt
⎛ dQ ⎞
0
...
555 – 20
...
15 = ⎜
⎟
⎝ dt ⎠
w1 (h1 + 0 + 0) +

dQ
= –0
...
11 MPa
t1 = 20°C

i
...
1622 kW loss by compressor

dW
= – 4
...
5M Pa
2

1

3
Cooles
3

n
n
(p2 V2 - p1 V1 ) =
(mRT2 − mRT1 )
n -1
n −1
1
...
0436 × 0
...
4
= 3
...
9854 – 4
...
165 kW
dt
dQ
For cooler
dt

∴
(b)

Page 46 of 265

First Law Applied to Flow Process

By: S K Mondal

Chapter 5
•

= m cP (t 2 − t1 )
= 0
...
005 × (111 – 25) kJ/s
= 3
...
8

Solution:

In water cooling tower air enters at a height of 1 m above the ground
level and leaves at a height of 7 m
...
Water enters at a height of 8 m and leaves at
a height of 0
...
The velocity of water at entry and exit are 3 m/s and 1
m/s respectively
...
Air temperatures are 30°C and 70°C at the entry and
exit respectively
...
25 kW
drives the air through the cooler
...
The values of cp of air and water are
1
...
187 kJ/kg K respectively
...
3
...
005 kJ/kg – K
p

w

V2 = 1 m/s, t2 = 50°C
w
w
w = w2 = 1 kg/s
0
...
187 kJ/kg – K
dW
= – 2
...
81
Or w ⎨1
...
187 (50 − 80) +

12 − 32 9
...
8 − 8) − 2
...
509 = –127
...
9346
∴
= 3
...
16 kg/s
w=
40
...
9

Air at 101
...
15 m2 cross-sectional area
...
18 MPa, 150°C through an opening of 0
...
The power output is 375 kW
...
Assume that air obeys the law
pv = 0
...
Take cp = 1
...

(Ans
...
23 kJ/kg)

Solution:

•

Volume flow rate at inlet (V)1 = V1A1 m3/s = 21 m3/s
•

p V1
101
...
304 kg/s
Inlet mass flow rate ( w1 ) = 1
=
R T1
0
...
304 × 0
...
325 kPa
1
t1 = 20°C
V1 = 140 m/s
A1 = 0
...
18 MPa = 180 kPa
2
t2 = 150°C
A2 = 0
...
66 m/s
A2
0
...
F
...
E
...
304 kg/s
∴

⎧
V 2 − V12 ⎫ dW
dQ
= w ⎨(h 2 − h1 ) + 2
⎬+
dt
2000 ⎭ dt
⎩
⎧
V 2 − V12 ⎫ dW
= w ⎨C p (t2 − t1 ) + 2
⎬+
2000 ⎭ dt
⎩
Page 48 of 265

First Law Applied to Flow Process

By: S K Mondal

Chapter 5

⎧
1712 − 1402 ⎫
= 25
...
005 (150 − 20) +
⎬ + 375 kW
2000
⎩
⎭
= 3802
...
76
= 150
...
304

Q5
...
The gas enters the
compressor at a temperature of 16°C, a pressure of 100 kPa, and an
enthalpy of 391
...
The gas leaves the compressor at a temperature
of 245°C, a pressure of 0
...
5 kJ/kg
...

(a) Evaluate the external work done per unit mass of gas assuming the
gas velocities at entry and exit to be negligible
...

(Ans
...
3 kJ/kg, 152
...
2 – 5345) kJ/kg
dm
= –143
...
e
...
2 kJ/kg
1

(b)

Q5
...
6 mPa = 600 kPa
h2 = 534
...
3 +
kJ/kg = (–143
...
6) kJ/kg
2000
= –152
...
e
...
One stream is supplied at the rate of 0
...
The other stream is
supplied at the rate of 0
...
At the exit from the engine the fluid leaves as two
Page 49 of 265

First Law Applied to Flow Process

By: S K Mondal

Solution:

Chapter 5

streams, one of water at the rate of 0
...
The engine develops a shaft power of 25 kW
...
Evaluate the enthalpy of the second exit stream
...
2402 kJ/kg)
dQ
∴
=0
dt
By mass balance

dW
= 25 kW
dt

w11 = 0
...
1 kg/s
h12 = 2569 kJ/kg
V12 = 120 m/s

w21 = 0
...
01 + 0
...
001 kg/s = 0
...
01 ⎜ 2952 +
⎟ + 0
...
001 × 420 + 0
...
522 + 257
...
42 + 0
...
722 = 0
...
2 kJ/kg
Q5
...
The engine has a
specific fuel consumption of 0
...
The net heat transfer rate from
the fuel-air stream to the jacket cooling water and to the surroundings is
35 kW
...
Compute the
increase in the specific enthalpy of the fuel air stream, assuming the
changes in kinetic energy and in elevation to be negligible
...
– 1877 kJ/kg mixture)
In 1 hr
...
3 × 26 = 7
...

∴
Mass flow rate of fuel vapor and air mixture

Page 50 of 265

First Law Applied to Flow Process

By: S K Mondal

Chapter 5
dW
= 26 kW
dt

wg=x kg/s
w1=15 x kg/s
o
t2=790 C

t1=30oC
wa=14 x kg/s
dQ
= – 35 kW
dt

w1 =

15 × 7
...
0325 kg/s
3600

Applying S
...
E
...

dQ
dW
w1 h1 +
= w1 h 2 +
dt
dt
dQ dW
or
w1 (h2 – h1) =
−
dt
dt
dQ dW
−
dt
∴
h 2 – h1 = dt
w1
−35 − 26
= –1877 kJ/kg of mixture
...
0325
An air turbine forms part of an aircraft refrigerating plant
...
The air leaves the turbine at a pressure
of 115 kPa, a temperature of 2°C, and a velocity of 150 m/s
...
Neglecting changes in
elevation, determine the magnitude and sign of the heat transfer per
unit mass of air flowing
...
005 kJ/kg K and the enthalpy
h = cp t
...
+ 7
...
13

Solution:

V12
V22
dW
dQ
+
+
h2 +
=
2000 dm
2000 dm
2
2
dQ (h − h ) + V2 − V1 + dW
or
=
2
1
2000
dm
dm

dW
= 54 kJ/kg
dm

h1 +

2

2

150 − 45
+ 54 kJ/kg
2000
= –56
...
2375 + 54 kJ/kg
= 7
...
01 − 58
...
14

p1 = 295 kPa
t1 = 58°C
V1 = 45 m/s

1

h1 = CPt
1
= 1
...
29 kJ/kg

2 p2 = 115 kPa
t2 = 2°C
z 1 = z2
V2 = 150 m/s
2 h2 = 2
...
15 MPa
0
...
5 m/s
Page

First Law Applied to Flow Process

By: S K Mondal

Chapter 5

Height above datum
10 m
2m
If the volume flow rate of the fluid is 40 m3/s, estimate the net energy
transfer from the fluid as work
...
60
...
F
...
E
...
15 M Pa
= 1150 kPa
V1 = 30 m/s
z1 = 10 m

1
p2 = 0
...
5 m/s
z2 = 2 m

2

2
datum

Flow rate = 40 m3/s ≡ 40 × 1000 kg/s = w (say)
∴

Or

⎛ 1150
302
9
...
5
9
...
3342 MW

Q5
...
18 kW power,
and three 100 W lamps
...
If each
person puts out heat at the rate of 630 kJ/h determine the rate at which
heat is to be removed by a room cooler, so that a steady state is
maintained in the room
...
1
...
7 kW
= +
dt
3600
dQelectic
3 × 100
kW = 0
...
18 × 2 +
dt
1000
dQ
= 1
...
36 kW
= w1 h1 − w2 h 2 +
=
∴
dt
45
dt
= 1
...
16

Air flows steadily at the rate of 0
...
85
m3/kg, and leaving at 4
...
9 bar and a specific
volume of 0
...
The internal energy of the air leaving is 88 kJ/kg
greater than that of the air entering
...

Calculate the power required to drive the compressor and the inlet and
outlet cross-sectional areas
...
45
...
057 m2, 0
...
F
...
E
...
4 [– 88 + 85 – 110
...
0076] – 0
...
357 – 0
...
416 kW [have to give to compressor]

dQ
= – 59 W
dt
w1 = 0
...
85 m3/kg
u1 = ?

w1 =
w2 =

A 2 V2
v2

A1 V1
v1
∴ A2 =

1

2

1

2

∴ A1 =

w2 = 0
...
5 m/s
p2 = 6
...
16 m3/kg
u2 = u1 + 88 kJ/kg

w1 v1
0
...
85
= 0
...
4 × 0
...
01422 m2
=
4
...


Chapter 6

Second Law of Thermodynamics
Some Important Notes

Regarding Heat Transfer and Work Transfer
•

Heat transfer and work transfer are the energy interactions
...


•

It is wrong to say 'total heat' or 'heat content' of a closed system, because heat or work is
not a property of the system
...


•

Work is said to be a high grade energy and heat is low grade energy
...


•

HEAT and WORK are not properties because their net change in a cycle is not zero
...


or

∫

R

dQ
=0
T

•

The more effective way to increase the cycle efficiency is to decrease T2
...
K
...
1

Solution:

An inventor claims to have developed an engine that takes in 105 MJ at a
temperature of 400 K, rejects 42 MJ at a temperature of 200 K, and
delivers 15 kWh of mechanical work
...
No)
Maximum thermal efficiency of his engine possible
200
ηm a x = 1 −
= 50%
400
∴
That engine and deliver output = η × input
= 0
...
5 MJ = 14
...


Q6
...
How does this compare
with resistance heating?
(Ans
...
4 kW)
COP =

desired effect
input

(COP) ref
...
P – 1
H
or
6=
W
H
So input (W) =
6
But motor efficiency 90% so
Electrical energy require (E) =

∴ (COP) H
...
= 6

W
H
=
0
...
9 × 6

= 0
...
52% of Heat (direct heating)
100
kW
H=
= 5
...
52 kW of work

Q6
...
666
...
1
...
is 5
So for each MJ removed from the cold body we need work
1MJ
= 200 kJ
5
For 200 kJ work output of heat engine hair η = 30%
200 kJ
= 666
...
3
Now
COP of H
...
= COP of Ref
...
E
...
3 MJ = 300 kJ
That will be the input to H
...

Q
∴ ( COP ) H
...
P
...
8 MJ
=

Q6
...
In process 1–2, 2
...
During process 2–1, 2
...

(a) Find the heat transfer in process 2–1
...
(a) – 708 kJ (b) Second law, W2–1 = 9348 kJ (c) Q2–1 = 0)

Solution:

From the first Law of thermodynamics
(a) For process 1–2
Q1–2 = E2 – E1 + W1–2
–732 = (E2 – E1) – 10080
[2
...
8 × 3600 kJ ]

–W

–Q

∴ E2 – E1 = 9348 kJ
For process 2–1

Q 21 = E1 – E 2 + W21

+W

= –9348 + 8640
= –708 kJ i
...
Heat flow out to the atmosphere
...
As Electric energy stored
in a battery is High grade energy so it can be completely converted to the
work
...
5

Solution:

Chapter 6

Q21 = –9348 + 9348 = 0 kJ

A household refrigerator is maintained at a temperature of 2°C
...
The door is opened 20 times a day, and the refrigerator
operates at 15% of the ideal COP
...
2
...

What is the monthly bill for this refrigerator? The atmosphere is at 30°C
...
Rs
...
80)
275
275
Ideal COP of Ref
...
82143
30 − 2
28
Actual COP = 0
...
4732
303 K
Heat to be removed in a day
Q1 = Q2 + W
(Q2) = 420 × 20 kJ
W
= 8400 kJ
R

∴

Q2

Work required = 5701
...
58385 kWh/day

275 K

Electric bill per month = 1
...
32 × 30 Rupees
= Rs
...
20
Q6
...
The heat pump is driven
by a reversible heat engine which takes in heat from a reservoir at 840°C
and rejects heat to a reservoir at 60°C
...
If the heat pump extracts 17 kJ/s
from the 5°C reservoir, determine
(a) The rate of heat supply from the 840°C source
(b) The rate of heat rejection to the 60°C sink
...
(a) 47
...
61 kW)

Solution:
COP of H
...


∴
∴

333
= 6
...
P
...
P
...
05454
WH
...

17
= 5
...
P
...
36 kW
5
...
E
...
36 = 33
...
E
...
7
1113

∴

WH
...
=

Page 58 of 265

278 K

1113 K

17 kW
WHP
H
...

W
Q3

30 kW
333 K

Q1
H
...

Q2

Second Law of Thermodynamics

By: S K Mondal
(a) ∴
∴

Chapter 6

W
= 0
...
61 kW
0
...
E
...
61 – 33
...
25
kW
(ii) For H
...
= 17 + 3
...
36 kW
∴ Total = 34
...
7

Solution:

A refrigeration plant for a food store operates as a reversed Carnot heat
engine cycle
...
If heat
is transferred from the cycle to the atmosphere at a temperature of 25°C,
calculate the power required to drive the plant
...
0
...
933
–

298 K

5 kW
=
W

∴

Q6
...
56 kW
8
...
The heat transfers from the
heat engine and from the heat pump are used to heat the water
circulating through the radiators of a building
...
Evaluate the ratio
of the heat transfer to the circulating water to the heat transfer to the
heat engine
...
1
...
E
...
27
Q1

Page 59 of 265

Second Law of Thermodynamics

By: S K Mondal

Chapter 6

Q2
= 0
...
73 Q1

T1
Q1
H
...


W = Q1 – Q2 = 0
...
P
...
08 Q1

T2
Q4
H
...


∴ Q2 + Q4 = (0
...
08) Q1 = 1
...
81 Q1
= 1
...
6
kJ are rejected at 0°C, determine the location of absolute zero on the
Celsius scale
...
– 270
...
9

Solution:

or
∴

20
a × 100 + b
a
=
= × 100 + 1
14
...
6986 × 10–3
b

For absolute zero, Q2 = 0
Q1
a ×100 + b
=
0
a×t+b
or
a×t+b=0
−b
1
or
t=
= −
= –270
...
6986 × 10 −3
Two reversible heat engines A and B are arranged in series, A rejecting
heat directly to B
...
4°C
...
143
...
5%, 80 kJ)

∴

Q6
...
4
694
T
694 − T
T − 277
...
4
or
or

2T – 277
...
27 K = 143
...
11

Solution:

Q1
HE
Q2
Q2
HE

277
...
36%
416
...
27
× 200 kJ = 119
...
4
× 119
...
94 kJ
Q1 =
416
...
4 K

A heat engine operates between the maximum and minimum
temperatures of 671°C and 60°C respectively, with an efficiency of 50% of
the appropriate Carnot efficiency
...
4°C to heat a block of flats in which the temperature is
to be maintained at 21
...
Assuming that a temperature difference of
11
...
Why is direct heating thermodynamically
more wasteful?
(Ans
...
79 kJ/kJ heat input)
273 + 60
333
Carnot efficiency (η) = 1 −
= 1−
= 0
...
323623 = 1 −

′
Q1
Q1

Page 61 of 265

Second Law of Thermodynamics

By: S K Mondal

Chapter 6

′
Q1
= 0
...
2
= 7
...
2 – 266
...
923 = 3
W
if Q3 = 1 kJ
Q3
1
=
∴
W=
3
...
923

∴

block
...
12

Solution:

= 0
...
6764 Q1 =
0
...
2549
Q1 =
(1 − 0
...
7877 kJ/kJ heat input to

An ice-making plant produces ice at atmospheric pressure and at 0°C
from water
...
Evaluate the
minimum electrical work in kWh required to produce 1 tonne of ice (The
enthalpy of fusion of ice at atmospheric pressure is 333
...

(Ans
...
11 kWh)
273
= 15
...
2
291 K
W
min

or

Wmin =

Q
1000 × 333
...
2
15
...
989 MJ = 6
...
13

W

Q2
R
Q1
273 K
0°C

A reversible engine works between three thermal reservoirs, A, B and C
...
The efficiency of the
engine is α times the efficiency of the reversible engine, which works
between the two reservoirs A and C
...
E
...
E
...
E
...
14

A reversible engine operates between temperatures T1 and T (T1 > T)
...
The second engine rejects energy at
temperature T2 (T2 < T)
...


Solution:

(a) If they produce same Amount and work
Then W1 = W2
or η1Q1 = η 2 Q 2
T ⎞
T ⎞⎛T ⎞
⎛
⎛
or ⎜1 − ⎟ ⎜ 1 ⎟ Q2 = ⎜1 − 2 ⎟ Q2
T1 ⎠ ⎝ T ⎠
⎝
T⎠
⎝
Q
Q
We know that 1 = 2
T
T1
T
or
Q1 = 1 Q2
T

Page 63 of 265

Second Law of Thermodynamics

By: S K Mondal

Chapter 6

T1
T
−1 = 1 − 2
T
T
T1 + T2
=2
or
T
T + T2
or
T= 1
2
i
...
, Arithmetic mean and T1, T2
or

(b) If their efficiency is same then
T
T
1−
= 1− 2
T
T1
or
∴
Q6
...


T1
Q1
H
...


W1

Q2
T
Q2
H
...


W2

Q3
T2

Two Carnot engines A and B are connected in series between two
thermal reservoirs maintained at 1000 K and 100 K respectively
...
Engine B takes in heat rejected by
engine A and rejects heat to the low-temperature reservoir
...

If engines A and B deliver equal work, determine
(d) The amount of heat taken in by engine B
(e) The efficiencies of engines A and B
(Ans
...
2 K, (c) 1148
...
3 kJ,
(d) 924 kJ, (e) 45%, 81
...
3K
Q2 =

Q1
1680 × 316
...
26 kJ

Q2
531
...
3
316
...
26) kJ
= 1148
...
26 – 168) kJ
= 363
...
45
(e) η A = 1 −
1000
100
= 0
...
16

Solution :

A heat pump is to be used to heat a house in winter and then reversed to
cool the house in summer
...
Heat transfer through the walls and roof is estimated to be 0
...

(a) If the outside temperature in winter is 5°C, what is the minimum
power required to drive the heat pump?
(b) If the power output is the same as in part (a), what is the maximum
outer temperature for which the inside can be maintained at 20°C?
(Ans
...
4°C)
(a) Estimated Heat rate
293 K
= 0
...
875 kJ/s
20°C
•
293
Q = 7875 kJ/s
COP =
= 19
...
P
...
875
= 0
...
53
•

(b) Given W = 403 W
•

Heat rate (Q1 ) = 0
...
17

T
R

•
Q1

293 K

Consider an engine in outer space which operates on the Carnot cycle
...
The rate at which heat is radiated is proportional to the
fourth power of the absolute temperature and to the area of the
radiating surface
...
18

Solution:

H
...


W

Q2
T2

W⎧
1
W ⎧ T2 ⎫
⎫
⎬ =
⎨
⎬
3
4 ⎨
σ ⎩ T1T2 − T24 ⎭
σ T2 ⎩ T1 − T2 ⎭

For minimum Area
∂A
=0
or
∂ T2
or
or

Q1

Q1
Q
W
= 2 =
T1
T2
T1 − T2
WT2
Q2 =
T1 − T2
WT2
= σ AT24
T1 − T2

∂
3
{T1T2 − T24 } = 0
∂ T2

3
T1 × 3 T22 − 4 T2 = 0
3T1 = 4T2
T2
3
=
proved
4
T1

It takes 10 kW to keep the interior of a certain house at 20°C when the
outside temperature is 0°C
...
Calculate the power required if the 10 kW heat flow
were supplied by operating a reversible engine with the house as the
upper reservoir and the outside surroundings as the lower reservoir, so
that the power were used only to perform work needed to operate the
engine
...
0
...
P
...
P
...


273 K
Q6
...

Suppose A is any refrigerator and B is reversible refrigerator and also assume

(COP)A > (COP) B
and

Q1A = Q1B = Q
Page 66 of 265

Second Law of Thermodynamics

By: S K Mondal
or
or
or

Chapter 6
Q1 A Q1B
>
WA
WB

Q
Q
>
WA WB
WA < WB

T1
Q2A
WA

Q2B
WB

R1
Q1A

Then
we
reversed
the
reversible refrigerator ‘B’ and
then
work
output
of
refrigerator ‘B’ is WB and heat
rejection is Q1B = Q (same)
So we can directly use Q to
feed for refrigerator and
Reservoir ‘T2’ is eliminated
then also a net work output
(WB – WA) will be available
...
e
...

So (COP) R ≥ (COP) A
Q6
...
21

Solution:

R2
Q1B

T2
T1 > T2 and T1 and T2 fixed

T1
WA

R

H
...

Q1A

WB

Q1B

T2

A house is to be maintained at a temperature of 20°C by means of a heat
pump pumping heat from the atmosphere
...
65 kW per unit of temperature difference
between the inside of the house and the atmosphere
...
For the same room temperature, the same heat loss rate,
and the same power input to the pump, what is the maximum
permissible atmospheric temperature?
(Ans
...
16
A solar-powered heat pump receives heat from a solar collector at Th,
rejects heat to the atmosphere at Ta, and pumps heat from a cold space
at Tc
...

Derive an expression for the minimum ratio Qh/Qc, in terms of the three
temperatures
...
2 kW 1 m2, what is the minimum collector area
required?
(Ans
...
25 kW, 131
...
E
...
4
= 120 m2
Required Area (A) =
0
...

R
Qc
Tc

Q6
...
If the
efficiency of the engine is 40% of the maximum possible and the COP of
the heat pump is 50% of the maximum possible, what is the temperature
of the reservoir to which the heat pump rejects heat? What is the rate of
heat rejection from the heat pump if the rate of heat supply to the
engine is 50 kW?
(Ans
...
5 K, 86 kW)

ηactual = 0
...
28
1000 ⎠

⎝
W = 0
...
72 Q1
Q3 = 2 Q2 + W = 1
...
72 Q1
0
...
5)
=
T − 300
6
...
143 = T × 0
...
58 K
Q3 = 1
...
E
...
P
...
23

A reversible power cycle is used to drive a reversible heat pump cycle
...
The
heat pump abstracts Q4 from the sink at T4 and discharges Q3 at T3
...

Q4 T4 (T1 − T2 ) ⎞
⎛
⎜ Ans
...
E
...
P
...
E
...
P
...

or

Q6
...


Solution:

Applying First Law of Thermodynamics
Q12 = E2 – E1 + W1
...


Chapter 7

Entropy
Some Important Notes

1
...


Sf – Si =

f

∫
i

⎛ dQ ⎞
=0
⎟
T ⎠rev
...

= (ΔS) irrev
...


4
...
(Isolated)

5
...


TdS = dH – Vdp

7
...


Clausius Inequality:

∫

Where K = Boltzmann constant
W = thermodynamic probability
...


General case of change of entropy of a Gas

p
V ⎫
⎧
S2 – S1 = m ⎨cv ln 2 + c p ln 2 ⎬
p1
V1 ⎭
⎩
Initial condition of gas p1 , V1, T1, S1 and
Final condition of gas p2 , V2, T2, S2

Page 70 of 265

Entropy

By: S K Mondal

Chapter 7

Questions with Solution P
...
Nag
Q7
...


On the basis of the first law fill in the blank spaces in the following table
of imaginary heat engine cycles
...

Cycle

(a)
(b)
(c)
(d)

Temperature
Source
327°C
1000°C
750 K
700 K

Rate of Heat Flow

Sink
Supply
27°C
420 kJ/s
100°C …kJ/min
300 K
…kJ/s
300 K
2500
kcal/h

Rejection
230 kJ/s
4
...
(a) Irreversible, (b) Irreversible, (c) Reversible, (d) Impossible)

Solution:
Cycle

Temperature

Rate of Heat Flow

Rate of
work

Efficiency

Remark

(a)

Source
327ºC

Sink
27ºC

Supply
420 kJ/s

Rejection
230 kJ/s

190kW

0
...
possible

(b)

1000ºC

100ºC

12000
kJ/km

4
...
7%
irrev
...
33 kJ/s

17
...
4%

(d)

700 K

Q7
...

possible
ηmax=57%
irrev
...
How much does the
entropy of 1 kg of ice change as it melts into water in each of the
following ways:
(a) Heat is supplied reversibly to a mixture of ice and water at 0°C
...

(Ans
...
2271 kJ/K)
1 × 335
Ice + Water
kJ/ K
Solution : (a) (ΔS) system = +
273
Q

= 1
...
3

Solution:

Two kg of water at 80°C are mixed adiabatically with 3 kg of water at
30°C in a constant pressure process of 1 atmosphere
...
187 kJ/kg K)
...
0
...
187 ln ⎜
⎟ + 3 × 4
...
05915 kJ/K
Q7
...
The
working fluid is water which, while receiving heat, evaporates from
liquid at 350°C to steam at 350°C
...
44
kJ/kg K
...
(a) 465
...
12 kJ/kg, (b) 0
...
44
(a)
(350 + 273)
or Q = 897
...
Q = 465
...
043 kg/s
(b)
P = mW or m =
=
W
465
...
5

A heat engine receives reversibly 420 kJ/cycle of heat from a source at
327°C, and rejects heat reversibly to a sink at 27°C
...
For each of the three hypothetical amounts of heat
rejected, in (a), (b), and (c) below, compute the cyclic integral of d Q / T
...
(a) Reversible, (b) Impossible, (c) Irreversible)
Solution:

(a)

(b)

(c)

Q7
...
35
T
600 300
∴ Cycle is Impossible

∫

dQ
420 315
−
= +
= – 0
...


∫

In Figure, abed represents a Carnot cycle bounded by two reversible
adiabatic and two reversible isotherms at temperatures T1 and T2 (T1 >
T2)
...
Prove that the efficiency of the oval cycle is
less than that of the Carnot cycle
...
7

Solution:

Water is heated at a constant pressure of 0
...
The boiling point is
164
...
The initial temperature of water is 0°C
...
3 kJ/kg
...
6
...
97
dT
= ∫ 1 × 4187 ×
T
273
p = 700 kPa
⎛ 437
...
187 ln ⎜
⎟ kJ/ K
⎝ 273 ⎠
T = 437
...
979 kJ/K
(ΔS)Eva pour

T
273
K

1 × 2066
...
97
= 4
...
697 kJ/kg – K

Q7
...
7 MPa, 20°C changes to 0
...
Process 1: a-2
consists of a constant pressure expansion followed by a constant volume
cooling, process 1: b-2 an isothermal expansion followed by a constant
pressure expansion, and process 1: c-2 an adiabatic

Page 74 of 265

Entropy

By: S K Mondal

Chapter 7

Expansion followed by a constant volume heating
...
Take cp = 1
...
718 kJ/kg K and assume the specific heats to be constant
...
287 T, where p is the pressure in kPa, v the specific volume
in m3/kg, and T the temperature in K
...
7 MPa = 700 kPa
v1 = 0
...
27306 m3/kg

Solution:

T1 = 293 K

p a = 700 kPa
v a = 0
...
27306 – 0
...
86 kJ/kg
Page 75 of 265

Entropy

By: S K Mondal

Chapter 7

u 2 – u a = –239 kJ/kg
Q1 – a =

Ta

∫c

P

dT

T1

= 1
...
865 kJ/kg
Qa – 2 =

T2

∫c

v

dT

Ta

= 0
...
766 kJ/kg
(ii) Δh = h2 – h1 = u 2 - u1 + p 2 v 2 – p1 v1
= 28
...
27306 – 700 × 0
...
246 kJ/kg
(iii) Q = Q2 + Q1 = 135
...
3275 kJ/kg – K

Q7
...
Assuming the specific heat of liquid water to
remain constant at 4
...

(Ans
...
02 J/K)

Solution:
273

S2 – S1 =

m cP dT
T
293

∫

1

273
kJ/ K
293
= –0
...
9694 J/K
− mL
S3 – S2 =
T
−0
...
271 J/K

293 K

= 0
...
2 × ln

3
T

273 K
4
268 K

S
Page 76 of 265

2

Entropy

By: S K Mondal

Chapter 7

268

m cP dT
268
⎛ 4
...
01 × ⎜
kJ/ K
⎟ × ln
T
273
⎝ 2 ⎠
273
= –0
...
10

∫

S4 – S1 = – 15
...
63 J/K

Calculate the entropy change of the universe as a result of the following
processes:
(a) A copper block of 600 g mass and with Cp of 150 J/K at 100°C is placed
in a lake at 8°C
...

(c) Two such blocks, at 100 and 0°C, are joined together
...
(a) 6
...
095 J/K, (c) 3
...
48 J/K
As unit of CP is J/K there for
∴
It is heat capacity
i
...

Cp = m c p

(ΔS) copper =

∫ mc

P

(ΔS) lake =

C p (100 − 8)

J/ K
281
150(100 − 8)
J/ K = 49
...
63 J/K

(b) Work when it touch water = 0
...
81 × 100 J = 588
...
6
=
J/ K = 2
...
09466 J/k = 2
...
638 J/K
+ ln
= 150 ln
373
273

{

Q7
...
Show that
the maximum work recoverable as the system is cooled to T0 is
⎡
T ⎤
W = Cv ⎢(T1 − T0 ) − T0 ln 1 ⎥
T0 ⎦
⎣

Solution:
For maximum work obtainable the process should be reversible
T0
dT
⎛T ⎞
(ΔS)body = ∫ Cv
= Cv ln ⎜ 0 ⎟
T
⎝ T1 ⎠
T1
Q−W
T0
(ΔS)cycle = 0
⎛T ⎞ Q−W
(ΔS)univ
...
E
...
12

If the temperature of the atmosphere is 5°C on a winter day and if 1 kg of
water at 90°C is available, how much work can be obtained
...
186 kJ/kg K
...
13

A body with the equation of state U = CT, where C is its heat capacity, is
heated from temperature T1 to T2 by a series of reservoirs ranging from
Page 78 of 265

Entropy

By: S K Mondal

Chapter 7

T1 to T2
...
Calculate the changes of entropy of
the body and of the reservoirs
...
14

A body of finite mass is originally at temperature T1, which is higher
than that of a reservoir at temperature T2
...
If the engine does work W, then it will reject heat Q–W to the
reservoir at T2
...


Solution:

If the body is maintained at constant volume having constant volume
heat capacity Cv = 8
...

(Ans
...
96 kJ)
Final temperature of the body will be T2
∴

S2 – S1 =

T2

∫ mc

v

T1

dT
⎛T ⎞
= m cv ln ⎜ 2 ⎟
T
⎝ T1 ⎠
[ cv = heat energy CV]

(ΔS) reservoir =

Q−W
T2

∴ (ΔS) H
...
= 0

or

Q−W
≥0
T2
T2 (S2 – S1) + Q – W ≥ 0

or

W ≤ Q + T2 (S2 – S1)

or

W ≤ [Q – T2 (S1 – S2)]

∴

Wmax = [Q – T2 (S1 – S2)]

∴

(ΔS) univ
...
4 ⎢373 − 303 + 303 ln ⎜
⎟⎥
⎝ 373 ⎠ ⎦
⎣
Page 79 of 265

Entropy

By: S K Mondal

Chapter 7

= 58
...
15

Each of three identical bodies satisfies the equation U = CT, where C is
the heat capacity of each of the bodies
...
If C = 8
...
756 kJ)

Solution:

U = CT
Therefore heat capacity of the body is C = 8
...
E
...
E
...
= C ln ⎜
⎟≥0
⎝ 540 × 250 × 200 ⎠

For minimum Tf
Tf3 = 540 × 250 × 200
∴ Tf = 300 K

∴
Q7
...
4(540 – 300) = 2016 kJ
Q1 – W1 = 8
...
4(300 – 200) = 840 kJ
∴ Q1 + Q2 – (W1 + W2) = 1260
or (W1 + W2) = 2016 – 1260 kJ = 756 kJ
Wmax = 756 kJ
In the temperature range between 0°C and 100°C a particular system
maintained at constant volume has a heat capacity
...
014 J/K and B = 4
...
What
is the maximum amount of work that can be transferred to the reversible
work source as the system is cooled from 100°C to the temperature of the
reservoir?
(Ans
...
508 J)
Solution:

Page 80 of 265

Entropy

By: S K Mondal

Chapter 7

Find temperature of body is 273 K

373 K

273

∴

Q =

∫

273
C v dT = AT + BT2 ]373

373

Q
2

2

= –A(100) + B( 273 – 373 ) J

H
...


= –28
...
08837 J/K

=

∫⎜
⎝

Q−W
; (ΔS)H
...
= 0
273
Q−W
≥0
(ΔS)univ = −0
...
125 + Q – W ≥ 0
W ≤ Q – 24
...
532 – 24
...
407 J
Wmax = 4
...
=
∴

or
or
or
or
Q7
...
18

A = 8
...
1 × 10-2 J/K2

If the bodies are initially at temperatures 200 K and 400 K and if a
reversible work source is available, what are the maximum and
minimum final common temperatures to which the two bodies can be
brought? What is the maximum amount of work that can be transferred
to the reversible work source?
(Ans
...
Find the
amount and direction of heat interaction with other reservoirs
...
Q2 = + 4
...
82 MJ)
Let Q2 and Q3 both incoming i
...
out from the system
∴
Q2 → +ve,
Q3 → +ve
Q3
Q2
5000
(ΔS) univ =
+
+
+ ( Δ S)H
...
+ ( Δ S)surrounds = 0
200 300 400

200 K

300 K

Q3

400 K

Q2

Q1 = 5 MJ

E

Or
or

W = 840 kJ
Q3 Q2 5000
+
+
+ 0+ 0 = 0
2
3
4
6 Q3 + 4 Q2 + 3 × 5000 = 0
Q3 + Q2 + 5000 – 840 = 0

Heat balance
or
4 Q3 + 4 Q2 + 16640 = 0
∴
(i) – (iii) gives
∴
∴
Q7
...
287 kJ/kg K, flows
steadily through an adiabatic machine, entering and leaving through
two adiabatic pipes
...
Determine which pressure and
temperature refer to the inlet pipe
...
A is the inlet pipe)
For the given temperature range, cp is given by
Cp = a ln T + b
Where T is the numerical value of the absolute temperature and a = 0
...
86 kJ/kg K
...
sB – s A = 0
...
A is the inlet pipe
...
287 × ln ⎜ ⎟
T
⎠
⎝5⎠
450

∫⎜
⎝

300

⎡ (ln T)2
⎤
⎛1 ⎞
= ⎢a
+ b ln T ⎥ − 0
...
287 ln ⎜ ⎟
2
450
⎝5⎠
or sB – s A = 0
...
20

Two vessels, A and B, each of volume 3 m3 may be connected by a tube of
negligible volume
...
7 MPa, 95 ° C, while vessel B
contains air at 0
...
Find the change of entropy when A is
connected to B by working from the first principles and assuming the

Page 83 of 265

Entropy

By: S K Mondal
Solution:

Chapter 7

mixing to be complete and adiabatic
...
8
...
0
...
287 × 368
= 19
...
7 MPa
700 kPa
368 K

350 kPa
478 K

Mass of gas ( m B ) =

Cp = 1
...
718 kJ/kg-K
R = 0
...
653842 kg
=
R TB
0
...
88335 × 0
...
66 kJ
u B = m B c v TB
Umixture
Or

= 7
...
718 × 478 kJ = 2626
...
6 K

If final pressure (pf)
∴
∴

∴

Q7
...
5372 × 0
...
6
kPa = 525 kPa
pf =
6

⎡
T
⎛ p ⎞⎤
(ΔS)A = m A ⎢c P ln f − R ln ⎜ f ⎟ ⎥ = 3
...
28795 kJ/K
TB
⎝ pB ⎠ ⎦
⎣
(ΔS)univ = (ΔS)A + (ΔS)B + 0 = 0
...
It is cooled reversibly by transferring
heat to a completely reversible cyclic heat engine until the block
reaches 20°C
...
Compute (i) the change in entropy for the block,

Page 84 of 265

Entropy

By: S K Mondal

Chapter 7

(ii) the change in entropy for the room air, (iii) the work done by the
engine
...

(Ans
...
5 J/K, 2
...
06 J/K
313
Q−W
(ΔS) air =
293
And Q = m c P (313 – 293) = 40000 J
5 × 400 × ln

313 K

5 kg

Q
H
...


As heat is reversibly flow then
(ΔS)Al + (ΔS) air = 0
W
or
–132
...
52 –
=0
293
or
W = 1
...
06 J/K
4000
(ΔS) air =
= 136
...
4587 J/K
Q7
...
What will the final temperature be if one
lets this system come to equilibrium (a) freely? (b) Reversibly? (c) What
is the maximum work which can be obtained from this system?

Solution:

(a)

Freely Tf =
(b)

T1 + T2
2

Reversible
Let find temperature be Tf
the (ΔS)hot =

Tf

∫C

T1

dT
T

= C ln
(ΔS)cold =

Tf

∫C

T2

Tf
T1

dT
⎛T ⎞
= C ln ⎜ f ⎟
T
⎝ T2 ⎠

∴ (ΔS)univ
...
E
...
23

Solution:

A resistor of 30 ohms is maintained at a constant temperature of 27°C
while a current of 10 amperes is allowed to flow for 1 sec
...

(Ans
...
The specific heat of the resistor is 0
...

(Ans
...
72 J/K)
As resistor is in steady state therefore no change in entropy
...

i 2 Rt
So (ΔS)atm =
Tatm

102 × 30 × 1
= 10 kJ/kg
300
If the resistor is insulated then no heat flow to
surroundings
So
(ΔS) surroundings = 0
=

And, Temperature of resistance (Δt)
102 × 30 × 1
= 333
...
01
∴
Final temperature (Tf) = 633
...
33
dT
∴
(ΔS) = ∫ m c
T
300
⎛ 633
...
01 × 0
...
725 J/K
⎝ 300 ⎠
(ΔS)univ = (ΔS)rev
...
725 J/K

Q7
...
By paddle-wheel work
transfer, the temperature of water is increased to 30°C
...
187 kJ/kg K, find the entropy
change of the universe
...
0
...
= 0

(ΔS)sys =

303

∫ mc

298

dT
T

303
= 0
...
13934 + 0 = 0
...
187 × ln
(ΔS)univ = ( ΔS)sys

∴

Q7
...
01 m
...
The rod is perfectly insulated along its
length and the thermal conductivity of copper is 380 W/mK
...

(Ans
...
985 W, 0
...
01 m
1m
K = 380 W/m – K

373 K
•

Q = kA

A = 7
...
854 × 10 −5 ×

100
W = 2
...
So at this
end
•

•

( Δ S)charge

Q
= −
373
•

And at the 273 K and from system Q amount of heat is rejected to the
surroundings
...
26

•

•

( Δ S)charge

Q
=
273
•

•

Q
Q
( Δ S)univ
...
00293 W/K
273 373
•

A body of constant heat capacity Cp and at a temperature Ti is put in
contact with a reservoir at a higher temperature Tf
...
Show that the entropy change of the universe is equal to

Page 87 of 265

Entropy

By: S K Mondal

Chapter 7
⎡ Ti − Tf
Cp ⎢
− ln
⎢ Tf
⎣

⎛
Ti − Tf
⎜1 +
⎜
Tf
⎝

⎞⎤
⎟⎥
⎟⎥
⎠⎦

Prove that entropy change is positive
...
–
Given ln (1 + x) = x –
{where x < 1}
2
3 4
Solution:
Final temperature of the body will be Tf

∴

(ΔS) resoier =

Tf

dT
⎛T ⎞
= C p ln ⎜ f ⎟
T
⎝ Ti ⎠
Ti
C p (Tf − T1 )

(ΔS) body = C p

∫

Tf

∴ Total entropy charge
T ⎤
⎡ T − Ti
+ ln f ⎥
(ΔS) univ = C p ⎢ f
Tf
Ti ⎦
⎣
T⎤
⎡ T − Ti
− ln i ⎥
= Cp ⎢ f
Tf ⎦
⎣ Tf
⎡ T − Ti
T − Tf
⎛
= Cp ⎢ f
− ln ⎜1 + i
Tf
⎝
⎣ Tf
Let

∴
∴

Tf

CP
Ti

⎞⎤
⎟⎥
⎠⎦

Tf − Ti
=x
as Tf > Ti
Tf
Tf − Ti
<1
Tf
(ΔS) in = CP {x – ln (1 + x)}

=
⎡
⎤
x
x3 x4
C p ⎢x − x +
−
+
+
...
α ⎥
3
4
5
⎣2
⎦
=
⎡ x 2 (3 − 2 x) x 4 (5 − 4 x)
⎤
Cp ⎢
+
+
...
27

(ΔS) univ is +ve

An insulated 0
...
2 kg water is in
equilibrium at a temperature of 20°C
...
05
kg of ice at 0°C in the calorimeter and encloses the latter with a heat
insulating shield
...
418 kJ/kg K and the latent heat of fusion of ice is 333
kJ/kg
...

(c) What will be the minimum work needed by a stirrer to bring back
the temperature of water to 20°C?
(Ans
...
68°C, (b) 0
...
84 kJ)

Solution:

Mass of ice = 0
...
75 × 0
...
2 × 4
...
05 + 0
...
187
× ( Tf – 273)
or 1
...
65 – 57
...
20935 Tf
or 337
...
1509 Tf

Wab = 0
...
75 kJ /kg-K
T1 = 293 K

or Tf = 277
...
68º C
(b) (ΔS)system

⎛ T ⎞
⎛ T ⎞
= 0
...
418 × ln ⎜ f ⎟ + 0
...
187 × ln ⎜ f ⎟
⎝ 293 ⎠
⎝ 293 ⎠
333 × 0
...
05 × 4
...
00275 kJ/K = 2
...

∴
W = C × (20 – 4
...
84 kJ
∴
Q7
...
36025

Show that if two bodies of thermal capacities C1 and C2 at temperatures
T1 and T2 are brought to the same temperature T by means of a
reversible heat engine, then
ln T =

C1lnT1 + C2lnT2
C1 + C2

Solution:
T

(ΔS) 1 =

∫C

1

T1

dT
⎛T⎞
= C1 ln ⎜ ⎟
T
⎝ T1 ⎠
T

(ΔS) 2 =

∫C

T2

2

dT
⎛T⎞
= C2 ln ⎜ ⎟
T
⎝ T2 ⎠

(ΔS)univ = (ΔS)1 + (ΔS)2
For reversible process for an isolated system (ΔS) since
...
E
...
29

C

1
2
⎛T⎞ ⎛T⎞
⎜T ⎟ ⎜T ⎟ = 1
⎝ 1⎠ ⎝ 2⎠
C
TC1 + C2 = T1C1 T2 2

W

Q–W

C1ln T1 + C2 ln T2
Proved
C1 + C2

T2

C2

Two blocks of metal, each having a mass of 10 kg and a specific heat of
0
...
A reversible refrigerator
receives heat from one block and rejects heat to the other
...

Mass = 10 kg
C = 0
...
30

Q + W = 10 × 10
...
87 kJ
Q = 10 × 0
...
87 kJ

A body of finite mass is originally at a temperature T1, which is higher
than that of a heat reservoir at a temperature T2
...
In this process there is a heat
flow Q out of the body
...

Page 90 of 265

Entropy

By: S K Mondal

Chapter 7

Solution:

Try please
...
31

A block of iron weighing 100 kg and having a temperature of 100°C is
immersed in 50 kg of water at a temperature of 20°C
...
45 and 4
...

(Ans
...
1328 kJ/K)
Let final temperature is tf ºC
∴
100 × 0
...
18 × (tf – 20)
100 – tf = 4
...
699
or
5
...
88
or
tf = 34
...
1732 K

Solution:

Q7
...
1732 ⎞
⎛ 307
...
45 ln ⎜
⎟ + 50 × 4
...
1355 kJ/K
36 g of water at 30°C are converted into steam at 250°C at constant
atmospheric pressure
...
2 J/g K and the latent heat of vaporization at 100°C is 2260 J/g
...
4619 kJ/kg K, and

Cp

= a + bT + cT2, where a = 3
...
195 × 10-3 K-1 and c = 0
...

Solution:

(Ans
...
8 J/K)

m = 36 g = 0
...
03143 kJ/K
mL
(ΔS) Vaporization =
T2
0
...
21812 kJ/K
523
dT
(ΔS) Vapor = ∫ m c p
T
373
523

= mR

a

∫ (T

T3

T

T2
T1

S

+ b + CT) dT

373

Page 91 of 265

Entropy

By: S K Mondal

Chapter 7
523

⎡
CT2 ⎤
= mR ⎢a ln T + bT +
⎥
2 ⎦ 373
⎣
523
C
⎡
⎤
= mR ⎢a ln
+ b × (523 − 373) + (5232 − 3732 ) ⎥
373
2
⎣
⎦
= 0
...
1 J/K

Q7
...
34

Solution:
Q7
...
In a time interval of 1s
(a) What is the change in entropy of the resistor?
(b) What is the change in entropy of the universe?
(Ans
...
167 J/K)
Try please
...
5 kg at an initial temperature of 260 K
melts at the pressure of 1 bar as a result of heat transfer from the
environment
...
Calculate the entropy
production associated with this process
...
4 kJ/kg, the specific heat of ice and water are 2
...
2 kJ/kg K
respectively, and ice melts at 273
...

(Ans
...
1514 kJ/K)
Try please
...
It is then heated reversibly at constant volume to state c
...
Heat is then rejected reversibly from the gas at constant volume till
it returns to state a
...
If Tb = 555 K and Tc
= 835 K, estimate Ta
...
4
...
Ta = b γ , 313
...
37

Solution:
Q7
...
4

d

V=C

a
S

γ = 1
...
286 K

Liquid water of mass 10 kg and temperature 20°C is mixed with 2 kg of
ice at – 5°C till equilibrium is reached at 1 atm pressure
...
Given: cp of water = 4
...
09 kJ/kg K and latent heat of fusion of ice = 334 kJ/kg
...
190 J/K)
Try please
...
The
initial temperature of the resistor is 10°C
...
85 J g K
...
(a) 0
...
173 J/K)
Try please
...

(a) Determine the heat absorbed and the increase in entropy of a mass m
of the substance when its temperature is increased at constant
pressure from T1 to T2
...

Given for copper: when T = 500 K, cp = 25
...
1 × 103 J/k mol K
...
⎜
2
T1
⎢
⎣
⎦⎥ ⎟
⎣
⎦
⎜
⎟
(b) 24
...
Ta
...
4 + 1

V=C
b

T

Ta =

Ta =

Q7
...
2 = a + b × 500
30
...
9
∴ b = 0
...
7 kJ/kg – K
⎡
⎤
⎛ 1200 ⎞
∴
S2 – S1 = ⎢21
...
007 (1200 − 500)⎥ kJ/ K = 23
...


Chapter 8

Availability & Irreversibility
Some Important Notes

1
...
E
...

= h1 − h 2 – T0 ( s1 − s2 )
(For steady flow system), it is (A1 – A2) as in steady state no change in volume is
CONSTANT VOLUME (i
...
change in availability in steady flow)

2
...


S

Availability function:
V2
A = h – T0s +
+ gZ
2
Availability = maximum useful work
For steady flow

Availability = A1 – A0 = (h1 – h0) – T0 ( s1 – s0 ) +

φ = u – T0s + p 0 V

V12
+ gZ
2

(∴V0 = 0, Z0 = 0)

For closed system
Availability = φ1 – φ0 = u1 – u 0 – T0 ( s1 – s 0 ) + p 0 (V1 – V0 )
Available energy is maximum work obtainable not USEFULWORK
...


Unavailable Energy (U
...
)
= T0 (S1 – S2)

5
...

Page 95 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8

6
...


7
...


for all processes

Wactual ⇒ dQ = du + d Wact
h1 +

this for closed system

dWact
V12
V2
dQ
+ g Z1 +
= h 2 + 2 + g Z2 +
this for steady flow
2
dm
2
dm

9
...


Gibb’s function, G = H – TS
•

11
...


13
...

⎝
T⎠
To Calculate dS
p
V ⎤
⎡
i) Use S2 – S1 = m ⎢cv ln 2 + cP l n 2 ⎥
p1
V1 ⎦
⎣

For closed system
TdS = dU + pdV
dT p
+ dV
or
dS = m c v
T
T
dT
dV
+ mR
= m cv
T
V
2
2
2
dT
dV
dS = m c v ∫
+ mR ∫
∫
T
V
1
1
1
For steady flow system
TdS = dH – Vdp
dT V
− dp
or
dS = m c p
T
T
2
2
2
dT
dp
- mR ∫
dS = m c p ∫
∫
T
p
1
1
1
But Note that

pV = mRT

V
mR
=
T
p
Page 96 of 265

Availability & Irreversibility

By: S K Mondal
And

Chapter 8

TdS = dU + pdV
TdS = dH – Vdp

Both valid for closed system only

14
...


m c p (T2 - T1 )

•

T0
•

Rate of Irreversibility (I) = T0 Sgen

Flow with friction
•

Decrease in availability = m RT0 ×

Δp
p1

Page 97 of 265

p, T2

Availability & Irreversibility

By: S K Mondal

Chapter 8

Questions with Solution P
...
Nag
Q8
...
(a) 11
...
25 kJ)

Solution:

Entropy change for this process
ΔS =

−100
kJ/ K
675

= 0
...
333 kJ
(a) Now maximum work obtainable
338 ⎞
⎛
′
Wmax = 100 ⎜1 −
⎟
625 ⎠
⎝
= 45
...
333 – 45
...
413 kJ
dQ
= ± constant
(b) Given
dT
Let dQ = ± mc P dT
∴ When source temperature is (675 – T) and
since temperature (288 + T) at that time if dQ
heat is flow then maximum
...


288 + T ⎞
⎛
∴ dWmax
...
E
...
887 mc P kJ

=

(675 – T)
Q1
W

H
...


m c p × 50 = 100 kJ
= 51
...
333 – 51
...
5603 kJ

Q 8
...
08 kJ/kg K) is cooled from 1300°C to 320°C
...
Determine the loss in available energy due to the above heat
transfer per kg of water evaporated (Latent heat of vaporization of
water at 260°C = 1662
...

(Ans
...
6 kJ)
Availability decrease of gas
Agas = h1 – h2 – T0 ( s1 – s2 )

⎛ T1 ⎞
⎟
⎝ T2 ⎠

= mc p ( T1 – T2 )

– T0 mc p ln ⎜

T ⎤
⎡
= m cP ⎢(T1 − T2 ) − T0 ln 1 ⎥
T2 ⎦
⎣
∴ T1 = 1573 K; T2 = 593 K; T0 = 303 K
= m × 739
...
5 1 −

303
533

}

= 717
...
08 × (1300 – 320) = 1 × 1662
...
5708 kg/ of water of evaporator
Agas = 1161
...
1 – 717
...
7 kJ
Page 99 of 265

Availability & Irreversibility

By: S K Mondal
Q 8
...
1 kJ/kg K)
...

(a) How much available energy per kg of gas is lost by throwing away
the exhaust gases?
(b) What is the ratio of the lost available energy to the engine work?
(Ans
...
58 kJ, (b) 0
...
1 ⎨(1073 − 303) − 303 ln ⎜
⎟⎬
⎝ 303 ⎠ ⎭
⎩
= 425
...
55
(b) r =
= 0
...
4

Solution:

A hot spring produces water at a temperature of 56°C
...
1 m3 of
water per min
...
19
...
1
329
kW
× 1000 × 4
...
559 kW
0
...

Find the available and unavailable energies of the heat added
...
0047 kJ/kg K
...
211
...
1 kJ)
Entropy increase
2066
dT
2066
ΔS = S2 – S1 = ∫ m c p
= 0
...
0047 × ln
= 0
...
5

329
287

= mc p ( T2 – T1 ) – T0 × 0
...
24 – 78
...
2 kJ
Heat input = m c p (T2 – T1) = 1250
...
086 kJ

Page 100 of 265

Availability & Irreversibility

By: S K Mondal
Q8
...
Determine the decrease in
available energy due to mixing
...
236 kJ)
m1 = 80 kg                                         2 = 50 kg
m
T1 = 100º = 373 K                           T2 = 60º C = 333 K
T0 = 288 K
m T + m2 T2
= 357
...
62

⎡
⎛ 373 ⎞ ⎤
= m cP ⎢(373 − 357
...
62 ⎠ ⎦
⎣
= 1088
...
62
T ⎞
⎛
Ain = ∫ m c p ⎜1 − 0 ⎟ dT
T⎠
⎝
333
⎡
⎛ 357
...
62 − 333) − 288 ln ⎜
⎟
⎝ 333 ⎠ ⎥
⎣
⎦
= 853
...
4 – 853
...
8 kJ

∴

Q8
...
2 MJ of
electrical energy
...

Let compressed air be considered for doing an equivalent amount of
work in starting the car
...
What is the volume of the tank that would be required to let the
compressed air have an availability of 5
...
287 T,
where T is in K, p in kPa, and v in m3/kg
...
0
...
2 MJ = 5200 kJ
Availability of compressed air
= u1 – u0 – T0 ( s1 – s0 )
A air

= m cv (T1 – T0) – T0 ( s1 – s0 )

( s1

– s0 ) = cv ln

W=

T0 R ln

p1
v
+ cp ln 1
p0
v0

= c p ln

T1
p
− R ln 1
p0
T0

p1
p0

⎛ 7000 ⎞
= 298 × 0
...
36 kJ/kg

Here T1 = T0 = 25º C = 298 K
Let atm

Given p1 = 7 MPa = 7000 kPa
Page 101 of 265

pr = 1 bar = 100 kPa

Availability & Irreversibility

By: S K Mondal

Chapter 8

5200
kg = 14
...
36
Specific volume of air at 7 MPa, 25ºC then
RT
0
...
012218 m3/kg
7000
p
∴ Required storage volume (V) = 0
...
8

Solution:

Required mass of air =

Ice is to be made from water supplied at 15°C by the process shown in
Figure
...
Determine the minimum work required to produce 1000 kg of ice
...
187 kJ/kg K, cp for ice = 2
...

(Ans
...
37 MJ)
Let us assume that heat rejection temperature is (T0)
(i) Then for 15ºC water to 0º C water if we need WR work minimum
...

When temperature is T if dT temperature decreases
dQ 2 = – mc p ice dT
∴
T0
⎞
− 1⎟
⎝T
⎠

∴

dWR = − m c p ice dT ⎛
⎜

∴

WRII = m c p ice

273

⎛ T0

∫⎜T
⎝

263

⎞
− 1 ⎟ dT = m c p ice
⎠

1
4
...
187 ⎡
273
⎤
= 1000 ×
⎢ T0 ln 263 − 10 ⎥
2 ⎣
⎦
273
⎡
⎤
− 10 ⎥ kJ
= 2093
...
9

Solution:

273
⎡
⎤
⎢T0 ln 263 − (273 − 263) ⎥
⎣
⎦

WR = (i) + (ii) + (iii)
= [1529
...
5ºC = 295
...
6 kJ = 33
...
4 MPa, 175°C
...

Calculate the availability in the initial and final states and the
irreversibility of this process
...

(Ans
...
6 kJ/kg, 222 kJ)
Given Ti = 175ºC = 448 K
Tf = 25ºC = 298 K
Vf = 1 m3
Vi = 1 m3
pi = 1
...
25 kPa
Calculated Data:
p 0 = 101
...
005 kJ/kg – K, cV = 0
...
287 kJ/kg – K
∴

Mass of air (m) =

pi Vi
1400 × 1
= 10
...
287 × 448
Page 103 of 265

Availability & Irreversibility

By: S K Mondal
∴
∴

Final volume (V0) =

Chapter 8
mRT0
p0

=

10
...
287 × 298
= 9
...
325

Initial availability
Ai = φ1 – φ0
= u1 – u0 – T0 ( s1 – s0 ) + p0 (V1 – V0)
V
p ⎫
⎧
= mc v (T1 - T0 ) - T0 ⎨mc p ln 1 + mc v ln 1 ⎬ + p0 (V1 - V0 )
V0
p0 ⎭
⎩
⎡
1
= m ⎢0
...
005 ln
9
...
718 ln
+ 101
...
1907) ⎥ kJ
101
...
58 kJ = 133
...
25 kPa and Tf = T0 ⎥
f
⎣
⎦
V
p ⎫
⎧
= 0 − T0 m ⎨cP ln f + cv ln f ⎬ + p0 (Vf − V0 )
V0
p0 ⎭
⎩
= (2065
...
92) kJ
= 1235
...
5 kJ/kg
Irreversibility = Loss of availability
= (1458
...
8) kJ = 222
...
10

Air flows through an adiabatic compressor at 2 kg/s
...
Compute
the net rate of availability transfer and the irreversibility
...

(Ans
...
1 kW and 21
...
7794 m3/s V f =
= 0
...
25 – 10
...
14 kW
i

Actual work required= m(h2 – h1 )
W = 2 × 251
...
5 kW

∴

Q8
...
– Wmin
...
5 – 481
...
36 kW

An adiabatic turbine receives a gas (cp = 1
...
838 kJ/kg K) at 7
bar and 1000°C and discharges at 1
...
Determine the
second law and isentropic efficiencies of the turbine
...

(Ans
...
956, 0
...
252

p1 = 7 bar = 700 kPa

T1
T

( c − cv ) T1
RT1
v1 =
= p
p1
p1
0
...
45828 m3/kg

∴

T2 = 938 K

T2
T2′

T0 = 298 K

RT2
= 1
...
09 × (1273 – 938) kW = 365
...
57584
+ 1
...
43828
= 0
...
055326
T2
T
1 × 1
...
055326
′
T2
T2
= 1
...
6 K
T2 =
=
1
...
05207 )
Page 105 of 265
= 1 × 0
...
5 bar = 150 kPa
v2 =

1

Availability & Irreversibility

By: S K Mondal

Chapter 8

Isentropic work = h1 − h′2 = m c p (T1 − T2′ )
= 3 × 1
...
6) kW = 415
...
15
= 87
...
75
Change of availability
ΔA = A1 – A2
= h1 – h2 – T0(S1 – S2 )
= mc P ( T1 – T2 ) + T0 ( S2 – S1 )
= 1 × 1
...
055326) kW= 381
...
15
= 95
...
64

∴ ηII =

Q8
...
5 bar
...
01 kg/s and the
efficiency of the compressor is 75%
...
Calculate
(a) The power required to drive the compressor
(b) The rate of irreversibility for the overall process (compressor and
cooler)
...
(a) 2
...
01 kg/s
RT
v1 = 1 = 0
...
5 bar = 550 kPa

288 K

T
1

S

For minimum work required
to compressor is isentropic
γ( p2 V2 − p1 V1 )
Wisentropic =
γ −1
γ −1
⎡
⎤
γ
⎛ p2 ⎞ γ
⎢
=
RT1 ⎜ ⎟
− 1⎥
⎢⎝ p1 ⎠
⎥
γ −1
⎣
⎦
0
...
4
1
...
287 × 288 ⎜
=
⎟
⎢⎝ 100 ⎠
⎥
0
...
55 kW/kg

Availability & Irreversibility

By: S K Mondal

Chapter 8

181
...
75
∴ Power required driving the compressor

Wact =
(a)

•

= m Wact = 2
...
55) = 60
...
85
60
...
25 K
1
...
25

=

∫

313

288 ⎞
⎛
1 × 1
...
25 ⎞ ⎫
= 1
...
21 − 313) − 288 ln ⎜
⎟ ⎬ kJ/kg
⎝ 313 ⎠ ⎭
⎩
= 65
...
85 + 65
...
15 kJ/kg
∴
Power loss due to irreversibility = 1
...
13

In a rotary compressor, air enters at 1
...
6 bar, 250°C
...
The atmosphere is at
1
...
Neglect the K
...
changes
...
19 kJ/kg, 0
...
1 bar = 110 kPa

p2

T1 = 294 K
p 2 = 6
...
287 ln ⎜
= ⎢1
...
064647 kJ/kg – K = 64
...
005 (523 – 294) – 293 × 0
...
2 kJ/kg
Page 107 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8

Actual work required (Wact ) = 230
...
064647 = 18
...
14

Solution:

In a steam boiler, the hot gases from a fire transfer heat to water which
vaporizes at a constant temperature of 242
...
5 MPa)
...
046
kJ/kg K over this temperature range
...
5 MPa is 1753
...
If the steam generation rate is 12
...
Take T0 = 21°C
...
(a) 22096 kW, (b) 15605
...
0 kW, (d) 20
...
6 × 1752
...
097 MW
•

If mass flow rate at gas is mg
•

Then mg cPg (1100 – 430) = 22097
or

•

mg = 31
...
15

Solution:

294 ⎞
⎛
•
mg cPg ⎜1 −
⎟ dT
T ⎠
⎝
703

∫

⎡
⎛ 1373 ⎞ ⎤
•
= mg cPg ⎢(1373 − 703) − 294 ln ⎜
⎟⎥
⎝ 703 ⎠ ⎦
⎣
= 15606 kJ/s = 15
...
497 MW
515
...
779 kW/K
An economizer, a gas-to-water finned tube heat exchanger, receives 67
...
0046 kJ/kg K, and 51
...
186 kJ/kg K
...
There are no changes of kinetic energy
and p0 = 1
...
Determine:
(a) Rate of change of availability of the water
(b) The rate of change of availability of the gas
(c) The rate of entropy generation
(Ans
...
2 kW, (b) 7079
...
73 kW/K)
T ⎞
⎛
(a) Rate of charge of availability of water = Q ⎜1 − 0 ⎟
⎝
T⎠
469
•
289 ⎞
⎛
= ∫ m w c p w dT ⎜1 −
⎟
T ⎠
⎝
402
469 ⎤
⎡
= 51
...
186 × ⎢(469 − 402) − 289 ln
kW
402 ⎥
⎣
⎦
Page 108 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8

= 4
...
5 × 1
...
0798 MW
∴ (c)

•

Rate of irreversibility (I) = 2
...
8808 kW/K
∴
Entropy generation rate Sgas =
T0
The exhaust gases from a gas turbine are used to heat water in an
adiabatic counter flow heat exchanger
...
The flow rates of the gas and water are
0
...
50 kg/s respectively
...
09 and 4
...
Calculate
the rate of exergy loss due to heat transfer
...

(Ans
...
5 kW)
Tgi = 260º C = 533 K
Tw i = 65ºC = 338 K
Tw o = 365
...
16

Solution:

•

•

mg = 0
...
09 kJ/kg – K

m w = 0
...
186 kJ/kg – K

To = 35º C = 308 K
To calculate Two from heat balance
•

•

mg cPg (Tgi − Tgo ) = m w cPw (Two − Twi )
∴

Two = 365
...
115 kW
⎝ 393 ⎠ ⎥
⎣
⎦
Rate of gain of availability of water
•
⎡
⎛ 365
...
7 − 338) − 308 ln ⎜
⎟ ⎥ = 7
...
916 kW

Q8
...
12 bar, 800 K flows steadily into a
heat exchanger which cools the gas to 700 K without significant pressure
drop
...
The mass flow rate
of air is twice that of the gas and the surroundings are at 1
...

Determine:
(a) The decrease in availability of the exhaust gases
...

(c) What arrangement would be necessary to make the heat transfer
reversible and how much would this increase the power output of
Page 109 of 265

Availability & Irreversibility

By: S K Mondal

Solution:

Chapter 8

the plant per kg of turbine gas? Take cp for exhaust gas as 1
...
05 kJ/kg K
...

(Ans
...
0731 kJ/kg K, (c) 38
...
08 ⎢(800 − 700) − 293 ln ⎜
⎟⎥
T ⎠
⎝
⎝ 700 ⎠ ⎦
⎣
700
= 65
...
F
...
F
...
5 K
Availability increases
521
...
05 × ⎢(521
...
257 kJ/kg
470 ⎥
⎣
⎦
•

∴ Sgas = 73
...
08 ln
700
⎛ 470 ⎞
= −2 m × 1
...
068673
470
∴
To = 503
...
08(800 – 700) = 108 kJ/kg
Q2 = 2m × 1
...
4 – 470) = 70
...
e
...
84 kJ/kg of gas flow
or

Q8
...
The rate of
flow of the products is 10 kg/s, and the products are cooled from 300°C to
Page 110 of 265

Availability & Irreversibility

By: S K Mondal

Solution:

Chapter 8

200°C, and for the products at this temperature cp = 1
...
The
rate of air flow is 9 kg/s, the initial air temperature is 40°C, and for the
air cp = 1
...

(a) What is the initial and final availability of the products?
(b) What is the irreversibility for this process?
(c) If the heat transfer from the products were to take place reversibly
through heat engines, what would be the final temperature of the
air?
What power would be developed by the heat engines? Take To = 300 K
...
(a) 85
...
68 kJ/kg, (b) 256
...
41 K, 353
...
09 (573 – 473) = 9 × 1
...
5 K

Or
(a)

Initial availability of the product
573 ⎤
⎡
= c p g ⎢(573 − 300) − 300 ln
300 ⎥
⎣
⎦
= 85
...
68 kJ/kg of product
300 ⎥
⎣
⎦
∴
Loss of availability = 46
...
5 ⎞ ⎤
= c p g ⎢(433
...
907 kJ/kg of air
⎝ 313 ⎠ ⎥
⎣
⎦
(b)

∴ Rate of irreversibility
•

I = (10 × 46
...
907 × 9) kW= 256
...
09 ln ⎜
10 × 1
...
4 = 399
...
005 × ln

or

•

•

∴ Q1 = mg c p g (Ti − Tf ) = 1090 kJ
•

Q2 = ma c p a (394
...
263 kJ
∴

Q8
...
74 kW output of engine
...
2 kJ of heat from a reservoir at 120°C
...
98 bar, 27°C
...

(Ans
...
5 kJ)
Maximum work from gas
= u1 – u2 – T0 ( s1 – s2 )
= m cv (T1 − T2 )
T
p ⎤
⎡
− T0 ⎢m cP ln 1 − mR ln 1 ⎥
T2
p2 ⎦
⎣
⎡
⎡
343
⎛ 3 ⎞⎤ ⎤
= 2 ⎢0
...
005 ln
− 0
...
07 kJ
Work done on the atmosphere = p0 (V2 – V1)

T
T ⎤
⎡
= 98 ⎢mR 0 − mR 1 ⎥
p2
p1 ⎦
⎣
T ⎤
⎡T
= 98 mR ⎢ 2 − 1 ⎥
⎣ p2 p1 ⎦
= 111
...
20

1

Q = 1
...
The compressor has an efficiency of 82%
...
For air, use

T2 s

⎛p ⎞
=⎜ 2⎟
T1 ⎝ p1 ⎠

γ −1/ γ

Where T2s is the temperature of air after isentropic compression and γ =
1
...
The compressor efficiency is defined as (T2s – T1) / (T2 – T1), where T2
is the actual temperature of air after compression
...
(a) 180
...
5 kJ/kg (c) 21 kJ/kg)
Solution:

Page 112 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8
2

p1 = 1 bar = 100 kPa
T1 = 30º C = 303 K

2S
T

p2 = 4 bar = 400 kPa
T2 = ? ηcom = 6
...
4
⎧
⎫
⎪
1
...
287 × 303 ⎪⎛ 400 ⎞ 1
...
92 kJ/kg
⎟
⎪
⎪
(1
...
92
= 180
...
82
′
∴ Extra work 32
...
21

Solution:

′
⎛ p′ ⎞ γ
T2
= ⎜ 2⎟
T1
⎝ p1 ⎠
′
32
...
6 K
Irreversibility (I) = (180
...
92) kJ/kg = 32
...
3 K

A mass of 6
...
Heat is transferred
to the air from a reservoir at 727°C
...
The environment is at 100 kPa, 17°C
...

(Ans
...
5, 621
...
005 m3
P1
Vo = 5
...
005 m3
m = 6
...
98 × 0
...
98 × 290

Page 113 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8

⎡
300
⎛ 200 ⎞ ⎤
−6
...
005 ln
− 0
...
005– 5
...
4 kJ

Final availability
T2
p ⎤
− R ln 2 ⎥ + p0 (V2 − V0 )
T0
p0 ⎦
⎣
= 6
...
718(600 – 290) – 6
...
287 ln ⎜
× ⎢1
...
005 − 5
...
5 kJ

Af = m

(b)

⎡

c v (T2 – T0) – mT0 ⎢cP ln

Maximum useful work
= u2 – u1 – T0 ( s2 – s1 ) + p 0 (V2 – V1)
T
p ⎤
⎡
= m cv (T2 − T1 ) − T0 m ⎢c p ln 2 − R ln 2 ⎥ + p0 (V2 − V1 )
T1
p1 ⎦
⎣
= 6
...
718(600 – 300) – 300
⎡
600
⎛ 400 ⎞ ⎤
× 6
...
005 ln
− 0
...
35 kJ

Heat transfer to the vessel
mRT
V
= m cv (T2 – T1) = 6
...
718 × (600 – 300) kJ

Q=

∫m c

v

dT

p=

= 1503
...
402 ⎜1 −
⎟
1000 ⎠
⎝
= 1067
...
22

Solution:

Air enters a compressor in steady flow at 140 kPa, 17°C and 70 m/s and
leaves it at 350 kPa, 127°C and 110 m/s
...
Calculate per kg of air
(a) The actual amount of work required
(b) The minimum work required
(c) The irreversibility of the process
(Ans
...
4 kJ, (b) 97
...
1 kJ)
Minimum work required
T2 = 127ºC = 400 K
T1 = 290 K
T0 = 280 K
2
2
V − V1
w = h 2 − h1 − T0 (s2 − s1 ) + 2
2000
T
V 2 − V12
p ⎤
⎡
= m cP (T2 − T1 ) − mT0 ⎢cP ln 2 − R ln 2 ⎥ + 2
T1
2000
p1 ⎦
⎣
Page 114 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8

400
350 ⎤
⎡
− 0
...
005(400 − 290) − 1 × 280 ⎢1
...
55 – 16
...
6 = 97
...
1102 − 702
kJ
2000

Actual work required
V 2 − V12
h 2 − h1 + 2
=
= (110
...
6) kJ = 114
...
86 kJ/kg
Q8
...
The environment is at 100 kPa, 17°C
...
(a) 159 kJ, (b) 109 kJ, (c) 50 kJ)
Maximum work output
V 2 − V22
w = h1 − h 2 − T0 (s1 − s2 ) + 1
2000
T1
p ⎫ V 2 − V22
⎧
= CP (T1 − T2 ) − T0 ⎨CP ln
− R ln 1 ⎬ + 1
T2
p2 ⎭
2000
⎩

{

= 1
...
005 ln

}

400
500 1502 − 702
− 0
...
41 kJ/kg
V12 − V22
= 100
...
8 = 109
...
109 kJ/kg
Calculate the specific exergy of air for a state at 2 bar, 393
...
15 K
...
287 kJ/kg K
...
72
...
24
Solution:

Q8
...
15
⎛ 2 ⎞⎤
− 0
...
15 − 293
...
15 ⎢1 × ln
293
...
28 kJ/kg
Calculate the specific exergy of CO2 (cp = 0
...
1889 kJ/kg K)
for a state at 0
...
15 K and for the environment at 1
...
15 K
...
– 18
...
8659 (268
...
15) − 293
...
8659 ln

p0 = 100 kPa
T0 = 293
...
15
70
− 0
...
15
100

p1 = 70 kPa, T1 = 68
...
772 kJ/kg

Q8
...
Because
of poor thermal insulation the brine temperature increases from 250 K at
the pipe inlet to 253 K at the exit
...
Take T0 = 293 K and cp = 2
...

(Ans
...
05 kW)

Solution:

1

2

T
...
85 kJ/kg – K
Entropy generation rate
•

•

Sgas = Ssys

•

Q
−
T0
•

•

= m (S2 − S1 ) −

m cP (253 − 250)
T0

•
3⎤
⎡ T
= m cP ⎢ ln 2 − ⎥ kW/ K
⎣ T1 T0 ⎦
= 0
...

T
S2 – S1 = c p ln 2
T1
•

∴

I = rate of energy degradation
= rate of exergy loss
•

To Sgen = 293 × 0
...
0548 kW

Q8
...
However, there is a loss of exergy
...
What is the rate of exergy loss due
to throttling?
Page 116 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8
i
i
p1 ⎞
⎛
⎜ Ans
...

•

= ( ΔS)sys + 0
•

= m(S2 − S1 )
•
⎛p ⎞
= m R ln ⎜ 1 ⎟
⎝ p2 ⎠
(as no heat interaction with surroundings)
TdS = dh – Vdp
dh
dp
V
mR
or dS =
−V
=
p
T
T
T
dp
dS = 0 − mR
p
2

or

S2 – S1 = − ∫ mR
1

∴

•
m kg/s

p1, T1

p2, T2

p
p
dp
= − mR ln 2 = mR ln 1
p
p1
p2
•

Irreversibility rate (I)
•

= T0 × Sgen

⎛p ⎞
= T0 × mR ln ⎜ 1 ⎟
⎝ p2 ⎠
⎛p ⎞
= mR T0 ln ⎜ 1 ⎟
⎝ p2 ⎠
Q8
...


Solution:

Air at 5 bar and 20°C flows into an evacuated tank until the pressure in
the tank is 5 bar
...

(a) What is the final temperature of the air?
(b) What is the reversible work produced between the initial and final
states of the air?
(c) What is the net entropy change of the air entering the tank?
(d) Calculate the irreversibility of the process
...
(a) 410
...
9 kJ/kg,
(c) 0
...
9 kJ/kg)
If m kg of air is entered to the tank then the enthalpy of entering fluid is equal to
internal energy of tank fluid
...
4 × 293 K = 410
...
29

v1 =

RT1
= 0
...
The
expansion and compression processes have a pressure ratio of 50
...

(Ans
...
965)

Solution:

Let Q1 amount of heat is in input
...
7 Q1
1000 ⎠
⎝
If there is no temperature differential between inlet and outlet then from Q1 heat
input Carnot cycle produce work
...
72549 Q1
1020 ⎠
⎝
W
0
...
965
∴
Second law efficiency ( ηII ) =
0
...
30

Solution:

Energy is received by a solar collector at the rate of 300 kW from a
source temperature of 2400 K
...

(Ans
...
80, 0
...
8
300
Page 118 of 265

Availability & Irreversibility

By: S K Mondal

Chapter 8

300 ⎞
⎛
(300 − 60) ⎜1 −
⎟
600 ⎠ = 0
...
31

For flow of an ideal gas through an insulated pipeline, the pressure
drops from 100 bar to 95 bar
...
5 kg/s and
has cp = 1
...
718 kJ/kg-K and if T0 = 300 K, find the rate of
entropy generation, and rate of loss of exergy
...
0
...
46 kW)

Solution:

1

2

•
m = 1
...
005 kJ/kg – K
cv = 0
...
5 × 0
...
022082 kW/K

•

Rate of loss of exergy = Irreversibility rate (I)
•

To Sgen = 300 × 0
...
6245 kW

Q8
...
Expansion takes place through a volume ratio of 9 according to
pv1
...
The surroundings are at 20°C, 1
...
Determine the loss of
availability, the work transfer and the heat transfer per unit mass
...
26 kl/kg-K and cv = 0
...

(Ans
...
1243 m3/kg (calculating)
T1 = 2500ºC = 2773 K
RT
∴ v1 = m 1 = 0
...
1 bar = 110 kPa

S
p2 = 279
...
11876 m3/kg
T2 = 1203
...
08 kJ/kg

W = 0
...
26 kJ/kg – K

1
...
38

n −1

T2
1
⎛v ⎞
= ⎜ 1⎟
= 0
...
38 = 1203
...
82(2773 – 1203
...
08 ln
− 0
...
2
⎝ 279
...
1243 – 1
...
2 + 1074 = –213
...
33

In a counterflow heat exchanger, oil (cp = 2
...
2 kJ/kg K) is heated from 290 K to
temperature T
...
Neglecting pressure drop, KE and PE effects and heat loss,
determine
(a) The temperature T
(b) The rate of exergy destruction
(c) The second law efficiency
Take T0 = I7°C and p0 = 1 atm
...
(a) 305 K, (b) 41
...
9%)
Solution:

440 K

c p = 2
...
2 kJ/kg – K
3200 kg/K

320 K

∴ T = 290 + 15 = 305 K

200 K
•

•

•

T0 = 17°C = 290 K
p0 = 1 m = 101
...
1 × ln 440 + 3600 × 4
...
039663 kW/K = 39
...
039663 kW

= 11
...
4 MJ/K
(c)

Availability decrease of oil
= A1 – A2 = h1 – h2 – T0 ( s1 – s2 )

T ⎤
•
⎡
= m0 c p0 ⎢(T1 − T2 ) − T0 ln 1 ⎥
T2 ⎦
⎣
800
440 ⎤
⎡
=
× 2
...
903 kW
Availability decrease of water
A1 – A2 = h1 – h2 – T0 ( s1 – s2 )

∴

T ⎤
•
⎡
= m w c p w ⎢(T1 − T2 ) − T0 ln 1 ⎥
T2 ⎦
⎣
3200
305 ⎤
⎡
=
× 4
...
4 kW
3600
290 ⎥
⎣
⎦
Gain of availability
1
...
85%
Loss for that
12
...


Chapter 9

Properties of Pure Substances
Some Important Notes

1
...


It is a Point
...
16 K
= 0
...


5 atm

And T = 216
...
45º C that so why
sublimation occurred
...
2 bar ≈ 225
...
15ºC ≈ 647
...
00317 m3/kg
At critical point
h fg = 0;

4
...


p = 0
...
587 mm of Hg

v fg = 0;

Sfg = 0

Mollier Diagram

⎡∵ TdS = dh − vdp⎤
⎢
⎥
⎢∴ ⎛ ∂ h ⎞ = T
⎥
⎜
⎟
⎢ ⎝ ∂ S ⎠p
⎥
⎣
⎦
∴ The slope of an isobar on the h-s co-ordinates is equal to the absolute saturation
temperature at that pressure
...

⎛∂h⎞
Basis of the h-S diagram is ⎜
⎟ =T
⎝ ∂ S ⎠P

Page 123 of 265

Properties of Pure Substances

By: S K Mondal
5
...


s = sf + x sfg

7
...


8
...


9
...
K
...
1

Complete the following table of properties for 1 kg of water (liquid,
vapour or mixture)

Solution:

p bar
a
b
c
d
e
f
g
h
i
j

tºC

v m3/kg

x/%

Superheat
0ºC
0
0
0
0
140
87
...
6
50
201
...
0563
35
25
...
0135
100º
0
...
42 0
...
6
1
...
27
10
320
0
...
8ºC
0
...
6
0
...
23
40
500
0
...
4ºC
0
...
203
100
Calculations: For (a) ………… For (b)
h = hf + x hf g
For (c) v = v f + x(v g − v f )
For (d) s = sf + x sf g

∴x=

h kJ/kg
2565
...
04
2608
...
3
3093
...
1
2635
...
3
2932
...
8

⇒ s= sf + x sf g

s − sf
= 0
...
258 +

20
( 0
...
258 ) ,
50

20
(3157
...
2) ) = 3093
...
123 +
(7
...
123 ) = 7
...
4646 − 0
...
8º C
0
...
425
38
...
4 +
(2960
...
4) = 2937
...
2 +

For (f)

s kJ/
kg – K
8
...
307
5
...
104
7
...
2235
6
...
090
6
...
2690

Page 125 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9

38
...
271 − 7
...
4400 = 0
...
462 – 0
...
7 + x × 2133, s = 1
...
1179 = 6
...
4
t = 262
...
111 +
(0
...
111),
50
12
...
5 +
(3023
...
5) = 2932
...
9
(6
...
545) = 6
...
545 +
50
s = 7
...
2

(a) A rigid vessel of volume 0
...
Evaluate the specific volume, temperature, dryness fraction,
internal energy, enthalpy, and entropy of steam
...
Show the
process on a sketch of the p–v diagram, and evaluate the pressure,
increase in enthalpy, increase in internal energy, increase in
entropy of steam, and the heat transfer
...

(Ans
...
86 m3/kg, 120
...
97, 2468
...
54 kJ/kg, 6
...
3 bar, 126 kJ/kg, 106
...
2598 kJ/kg K, 106
...
86 m3
= 0
...
Volume = 0
...
2º C
v − vf
→ v = v f + x(v g − v f )
∴x=
vg − v f

(a) → Specific volume = Volume/mass =

0
...
001061
= 0
...
885 − 0
...
86 = 2472 kJ/kg
→ Here h = h f + x h f g = 504
...
97172 × 2201
...
5301 + 0
...
5967 = 6
...
86 m3/kg

Page 126 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9

S
p

V
0
...
86
+ (2 + 0 − 2) = 2
...
885 − 0
...
86 m3/kg
0
...
8 − 120
...
2 = 121
...
23 K
0
...
0641 = 2
...
23
From Molier diagram ps = 2
...
095
∴
Δh = 127 kJ/kg, Δs = 0
...
5 kJ/kg
TS =

Q9
...
Find the change in volume,
enthalpy, internal energy and entropy
...
2
...
5 kJ, 26047
...
842 kJ/K)

Solution:

2
30°C

p

1

2

T
1

m = 10 kg
V

At state (1)
p1 = 10 bar = 1000 kPa
T1 = 45ºC = 318 K

S
ΔS

At state (2)
p2 = p1 = 10 bar
T2 = 300ºC
For Steam Table
Page 127 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9

v1 = 0
...
4 kJ/kg
u1 = h1 − p1 v1 = 187
...
693 kJ/kg – K

∴

Change in volume

Enthalpy change
Internal Energy change
Entropy change
Q9
...
258 m3/kg
u2 = 2793
...
2 kJ/kg
s2 = 7
...
57 m3
= m(h 2 − h1 ) = 28
...
0581 MJ
= m ( s2 – s1 ) = 64
...
At a section downstream where the
pressure is 3 bar, the quality is to be 95%
...

(Ans
...
67 kg/h)

Solution:

3
1

2

m1
p1 = 5 bar
= 500 kPa
T1 = 300°C
h1 = 3064
...
5°C
h2 = 561
...
95 × 2163
...
44 kJ/kg

T3 = 40º C
h 3 = 167
...
2 − 2616
...
44 − 167
...
23 kg/hr
...
5

A rigid vessel contains 1 kg of a mixture of saturated water and
saturated steam at a pressure of 0
...
When the mixture is heated,
the state passes through the critical point
...

(Ans
...
003155 m3, (b) 0
...
0018 kg,
(c) 233
...
46 kJ/kg)
Solution:
3
30 bar

2

p

p1 = 1
...
So the volume of
the vessel is critical volume of water = 0
...
00317 − 0
...
159 − 0
...
0018282 kg
∴ Mass of water = 0
...
e
...
8ºC
...
00317 − 0
...
029885
0
...
001216

∴ Q = (1008
...
029885 × 1793
...
1 + 0
...
2) – 3000 (0
...
029885 (0
...
001216)) + 150(0
...
001828 (1
...
0018282))
= 581
...
6

A rigid closed tank of volume 3 m3 contains 5 kg of wet steam at a
pressure of 200 kPa
...
Determine the final pressure and the heat transfer to the
tank
...
304 kPa, 3346 kJ)

Page 129 of 265

Properties of Pure Substances

By: S K Mondal
Solution:

Chapter 9

V1 = 3 m3
m = 5 kg
∴

∴

3
= 0
...
6 − 0
...
67758
vg − v f
(0
...
001061)

v1 =

h1 = h f + x1 h fg = 504
...
67758 × 2201
...
5 kJ/kg
u1 = h1 − p1 v1 = 1996
...
6 = 1876
...

∴
v g2 = 0
...
606 m3/kg
vg = 0
...
006
= 303
...
019

∴
∴
∴
Q9
...
6 m3

u2 = h2 − p2 v 2 = 2725 – 303
...
6 = 2543 kJ/kg
Heat supplied Q = m(u2 − u1 ) = 3333 kJ

p2 = 300 ×

Steam flows through a small turbine at the rate of 5000 kg/h entering at
15 bar, 300°C and leaving at 0
...
The steam enters
at 80 m/s at a point 2 in above the discharge and leaves at 40 m/s
...
How much error
would be made if these terms were neglected? Calculate the diameters of
the inlet and discharge tubes
...
765
...
44%, 6
...
9 cm)
5000
•
kg/s
m = 5000 kg/hr =
3600
p1 = 15 bar
t1 = 300º C

1

2m
Z0
∴ From Steam Table
h1 = 3037
...
1 bar
(100 − 4)
x2 =
= 0
...
8º C
Page 130 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9

v1 = 0
...
8 + 0
...
8 = 2489 kJ/kg
∴
V2 = 40 m/s, Z2 = Z0 m

v 2 = 14
...
81(2) ⎤
(3037
...
45 kW
If P
...
and K
...
is neglected the

∴

•

∴

W′ = m(h1 − h 2 ) = 762
...
44%
W
•

Area at inlet (A1) =

mv1
= 0
...
34 cm2
V1
∴ d1 = 6
...
8

Solution :

mv 2
= 0
...
9 cm

A sample of steam from a boiler drum at 3 MPa is put through a
throttling calorimeter in which the pressure and temperature are found
to be 0
...
Find the quality of the sample taken from the
boiler
...
0
...
1 MPa = 1 bar
t 2 = 120º C
20
(2776
...
2)
h 2 = 2676
...
3 kJ/kg
1
2
∴
h1 = h 2
h
∴
h1 = 2716
...
9

h1 − h f1
h fg1

S

2716
...
4
= 0
...
9

It is desired to measure the quality of wet steam at 0
...
The quality
of steam is expected to be not more than 0
...

(a) Explain why a throttling calorimeter to atmospheric pressure will
not serve the purpose
...
0
...
Then minimum temperature required
t = tsat + 5ºC = 100 + 5 = 105º C
Then Enthalpy required
5
(2776
...
5 MPa = 5 bar dryness fraction is < 0
...
9 hfg = 640
...
9 × 2107
...
76 kJ/kg
So it is not possible to give 5º super heat or at least saturation i
...
(2676 kJ/kg) so
it is not correct
...
1
∴x=
= 0
...
4

Q9
...
55 kg
Mass of steam condensed after passing through the throttle valve –4
...

(Ans
...
85)
p1 = 15 bar = p2
T1 = 198
...
3 kJ/kg

1

2

1

2

1 bar
t = 120°C

3

3
4
...
55 kg
h 2 = h 2f + x 2 × h fg2 = 844
...
2
∴
∴

x 2 = 0
...
2

Page 132 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9

x 2 × 4
...
96216 × 4
...
85
=
4
...
55
4
...
55
h1 = h f 1 + x h fg1 = 844
...
85 × 1945
...
6 kJ/kg
=

But at 1 bar minimum 5º super heat i
...
105ºC enthalpy is 2686 kJ/kg
So it is not possible to calculate only by throttling calorimeter
...
11

Solution:

Steam from an engine exhaust at 1
...
The calorimeter has
two 1 kW heaters and the flow is measured to be 3
...
Find the
quality in the engine exhaust
...
0
...
921)
30
(2776
...
2) = 2736
...
2 +
50
•

•

•

m h1 = m h 2 − Q

1 1
...
4 kg
•
m =
5 mm
= 0
...
25 bar: from Steam Table
At 1
...
4 kJ/kg
At 1
...
2 kJ/kg
At 1
...
3 kJ/kg;
If dryness fraction is x
Then 2560 = 444
...
9441
If outlet temperature is 105º C then
h2 = 2686 kJ/kg

hfg = 2244
...
8 kJ/kg
hfg = 2241 kJ/kg

(then from problem 9
...
53 kJ/kg

Then if dryness fraction is x2 then
∴ x 2 = 0
...
3 + x 2 × 2241
Q9
...

The steam flow rate is 1 kg/s
...
Neglect the velocity of steam
at the inlet to the nozzle
...
The cooling water enters the condenser at 25°C and
leaves at 35°C
...

(Ans
...
0101 m2, 47
...
6 kJ/kg
t 2 = 45
...
925 kJ/kg-K
s2 = 6
...
8 kJ/kg
v 2 = 12
...
1 bar

∴
∴
∴

t1 = 250°C m = 1 kg/s
V1 = 0
V2
s1 = s2 = 0
...
501
∴ x = 0
...
8 + 0
...
8 = 2193
...
13

Solution:

mv 2
= 100
...
8 – 191
...
187 (35 – 25)
m = 47
...
The exponent n has the value 1
...
Find
the final specific volume, the final temperature, and the heat transferred
per kg of fluid
...
206 m3/s
h1 = 2827
...
15
∴
v 2 = 1 1 = ⎜ 1 ⎟
...
206 = 1
...
694 m
∴
1
...
001043 + x (1
...
001043)
∴
x = 0
...
6º C
= u1 − u2
= (h1 − h 2 ) − (p1 v1 − p2 v 2 )
Page 134 of 265

Properties of Pure Substances

By: S K Mondal
[h 2 = h f 2

Chapter 9

= (2827
...
8) – (1000 × 0
...
5256)
= 323
...
5 + 0
...
9
= 2450
...
27 kJ/kg
Work done (W) = 1 1
n −1
∴
From first law of thermo dynamics
Q2 = (u2 − u1 ) + W1 – 2
= (–323
...
27) = 32
...
14

Solution:

Two streams of steam, one at 2 MPa, 300°C and the other at 2 MPa, 400°C,
mix in a steady flow adiabatic process
...
Evaluate the final
temperature of the emerging stream, if there is no pressure drop due to
the mixing process
...
Determine the exit
velocity of the stream and the exit area of the nozzle
...
340°C, 0
...
77 cm2)
p2 = 2 MPa = 20 bar
t2 = 400º C
•

m2 = 2 kg/min
h2 = 3247
...
127 kJ/kg-K

1
3

p1 = 2 MPa = 20 bar
t1 = 300°C
•
m 1 = 3 kg/min

2

For Steam table
h1 = 3023
...
766 KJ/kgK

•
•
•
m 3 = m 1 + m2 = 5 kg/min
p = 20 bar
3
40
s3 = 6
...
956 – 6
...
918 kJ/kg – K

For adiabatic mixing process
•

•

•

m1 h1 + m2 h 2 = m3 h 3
∴

h3 = 3113
...
14 − 3023
...
5
Rate of increase of the enthalpy of the universe

∴

•

Final temperature (t) = 300 +

•

•

•

sgen = m3 S3 − m1 S1 − m2 S2 = 0
...
918 = 0
...
870
∴ x = 0
...
3 + 0
...
9 = 1937
...
14 − 1937
...
3 m/s
Page 135 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9
•

Q9
...
0054 m2 = 54 cm2
V
Boiler steam at 8 bar, 250°C, reaches the engine control valve through a
pipeline at 7 bar, 200°C
...
1 bar, 0
...
Determine per kg of steam
(a) The heat loss in the pipeline
(b) The temperature drop in passing through the throttle valve
(c) The work output of the engine
(d) The entropy change due to throttling
(e) The entropy change in passing through the engine
...
(a) 105
...
35 kJ/kg,
(d) 0
...
3657 kJ/kg K)
∴
From Steam Table
h1 = 2950
...
8
h3 = 2844
...
3 kJ/kg
∴

Heat loss in pipe line = (h1 − h 2 ) = 105
...
8º C
hg = 2747
...
4
2855
...
8
× (200 − 151
...
4 − 2747
...
74 = 195
...
74º C (drop)

(c)

p4 = 0
...
9
t4 = 45
...
8 – 2345
...
48 kJ/kg

(d)

∴
(e)

From Steam Table
s2 = 6
...
26 − 151
...
059 − 6
...
8192 +
(200 − 151
...
1494 kJ/kg – K

= 7
...
9 sfga = 0
...
9 × 7
...
4 kJ/kg – K
ΔS = s4 – s3 = 0
...
16

Chapter 9

Tank A (Figure) has a volume of 0
...
The valve is
then opened, and the tanks eventually come to the same pressure,

Which is found to be 4 bar
...
What is the volume of tank B?
(Ans
...
89 m3)
Solution:

t1= 200°C
From Steam table
pa = 15
...
157 – 10–3
V g = 0
...
1 m3

B

Initial

Initial volume of liquid =

10
× 0
...
643 kg
Initial mass of steam = (mg)
90
× 0
...
70666 kg
= 100
0
...
3497 kg
After open the valve when all over per = 4 bar at 200ºC
Then sp
...
534 m3/kg
∴
Total volume (V) = 9
...
534 m3 = 4
...
8927 m3
Q9
...
5 MPa and 250°C, when it undergoes the following
processes:
(a) It is confined by a piston in a cylinder and is compressed to 1 MPa
and 300°C as the piston does 200 kJ of work on the steam
...
Changes in K
...
and P
...
are negligible
...
Then 200 kJ of
shaft work is transferred to the steam, so that its final condition is 1
MPa and 300°C
...
(a) –130 kJ (b) – 109 kJ, and (c) – 367 kJ)

Page 137 of 265

Properties of Pure Substances

By: S K Mondal
Solution:

Initially:
t i = 250ºC
∴ From Steam Table
u i = 2729
...
474 m3/kg

(a)

Chapter 9
pi = 0
...
7 kJ/kg

After compression
p = 1 mPa = 10 bar
T = 300ºC
∴ From S
...
u = 2793
...
2 kJ/kg
and
Winput = 200 kJ
∴ From first law of thermodynamics
Q1 – 2 = m(u2 − u1 ) + W1 – 2
[(2793
...
5) – 200] kJ
= [63
...
3 kJ
i
...
heat rejection to atm
...
2 – 2960
...
5 kJ/kg
[heat rejection]

(c)

–W

Energy of the gas after filling
E1 = u1 kJ/kg = 2729
...
2 kJ/kg
∴ ΔE = E2 – E1
= (2793
...
5) kJ/kg
= 63
...
474 kJ/kg = –237 kJ/kg
∴
From first law of thermodynamic
Q = ΔE + W1 + W2
= (63
...
3 kJ/kg

Page 138 of 265

Properties of Pure Substances

By: S K Mondal
Q9
...
The
valve and the pipeline are well insulated
...
008 kg/s while the coil takes 3
...
The main
pressure is 4 bar, and the pressure and temperature of the steam
downstream of the coil are 2 bar and 160°C respectively
...

(a) Evaluate the quality of steam in the main
...

(Ans
...
97, (b) Not suitable)
(a)

2

2
1

3

3

1

p2 = 4 bar

3
...
8993 kW
1000
p3 = 2 bar; t 2 = 160ºC

t 2 = 143
...
8 +

•

m2 = 0
...
5 − 2768
...
14 kJ/kg

From steady flow energy equation
•

•

•

•

m h 2 + Q = m h3 + 0 : h2 = h3 −

Q
•

m

If dryness fraction of steam x then
h2 = hf2 + x hfg2
or
2676
...
7 + x × 2133

= 2676
...
9714

(b) For throttling minimum enthalpy required 2686 kJ/kg if after throttling 5ºC
super heat and atm
...

Q9
...
Tank A has a
volume of 0
...
5 bar, 200°C
...
35 m3 and contains steam at 6 bar with a quality of 90%
...
If there
is no heat transfer during the process, what is the final pressure?
Compute the entropy change of the universe
...
322
...
1985 kJ/K)
From Steam Table
from Steam Table
t B = 158
...
Enthalpy (h A ) = 2872
...
Enthalpy (h B )
Sp
...
193 m
Page 139 of 265

Properties of Pure Substances

By: S K Mondal

Chapter 9
= (670
...
9 × 2085) = 2547 kJ/kg

A

B

VA = 0
...
5 bar
tA = 200°C

VB = 0
...
9

Sp
...
2 kJ/kg =
Sp
...
(v B ) = v Bf + x ( v Bg – v Bf ) = 0
...
entropy (s) = 7
...
in energy (uB ) = uf + x × ufg = 2376
...
entropy (sB ) = 6
...
4 – 600 × 0
...
74 kJ/kg
ufg = hfg – pfB ( v g – v f )

Q9
...
2341 kg
vB

= 1896
...
61242 kg
vA
∴
From First Law of thermodynamics
U1 = U2
∴ m A u A + mB uB = (m A + m B ) u
∴ u = 2469
...
5686 m3/kg
And sp
...
3 m and a 0
...
The vessel contains water at 25°C with a quality of 1%
...

Considering the vessel and water together as a system, calculate the
heat transfer during this process
...
7 g/cm3
and its specific heat is 0
...

(Ans
...
82 kJ)
4
Volume of water vapour mixture (V) = π d3 = 0
...
volume = π d 3 = 0
...
0146117 m3
∴
Mass = 39
...
3 m
0
...
3 +
× 2 m = 0
...
001003 +

Chapter 9
p1 = 0
...
36) = 0
...
3 = 129
...
95 kJ/kg
0
...
26023 kg
Mass of water and water vapour =
0
...
volume vg = 0
...
9 +

For Steam Table
At 4
...
441
At 4
...
423
0
...
434603
(pf ) = 4
...
2 ×
= 4
...
441 − 0
...
07
h f = 2739
...
9 − 2739
...
55 kJ/kg
Then
0
...
26023(2555 – 127
...
58 kJ
Heat required for A1
= 39
...
896 × (146 – 25)
= 4277
...
76 kJ

Q9
...
75 quality, flowing
also with negligible velocity at the rate of 5 kg/min
...
Determine
(a) The state of steam after mixing
(b) The state of steam after throttling
(c) The increase in entropy due to throttling
(d) The velocity of steam at the exit from the nozzle
(e) The exit area of the nozzle
...
E
...

(Ans
...
975 dry, (b) 5 bar, 0
...
2669 kJ/kg K,
(d) 540 m/s, (e) 1
...
6 kJ/kg
h 2 = 762
...
75 × 2013
...
8 kJ/kg
3 × 2942
...
8
∴
h3 =
8
= 2524 kJ/kg
Page 141 of 265

Properties of Pure Substances

By: S K Mondal
•
m 1 = 3 kg/min
1
10 bar
250°C

Chapter 9

3

4

1

5
5 bar

1
10 bar
x2 = 0
...
8°c

(a)

h3 = 762
...
6 or x 3 = 0
...
1 + x 4 × 2107
...
89395
s4 = 1
...
9588 = 6
...
2933 = 1
...
5967
x 5 = 0
...
7 + 0
...
6 = 2378
...
1382 + 0
...
4446 = 6
...
2933 – 6
...
18185 kJ/kg – K

(d)

V=

(e)

A=
⇒

Q9
...
4) = 540 m/s

mv
V
m × x 5
...
885
8
0
...
851 ×
m = 1
...
At a reduced load, as the governor takes action,
the pressure of steam is reduced to 59 bar by throttling before it is
admitted to the turbine
...

(Ans
...
65 kJ/kg
h 2 = 3167
...
4945 kJ/kg-K
t 2 = 396
...
6
(6
...
333)
50
= 6
...
032356 kJ/kg – K

s2 = 6
...
14
Q9
...
8 to pressure 3 bar
...
5°C
...
Draw the T–s and h–s diagrams
...
– 59 kJ/kg, 0
...
588 kJ/kg K during heating)

Solution:

p1 = 7 bar
t1 = 165º C
For Steam Table
h1 = h f + 0
...
8 × sfg

= 5
...
5°C

x = 0
...
5º C
h 2 = 561
...
8 × 2163
...
6716 + 0
...
3193
S
= 5
...
6 kJ/K
s3 = 7
...
16452 kJ/kg-K

Enthalpy charge in heating = h3 − h 2 = 573
...
38396 kJ/kg – K
Page 143 of 265

 

Page 144 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

10
...


As p → 0, or T → ∞, the real gas approaches the ideal gas behaviour
...
3143 kJ/kmole-K

2
...


4
...


γ −1
γ

=

⎛ v1 ⎞
⎜v ⎟
⎝ 2⎠

γ −1

For isentropic process

RT1
u2 − u1 =
γ −1
(i)

For closed system

γ −1
γ −1
⎡
⎤
⎡
⎤
γ
⎛ p2 ⎞ γ
⎛ p2 ⎞ γ
⎢
⎢
⎥; h − h =
(RT1 ) ⎜ ⎟
− 1⎥
−1
2
1
⎢⎝ p1 ⎠
⎥
⎢⎜ p1 ⎟
⎥
γ −1
⎣
⎦
⎣⎝ ⎠
⎦
2
p v −p v
∫ pd v = 1 1γ − 12 2
1

2

6
...


For minimum work in multistage compressor, p2 =

For steady flow

∫ v dp

=

p2 p3
=
p1 p2
(ii) Equal discharge temperature (T2 = T3 )

(i)

Equal pressure ratio i
...


Page 145 of 265

p1 p3

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

p3
p2

p1
T

2

3

2′

And (iii) Equal work for the two stages
...


9
...


(iii)

Virial Expansions:
pv
= 1 + B′p + C′p2 + D′p3 + ……………
RT
pv
B C
D
Or
= 1 + + 2 + 3 +
...


Chapter 10

vc
8 pc v c
; values of Z at critical point 0
...

a
bR
μ = x1 μ1 + x 2 μ2 +
...

12
...
mc R c
m1 + m2 +
...
m c uc
um = 1 1
;
m1 + m2 +
...
mc cPc
m1 + m2 +
...
m c h c
m1 + m 2 +
...
mc cv c
m1 + m2 +
...
+ m c R c ln c ⎥
p
p
p⎦
⎣
Gibbs function G = RT ∑ n x ( φk + ln p + ln x k )

13
...
K
...
10·1
Solution:

What is the mass of air contained in a room 6 m × 9 m × 4 m if the
pressure is 101
...
256 kg)
Given pressure (p) = 101
...
287 kJ/kg – K, Gas constant mass is m kg

∴
Q
...
2

(c)

Solution:

m=

pV
RT

=

101
...
9 kg
0
...
It is changed to 12 MPa at room
temperature (27°C)
...

(b) Explain how the actual cylinder contains nearly 15 kg of gas
...
25 m
(H) = 80 cm = 0
...
03927 m3
4
Gas pressure
(p) = 12 MPa = 12000 kPa
Page 147 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

Temperature
(a)

(T) = 27º C = 300 K

Mass of gas filled in the cylinder
⎡
⎤
pV
R 8
...
51964 ⎥
⎢ Here R = Gas constant =
RT
M
16
⎣
⎦
= 3
...
At
that pressure it is not a gas it is liquid form in atmospheric temperature so
its weight is amount 14 kg
...
10
...
03927
=
= 375 K = 102º C
mR
3
...
51964

A certain gas has cP = 0
...
653 kJ/kg K
...


Solution:

Gas constant, R = c p – c v = (0
...
26 kJ/kg – K
If molecular weight,( M )kJ/kg – mole
R
8
...
98 kJ/kg – mole
Then R = MR
∴M=
R
0
...
10
...
26
...


Solution:

Gas constant of acetylene (C2 H2 ) (R) =

R
8
...
3198 kJ/kg – K
M
26

As adiabatic index ( γ ) = 1
...
55 kJ/kg – K
γ −1
R
cv =
= 1
...
10
...

γ
Mono-atomic: c p =
R = 20
...
47 kJ/kg – mole – K
γ −1
γ
cp =
R = 29
...
79 kJ/kg – mole – K
γ −1

Page 148 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal
cp =

Polyatomic:

Chapter 10
γ
R = 33
...
94 kJ/kg – mole – K
Q
...
6

Solution:

A supply of natural gas is required on a site 800 m above storage level
...
1 bar from storage is pumped steadily to a point on
the site where its pressure is 1
...
If the work transfer to the gas at the pump is 15 kW,
find the heat transfer to the gas between the two points
...
E
...
33 (g = 9
...

(Ans
...
9 kW)
Given:
At storage
(p1 ) = 1
...
2 bar = 120 kPa

T3 = 288 K
•

(V 3 ) = 1000 m3/m =

Flow rate
Gas constant (R) =

5 3
m /s
18

R
= 0
...
22273 kg/s
RT3

2

•

⎛ dW ⎞
Pump work ⎜
⎟ = –15 kW
⎝ dt ⎠
∴
From steady flow energy equation
dQ
dW
•
•
m(h1 + 0 + gZ1 ) +
= m( h3 + 0 + g Z3 ) +
dt
dt
(Z3 − Z1 ) ⎤ dW
dQ
• ⎡
∴
= m ⎢(h3 − h1 ) + g
+
dt
1000 ⎥
dt
⎣
⎦
Δ Z ⎤ dW
• ⎡
= m ⎢c P (T3 − T1 ) + g
+
1000 ⎥
dt
⎣
⎦
9
...
22273 ⎢2
...
7 kJ/s = 63
...
10
...
0943 kJ/kg
γ −1

800m

P
1

A constant volume chamber of 0
...
Heat is transferred to the air until the temperature is 100°C
...


Page 149 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal
Solution:

Chapter 10

Constant volume (V) = 0
...
95 kPa
V
Work done = ∫ pdV = 0
p1 =

∫ du + ∫ dW = ∫ dW = m c ∫ dT = m c
Change in internal Energy = ∫ du = 68
...
475 kJ

Heat transferred Q =

v

P

Change in Entropy =

∫ d s = s2 –

2

v

(T2 – T1 ) = 68
...
718 × ln
265
...
2114 kJ/kg – K
= m c v ln

Q
...
8

One kg of air in a closed system, initially at 5°C and occupying 0
...
There
is no work other than pdv work
...


Solution:

T1 = 278 K
V1 = 0
...
95 kPa
T2 = 100º C = 373 K
p2 = 265
...
40252 m3
p2
(a) Work during the process
2

(W12) =

∫ p dV

= p(V2 − V1 ) = 27
...
476 kJ

V

(c) Entropy change of the gas
V
p
s2 – s1 = mc p ln 2 + mc v ln 2
V1
p1
v
= m c p ln 2 = 0
...
10
...
1 m 3 of hydrogen initially at 1
...
1 MPa
...

Page 150 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal
Solution:

Chapter 10

V1 = 0
...
2 MPa = 1200 kPa
T1 = 473 K

1
1

p

2

T

2
S

V

R
8
...
157 kJ/kg – K
M
2
p V
m = 1 1 = 0
...
1 MPa = 100 kPa
Heat transferred (Q) = Δu + ΔW
mRT2
3
∴
V2 =
= 1
...
2 ⎞
= 4
...
1 ⎠
= 4886 kJ
2

Work done (W) =

∫ p dV

= 4886 kJ

1

Entropy change, s2 – s1 = mc p ln

V2
p
+ mc v ln 2
V1
p1

⎡
⎛ 1
...
06103 ⎢14
...
4 ln ⎜
⎟⎥
⎝ 0
...
6294 kJ/kg – K
For H2 diatomic gas (γ = 1
...
4 kJ/kg – K
cp =
R = 14
...
10
...
5 MPa, 15°C to 0
...
Find the final temperature, and
per kg of air, the change in enthalpy, the heat transferred, and the work
done
...
5 MPa = 500 kPa
T1 = 15ºC = 288 K
Let mass is 1 kg
∴
v1 = 1 × R × T1 = 0
...
2 MPa = 200 kPa
∴

γ
γ
p1 v1 = p2 v 2 :

1

1
γ

⎛p ⎞
v2 = v1 × ⎜ 1 ⎟ = 0
...
33 kJ/kg

2
S

The Heat transferred (Q) = 0
The work done
2
p v − p2 v 2
(W) = ∫ p d v = 1 1
γ −1
1
= 47
...
10
...
Neglect velocity and elevation changes
...
e
...
4 kJ/kg
2

(W) = − ∫ v dp =

Shaft work

1

γ −1
γ

valid
...
33 kJ/kg
γ −1

Heat transferred:
h1 + 0 + 0 +

∴

Q
...
12

dQ
dW
= h2 + 0 + 0 +
dm
dm
dQ
dW
= (h 2 − h1 ) +
= –66
...
33 = 0
dm
dm
[As it is reversible adiabatic so dQ = 0]

The indicator diagram for a certain water-cooled cylinder and piston air
compressor shows that during compression pv1
...
The
compression starts at 100 kPa, 25°C and ends at 600 kPa
...
8553 m3/kg
∴ v1 =
p1
p2 = 600 kPa
1
n

⎛p ⎞
v 2 = v1 ⎜ 1 ⎟ = 0
...
3

= constant

p
1

n −1
n

∴

⎛p ⎞
= 451 K
T2 = T1 × ⎜ 2 ⎟
⎝ p1 ⎠
dQ
dW
h1 + 0 + 0 +
= h2 + 0 + 0 +
dm
dm
dQ
dW
= (h 2 − h1 ) +
dm
dm
n[p1 v1 − p2 v 2 ]
dW
=
n −1
dm
= –189
...
774
= 153
...
774
= –36 kJ/kg
[Heat have to be rejected]

∴

Q
...
13

V

An ideal gas of molecular weight 30 and γ = 1
...
5
m 3 at 100 kPa and 77°C
...
25 = constant to a pressure of 3 MPa
...


Solution:

R
= 0
...
3; n = 1
...
9238 kJ/kg – K
∴ cv =
γ −1
R
cP = γ
= 1
...
5 m3; T1 = 350 K
p2 = 3 MPa = 3000 kPa
R=

2
p

1

⎛ p ⎞n
V2 = V1 ⎜ 1 ⎟ = 0
...
5464 kg
RT1
pV
∴ T2 = 2 2 = 691 K
mR
∴
Page 153 of 265

1
V

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10
2

Work done (W1 – 2) =

∴ p1 V

n
1

∫ pdV

1
n
2

n

= p V = p2 V
= p1 V1n

2

p Vn
dV
= 1 1
∫ Vn − n + 1
1

1 ⎤
⎡ 1
⎢ Vn − 1 − Vn − 1 ⎥
1
⎣ 2
⎦

p2 V2 − p1 V1
p V − p2 V2
= 1 1
1−n
n −1
100 × 1
...
09872
kJ = –584
...
25 − 1

=

Heat transfer Q = u2 – u1 + W1 – 2
= mc v (T2 – T1 ) + W1 – 2
= [1
...
9238 (691 – 350) – 584
...
5 kJ
p
V ⎤
⎡
ΔS = S2 – S1 = ⎢mc v ln 2 + mc P ln 2 ⎥
p1
V1 ⎦
⎣
= – 0
...
10
...
15 m 3 to a temperature of 550 K
and a volume of 0
...

If the air expands according to the law, pv n = constant, between the
same end states, calculate the heat given to, or extracted from, the air
during the expansion, and show that it is approximately equal to the
change of entropy multiplied by the mean absolute temperature
...


Q
...
15

0
...
For the total path, find the work
transfer, the heat transfer, and the change of entropy
...


Q
...
16

An ideal gas cycle of three processes uses Argon (Mol
...
40) as a
working substance
...
014 m 3 , 700 kPa, 280°C to 0
...
Process 2-3 is a reversible
isothermal process
...
Sketch the cycle in the p-v and T-s planes, and find
(a) the work transfer in process 1-2, (b) the work transfer in process 2-3,
and (c) the net work of the cycle
...
67
...


Q
...
17

A gas occupies 0
...
It is expanded in the nonflow process according to the law pv1
...
Sketch the process on the p-v and T-s diagrams, and
calculate for the whole process the work done, the heat transferred, and
the change of entropy
...
047 and c V = 0
...

Solution:

Try please
...
10
...
5 kg of air at 600 kPa receives an addition of heat at constant volume
so that its temperature rises from 110°C to 650°C
...
32
...
Calculate (a) the change of entropy during each of the three
stages, (b) the pressures at the end of constant volume heat addition and
at the end of expansion
...

Try please
...
10
...
5 kg of helium and 0
...
Find (a) the volume of the mixture, (b) the partial
volumes of the components, (c) the partial pressures of the components,
(d) the mole fractions of the components, (e) the specific heats cP and cV
of the mixture, and (f) the gas constant of the mixture
...


Q
...
20

A gaseous mixture consists of 1 kg of oxygen and 2 kg of nitrogen at a
pressure of 150 kPa and a temperature of 20°C
...


Solution:

Try please
...
10
...
1 m 3
...
The pressure in one compartment is 2
...
The diaphragm is ruptured so that
the air in both the compartments mixes to bring the pressure to a
uniform value throughout the cylinder which is insulated
...


Solution:

Try please
...
10
...
Part (a) contains oxygen and has a volume of 0
...
2 m 3 and contains nitrogen, while (c) is 0
...
All three parts are at a pressure of 2 bar and a temperature of 13°C
...
The vessel may be taken as being
completely isolated from its surroundings
...
0
...
0783, 155 of 265 2 bar; 0
...
1429, 0
...
)
Page 0
...
1 m 3

0
...
05 m 3

N2

O2

W2

p = 2 bar = 200 kPa
T = B° C = 286 K
After mixing temperature of the mixture will be same as before 13ºC = 286 K and
also pressure will be same as before 2 bar = 200 kPa
...
1 + 0
...
05) = 0
...
1
ma =
=
∴
kg = 0
...
3143
Ra T
× 286
32
pVa
200 × 0
...
471 kg
8
...
05
mc =
=
kg = 0
...
319
Rc T
× 286
44
v ⎤
T
p
⎡ p
∴
ΔS = S2 – S1 = mc P ln 2 − mR ln 2
Here T2 = T1 so ⎢∵ 2 = 1 ⎥
T1
p1
⎣ p1 v 2 ⎦
(S2 − S1 )O2 = mO2 R O2 ln

Vo
8
...
35 ⎞
× ln ⎜
= 0
...
1 ⎠

= 0
...
3143
⎛ V ⎞
⎛ 0
...
471 × 32 × ln ⎜ 0
...
078267 kJ/K
⎝
⎠
⎝ Vn2 ⎠
8
...
35 ⎞
(S2 − S1 )CO2 = mCO2 RCO2 ln ⎜
⎟ = 0
...
05 ⎟ = 0
...
1
Mole fraction of
O2 (x O2 ) =
0
...
2
Mole fraction of
N 2 (x N2 ) =
0
...
05
Mole fraction of
CO2, x O2 =
0
...
1
× 200 = 57
...
35
0
...
29 kPa
Partial pressure of
N 2 ; (pN2 ) = x N 2 × p =
0
...
05
Partial pressure of
× 200 = 28
...
35

( )

Page 156 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

Q
...
23

A Carnot cycle uses 1 kg of air as the working fluid
...
The maximum
pressure of the cycle is 1 MPa and the volume of the gas doubles during
the isothermal heating process
...


Solution:

Try please
...
10
...

Sketch the cycle on the p-v and T'-s diagrams
...

3
(Ans
...
378))
For process 1 – 2 constant volume heating
Q1 – 2 = Δu + pdv
= mc v ΔT + pdv

Solution:

= mc v ΔT = mc v (T2 – T1 )

2
2

T
1

p

3

3

1
V

S
Q2 – 3 = 0 as isentropic expansion
...
37778
3
(8 − 1)
Q
...
25

Solution :

Using the Dietetic equation of state
⎛
RT
a ⎞
P=

...
⎟
v v
⎝
⎠
(c) Show that
a
TB =
bR
Try please
...
10
...


Solution :

Try please
...
10
...
5 kg moles of steam at 236
...
and 776
...

At this volume and the given pressure, what would the temperature be
in K, if steam behaved like a van der Waals gas?
The critical pressure, volume, and temperature of steam are 218
...
,
57 cm 3 /g mole, and 647
...


Solution :

Try please
...
10
...
Vessel a contains air at 7 bar, 95°C while B
Page 158 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal
Solution:

Chapter 10

contains air at 3
...
Find the change of entropy when A is
connected to B
...

(Ans
...
975 kJ/kg K)
VA = VB = 3m3
pA = 7 bar = 700 kPa
TA = 95ºC = 368 K
pB = 3
...
883 kg
RTA

mB =

In case of Adiabatic mixing for closed system
Internal energy remains constant
...
TA + m B c v
...
6 K
m A + mB
After mixing partial for of A
⎡ Total pressure
⎤
⎢
⎥
⎢∴ p = mRT = 525
...
1 kPa
pAf = A
V
m RT
pBf = B
= 145
...
0957 kJ/K
p
T
− m B R ln Bf
ΔsBf − sB = m B c P ln
TB
pB
= 0
...
62 kJ/K
∴

Q
...
29

T=

pB VB
= 7
...
5 bar
TB = 205°C

An ideal gas at temperature T1 is heated at constant pressure to T2
n

and

then expanded reversibly, according to the law pv = constant, until the
temperature is once again T1 What is the required value of n if the
changes of entropy during the separate processes are equal?
⎛
⎛
2γ ⎞ ⎞
⎜ Ans
...
10
...
142 m 3 capacity, at a pressure and temperature of 23
...
A valve is opened momentarily and the pressure falls
immediately to 6
...
Sometimes later the temperature is again 18°C
and the pressure is observed to be 9
...
Estimate the value of specific
heat ratio
...
1
...
142
= 8
...
3143
R SO2 T
× 291
64

Mass of SO2 after closing the valve
Page 160 of 265

R SO2 = 0
...
142
= 3
...
1 × 0
...
9 × 0
...
65 K
As valve is opened momentarily term process is adiabatic
So

or

or
∴

Q
...
31

Solution :

T2
⎛p ⎞
= ⎜ 2⎟
T1
⎝ p1 ⎠

γ −1
γ

220
...
9 ⎞
or
= ⎜
⎟
291
⎝ 23
...
65 ⎞
ln ⎜
⎟
1⎞
⎛
⎝ 299 ⎠ = 0
...
9 ⎞
⎝
ln ⎜
⎟
⎝ 23
...
22903 = 0
...
297

A gaseous mixture contains 21% by volume of nitrogen, 50% by volume of
hydrogen, and 29% by volume of carbon-dioxide
...
(At 10ºC, the c p values
of nitrogen, hydrogen, and carbon dioxide are l
...
235, and 0
...
)
A cylinder contains 0
...
The gas
undergoes a reversible non-flow process during which its volume is
reduced to one-fifth of its original value
...
2 = constant, determine the work and heat transfer in magnitude and
sense and the change in entropy
...
19
...
423 kJ/kg K, 1
...
24 kJ, – 0
...
42334 kJ/kg – K
=

c p Mix =

m N2 C pN + m H2 C pH + mCO2 C pCO
2

2

2

m N2 + m H2 + mCO2
Page 161 of 265

[mN2 + m H2 + mCO2 = 1964]

Properties of Gases and Gas Mixtures

By: S K Mondal
=

Chapter 10
21 × 28 × 1
...
235 + 0
...
5738 kJ/kg – K
588 + 100 + 1276

R
= 0
...
235 −
= 10
...
828 −
= 0
...
039 −

∴

∴

588 × 0
...
078 + 1276 × 0
...
1505 kJ/kg – K
588 + 100 + 1276
c v mix = c P mix – Rmix = 1
...
42334 = 1
...
368

cv mix

Given
p1 = 1 bar = 100 kPa
⇒

p2 = 690 kPa (Calculated)
v
V2 = 1 = 0
...
5 K (Calculated)

V2 = 0
...
2
p1
⎝ v2 ⎠
p2 = 100 × 51
...
5 K

1
V

⎡
dV ⎤
⎢∵ W = ∫ pdV = C ∫ 4 ⎥
1
1 V ⎦
⎣
2

2

100 × 0
...
017
1
...
15 kJ
[i
...
work have to be given to the system)
=

– W

Q = u2 − u1 + W
m=

p1 V1
= 0
...
7748 – 16
...
3752 kJ
[i
...
Heat flow through system]
Page 162 of 265

– Q

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

Charge of entropy
⎛T ⎞
⎛p ⎞
ΔS = S2 – S1 = mc P ln ⎜ 2 ⎟ − m R ln ⎜ 2 ⎟
⎝ T1 ⎠
⎝ p1 ⎠
⎡
⎛ 390
...
5738 ln ⎜
⎟ − 0
...
022062 kJ/K = –22
...
10
...
In an adjacent compartment is one mole of an ideal
gas at temperature 2Tand pressure p
...
Show that the entropy increase due to the mixing process is
given by
⎛ 27
32 ⎞
γ
+
R ⎜ ln
ln
⎟
4 γ − 1 27 ⎠
⎝
Provided that the gases are different and that the ratio of specific heat γ
is the same for both gases and remains constant
...

γ −1
4
27 ⎥
⎣
⎦

Q
...
33

n1 moles of an ideal gas at pressure p1 and temperature T are in one
compartment of an insulated container
...
When the partition is removed, calculate (a) the final
pressure of the mixture, (b) the entropy change when the gases are
identical, and (c) the entropy change when
...

Prove that the entropy change in (c) is the same as that produced by two
independent free expansions
...


Q
...
34

Assume that 20 kg of steam are required at a pressure of 600 bar and a
temperature of 750°C in order to conduct a particular experiment
...

Predict if this is an adequate storage capacity using:
(a) The ideal gas theory,
(b) The compressibility factor chart,
(c) The van der Waals equation with a = 5
...
03042 litres/gmol for steam,
(d) The Mollier chart
(e) The steam tables
...


Solution:

Try please
...
10
...
70
m 3 at 75°C, using the Beattie-Bridgeman equation of state
...
Which is more accurate and why?
For CO2 with units of atm, litres/g mol and K, A o = 5
...
07132, Bo =

0
...
07235, C * 10-4 = 66
...

Solution:

Try please
...
10
...
25 =
Page 164 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

constant
...
5p1 v1 ⎢1 – ⎜ 2 ⎟ ⎥
⎢ ⎝ p1 ⎠ ⎥
⎣
⎦
Take γ for air as 1
...

Solution:

Using S
...
E
...

⎡ V2
⎤
Q − W + Δ ⎢ 2 + g Z ⎥ = h2 – h1
⎣ 2
⎦
or

Q – W = mc p (T2 − T1 )
=

γ
mR(T2 − T1 )
γ −1

∴ p1 v1 = mRT1

p2 v 2 = mRT2

γ p1 v1 ⎡ p2 v 2
⎤
⎢ p v − 1⎥
γ −1 ⎣ 1 1
⎦
n −1
⎡
⎤
γ
⎛p ⎞ n
=
− 1⎥
p1 v1 ⎢⎜ 2 ⎟
⎢⎝ p1 ⎠
⎥
γ −1
⎣
⎦
=

∴ Q → 0 and as

Here adiabatic process

So

n −1
⎡
⎤
γ
⎛ p2 ⎞ n ⎥
⎢1 −
× p1 v1
W=
⎢ ⎜ p1 ⎟
⎥
γ −1
⎣ ⎝ ⎠
⎦

γ = 1
...
25
1
⎡
⎤
⎛ p2 ⎞ 5 ⎥
W = 3
...
10
...
5 m 3 /kg
...
4
...
(b) What is the change in entropy
when the gas is expanded to pressure 100 kPa according to the law pv1
...
5 = const
...
(a) 1
...
833 kJ/kg K, (b) 0
...
039 kJ/kg K
(d) Entropy increases when n < γ and decreases when n > γ )
p1 = 200 kPa
Given
T1 = 300 K
v1 = 0
...
4
(a) Gas constant( R) =

p1 v1
200 × 0
...
33333 kJ/kg – K
300
T1
Page 165 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

∴ Super heat at constant Pressure
γ
1
...
33333 = 1
...
4 − 1
γ −1
c V = c p – R = 0
...
3
v 2 = v1 × ⎜ 1 ⎟ = 0
...
85218 ⎞
⎛ 100 ⎞
= 1
...
83333 × ln ⎜
⎟ kJ/kg − K
⎝ 0
...
044453 kJ/kg – K = 44
...
5 = C
...
5
v2 = v1 × ⎜ 1 ⎟ = 0
...
7937 ⎞
⎛ 150 ⎞
s2 – s1 = 1
...
83333 ln ⎜
⎟ kJ/kg − K
⎝ 0
...
03849 kJ/kg – K
(d) n > γ is possible if cooling arrangement is used and ΔS will be –ve

Q
...
38

Solution:

(a) A closed system of 2 kg of air initially at pressure 5 atm and
temperature 227°C, expands reversibly to pressure 2 atm following
the law pv1
...
Assuming air as an ideal gas, determine the
work done and the heat transferred
...
193 kJ, 72 kJ)
(b) If the system does the same expansion in a steady flow process, what
is the work done by the system?
(Ans
...
625 kPa
1
T1 = 277º C = 550 K
p2 = 2 atm = 202
...
9 K
⎝ p1 ⎠
p V − p2 V2
mR(T1 − T2 )
=
W1 – 2 = 1 1
n −1
n −1
2 × 0
...
9)
=
= 211
...
25 − 1
Reversible polytropic process
Heat transfer
Q1 – 2 = u2 − u1 + W1 – 2
= mc v (T2 − T1 ) + W1 – 2
Page 166 of 265

2
V

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

= 2 × 0
...
9 – 550) + W = –132
...
46 = 79
...
10
...
33 kJ
n −1
n −1

Air contained in a cylinder fitted with a piston is compressed reversibly
according to the law pv1
...
The mass of air in the cylinder is 0
...
The initial pressure is 100 kPa and the initial temperature 20°C
...
Determine the work and the
heat transfer
...
– 22
...
6 kJ)
It is a reversible polytropic process
p2 = 1345
...
1 kg
p1 = 100 kPa
T2 = 492
...
010511 m3
P1
= 0
...
25
⎛V ⎞
2
∴
p2 = p1 ⎜ 1 ⎟ = 100 × 81
...
084091 − 1345
...
010511
=
1
...
93 kJ
Q1 – 2 = u2 − u1 + W1 – 2

∴ W1 – 2 =

V

= mc v (T2 − T1 ) + W1 – 2
= 0
...
718 × (492
...
93
= –8
...
10
...
Initially
the cylinder contains 0
...
5 bar, 20°C
...

Determine: (a) the polytropic index n, (b) the final volume of air, (c) the
work done on the air and the heat transfer, and (d) the net change in
entropy
...
(a) 1
...
1676 m3 (c) –95
...
5 kJ, (d) 0
...
5 bar = 150 kPa
T1 = 20ºC = 293 K
V1 = 0
...
89189 kg
RT1
p2 = 6 bar = 600 kPa
T2 = 120º C = 393 K

2

∴m=

T
⎛p ⎞
∴ 2 = ⎜ 2⎟
T1
⎝ p1 ⎠

p

n −1
n

1
V

⎛T ⎞
ln ⎜ 2 ⎟
1⎞
⎛
⎝ T1 ⎠ = 0
...
2687
(a) The polytropic index, n = 1
...
189 × 0
...
16766 m3
(b) Final volume of air (V1) =
=
600
p2
2

(c)

=

W1 – 2

∫ pdV
1

=

=

p1 V1 − p2 V2
n −1

150 × 0
...
16766
kJ = –95
...
2687 − 1

Q1 – 2 = u2 − u1 + W1 – 2
= mc v (T2 − T1 ) + W1 – 2
= 0
...
718(393 – 293) + W1 – 2
= –31
...
10
...
16766 ⎞ ⎤
= 0
...
718 ln ⎜
⎟ + 1
...
091663 kJ/K
⎝ 150 ⎠
⎝ 0
...
9169 + 2
...
974 * 10-8 T2 kJ/kg K
Determine the change in internal energy and that in entropy of air when
it undergoes a change of state from 1 atm and 298 K to a temperature of
2000 K at the same pressure
...
1470
...
1065 kJ/kg K )
p1 = p2 = 1 atm = 101
...
9169 + 2
...
974
× 10–3 T2 kJ/kg – K
Δu = u2 − u1 =

∫m c

v

dT

Page 168 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal
=
=

Chapter 10

∫ m(c − R) dT
∫ mc dT − mR ∫ dT
P

P

2000

= 1×

∫

2

T
(0
...
577 × 10 −4 T − 3
...
287

∫

dT kJ / kg

S

298

= (1560
...
96 – 105
...
47) kJ/kg
= 1470
...
577 × 10 −4 (2000 − 298)
∴ s2 – s1 = 0
...
974 × 10 −8 × ⎜
⎟
2
⎝
⎠

= 2
...
10
...
25 m 3 to 0
...
32 = const
...

(a) Draw the p-v and T-s diagrams
...
(b) 262
...
63 kPa,
(c) 57
...
4 kJ, (d) 0
...
25 m3
T1 = 100ºC = 373 K

p1

1

2

T

p
2
V
∴

m=

p2

1

p1 v1
= 0
...
5643 kg
RT1
Page 169 of 265

S

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

8
...
29694 kJ/kg
28
n
⎛ v1 ⎞
p2 = p1 × ⎜ ⎟ = 58
...
75 m3
n −1

⎛p ⎞ n
= 262
...
25 − 58
...
75
=
W= 1 1
= 57
...
32 − 1)
Q = u2 − u1 + W = mc v (T2 − T1 ) + W
= 0
...
7423 (262
...
7423
γ −1
γ= 1+

= 11
...
10
...
4
R =
× 0
...
04 kJ/kg – K
1
...
4
5

V
p ⎤
⎡
Δs = s2 – s1 = m ⎢cP ln 2 + cV ln 2 ⎥
V1
p1 ⎦
⎣
⎡
⎛ 0
...
633 ⎞ ⎤
= 0
...
04 × ln ⎜
⎟ + 0
...
25 ⎠
⎝ 250 ⎠ ⎦
⎣
= 0
...
66 +
0
...
If the initial pressure of the gas is 1 atm,
calculate the final pressure, the heat transfer, the work done and the
change in entropy
...
2
...
5 kJ/kg, 2
...
325 kPa
R =
16
T1 = 27ºC = 300 K
= 0
...
5385
p1
T2 = 500ºC = 773 K
= 1
...
577 atm
(ii) Heat transfer Q =

∫ mc

v

dT

= m ∫ [c P − R]dT

S
Page 170 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

773

= 1×

∫ (1
...
8675 × 10

−3

− 0
...
58411(773 − 300) + 3
...
66 0
...
1038 + 3
...
7 kJ/kg

cP =

2

(iii)

Work done =

∫ pdV = 0
1

∴

Tds = du = mc v dT

dT
dT
= m(c p − R)
T
T
2
773
⎛ 1 × 0
...
8675 × 10 −3 T ⎞
dS = ∫ ⎜
∴
⎟ dT
∫
T
⎝
⎠
1
300
773
s2 – s1 = 0
...
8675 × 10 −3 (773 − 300) = 2
...
10
...
25 = const
...
9 m 3 to a final volume of 0
...
Determine the final pressure and the change of entropy per kg of
air
...
1
...
0436 kJ/kg K)
p1 = 1 bar
V1 = 0
...
6 m3
1
...
66 bar
p2 = p1 ⎜
⎟
⎝ V2 ⎠

2

2
T

p

1

1
V

S

V
p ⎞
⎛
Δs = s2 − s1 = ⎜ c p ln 2 + cv ln 2 ⎟
V1
p1 ⎠
⎝
⎛ 0
...
66 ⎞
= 1
...
718 × ln ⎜
⎝ 1 ⎠
⎝ 0
...
043587 kJ/kg – K

Q
...
45

In a heat engine cycle, air is isothermally compressed
...
Draw the cycle on p-v and T'-s coordinates
...
Determine the pressure ratio and
the cycle efficiency if the initial temperature is 27°C and the maximum
temperature is 327°C
...
13
...
4%)
Heat addition (Q1) = Q2 – 3 = mc p (T3 − T2 )

p2

p

2

3

p1

3

W

T
2
1

1
S

V

⎛p ⎞
Heat rejection (Q2) = mRT1 ln ⎜ 2 ⎟
⎝ p1 ⎠
⎛p ⎞
RT1 ln ⎜ 2 ⎟
Q
⎝ p1 ⎠
∴
η= 1− 2 = 1−
C p (T3 − T2 )
Q1

p2
=r
p1

∴

⎛p ⎞
ln ⎜ 2 ⎟
γ −1
⎝ p1 ⎠
= 1−
γ ⎛ T3
⎞
⎜ T − 1⎟
⎝ 1
⎠

Here,

T3
⎛p ⎞
= ⎜ 3⎟
T1
⎝ p1 ⎠

cp =
= 1−

⇒
And

γ −1
γ

∴

(r

Proved

− 1)

If initial temperature (T1) = 27ºC = 300 K = T2
T3 = 327ºC = 600 K
γ

∴

ln r
γ −1
γ

1
...
4 − 1
r= ⎜ 3⎟
= ⎜
= 11
...
4 − 1)
ln (11
...
4 − 1
(1
...
314) 1
...
30686

γR
γ −1

γ −1
γ

= r

γ −1
γ

Properties of Gases and Gas Mixtures

By: S K Mondal
Q
...
46

Solution:

Chapter 10

What is the minimum amount of work required to separate 1 mole of
air at 27°C and 1 atm pressure (assumed composed of 1/5 O2 and 4/5 N2 )
into oxygen and nitrogen each at 27°C and 1 atm pressure?
( Ans
...
0064 kg
5
4
N2 = mole = 0
...
3143
8
...
0224 ×
× 300 ln ⎜ ⎟
= 0
...
248 kJ = 1248 J
A closed adiabatic cylinder of volume 1 m 3 is divided by a partition into
two compartments 1 and 2
...
6 m 3 and
contains methane at 0
...
4 m 3 and contains propane at 0
...
The partition is removed
and the gases are allowed to mix
...

(b) What are the molecular weight and the specific heat ratio of the
mixture?
The mixture is now compressed reversibly and adiabatically to 1
...
Compute
(c) the final temperature of the mixture,
(d) The work required per unit mass, and
(e) The specific entropy change for each gas
...
10
...
72 and 74
...

(Ans
...
8609 kJ/K, (b) 27
...
193 (c) 100
...
255 kJ/kg K)
After mixing
pf = 400 kPa
Tf = 313 K

1

But partial pressure of (p1f )
CH4 =
∴

0
...
4 × 400 = 160 kPa
Page 173 of 265

V1 = 0
...
4 m3
p2 = 400 kPa
T2 = 313 K
C3 H8

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

⎡
T
p ⎤
( ΔS)CH4 = mCH4 ⎢c PCH ln 2 − R ln 2 ⎥
4
T1
p1 ⎦
⎣
⎛p ⎞
= mCH4 RCH4 ln ⎜ i ⎟
⎝ pf ⎠

(a)

=

⎛p ⎞
p1 V1
× RCH4 ln ⎜ i ⎟
RCH4 T1
⎝ pf ⎠

⎛p
p1 V1
× RCH4 ln ⎜ i
⎜ pf
T1
⎝ 1
p2 V2
pi
=
× ln
T2
pf2
=

( ΔS)C3H8

⎞
⎟
⎟
⎠

(ΔS) Univ = ( ΔS)CH4 + ( ΔS)C3H8
400 × 0
...
4 ⎛ 400 ⎞
ln ⎜
ln ⎜
⎟+
⎟ kJ /K
313
313
⎝ 240 ⎠
⎝ 160 ⎠
= 0
...
6 × 16 + 0
...
2
x
x
n1c p1 + n 2 c p2
0
...
72 + 0
...
56
=
= 51
...
3143

∴
∴

Q
...
48

c v mix = c P mix – R = 42
...
256
= 1
...
9417

An ideal gas cycle consists of the following reversible processes: (i)
isentropic compression, (ii) constant volume heat addition, (iii)
isentropic expansion, and (iv) constant pressure heat rejection
...

An engine operating on the above cycle with a compression ratio of 6
starts the compression with air at 1 bar, 300 K
...
5, calculate the efficiency and the
m
...
p
...
Take
γ = 1
...
718 kJ/kg K
...
0
...
5322 bar)

Solution:

Q2 – 3 = u3 − u2 + pdV = mc v (T3 − T2 )
Page 174 of 265

Properties of Gases and Gas Mixtures

By: S K Mondal

Chapter 10

Q1 – 4 = mc p (T4 − T1 )
∴

η= 1−

m c p (T4 − T1 )
m c v (T3 − T2 )

⎛ T − T1 ⎞
= 1 − γ⎜ 4
⎟
⎝ T3 − T2 ⎠
γ −1

∴

⎛v ⎞
T2
γ
= ⎜ 1⎟
= rk − 1
T1
v2 ⎠
⎝
γ
T2 = T1 × rk − 1
γ −1

⎛p ⎞ γ
T3
= ⎜ 3⎟
T4
⎝ p4 ⎠
p3
∴
=9
p2
p v
p2 v 2
= 3 3
T2
T3
T
∴ 3 = (a × r)
T4

⎛p ⎞
= ⎜ 3⎟
⎝ p1 ⎠

V=C
3
Q1
W

γ −1
γ

γ

γ −1
γ

⎛ ⎛ V ⎞γ 1 ⎞
= ⎜⎜ 2 ⎟ × ⎟
⎜ ⎝ V1 ⎠ a ⎟
⎝
⎠
−
= (rk γ a −1 )
1
T4 = rk − γ
...
a

γ −1
γ

⎛p
p ⎞
= ⎜ 1 × 2⎟
⎝ p2 p3 ⎠

η= 1−

S

T

Q1
2

γ

...
rk − 1

1

Q2
S

1
γ

γ(a T1 − T1 )
γ
γ
(aT1rk − 1 − T1rk − 1 )
1

[ γ(a γ − 1)]
= 1 − γ −1
Proved
...
5,

Q2

3

= a
...
T3

1
−1
γ

1

γ −1
γ

r−1
r

1−γ
γ

W

p
T
∴ 3 = 3 = a,
p2
T2

⎛p ⎞
= ⎜ 1⎟
⎝ p3 ⎠

p=C

2

T

v ⎞
p2
= 1 ⎟ = aγ
p1
v2 ⎠
γ
∴ T3 = aT2 = aT1rk − 1
⎛p ⎞
T4
= ⎜ 4⎟
T3
⎝ p3 ⎠

γ −1
γ

rk = 6,
γ = 1
...
4(2
...
4 − 1)
∴ η = 1 − 1
...
57876
6
(2
...
718 × 300 × 60
...
5 – 1)
= 661
...
9 m kJ
1

For V4 = ; T4 = 2
...
4 × 300 = 577
...
287 × 577
...
6567 m m3
m R T1
m × 0
...
861m m3
∴
V4 – V1 = 0
...
e
...
is pm then
pm ( V4 – V1 ) = W
382
...
7957 m
= 481
...
8121 bar

Q10
...
50

Solution:
Q10
...
Show that for a reversible adiabatic
process pv y = constant, where
γ = (b + 1)/b
...

(a) Show that the slope of a reversible adiabatic process on p-v
coordinates is
dp
1 cp
1 ⎛ ∂v ⎞
=
wherek = − ⎜
⎟
dv kv c v
v ⎝ ∂p ⎠ T
(b) Hence, show that for an ideal gas, pv γ = constant, for a reversible
adiabatic process
...

A certain gas obeys the Clausius equation of state p (v – b) = RT and has
its internal energy given by u = c v T
...

γ

Solution:

Try please
...
52

Solution:
Q10
...
Show
that the ratio of the slope of the adiabatic curve to the slope of the
isothermal curve is equal to γ
...
6
...

Try please
...
54

Solution:
Q10
...
Determine:
(a) the final temperature of the two gases and
(b) the change of entropy due to this energy exchange
...

The pressure of a certain gas (photon gas) is a function of temperature
only and is related to the energy and volume by p(T) = (1/3) (U/V)
...

(a) Find expressions for work and heat of reversible isothermal and
adiabatic processes
...

(c) Determine the efficiency of the cycle in terms of pressures
...

The gravimetric analysis of dry air is approximately: oxygen = 23%,
nitrogen = 77%
...
3 kg air to produce
...
(a) 21% O2 , 79% N 2 , (b) 0
...
84 m3 /kg, (f) 1
...
71875: 2
...
75 × 100
:
(0
...
75)
2
...
72: 79
...
71875 ×

Properties of Gases and Gas Mixtures

By: S K Mondal
(b)

Chapter 10

Let total mass = 100 kg
∴
O2 = 23 kg, N2= 77 kg
∴

R=

23 × R O2 + 77 × R N2
23 + 77

’

8
...
3143
+ 77 ×
32
28
=
23 + 77
= 0
...
7928 × 28 = 28
...
2072 × 101
...
995 kPa
Partial pressure of N2 = x N 2 × p = 0
...
325 kPa = 80
...
2884 × 288 3
=
m / kg = 0
...
325
ρ
ρ =   ρ1 + ρ2
pN 2
pO2
1
1
1
+
=
+
=
v
v1 v 2
R O2 × 288 R N2 × 288

Sp
...
2072 × 101
...
7928 × 101
...
3143 × 2
...
3143 × 288
3/kg
v = 0
...
3 kg of air O2 = 2
...
23 kg = 0
...
3 × 0
...
771 kg = 63
...
25 ×
kg = 2
...
024 – 0
...
495 kg

Page 178 of 265

Thermodynamic Relations

By: S K Mondal

Chapter 11

11
...
If a relation exists among the variables x, y and z, then z may be expressed as a
function of x and y, or
⎛ ∂z ⎞
⎛ ∂z ⎞
dz = ⎜ ⎟ dx + ⎜ ⎟ dy
⎝ ∂x ⎠ y
⎝ ∂y ⎠ x

then dz = M dx + N dy
...
Differentiating M partially with respect to y, and N
with respect to x
...
∂y
⎝
⎠x
∂2 z
⎛ ∂N ⎞
⎜ ∂x ⎟ = ∂y
...

Theorem 2
...
Similarly any one of x, y and z may be regarded to
be a function of f and any one of x, y and z
...
Among the variables x, y, and z any one variable may be considered as a function
of the other two
...
The following relation holds good
⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞
⎜ ∂V ⎟ ⎜ ∂T ⎟ ⎜ ∂p ⎟ = −1
⎝
⎠T ⎝
⎠p ⎝
⎠v

Maxwell’s Equations

A pure substance existing in a single phase has only two independent variables
...

For a pure substance undergoing an infinitesimal reversible process
(a) dU = TdS - pdV
(b) dH = dU + pdV + VdP = TdS + Vdp
(c) dF = dU - TdS - SdT = - pdT - SdT
(d) dG = dH - TdS - SdT = Vdp - SdT
Since U, H, F and G are thermodynamic properties and exact differentials of the type
dz = M dx + N dy, then
⎛ ∂M ⎞
⎛ ∂N ⎞
⎜
⎟ =
∂y ⎠ x ⎜ ∂x ⎟ y
⎝
⎠
⎝

Applying this to the four equations

Page 180 of 265

Thermodynamic Relations

By: S K Mondal

Chapter 11

⎛ ∂T ⎞
⎛ ∂p ⎞
⎜ ∂V ⎟ = − ⎜ ∂S ⎟
⎝
⎠s
⎝
⎠v
⎛ ∂T ⎞
⎛ ∂V ⎞
⎜ ∂P ⎟ = ⎜ ∂S ⎟
⎝
⎠s ⎝
⎠p
⎛ ∂p ⎞
⎛ ∂S ⎞
⎜ ∂T ⎟ = ⎜ ∂V ⎟
⎝
⎠V ⎝
⎠T
⎛ ∂S ⎞
⎛ ∂V ⎞
⎜ ∂T ⎟ = − ⎜ ∂p ⎟
⎝
⎠P
⎝
⎠T

These four equations are known as Maxwell’s equations
...


Then                S = S ( T, V )

⎛ ∂S ⎞
⎛ ∂S ⎞
            or        dS = ⎜
dT + ⎜
⎟ dV
∂T ⎟ V
⎝
⎠
⎝ ∂V ⎠ T
multiplying both side by T

Since
and
∴

⎛ ∂S ⎞
⎛ ∂S ⎞
TdS = T ⎜
⎟ dT + T ⎜ ∂V ⎟ dV
⎝ ∂T ⎠ V
⎝
⎠T
⎛ ∂S ⎞
T⎜
⎟ = CV , heat capacity at constant volume
⎝ ∂T ⎠ V
⎛ ∂p ⎞
⎛ ∂S ⎞
⎜ ∂V ⎟ = ⎜ ∂T ⎟ by Maxwell 's equation
⎠V
⎝
⎠T ⎝
⎛ ∂p ⎞
TdS = CV dT + T ⎜
⎟ dV
⎝ ∂T ⎠ V

dividing both side by T
dS = CV

dT ⎛ ∂p ⎞
dV proved
+
T ⎜ ∂T ⎟ V
⎝
⎠

(ii) Derive:

⎛ ∂V ⎞
TdS = CpdT − T ⎜
⎟ dp
⎝ ∂T ⎠ p

[IES-1998]

Let entropy S be imagined as a function of T and p
...


(iii) Derive:

TdS = C V dT + T

k Cv dp Cp
β
dV = CpdT − TVβ dp =
dV
+
k
β
βV
[IES-2001]

We know that volume expansivity (β) =

1 ⎛ ∂V ⎞
V ⎜ ∂T ⎟ p
⎝
⎠

and

isothermal compressibility (k) = −

∴

1 ⎛ ∂V ⎞
V ⎜ ∂p ⎟ T
⎝
⎠

From first TdS equation

⎛ ∂p ⎞
TdS = CV dT + T ⎜
⎟ dV
⎝ ∂T ⎠ V
⎛ ∂V ⎞
⎜ ∂T ⎟
β
⎝
⎠p
⎛ ∂V ⎞ ⎛ ∂p ⎞
=−
= −⎜
⎟ ⋅⎜
⎟
k
⎛ ∂V ⎞
⎝ ∂T ⎠ p ⎝ ∂V ⎠ T
⎜ ∂p ⎟
⎝
⎠T

As

⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂p ⎞
⎜ ∂T ⎟ ⋅ ⎜ ∂p ⎟ ⋅ ⎜ ∂V ⎟ = − 1
⎝
⎠p ⎝
⎠T
⎠V ⎝

∴

⎛ ∂V ⎞ ⎛ ∂p ⎞
⎛ ∂p ⎞
−⎜
⎟ ⋅ ⎜ ∂V ⎟ = ⎜ ∂T ⎟
⎝ ∂T ⎠ p ⎝
⎠T ⎝
⎠V

or

β ⎛ ∂p ⎞
=
k ⎜ ∂T ⎟ V
⎝
⎠

β
⋅ dV
k
From second TdS relation
∴

TdS = C V dT + T ⋅

proved

Page 182 of 265

Thermodynamic Relations

By: S K Mondal

Chapter 11

⎛ ∂V ⎞
TdS = CpdT − T ⎜
⎟ dp
⎝ ∂T ⎠ p
as
∴
∴

β=

1 ⎛ ∂V ⎞
V ⎜ ∂T ⎟ p
⎝
⎠

⎛ ∂V ⎞
⎜ ∂T ⎟ = Vβ
⎝
⎠p
TdS = CpdT − TVβ dp

proved

Let S is a function of p, V
∴
S = S(p, V)
∴

⎛ ∂S ⎞

⎛ ∂S ⎞

dS = ⎜ ⎟ dp + ⎜ ⎟ dV
⎝ ∂V ⎠ p
⎝ ∂p ⎠ V

Multiply both side by T
⎛ ∂S ⎞
⎛ ∂S ⎞
TdS = T ⎜ ⎟ dp + T ⎜
⎟ dV
⎝ ∂V ⎠ p
⎝ ∂p ⎠ V
or

⎛ ∂S ∂T ⎞
⎛ ∂S ∂T ⎞
⋅
⋅
TdS = T ⎜
dp + T ⎜
⎟ dV
∂T ∂p ⎟ V
⎝ ∂T ∂V ⎠ p
⎝
⎠

or

⎛ ∂S ⎞ ⎛ ∂T ⎞
⎛ ∂S ⎞ ⎛ ∂T ⎞
⋅
TdS = T ⎜
dp + T ⎜
⎟ ⋅⎜
⎟ dV
∂T ⎟ V ⎜ ∂p ⎟ V
⎝
⎠ ⎝
⎝ ∂T ⎠ p ⎝ ∂V ⎠ p
⎠
⎛ ∂S ⎞
Cp = T ⎜
⎟
⎝ ∂T ⎠ p

∴

and

⎛ ∂S ⎞
CV = T ⎜
⎟
⎝ ∂T ⎠ V

⎛ ∂T ⎞
⎛ ∂T ⎞
TdS = Cv ⎜
⎟ dp + Cp ⎜ ∂V ⎟ dV
⎝
⎠p
⎝ ∂p ⎠ V
β ⎛ ∂p ⎞
=
k ⎜ ∂T ⎟ V
⎝
⎠

From first

or

k ⎛ ∂T ⎞
=
β ⎜ ∂p ⎟ V
⎝
⎠

k
⎛ ∂T ⎞
dp + Cp ⎜
⎟ dV
β
⎝ ∂V ⎠ p

∴

TdS = Cv

∴

β=

∴

1
⎛ ∂T ⎞
⎜ ∂V ⎟ = βV
⎝
⎠p

∴

TdS =

1 ⎛ ∂V ⎞
V ⎜ ∂T ⎟ p
⎝
⎠

Cv k dp Cp
+
dV
β
βV

proved
...



...
It indicates the following important
facts
...
(Cp – Cv) is always
⎝ ∂T ⎠ p
⎝
⎠T
positive
...

(b) As T → 0 K ,C p → Cv or at absolute zero, Cp = Cv
...
g for water at 4ºC, when density is maximum
...
Cp = Cv
...


From Tds relations

Page 185 of 265

Thermodynamic Relations

By: S K Mondal

Chapter 11

⎛ ∂V ⎞
⎛ ∂p ⎞
TdS = CpdT − T ⎜
⎟ dP = CV dT + T ⎜ ∂T ⎟ dV
⎝ ∂T ⎠ p
⎝
⎠V
∴

or

(C

p

⎛ ∂V ⎞
⎛ ∂p ⎞
− Cv ) dT = T ⎜
⎟ dP + T ⎜ ∂T ⎟ dV
⎝ ∂T ⎠ p
⎝
⎠v

⎛ ∂V ⎞
⎛ ∂p ⎞
T⎜
T⎜
∂T ⎟ p
∂T ⎟ V
⎝
⎠
⎠ dV − − − i
dT =
dP + ⎝
()
Cp − CV
Cp − CV

S in ce T is a function of ( p, V )
T = T ( p, V )
∴

⎛ ∂T ⎞
⎛ ∂T ⎞
dT = ⎜
⎟ dp + ⎜ ∂V ⎟ dV
⎝
⎠p
⎝ ∂p ⎠ V

− − − ( ii )

Compairing ( i ) & ( ii ) we get
⎛ ∂V ⎞
T⎜
⎟
⎝ ∂T ⎠ p ⎛ ∂T ⎞
=⎜
⎟
Cp − CV
⎝ ∂p ⎠ V

as

⎛ ∂V ⎞ ⎛ ∂p ⎞
Cp − CV = T ⎜
⎟ ⋅⎜
⎟
⎝ ∂T ⎠ p ⎝ ∂T ⎠ V
dU = dQ − pdV

∴

and

⎛ ∂p ⎞
T⎜
⎟
⎝ ∂T ⎠ V = ⎛ ∂T ⎞
⎜ ∂V ⎟
Cp − CV
⎝
⎠p

dU = TdS − pdV

∴

or
or

⎛ ∂U ⎞
⎛ ∂S ⎞
⎜ ∂V ⎟ = T ⎜ ∂V ⎟ − p
⎝
⎠T
⎝
⎠T
⎛ ∂U ⎞
⎛ ∂S ⎞
⎜ ∂V ⎟ + p = T ⎜ ∂V ⎟
⎝
⎠T
⎝
⎠T

From Maxwell 's Third relations
⎛ ∂p ⎞
⎛ ∂S ⎞
⎜ ∂T ⎟ = ⎜ ∂V ⎟
⎝
⎠V ⎝
⎠T
∴

⎧
⎛ ∂V ⎞ ⎛ ∂p ⎞
⎛ ∂U ⎞ ⎫ ⎛ ∂V ⎞
Cp − CV = T ⎜
⋅
= ⎨p + ⎜
⎟ ⎬⎜
⎟
∂T ⎟ p ⎜ ∂T ⎟ V ⎩
⎝
⎠ ⎝
⎠
⎝ ∂V ⎠ T ⎭ ⎝ ∂T ⎠ p

(vi) Prove that
Joule – Thomson co-efficient

T2 ⎡ ∂ ⎛ V ⎞ ⎤
⎛ ∂T ⎞
μ=⎜
⎜ ⎟
⎟ =
∂p ⎠h Cp ⎢ ∂T ⎝ T ⎠ ⎥ p
⎝
⎣
⎦

[IES-2002]

The numerical value of the slope of an isenthalpic on a T – p diagram at any point is called
the Joule – Kelvin coefficient
...
During the
evaporation, the pr
...


sg − s f
⎛ dp ⎞
⎜ dT ⎟ = v − v
⎝
⎠sat
g
f

∴

sg − sf = sfg =
or

h fg
T

h fg
⎛ dp ⎞
⎜ dT ⎟ =
⎝
⎠sat T ( v g − v f )

→ It is useful to estimate properties like h from other measurable properties
...
e
...

At very low pressure   g ≈ v f g as v f very small
v
pv g = RT
∴

or

vg =

RT
p

h fg
h fg
h ⋅p
dp
=
=
= fg 2
dT T ⋅ v g T ⋅ RT
RT
p

or

dp h fg dT
=
⋅
p
R T2

or

⎛ p ⎞ h fg ⎛ 1
1 ⎞
ln ⎜ 2 ⎟ =
⎜ − ⎟
R ⎝ T1 T2 ⎠
⎝ p1 ⎠

→ Knowing vapour pressure p1 at temperature T1, we may find out p2 at temperature T2
...
The Joule–Thomson coefficient μJ is defined as

⎛ ∂T ⎞

μJ = ⎜
⎟
⎝ ∂p ⎠h
Like other partial differential coefficients introduced in this section, the Joule–Thomson
coefficient is defined in terms of thermodynamic properties only and thus is itself a property
...

A relationship between the specific heat cp and the Joule–Thomson coefficient μJ can be
established to write

⎛ ∂T ⎞ ⎛ ∂p ⎞ ⎛ ∂h ⎞
⎜ ∂p ⎟ ⎜ ∂h ⎟ ⎜ ∂T ⎟ = − 1
⎠p
⎝
⎠ h ⎝ ⎠T ⎝
The first factor in this expression is the Joule–Thomson coefficient and the third is cp
...
The following expression results:

cp =

⎤
1 ⎡ ⎛ ∂v ⎞
T⎜
− v⎥
⎢
μJ ⎣ ⎝ ∂T ⎟ p ⎦
⎠

allows the value of cp at a state to be determined using p–v–T data and the value of the Joule–
Thomson coefficient at that state
...

The numerical value of the slope of an isenthalpic on a T-p diagram at any point is called the
Joule-Kelvin coefficient and is denoted by μJ
...
The region inside the inversion curve where μJ is positive is called the
cooling region and the region outside where μJ is negative is called the heating region
...
Two application of the equation are given below(a) For an ideal gas, p =

nRT
V
nR p
⎛ ∂p ⎞
∴⎜
=
=
V
T
∂T ⎟V
⎝
⎠
p
⎛ ∂U ⎞
∴⎜
⎟ = T
...

⎛ ∂U ⎞ ⎛ ∂p ⎞ ⎛ ∂V ⎞
⎜
⎟ ⎜
⎟ ⎜
⎟ =1
⎝ ∂p ⎠T ⎝ ∂V ⎠T ⎝ ∂U ⎠T
⎛ ∂U ⎞ ⎛ ∂p ⎞
⎛ ∂U ⎞
⎜
⎟ ⎜
⎟ = ⎜ ∂V ⎟ = 0
⎠T
⎝ ∂p ⎠T ⎝ ∂V ⎠T ⎝
⎛ ∂U ⎞
⎛ ∂p ⎞
since ⎜
⎟ ≠ 0, ⎜ ∂p ⎟ = 0
⎝ ∂V ⎠T
⎝
⎠T

U does not change either when p changes at T = C
...

Another important point to note is that for an ideal gas
⎛ ∂p ⎞
pV = nRT and T ⎜
⎟ −p=0
⎝ ∂T ⎠v

Page 189 of 265

Thermodynamic Relations

By: S K Mondal

Chapter 11

Therefore
dU = Cv dT
holds good for an ideal gas in any process (even when the volume changes)
...
shown in above that the
enthalpy of an ideal gas is not a function of either volume or pressure
...
e ⎜
⎟ = 0 and ⎜
⎟ = 0⎥
⎝ ∂V ⎠T
⎣ ⎝ ∂p ⎠T
⎦

but a function of temperature alone
...
)
However, for any other substance the relation dH = Cp dT holds good only when the pressure
remains constant or dp = 0
...
The energy density (u), defined as the ratio of energy to
volume, is a function of temperature only, or
u=

U
= f (T )only
...

since
...

T du u
−
3 dT 3
du
dT
∴
=4
u
T
u=

Page 190 of 265

Thermodynamic Relations

By: S K Mondal

Chapter 11

or
ln u = ln T4 + lnb
or
u = bT4
where b is a constant
...

Since U = uV = VbT 4

and

⎛ ∂U ⎞
3
⎜ ∂T ⎟ = Cv = 4VbT
⎝
⎠V
1 du 4
⎛ ∂p ⎞
3
⎜ ∂T ⎟ = 3 dT = 3 bT
⎝
⎠V

From the first TdS equation
⎛ ∂p ⎞
TdS = Cv dT + T ⎜
⎟ dV
⎝ ∂T ⎠v
4
= 4VbT 3 dT + bT 4
...

Q=

4
bT 4 ΔV
3

For a reversible adiabatic change of volume
4
bT 4 dV = −4VbT 3 dT
3
dV
dT
or
= −3
V
T
3
or VT = const

If the temperature is one-half the original temperature
...


Gibbs Phase Rule
Gibbs Phase Rule determines what is expected to define the state of a system

F=C–P+2
F = Number of degrees of freedom (i
...
, no
...
g
...
Therefore, F = 2
•
•
•
•
•

To determine the state of the nitrogen gas in a cylinder two properties are adequate
...
If any one property is specified it is sufficient
...
The triple point is uniquely defined
...
dv ⎞
(c ) ∫ ⎜
+
⎟
v ⎠
⎝ T

⎛ dT v
...

Thus

∫ pdv and ∫ vdp are not exact differentials and thus not properties
...
If
the differential is of the form Mdx + Ndy, then the test for exactness is

⎡ ∂M ⎤ ⎡ ∂N ⎤
⎢ ∂y ⎥ = ⎢ ∂x ⎥
⎣
⎦x ⎣ ⎦ y
Now applying above test for
2
R
⎛ dT p
...
dv ⎞
⎟ is not a point function and hence not a property
...
dp ⎞ ⎡ ∂ (1/ T ) ⎤
⎡ ∂ ( −v / T ) ⎤
⎡ ∂ (− R / P) ⎤
And for ∫ ⎜
−
⎟⎢
⎥ = ⎢ ∂T ⎥ = ⎢ ∂T
⎥ or 0 = 0
T ⎠ ⎣ ∂p ⎦T ⎣
⎦P ⎣
⎦P
⎝ T

∫⎜ T
⎝

+

Thus

∫⎜ T
⎝

⎛ dT

−

v
...
Vapour Power Cycles
Some Important Notes
A
...
F
...
E
...

For reversible Adiabatic Compression Tds = dh – vdp where ds = 0
∴
dh = vdp as v = constant
Δh = vΔp
or
h 4 − h3 = v(p1 − p2 ) = WP

(iv)
B
...


Steam rate =

D
...


Chapter 12

About Turbine Losses: If there is heat loss to the surroundings, h2 will decrease,
accompanied by a decrease in entropy
...
(figure in below)
...

F
...


Mean temperature of heat addition:

Q1 = h1 − h 4 s = Tm (s1 − s4 s )

∴

Tm =

h1 − h 4 s
s1 − s4 s

1
5
T
↑

Tm

4s
2s

3
→S

Page 194 of 265

Vapour Power Cycles

By: S K Mondal
H
...
For calculation of m2
...


For Binary vapour Cycles:

m kg
a
1

d

b

c

T

5

1 kg 6

4
3

2
S

WT = m (ha – hb) + (h1 – h2) kJ/kg of steam
WP = m (hd – hc) + (h4 – h3) kJ /kg of steam
Q1 = m (ha – hd) + (h1 – h6) + (h5 – h4) kJ /kg of steam
...
e
...


Chapter 12

Efficiency of Binary vapour cycle:
1 – η = (1 − η1 ) (1 − η2 )
...


Overall efficiency of a power plant
ηoverall = ηboiler × ηcycle × ηturbine (mean) × ηgenerator

Questions with Solution P
...
Nag
Q
...
1

for the following steam cycles find
(a) WT in kJ/kg

(b) Wp in kJ/kg,

(d) cycle efficiency,
(c) Q1 in kJ/kg,
(e) steam rate in kg/kW h, and
(f) moisture at the end of the turbine process
...

Boiler Outlet

Type of Cycle

Condenser Pressure

10 bar, saturated 1 bar

Ideal Rankine Cycle

-do-

-do-

Neglect Wp

-do-

-do-

-do-

0
...
1 bar
Boiler Outlet

Condenser Pressure

10 bar, saturated 0
...
1 kJ/kg

2
S

s1 = 6
...
5865 = 1
...
3594 – 1
...
8724
h 2 = 417
...
8724 × 2258 = 2387
...
46 kJ/kg

∴
(a)
(b)
(c)
(d)

(e)
(f)

h4 = h3 + WP
WP = 1
...
94 kJ/kg
h4 = 418
...
1 – 2387
...
8 kJ/kg
WP = 0
...
1 – 418
...
7 kJ/kg
W
W − NP
390
...
94
Cycle efficiency (η) = net = T
=
2359
...
52%
3600
3600
Steam rate =
kJ / kWh =
= 9
...
8 − 0
...
1276 ≅ 12
...
12
...
Steam wells are drilled to tap this steam supply which is
available at 4
...
The steam leaves the turbine at 100 mm
Hg absolute pressure
...
75
...
If the unit produces 12
...
5 bar
T1 = 175ºC
From super heated STEAM TABLE
...
8
s = 6
...
5
s = 7
...
7
s = 6
...
4
s = 7
...
8 + (2860
...
8)
2
⎛ 175 − 152 ⎞
h = 2748
...
4 − 2748
...
7 kJ/kg
= 2800 kJ/kg
1
23
s = 6
...
1706 − 6
...
8213 + (7
...
8213)
2
48
= 7
...
9353
∴
at 4
...
7 + 2800
h1 =
= 2803
...
0503 + 6
...
9928 kJ/kg – K
2
Pressure 100 mm Hg
100
=
m × (13
...
81 m/s2
1000
= 0
...
342 kPa
Here also entropy 6
...
T
...
83
sf = 0
...
7549 hf = 225
...
8

sg = 8
...
0085

hfg = 2373
...
342 kPa [Interpolation]
⎛ 15 − 13
...
6493 + ⎜
⎟ (0
...
6493) = 0
...
342 ⎞
⎟ (8
...
1502) = 8
...
9928 = 0
...
1032 – 0
...
85033
At 13
...
1502 + ⎜
⎝ 15 − 10
∴
∴
∴

Vapour Power Cycles

By: S K Mondal

Chapter 12

⎛ 15 − 13
...
83 + ⎜
⎟ (225
...
83) = 203
...
342 ⎞
h fg = 2392
...
1 – 2392
...
3 kJ/kg
⎝ 15 − 10 ⎠
h2s = hf + x hfg = 203
...
85033 × 2386
...
3 kJ/kg

ηisentropic =
∴

h1 − h 2′
h1 − h 2s

h1 − h 2′ = ηisentropic × (h1 – h2s)
h′2 = h1 – ηisentropic (h1 – h2s)

∴

= 2803
...
75 (2803
...
3) = 2375 kJ/kg
...
4 – 2373) %
= 428
...
36
≈ 0
...
28%
∴
Efficiency of the plant = T =
2803
...
WT = 12
...
5 × 103
= 29
...
36

Q
...
3

A simple steam power cycle uses solar energy for the heat input
...
It then evaporates in the boiler at this pressure, and enters the
turbine as saturated vapour
...
The flow rate is 150 kg/h
...
58 kW/ m 2
...
(c) 2
...
2 m 2 )

Solution:

From Steam Table

T1 = 120
...
23 K
h1 = 2706
...
1271 kJ/kg – K

2 bar
1
T

4
2s 2

3

At 40°C saturated pressure 7
...
57

S
hfg = 2406
...
5725

sg = 8
...
9 × 2406
...
6 kJ/kg
For h2s if there is dryness fraction x
7
...
5725 + x × (8
...
5725)
∴
x = 0
...
57 + 0
...
7 = 2220
...
7 − 2333
...
72%
2706
...
4
(b)
Net work output WT = h1 – h2 = 373
...
55 kW i
...
(WT − WP ) × m
Pump work, WP = v ( p1 – p2 )

∴
∴
∴

= 1
...
384) kJ/kg = 0
...
57 kJ/kg,
ha = 167
...
7 – 167
...
1 − 0
...
69 %
2539
Q1
•

Q1 × m
Required area A =
collection picup
2539 × 150
= 182
...
58 × 3600

Q
...
4

Solution :

In a reheat cycle, the initial steam pressure and the maximum
temperature are 150 bar and 550°C respectively
...
1 bar and the moisture at the condenser inlet is 5%, and
assuming ideal processes, determine (a) the reheat pressure, (b) the
cycle efficiency, and (c) the steam rate
...
13
...
6%,2
...
6 kJ/kg
s1 = 6
...
1 bar

T = 45
...
8 kJ/kg

h fg = 2392
...
8 + 0
...
8 = 2465 kJ/kg
sf = 0
...
501

s4 = sf + x sfg = 0
...
95 × 7
...
775 kJ/kg – K
From Molier Diagram at 550°C and 7
...
25 bar
From S
...
at 10 bar 550°C
s = 7
...
7045
⎛ p − 10 ⎞
∴
7
...
8955 + ⎜
⎟ (7
...
8955)
⎝ 15 − 10 ⎠
–0
...
0382)
∴
p – 10 = 3
...
15 bar
∴
from Molier Dia
...
8 kJ/kg
WP = v 5 ( p1 – p3 ) = 0
...
505 kJ/kg
h6 = h5 + WP = 193
...
6 kJ/kg
WP = 1
...
5 kJ/kg
Q = (h1 – h6) + (h3 – h2) = 4040
...
5
ηcycle = net =
× 100 % = 43
...
3
Q

Steam rate =

Q
...
5

3600
3600
kg / kWh = 2
...
5
WT − WP

In a nuclear power-plant heat is transferred in the reactor to liquid
sodium
...
The steam leaves this heat exchanger as
saturated vapour at 55 bar, and is then superheated in an external gasfired super heater to 650°C
...
The turbine efficiency is 75% and the condenser
temperature is 40°C
...

Page 201 of 265

Vapour Power Cycles

By: S K Mondal
Solution:

Chapter 12

From Steam Table at 55 bar saturated state
h9 = 2789
...
T
...
5
h = 3900
...
3
s = 7
...
5122
s = 7
...
4
s = 7
...
2
s = 7
...
3
s = 7
...
8 kJ/kg
s1 = 7
...
3406
At 200°C
at 250°C if temp is t
s = 7
...
379
h = 2860
...
2
⎛ 7
...
171 ⎞
Then h2 = 2860
...
2 − 2860
...
379 − 7
...
7
hf = 167
...
573

sfg = 7
...
573 + x × 7
...
3406
∴
x = 0
...
6 + 0
...
7 = 2287 kJ/kg
h4 = hf = 167
...
001010 × (400 – 7
...
397 kJ/kg ≈ 0
...
7 kJ/kg

[at 4 bar saturated liquid]
Page 202 of 265

Vapour Power Cycles

By: S K Mondal

Chapter 12

WP6 − 7 = v 6 ( p1 – p2 ) = 0
...
53 kJ/kg
∴

h7 = h 6 + WP6 − 7 = 610
...
1622 kg
(1 – m) h5 + mh2 = h5
∴
WT = [(h1 – h2) + (1 – m) (h2 – h3) × 0
...
8 kJ/kg ;
Wnet = WT − WP4 − 5 − WP6 − 7 = 1043
...
638 kg/s
1049
...
638(2789
...
23) = 167
...
638(3779
...
9) = 75
...
12
...
12
...
12
...
12
...
p
...
It is reheated at constant pressure to 400°C and then
expands in a l
...
turbine to 40°C
...
p
...
Assume all
ideal processes
...

Try please
...
08 bar
...
5 bar and 0
...

Calculate the thermal efficiency of the plant, neglecting pump work
...

The net power output of the turbine in an ideal reheat-regenertive cycle
is 100 MW
...
P
...
After
...
07 bar
...
P
...
(d) If there is a
10°c rise in the temperature of the cooling 'water, what is the rate of
flow of the cooling water in the condenser? (e) If the velocity of the
steam flowing from the turbine to the condenser is limited to a
maximum of 130 m/s, find the diameter of the connecting pipe
...

A mercury cycle is superposed on the steam cycle operating between the
boiler outlet condition of 40 bar, 400°C and the condenser temperature
Page 203 of 265

Vapour Power Cycles

By: S K Mondal

Chapter 12

of 40°C
...
2 bar is used to
impart the latent heat of vaporization to the water in the steam cycle
...

Compute (a) kg of mercury circulated per kg of water, and (b) the
efficiency of the combined cycle
...
5

72
...
0

0
...
5167

0
...
3

38
...
55

0
...
6385

80
...
0333
77
...
163

Solution:

Try please
...
12
...
At an evaporation rate of
1,000,000 kg/h for the mercury, its specific enthalpy rises by 356 kJ/kg in
passing through the boiler
...
The mercury gives up 251
...
The overall boiler efficiency is 85%
...
The steam auxiliaries require 5% of the energy generated by
the units
...

Try please

Solution:
Q
...
11

A sodium-mercury-steam cycle operates between l000°C and 40°C
...
Mercury boils at 24
...
141 bar
...
Steam is formed at 30 bar and is superheated in the sodium
boiler to 350°C
...
0 8 bar
...
Find (a)
the amounts of sodium and mercury used per kg of steam, (b) the heat
added and rejected in the composite cycle per kg steam, (c) the total
work done per kg steam
...

For mercury, at 24
...
78 kJ/kg

sg = 0
...
141 bar, s j =0
...
64kJ/kg K, h j =36
...
77 kJ/kg

For sodium, at 1000°C, hg = 4982
...
85 kJ/kg
At 670°C, hf = 745
...

Try please
...
12
...
12
...
12
...
12
...
Find the steam condition
required at the inlet to the turbine
...

A 10,000 kW steam turbine operates with steam at the inlet at 40 bar,
400°C and exhausts at 0
...
Ten thousand kg/h of steam at 3 bar are to
be extracted for process work
...
Find the boiler capacity required
...

A 50 MW steam plant built in 1935 operates with steam at the inlet at 60
bar, 450°C and exhausts at 0
...
It is
proposed to scrap the old boiler and put in a new boiler and a topping
turbine of efficiency 85% operating with inlet steam at 180 bar, 500°C
...
The flow rate is just sufficient to produce
the rated output from the old turbine
...
What is the additional power developed?
Try please
...
The
condensate from the heating system is returned to the boiler plant at
65°C, and the heating system utilizes for its intended purpose 90% of the
energy transferred from the steam it receives
...
(a) What fraction of the energy supplied to the steam plant serves a
useful purpose? (b) If two separate steam plants had been set up to
produce the same useful energy, one to generate heating steam at 2 bar,
and the other to generate power through a cycle working between 20
bar, 400°C and 0
...
91
...
5%)
From S
...
at 20 bar 400°C
h1 = 3247
...
127 kJ/kg – K
1
20 bar

T

3
65°C

At 2 bar

2 bar

4

Q0

2

S
Page 205 of 265

Vapour Power Cycles

By: S K Mondal

Chapter 12

sf = 1
...
5967
sg = 7
...
3 kJ/kg
At 2 bar saturated temperature is 120
...
7 – 4
...
2 – 65) = 273
...
001 × (2000 – 200) = 1
...
4 kJ/kg
∴ Heat input (Q) = h1 – h4 = (3247
...
4) = 2972
...
6 – 2706
...
7 kJ/kg
= 378
...
3 – 273
...
9 = 2189
...
9 – 1
...
1 kJ/kg
∴ Fraction at energy supplied utilized

Wnet + Q0
377
...
4
=
× 100%
2972
...
35%

=

(b) At 0
...
559, sfg = 7
...
559 + x × 7
...
127
∴
x = 0
...
4 + 0
...
1 = 2214
...
4 kJ/kg
∴
∴

WP = 0
...
4 kJ/g
...
7 = 723
...
24 kJ/kg

Here heat input for power = (h1 – h4) = 3082
...
kJ/kg
For same 377
...
52285 kg of water
So heat input = 1611
...
4
kJ = 2432
...
9
377
...
4
× 100%
∴ Fraction of energy used =
1611
...
7
= 63
...
12
...
p
...
841
...
Saturated vapour leaves the separator and is
Page 206 of 265

Vapour Power Cycles

By: S K Mondal

Solution:

Chapter 12

expanded isentropically to 0
...
p
...

The condensate leaving the condenser at 0
...
Calculate the cycle efficiency and turbine
outlet quality taking into account the feed pump term
...

(Ans
...
5%, 0
...
716)
Form Steam Table at 30 bar saturated
h1 = 2802
...
1837

From Molier diagram
h2 = 2300 kJ/kg
pr = 2
...
T
...
5 kJ/kg, sg = 7
...
2°C
hf = 551
...
04 bar S
...
423 kJ/kg, sfg = 8
...
5 kJ/kg, hfg = 2432
...
7 bar
4

S
∴
If dryness fraction is x the
7
...
423 + x × 8
...
8186
∴
h4 = hf + x hfg = 2113 kJ/kg
WP5– 6 = 0
...
001071(3000 – 280) = 2
...
3 kJ/kg
h2 – 2380 kJ/kg
h3 – 2721
...
5 kJ/kg
h6 = 124
...
4 kJ/kg
h8 = 554
...
841 = 0
...
98 kJ/kg ≈ 3 kJ/kg
Page 207 of 265

Vapour Power Cycles

By: S K Mondal

Chapter 12

∴ Wnet = 931 kJ/kg
Heat supplied (Q) = m(h1– h8) + (1 – m) (h1 – h6) = 2609
...
68% with turbine exhaust quality 0
...
5
If No separation is taking place, Then is quality of exhaust is x
⇒ x = 0
...
1837 = 0
...
052
∴η=

∴

h4 = hf + x × hfg = 1862 kJ/kg

∴

WT = h1-h4 = 941
...
28 kJ/kg
∴
Heat input, Q = h1 – h6 = 2677
...
28
∴
η=
× 100% = 35%
2677
...
12
...
Steam enters the h
...
turbine at 80 bar, 500°C and expands till it
becomes saturated vapour
...
p
...
07 bar
...
p
...

Neglect pump work
...
6
...
4 kg/s, 43
...
T of 80 bar 500°C
h1 = 3398
...
724 kJ/kg – K
s2 = 6
...
6 bar so
Reheat pr
...
6 bar

1 80 bar 500°C
3
400°C
8

T

7

6

m kg

(1 – m) kg

2

0
...
5 kJ/kg
h3 = 3270
...
6(3268
...
3) = 3269
...
708 + 0
...
635 – 7
...
6643 kJ/kg – K
At 0
...
4,

hfg = 2409
...
559,

Chapter 12

sfg = 7
...
6642 = 0
...
717 ⇒ x = 0
...
4 + 0
...
1 = 2381
...
8 kJ/kg ≈ h8
h5 = 163
...
2016 kg/kg of steam at H
...
7984
WT = h1 – h2 + (1 – m) (h3 –h4) = 1347
...
5 kJ/kg at H
...
6
...
P =

(c) Cycle efficiency (η) =
Q
...
18

80 × 103
kg/s = 59
...
6

W
1347
...
21%
3118
...
The steam leaves boiler B at
30 bar, 320°C and expands in the H
...
turbine to 2 bar, the efficiency of
the H
...
turbine being 75%
...
The dry steam enters the L
...
turbine at 2 bar
and expands to the condenser pressure 0
...
P
...
The drainage from the

Page 209 of 265

Vapour Power Cycles

By: S K Mondal

Solution:

Chapter 12

Separator mixes with the condensate from the process heater and the
combined flow enters the hotwell H at 50°C
...
A pump extracts the condensate from condenser C
and this enters the hot well at 38°C
...
Also calculate, as percentage of heat
transferred in the boiler, (a) the heat transferred in the process heater,
and (b) the work done in the turbines
...


Q
...
19

In a combined power and process plant the boiler generates 21,000 kg/h
of steam at a pressure of 17 bar, and temperature 230 °C
...
56 kW, the steam
leaving the process heater 0
...
5 bar
...
P
...
5 bar
...
P
...
5 kW
...
3 bar, and the steam is 0
...
Draw a line
diagram of the plant and determine (a) the steam quality at the exhaust
from the H
...
turbine, (b) the power developed by the H
...
turbine, and
(c) the isentropic efficiency of the H
...
turbine
...
(a) 0
...
8
2923
...
455
6
...
8 +
(2623
...
8) = 2872
...
709 − 6
...
6074 kJ/kg
s = 6
...
4°C
250°C
Page 210 of 265

Con

Vapour Power Cycles

By: S K Mondal

Chapter 12

h = 2797
...
5
s = 6
...
545
∴
at 230°C
17
...
2 +
(2902
...
2) = 2846
...
6
17
...
545 − 6
...
434 kJ/kg
s = 6
...
6
∴
at 17 bar 230°C
2
h1 = 2872
...
5 − 2872
...
2 kJ/kg
5
2
s1 = 6
...
434 − 6
...
5381 kJ/kg
5
h2 = 871
...
957 × 1921
...
7 kJ/kg ≈ h3
h4 = ?
∴
Mass flow through process heater
132
...
97597 kg/s = 3513
...
5 kJ/kg = 4
...
5 h4 + 3513
...
(i)
h6 = 289
...
912 × 2336
...
8 kJ/kg
•

WT = m(h5 − h 6 )
WT

⎛ 1337
...
8 ⎟ = 2649
...
7 kJ/kg

•

At 3–5 bar hg = 2731
...
7 = 584
...
4 ⇒ x = 0
...
5
(2862
...
7) kJ/kg = 1095 kW
=
3600
At 3
...
7273, sfg = 5
...
5381 = 1
...
2119
x = 0
...
3 + 0
...
4 = 2566
...
2 − 2636
...
= 1
=
× 100% = 76
...
2 − 2566
...
12
...
6 MW and the heating load is
1
...
Steam is generated at 40 bar and 500°C and is expanded
isentropically through a turbine to a condenser at 0
...
The heating
load is supplied by extracting steam from the turbine at 2 bar which
condensed in the process heater to saturated liquid at 2 bar and then
pumped back to the boiler
...

(Ans
...
07 t/h, (b) 71
...
607 MW)
From steam table at 40 bar 500°C
Page 211 of 265

Vapour Power Cycles

By: S K Mondal

Chapter 12

h1= 3445
...
090 kJ/kg

T

5

1
1 kg

7
6
4

Q0

2

m kg

(1 – m) kg

(1 – m )k g
3

S

→ at 2 bar

sf = 1
...
5967
∴
∴

7
...
5301 + x × 5
...
9934
h2 = 504
...
9934 × 2201
...
8 kJ/kg
→ at 0
...
521, sfg = 7
...
809 ⇒ x = 0
...
5 + 0
...
9 = 2183
...
5 kJ/kg
h6 = 504
...
001006 × (4000 – 6) = 4 kJ/kg
h5 = h4 + WP = 155
...
001061 × (4000 – 100) = 4 kJ/kg

so
h7 = h6 + WP = 508
...
8 – 504
...
1 kJ/kg
For WT = (h1 – h2) + (1 – m) (h2 – h3)
= 753
...
5 – 508 m
∴ Wnet = WT – WP4 −5 (1 − m) − m WP6 −7
= (1257
...
5 – 508m) = 5600
wm × 2187
...
668 kg/s = 16
...
11391 kg/kg of the generation
(a)

Steam generation capacity of boiler = 16
...
(i)

...
12
...
169 MW
Heat rejection to the condenser
= (1 – m) (h3 – h4) = 8
...
5 bar
...
07 bar and the condition line may be assumed to be
straight (the condition line is the locus passing through the states of
steam leaving the various stages of the turbine)
...
4 kW, estimate the maximum
steam flow through the high and low pressure stages
...

(Ans
...
543 and 1
...
743 + 6
...
6625 kJ/kg
s1 =
2
3115
...
5
h1 =
= 3103
...
7273
at 3
...
2119
∴
if condition of steam is x1
6
...
7273 + x1 x 5
...
9469
∴ h2 = 584
...
9469 × 2147
...
7 kJ/kg
At 0
...
559
sfg = 7
...
6625 = 0
...
717

⇒ x 2 = 0
...
5 bar

m kg

5

2
0
...
4 + 07909 × 2409
...
8 kJ/kg
h4 = 163
...
3 kJ/kg
WP4 − 5 = 0
...
5 kJ/kg
Page 213 of 265

Vapour Power Cycles

By: S K Mondal
∴

Chapter 12

h5 = h4 + WP4 − 5 = 166
...
001079 (3500 – 350) = 3
...
7 kJ/kg

Let boiler steam generation rate = w kg/s
∴
WT = w [(h1 – h2) + (1 – m) (h2 – h3)]
Wnet = w [(h1 – h2) + (1 – m) (h2 – h3) – (1 – m) 3
...
4] kW
= w [486
...
4 – 542 m]
= w [1031
...
4) kW
Here w [1031
...
4 = 1400

...
3311 kg/s
;
m = 0
...
P
∴
AT H
...
3311 kg/s
At L
...
643 kg/s
Q
...
22

Solution :

Geothermal energy from a natural geyser can be obtained as a
continuous supply of steam 0
...
This is utilized in a mixed-pressure cycle to augment the
superheated exhaust from a high pressure turbine of 83% internal
efficiency, which is supplied with 5500 kg/h of steam at 40 bar and 500 °c
...
10 bar in a low pressure turbine of 78% internal
efficiency
...

(Ans
...
3 kJ/kg
s1 = 7
...
7 kJ/kg, hfg = 2201
...
5301 kJ/kg, sfg = 5
...
090 = 1
...
5967
∴
x1 = 0
...
7 + 0
...
6 kJ/kg = 2691
...
= 1
h1 − h 2s
∴
h2 – h1 – ηin (h1 – h2s)
= 3445
...
83 (3445
...
8) = 2819
...
31 kJ/kg – K
From molier diagram
Adiabatic mixing
h5 = 504
...
87 × 2201
...
3 kJ/kg from molier dia at 2 bar 2688
...
7 + x 2 × 2201
...
3 ⇒ x 3 = 0
...
5301 + 0
...
5967 = 7
...
1 bar
sf = 0
...
501
∴ x4 × 7
...
649 = 7
...
8575
∴
h4s = 191
...
8575 × 2392
...
6 kJ/kg
5500
(h1 − h 2 ) = 955
...
P =
3600
8200
(h3 − h 4 s ) × 0
...
08 kW
WTL
...
6 kW
5500
WP =
× 0
...
16 kW
3600
∴

Wnet = 1739
...
8 kJ/kg, h6 = h5 + WP = 195
...
12
...
5 kW
3600

1739
...
04%
4964
...
A binary cycle is proposed,
using Refrigerant 12 as the low temperature fluid, and water as the high
temperature fluid
...
06 bar, at which pressure it is
condensed by the generation of dry saturated refrigerant vapour at 30°C
from saturated liquid at -40°C
...
Determine the mass ratio of R-12 to water and the
efficiency of the cycle
...

(Ans
...
86; 44
...
)
Page 215 of 265

Vapour Power Cycles

By: S K Mondal
Solution :

Chapter 12

at 80 bar 500°C
h1 = 3398
...
724 kJ/kg – K
80 bar
1
1 kg
0
...
06 bar if quality is x, then
6
...
521 + x 2 × 7
...
79434
∴
h2s = 151
...
79434 × 2415
...
5 kJ/kg
h3 = 151
...
001006 (8000 – 6) = 8 kJ/kg
∴
h4 = 159
...
3 – 2070
...
81 = 1075
...
5 kJ/kg
Q1 = h1 – h4 = 3238
...
3 kJ/kg

For R-12
at 30°C saturated vapour
ha = 199
...
45 bar sg = 0
...
7274, ∴ if dryness x b then

sf = 0,
xb × 0
...
6854
∴
∴
hC = 0

⇒ x b = 0
...
0 kJ/kg
hb – 0
...
24 kJ/kg

⎛ 0
...
77 ⎞
−3
WP = Vc (pa – pc ) = ⎜
⎟ × 10 (745 – 64
...
4868 kJ/kg
2
⎝
⎠
∴
hb = hC + WP = 0
...
6 – 0
...
11 m = 2171
...
905 kg of R-12/kg of water

Power output WTR = m (ha – hb) × η
= 10
...
6 – 159
...
83 = 365
...
8 kJ/kg of steam
Woutput = Wnet H2 O + Wnet R12
= (1067
...
8) = 1432
...
32
η=
× 100% = 44
...
8
Heat input

∴

Q
...
24

Solution :

Chapter 12

Steam is generated at 70 bar, 500°C and expands in a turbine to 30 bar
with an isentropic efficiency of 77%
...
The mixture then expands with
an isentropic efficiency of80% to 0
...
At a point in the expansion
where me pressure is 5 bar, steam is bled for feedwater heating in a
direct contact heater, which raises the feed water to the saturation
temperature of the bled steam
...
Assume that the L
...

expansion condition line is straight
...
0
...
9%)
From Steam Table 70 bar 500°C
h1 = 3410
...
798 kJ/kg – K
s1 at 30 bar 400°C

h′3 = 3230
...
921 kJ/kg
′
From Molier diagram h2s = 3130 kJ/kg
∴
h2 = h1 – ηisentropic × (h1 – h2s)
= 3410
...
77 (3410
...
5 kJ/kg
For adiabatic mixing
1 × h2 + 2 × h′3 = 3 × h3

∴
∴

h3 = 3218
...
875 kJ/kg
(From Molier diagram)
h4′ = 2785 kJ/kg
h5s = 2140 kJ/kg
h5 = h3 – η(h5 – h5s) = 3218
...
80 (3218
...
06 bar
5s

S
From S
...
P
...
5 + 0
...
5 kJ/kg
m × 2920 + (3 – m) × 152 = 3 × 640
...
529 kg
hg = 640
...
P = 1 × (h1 – h2) = (3410
...
5) kJ/kg = 215
...
P = 3 (h3 – h4) + (3 – m) (h4 – h5)
= 3 (3218
...
529) (2920 – 2355
...
5 kJ/kg of steam H
...
001(3000 – 500)
+ 1 × 0
...
74 kJ/kg of H
...
8 + 2290
...
74) kJ/kg & H
...
6 kJ/kg of H
...
6 kJ/kg
h9 = h8 + WP8 − 9

= (3410
...
6) + 2 (3230
...
6)
= 7940
...
P steam
∴

Q
...
25
Solution:

ηcycle =

= 642
...
6
× 100% = 31
...
3

An ideal steam power plant operates between 70 bar, 550°C and 0
...
It has seven feed water heaters
...

Try please
...
12
...
p
...
It is reheated at constant pressure to 400°C and then
expands in a I
...
turbine to 40°C
...
67%, find (a) the reheat pressure, (b) the
pressure of steam at inlet to the h
...
turbine, (c) the net work output per
kg, and (d) the cycle efficiency
...

(Ans
...
8%)

Solution:

From S
...
at 40°C, 14
...
8533

1 p
1
3

T
6
5

p2

2
p3
4

S
Page 218 of 265

Vapour Power Cycles

By: S K Mondal

Chapter 12

p3 = 0
...
6 kJ/kg; hfg = 2406
...
6 + 0
...
7
= 2221
...
573 kJ/kg – K
sfg = 7
...
573 + 0
...
685 = 7
...
1306 From Steam Table
Pr = 20 bar,
At 20 bar saturation
h2 = 2797
...
6 kJ/kg
S2 = 6
...
3366 kJ/kg – k
From Steam Table
Pr = 200 bar
∴
h1 = 3393
...
6 kJ/kg
WP = 0
...
38) kJ/kg
∴
h6 = h5 + W = 187
...
7 kJ/kg

(c)

∴

Wnet = WT – WP = 1602
...
5 – 187
...
6 – 2797
...
3 kJ/kg
(d)
Q
...
27

Solution:

∴

η=

1602
...
83 %
3656
...
The condenser pressure is 0
...
8778
...
7592
...

(Ans
...
51%, (d) l
...
1 bar (saturated S
...
)
sf = 0
...
501 kJ/kg – K
∴

s4 = s3 = 0
...
8778 × 7
...
233 kJ/kg – K
s4 " = s1 = 0
...
7592 × 7
...
344 kJ/kg – K

h4 = 191
...
8778 × 2392
...
2 kJ/kg
From super heated steam turbine at 500°C 7
...
5 kJ/kg, h2 = 2892
...
1 bar

5

4′

4

S

At 500°C and 6
...
6 kJ/kg – K
h5 = 191
...
001010 (15000 – 10) =15
...
94 kJ/kg
∴
WT =(h1 – h2) + (h3 – h4) = 1580
...
46 kJ/kg
Q1 = (h1 – h6) + (h3 – h2) = 3665
...
45
∴
η=
≈ 42
...
86
Q
...
28

Solution:

In a cogeneration plant, steam enters the h
...
stage of a two-stage
turbine at 1 MPa, 200°C and leaves it at 0
...
At this point some of
the steam is bled off and passed through a heat exchanger which it
leaves as saturated liquid at 0
...
The remaining steam expands in
the I
...
Stage of the turbine to 40 kPa
...
Calculate the required mass flow rate of steam
into the h
...
stage of the turbine
...
80
...
2
...
T at 1 MPa 200°C
i
...
9 kJ/kg
s1 = 6
...
6716 sfg = 5
...
694 = 1
...
3193
x′ = 0
...
4 + 0
...
9 kJ/kg
h2 = h1 – η(h1 – h2s) = 2827
...
8 (2827
...
0) kJ/kg
= 2648
...
6716 + x2′ × 5
...
3 [2648
...
4 + x2 × 2163
...
8042 kJ/kg – K
If at 3s condition of steam is x3 then 40 kPa = 0
...
8697
6
...
0261 + x 3 × 6
...
7 + 0
...
2 = 2334
...
3 kJ/kg
h4 = 317
...
001 × (1000 – 40) ≈ 1 kJ/kg

∴

h5 = 318
...
1 kJ/kg, WP6 − 7 = 0
...
7 kJ/kg
h7 = 562
...
7
= (429
...
1 m) kJ/kg of steam of H
...
3
If mass flow rate of w then
w (429
...
1 m) = 1000
mw × 2087
...
4678 kg/s

∴

required mass flow rate in H
...
4678 kg/s

∴

Page 221 of 265

 

Page 222 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13

13
...
Compression ratio, (rc)
rc =

Volume at the begining of compression (V1 )
Volume at the end of compression (V2 )
∴

rc =

2
p

V1
bigger term
=
V2
smaller term

1
V

2
...
Cut-off ratio, ρ =

V

volume after heat addition (v 2 )
volume before heat addition (v1 )

(For constant Pressure heating)
∴

ρ=

V2
V1

bigger term
=
smaller term

Relation

p
1

rc = re
...
Constant volume pressure ratio,
∝=

Q

Pressure after heat addition
Pressure before heat addition

[For constant volume heating)

Page 223 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13
2

p

bigger term

2
α = p = smaller term
1

∴

Q

p

1
V

5
...


Q1

2

3

pV = C

1

Q2

4

V

Carnot cycle: The large back work (i
...


7
...


8
...


9
...


10
...


Otto cycle (1876)

η =1−

1
γ −1
c

r

V=

3
p

T

4
2

1

2

3

V=C
4

1

V

S
1

For Wmax;

C

2( γ − 1)
T
rc = ⎛ max ⎞
⎜T ⎟
⎝ min ⎠

Page 224 of 265

Gas Power Cycles

By: S K Mondal
Diesel cycle (1892)

p

V

=

C

b
...
γ(ρ − 1)

C
...
With same compression ratio and heat rejection

4

∴    ηotto > ηDual > ηDiesel

p

3

Q1

5

6

2

7
1
V

Page 225 of 265

Q2

Gas Power Cycles

By: S K Mondal
b
...


Brayton cycle

η= 1−

1
γ −1
c

r

1

=1−

γ −1
γ
P

r

p=C
3
2

T

p=C

4
1
S

∴

Brayton cycle efficiency depends on either compression ratio ( rc ) or Pressure ratio

rp
* For same compression ratio

⎡ηOtto = ηBrayton ⎤
⎣
⎦
γ

⎛ T ⎞ ( γ − 1)
a
...


=1−

For Maximum work
γ

(i)

⎛ T ⎞ 2( γ − 1)
( rp ) opt
...
= ⎜ ηT ηC max ⎟
Tmin ⎠
⎝

Question and Solution (P K Nag)
In a Stirling cycle the volume varies between 0
...
06 m3 , the
maximum pressure is 0
...
The working fluid is air (an ideal gas)
...
(b) Find the
efficiency and the work done per cycle for the cycle with an ideal
regenerator, and compare with the Carnot cycle having the same
isothermal heat supply process and the same temperature range
...
(a) 27
...
7 kJ/kg, (b) 32
...
1

Solution:
Given V1 = 0
...
03 m3 = V3
p3 = 200 kPa

3
Q1

T1 = T2 = 270°C = 543 K
T3 = T4 = 540°C = 813 K

p
2

p3 V3
200 × 0
...
025715 kg
R T3
0
...
025715 × 0
...
985 kJ

Here m =

⎛V ⎞
pdV = m R T3 ln ⎜ 4 ⎟
3
⎝ V3 ⎠
pV = mRT = C
3
⎛V ⎞
W1 – 2 = ∫ pdV = m RT1 ln ⎜ 1 ⎟
1
⎝ V2 ⎠
∴p=

∫

4
T=C
1

∴ Heat addition Q1 = Q2 – 3 = m c v (T3 – T2)

W3 – 4 =

T=C

4

mRT
V

⎛V ⎞
m(RT3 − RT1 ) ln ⎜ 1 ⎟
⎝ V2 ⎠ =
∴η=
4
...
025715 × 0
...
7%
4
...
3812 kJ = 53
...
21%
813
Q13
...
The pressure at the beginning of isothermal
compression is 1
...
Determine (a) the compressor and turbine work
per kg of air, and (b) the cycle efficiency
...
(a) wT = 465 kJ/kg, wC = 121
...
738)

Solution:

Given T1 = T2 = 288 K
T3 = T4 = 1100 K
p1 = 1
...
3 kPa
RT1
= 0
...
82%
1100

V1
T ⎛p ⎞ ⎛p ⎞
= 1 × ⎜ 2 ⎟= ⎜ 2 ⎟
V2
T2 ⎝ p1 ⎠ ⎝ p1 ⎠

= 0
...
005 (1100 – 288) kJ/kg = 602
...
5

p 2 (V3 – V2)

An engine equipped with a cylinder having a bore of 15 cm and a stroke
of 45 cm operates on an Otto cycle
...

(Ans
...
4%)

Page 228 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13

Solution:
V2 = 2000 cm3 = 0
...
V
...
002 +

3

π × 0
...
45 = 0
...
9761
V2
1
ηair std = 1 − γ − 1 = 47
...
10

4

2

1
V1

V2
SV

VCL

Two engines are to operate on Otto and Diesel cycles with the following
data: Maximum temperature 1400 K, exhaust temperature 700 K
...
1 MPa, 300 K
...

(Ans
...
656, p max = 2
...
456, p max = 1
...
45 kJ/kg, η = 60
...
861 m3/kg
v1 =
p1

∴

⎛p ⎞
T3
= ⎜ 3⎟
T4
⎝ p4 ⎠

γ −1
γ

⎛v ⎞
= ⎜ 4⎟
⎝ v3 ⎠

γ −1

V=C

3
3

V=C
T

Q1

p

4

2

2
4

Q2
1

1
S

∴
∴

⎛ 1400 ⎞ ⎛ v1 ⎞
⎜ 700 ⎟ = ⎜ v ⎟
⎝
⎠ ⎝ 2⎠

v2 =

v1
1
γ −1

=

γ −1

0
...
9
2
= 0
...
657)0
...
5 kPa
p1
⎝ v2 ⎠
T
1400
⇒ p3 = 3 × p2 =
× 1131
...
64 MPa
600
T2

v
rc = 1 = 2 γ − 1 = 5
...
718 [(1400 – 600) – (700 – 300)] kJ/kg = 287
...

Q1 − Q2
287
...
5 ≈ 50%
=
0
...
861 m3/kg

p1 = 100 kPa
T3 ⎛ v 4 ⎞
= ⎜ ⎟
T4 ⎝ v 3 ⎠

∴

γ −1

0
...
4 = 22
...
5 = 0
...
287 × 1400
= 2639
...
1522
V3

2

3

p

4
1

∴

p 2 = p3

∴

T2 ⎛ p2 ⎞
= ⎜ ⎟
T1 ⎝ p1 ⎠

V
γ −1
γ

⎛v ⎞
= ⎜ 1⎟
⎝ v2 ⎠

γ −1

1

∴ T2 = 764 K
1

⎛ p ⎞γ
v
2639
...
4
rc = 1 = ⎜ 2 ⎟ = ⎛
⎜ 100 ⎟ = 10
...
64 MPa
Q1 = Q2 – 3 = CP (T3 – T2) = 1
...
84 kJ/kg
Page 230 of 265

Gas Power Cycles

By: S K Mondal

Q2 = Q4 – 1 = Cv (T4 –T1) = 0
...
2 kJ/kg
W = Q1 – Q2 = 351
...
11

Chapter 13

W
351
...
84
Q1

An air standard limited pressure cycle has a compression ratio of 15 and
compression begins at 0
...
The maximum pressure is limited
to 6 MPa and the heat added is 1
...
Compute (a) the heat
supplied at constant volume per kg of air, (b) the heat supplied at
constant pressure per kg of air, (c) the work done per kg of air, (d) the
cycle efficiency, (e) the temperature at the end of the constant volume
heating process, (f) the cut-off ratio, and (g) the m
...
p
...


(Ans
...
5%, (e) 1252 K, (f) 2
...
21 MPa)
Solution:

rc =

v1
= 15
v2

p1 = 100 kPa
∴ v1 =

T1 = 40°C = 313 K

3
p

2

RT1
= 0
...
4 – 1 ⇒ T2 = 924
...
4 ⇒ p 2 = 4431 kPa
p1
⎝ v2 ⎠

∴
∴
∴

p V
p
p2 V2
6000
× 924
...
8 k = 2684
...
12842 m3/kg
...
1773 ∴ T5 = 1233 K
T5

(a) Heat supplied at constant volume = Cv (T3 – T2) = 235 kJ/kg
(b) Heat supplied at constant Pressure = (1675 – 235) = 1440 kJ/kg
(c) Work done = Q1 – Q2 = 1675 – Cv (T5 – T1) = 1014
...
44
× 100% = 60
...
12842
=
= 2
...
05988
v3

[∴ v3 =
(g) m
...
p
...
13

V2 ) = W

RT3
= 0
...
44
= 1209
...
2099 MPa
v
v1 − 1
15

Show that the air standard efficiency for a cycle comprising two
constant pressure processes and two isothermal processes (all
reversible) is given by

η=

(T1 − T2 ) ln ( rp )

(γ −1) / γ

(γ −1) / γ

T1 ⎡1 + ln ( rp )
− T2 ⎤
⎢
⎥
⎣
⎦
Where T1 and T2 are the maximum and minimum temperatures of the
cycle, and

rp is the pressure ratio
...

3
⎝ V4 ⎠
⎝ V3 ⎠
⎛V ⎞
∴
Wnet = W1 – 2 + W3 – 4 = R(T1 − T3 ) ln ⎜ 2 ⎟
⎝ V1 ⎠
= R (T1 – T2) ln rP
...

γ −1
Total heat addition (Q1)

η =

Multiply

= Q12 + const Pr
...
14

Chapter 13

Obtain an expression for the specific work done by an engine working
on the Otto cycle in terms of the maximum and minimum

r

Temperatures of the cycle, the compression ratio k , and constants of the
working fluid (assumed to be an ideal gas)
...
rc−( r − 1)
Then

γ −1

⎛v ⎞
= ⎜ 2⎟
⎝ v1 ⎠

= rcγ − 1

γ −1

= rc−( γ − 1) Let rcγ − 1

3

T
= 3
x

p

T
⎡
⎤
W = Cv ⎢ T3 − T1x − 3 + T1 ⎥
x
⎣
⎦
dW
For maximum W,
=0
dx
T
⎡
⎤
∴ Cv ⎢0 − T1 + 3 + 0 ⎥ = 0
2
x
⎣
⎦
T
∴ x2 = 3
T1
∴ rcγ − 1 =

T3
=
T1

Q1

4
Q2

2
1

Tmax
Tmin
1

1

⎛ T ⎞ 2(1 − γ )
⎛ T ⎞ 2( γ − 1)
Proved
...
15
A dual combustion cycle operates with a volumetric compression ratio rk
= 12, and with a cut-off ratio 1
...
The maximum pressure is given by
pmax = 54 p1 ' where p1 is the pressure before compression
...
35, show that the m
...
p
...


(Ans
...
2 p1 ' T3 = 1490 K,

T4 = 2410 K, T5 = 1200 K, (b) η = 0
...
615
v3

pv1
...
35

p max = p3 = p 4 = 54 p1
4

3

3
T

2

4

p

5

5

2

1

1
V

S
∴
And

⎛v ⎞
T2
= ⎜ 1⎟
T1
⎝ v2 ⎠

⎛v ⎞
p2
= ⎜ 1⎟
p1
⎝ v2 ⎠
p
p2
= 3
T3
T2

n −1

∴ T2 = T1 × (12 )

(1
...
3862 T1

n

∴ p2 = p1 × (12)1
...
635 p1
∴ T3 =

p3
54p1
× T2 =
× 2
...
5 T1
T2
28
...
615
v1 = 0
...
615 T3 = 1
...
5 T1 = 7
...
6019 T1
W = [Cv (T3 – T2) + CP (T4 – T3) – Cv (T5 – T1) = 2
...

Page 235 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13
p m (v1 – v2) = W

2
...
4308 T1
= 9
...
4308 T1
× 100 % = 56
...
299 T1
(a)
T1 = 335 K, T2 = 799
...
5 K, T4 = 2434
...
6 K
...
16

Recalculate (a) the temperatures at the cardinal points, (b) the m
...
p
...
15 is a Diesel
cycle with the same compression ratio and a cut-off ratio such as to give
an expansion curve coincident with the lower part of that of the dual
cycle of Problem 13
...

(Ans
...
82 p1 , (c) η = 0
...
615
v2

∴ T3 =
Then

2

v3
× T2 = 1
...
4 = 1291 K
v2

⎛v ⎞
T2
= ⎜ 1⎟
T1
⎝ v2 ⎠

3

p
4

n −1

∴ T2 = T1 (12 )

1
...
4 K

n

⎛v ⎞
p
But 2 = ⎜ 1 ⎟
p1
⎝ v2 ⎠
Continue to try…
...
19

Solution:

In a gas turbine plant working on the Brayton cycle the air at the inlet is
at 27°C, 0
...
The pressure ratio is 6
...
The turbi- ne and compressor efficiencies are each
80%
...

(Ans
...
4 kJ/kg, (b) 351
...
43 kJ/kg,
(d) 16
...
25
p4
= 6
...
861
p3

v3 = 0
...
70243)0
...
4 K

∴ 0
...
23255

∴ T4 = 558
T2s = 635
...
68808
T2 = 723 K

T1 − T2
⇒ T1 – T2 = 350
T1 − T2s
T2 = T1 – 350 = 723 K
η=

∴

(a) Compressor work (Wc) = (h4 – h3) = Cp(T4 – T3) = 259
...
75 kJ/kg
(c) Heat supplied (Q1) = Cp(T1– T4) = 517
...
86%
Q1

(d) Cycle efficiency (η) =

(e) Turbine exhaust temperature (T2) = 723 K
Q13
...
What will be the improvement in cycle efficiency and output if the
turbine process Is divided into two stages each of pressure ratio 3, with
intermediate reheating to 727°C?
(Ans
...
3%, 30
...
8 K

γ− 1
γ

⎛p ⎞
= ⎜ 1⎟
⎝ p2 ⎠

γ −1
γ

γ −1
γ

× T1 = 562 k

⎛1 ⎞
= ⎜ ⎟
⎝9⎠

γ −1
γ

∴ T4 =

T3
9

Page 238 of 265

γ −1
γ

= 533
...
6 K

p1

6730
...
6 K

= 730
...
22 kJ/kg
Q = h3 – h2 = CP (T3 – T2) = 440
...
62 %
W = (h3 – h4) + (h5 – h6) – (h2 – h1)
= CP [(T3 – T4) + (T5 – T6) – (T2 – T1)] = 278
...
94 kJ/kg

∴

η = 39
...
13 − 46
...
07 %
46
...
18 − 205
...
6%
205
...
28

Obtain an expression for the specific work output of a gas turbine unit
in terms of pressure ratio, isentropic efficiencies of the compressor and
turbine, and the maximum and minimum temperatures, T3 and T1 •
Hence show that the pressure ratio
⎛
T ⎞
rp = ⎜ηT ηC 3 ⎟
T1 ⎠
⎝

rp

for maximum power is given by

γ / 2(γ −1)

Page 239 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13

If T3 = 1073 K, T1 = 300 K, ηC = 0
...
8 and γ = 1
...

(Ans
...
263, 100 kJ/kg, 17
...
Work

⎤
⎛ γ− 1
⎞⎥
γ
⎜ rp − 1 ⎟ ⎥ kJ/ kg
⎜
⎟
⎝
⎠⎥
⎦

dW
=0
dx

⎡η T
T ⎤
dW
= CP ⎢ T 2 3 − 1 ⎥ = 0
dx
ηC ⎦
⎣ x

∴

∴

x2 = ηT ηC

∴

x=

T3
TT1

ηT ηC

Tmax
Tmin
γ

⎛
T ⎞ 2( γ− 1)
Proved
...
8, η7 = 0
...
4 then
1
...
4 – 1)
= 4
...
= ⎛ 0
...
8 × 1073 ⎞
⎜
⎟
300 ⎠
⎝
γ− 1
γ
(rp )opt = x = 1
...
005 ⎢0
...
513 − 1) ⎥ kJ/kg
⎟−
1
...
08
⎝
⎣
⎦
= 99
...
005 (1073 – 492
...
5 kJ/kg

= 492
...
18
× 100% = 17%
583
...
29

A gas turbine plant draws in air at 1
...
5
...

Compression is conducted in an uncooled rotary compressor having an
isentropic efficiency of 82%, and expansion takes place in a turbine with
an isentropic efficiency of 85%
...

For an air flow of 40 kg/s, find (a) the overall cycle efficiency, (b) the
turbine output, and (c) the air-fuel ratio if the calorific value of the fuel
used is 45
...

(Ans
...
4%, (b) 4272 kW, (c) 115)

Solution:

p1 = 101
...
5 kPa
p1
T4 = 750°C = 1023 K
γ−1

T
⎛p ⎞ γ
∴ 2s = ⎜ 2 ⎟ ⇒ T2s = 460
...
6K
T
⎛p ⎞
∴ 5s = ⎜ 5 ⎟
T4
⎝ p4 ⎠

γ− 1
γ

⎛p ⎞
= ⎜ 1⎟
⎝ p2 ⎠

γ− 1
γ

⎛ 1 ⎞
= ⎜
⎟
⎝ 5
...

46 1 kg

(1 + m) kg

2
499
...
7 K

5s
628
...
5 ⎠

ηT =

1
...
4

= 628
...
3 K
T4 − T5s
Page 242 of 265

Gas Power Cycles

By: S K Mondal
∴

Chapter 13

T5 = 687
...
7Cp (T5 – T2) = 132
...
33 and
CpT3 = 132
...
33 m + CpT2 = 634
...
33 m
Heat addition (Q1) = Cp (T4 – T3) = CpT4 – CpT3
= 393
...
33m = m × 45
...
68 × 10–3 kJ/kg of air
∴
Q1 = 392
...
00868 × 1
...
7) kJ/kg of air 340 kJ/kg
Wc = (h2 – h1) = Cp (T2 – T1) = 1
...
6 – 283)
= 217
...
32 kJ/kg
122
...
16%
(a)
η=
392
...
32 kJ/kg of air
= 4893 kW
1 kg air
= 115
...
00868 kg of fuel
A gas turbine for use as an automotive engine is shown in Fig
...
43
...
30

the turbine to drive the compressor
...
Consider air as the
working fluid, and assume that all processes are ideal
...


Page 243 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13

Solution :

Try please
...
31

Repeat Problem 13
...

Try please
...
32

An ideal air cycle consists of isentropic compression, constant volume
heat transfer, isothermal expansion to the original pressure, and
constant pressure heat transfer to the original temperature
...
Determine
the cycle efficiency and the m
...
p
...
51
...
45 bar)

Solution:
V=C

3
Q1′
2

3

p

T=C

T

Q1′
2

T=C

4

Q2
S=C

r

pV = C
1

1

Q2 4

S

V

∴

p=C

Q1 ′′

Q1 ′′

Compression ratio, rc =

V1
V2

V4
V
= 4
V3
V2
Heat addition Q1 = Q1′ + Q1′′
= constant volume heat addition
(Q1′ + constant temperature heat addition Q1′′)
Heat rejection, Q2 = Cp (T4 – T1)

Expansion ratio, re =

∴ T2 – T1
...
re

γ −1
γ

γ −1

⎛v ⎞
= rcγ − 1
= ⎜ 1⎟
v2 ⎠
⎝
v
and v2 = 1 and p 2 = p1 rcγ
rc

T
⎛p ⎞
Hence 2 = ⎜ 2 ⎟
T1
⎝ p1 ⎠

∴

p
p
v4
= 3 = 3 = re
p4
p1
v3

Page 244 of 265

Gas Power Cycles

By: S K Mondal
∴
∴

Chapter 13

p
p2
= 3
T2
T3

or

η= 1−

p3
r
r
× T2 = e × T1
...
e − T1 ⎟
rc
⎝
⎠
= 1−
re
r
⎛
⎞
Cv ⎜ T1
...
T1 e In re
rc
rc
⎝
⎠
⎛r
⎞
γ ⎜ e − 1⎟
rc
⎝
⎠
= 1−
re
re
⎛
γ −1 ⎞
⎜ r − rc ⎟ + ( γ − 1) r In re
c
⎝ c
⎠
γ(re − rc )
= 1−
γ
(re − rc ) + ( γ − 1) re l n re
∴

η= 1−

γ[re − rc ]
(re − r ) + ( γ − 1) re l n re
γ
c

Given p1 = 1 bar = 100 kPa
T1 = 40°C = 313 K

rc =8 and p3 = 100 bar = 10000 kPa
p3
= 100
p1
1
...
4
(100 − 8 ) + (1
...
8
= 1−
265
...
51548 = 51
...
re ∴ re =

⇒ T3 = T1 ×

re
313 × 100
=
= 3912
...
718 (3912
...
287 × 3912
...
5 kJ/kg
∴ p m (V4 – V2) = W

∴ v 4 = 100 v 2

∴ p m (100 –1) v 2 = W
∴ pm (99) ×

v1
=W
8
Page 245 of 265

v2 =

v1
rc

Gas Power Cycles

By: S K Mondal

Chapter 13

8W
= 346
...
461 bar
( v 4 – V3) = 40 58

∴ pm =
∴ pm

∴ pm =

Q13
...
89831 kJ/kg
p1

4058
= 365 bar
v4
v4 −
100

Show that the mean effective pressure, pm ' for the Otto cycle is
Given by

(p

3

pM =

1 ⎞
γ ⎛
− p1 rk ⎜1 − γ-1 ⎟
⎝ rk ⎠
( γ − 1)( rk −1)

)

Where p3 = pmax ' p1 = pmin and rk is the compression ratio
...
rc

γ −1

= rcγ − 1

3
p

γ

PV = C
Q1

2

pv = C
γ

pv = C

v2  =

V

v1
rc

p3
p
= 2
T3
T2
∴ T3 = T2 ×

p3
p
r γ − 1 × p3
T ×p
= T2 × 3 γ = T1 c
= 1 3
γ
rc p1
p2
p1 rc
p1 rc
γ −1

γ −1

T3
⎛v ⎞
⎛v ⎞
= ⎜ 4 ⎟ = ⎜ 1 ⎟ = rcγ− 1
T4
⎝ v2 ⎠
⎝ v3 ⎠
T3
T1 p3
T p
= 1 3
∴ T4 = γ − 1 =
γ -1
rc
rc p1 × rc
rcγ p1
W = Q1 – Q2
= Cv (T3 – T2) – Cv (T4 – T1)
pm (V1 – V2) =W
∴
∴ pm =

4

γ

p2 = p1 × rcγ

Cv [(T3 − T2 ) − (T4 − T1 )]
V1 − V2

Page 246 of 265

Q2
1

Gas Power Cycles

By: S K Mondal

Chapter 13

T p
⎡T p
⎤
cv ⎢ 1 3 − T1 rcγ - 1 − 1 3 + T1 ⎥
γ
rc p1
⎣ rc p1
⎦
=
v1
v1 −
rc
p3
⎡
⎤
p3 − p1 rcγ − γ − 1 + p1 rc ⎥
cV T1 ⎢
rc
⎢
⎥
=
V1 p1 ⎢
(rc − 1)
⎥
⎣
⎦
R
⎡
⎢ cV = γ − 1
⎢
⎢∵ p1 V1 = RT1
⎣
RT1
=
V1 p1

p3
+ ( p3 – p1 rcγ )]
rcγ − 1
( γ − 1) (rc − 1)

[( p3 − p1 rcγ ) −

1 ⎞
⎛
( p3 − p1 rcγ ) ⎜1 − γ − 1 ⎟
rc
⎝
⎠ Proved
=
( γ − 1)(rc − 1)
Q13
...
Derive expressions for net work output per kg of air and
corresponding efficiency of the cycle in terms of the maximum and the
minimum temperatures
...

2
T
(Ans
...
14
max

(

)

(Wnet )max = 236
...
469 )

T1 = Tmin

Solution:

T4 = Tmax
∴
∴

T2
⎛p ⎞
= ⎜ 2⎟
T1
⎝ p1 ⎠
T2 = T1 x

γ− 1
γ

γ −1

γ− 1

= rp γ = x (say)
γ −1

T5 ⎛ p5 ⎞ γ
1
⎛p ⎞ γ
= ⎜ ⎟ = ⎜ 1⎟
=
x
T4 ⎝ p4 ⎠
⎝ p2 ⎠
T
∴
T5 = 4
x
For regeneration 100% effective number
Cp (T5 – T2) = Cp (T3 – T2)
T
∴
T3 = T5 = 4
x
WT = h4 – h5 = Cp (T4 – T5)
Page 247 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13
T ⎞
⎛
= Cp ⎜ T4 − 4 ⎟
⎝
x ⎠
p2

4
Q1
3
T

p1

2
5
Q2

1

S

And

1⎞
⎛
= Cp T4 ⎜1 − ⎟
⎝
x⎠

Wc = h2 – h1
= Cp (T2 – T1)
= Cp T1 (x – 1)

⎡ ⎛
⎤
1⎞
Wnet = WT – WC = Cp ⎢T4 ⎜1 − ⎟ − T1 (x − 1) ⎥
x⎠
⎣ ⎝
⎦
For Maximum Net work done
∂ Wnet
1
= 0 ∴ T4 × 2 − T1 = 0
∂x
x
Tmax
T4
=
∴
x2 =
Tmin
T1

∴

∴

x=

Tmax
Tmin
γ

⎛ T ⎞ 2( γ − 1)
Heat addition ∴ (   p ) opt
...
=
=
1⎞
Q1
⎛
T4 ⎜1 − ⎟
x⎠
⎝
= 1−

T
T4
T1
×x = 1− 1 ×
= 1−
T4
T4
T1

Page 248 of 265

Tmin
Tmax

Gas Power Cycles

By: S K Mondal

Chapter 13

Wopt
...
= 1 −

Tmin = 30°C = 303K

1
...
4–1)

= 9
...
9%
Tmax

Wopt
...
005 ( 1073 − 303)2 = 236
...
40

Show that for the Sterling cycle with all the processes occurring
reversibly but where the heat rejected is not used for regenerative
heating, the efficiency is giver: by
⎛ T1
⎞
− 1 ⎟ + (γ − 1) ln r
⎜
T
⎠
η =1 − ⎝ 2
⎛ T1
⎞
T1
⎜ − 1 ⎟ + (γ − 1) ln r
T2
⎝ T2
⎠
Where r is the compression ratio and T1 / T2 the maximum to minimum
temperature ratio
...
307 kJ/kg K,
c p =
...
50 kJ/kg K) with a pressure and temperature prior to isothermal
...
55
MPa and heat supplied during the constant volume heating of 9300
kJ/kg
...
41

Helium is used as the working fluid in an ideal Brayton cycle
...
The gas is
heated to l000 °C before entering the turbine
...
Determine:
(a) the temperatures at the end of compression and expansion, (b) the
heat supplied, the heat rejected and the net work per kg of He, and (c) the
cycle efficiency and the heat rate
...
1926 kJ/kg K
...
(a) 4 65
...
2 K, (b) 4192
...
2, 1491
...
3557, 10,121kJ/kWh)
T2
⎛p ⎞
= ⎜ 2⎟
T1
⎝ p1 ⎠

Solution:

∴

γ −1
γ

=

60
20

Cp = 5
...
0786
c v = c p – R = 3
...
1926
= 1
...
114

∴

γ −1
= 0
...
7 K

p1
4

820 K

Q2
1 300 K

S
∴

⎛ 60 ⎞
T2 = T1 × ⎜ ⎟
⎝ 20 ⎠

γ −1
γ

= 465
...
7K
End of expansion temperature T4 = 820K

(b)

Heat supplied (Q1) = h3 – h2 = CP (T3 –T2) = 4192 kJ/kg
Heat rejected (Q2) = h4 – h1 = CP (T4 –T1) = 2700 kJ/kg
Work,
W = Q1 – Q2 = 1492 kJ/kg

W
1492
× 100% = 35
...
356
η

(c)

Q13
...
The pressure ratio across the compressor is 6 to 1 and the
temperature at the turbine inlet is 1400 K
...
Assuming that the efficiency
of the compressor and turbine are both 85% and that the nozzle
efficiency is 95%, determine the pressure at the nozzle inlet and the
velocity of the air leaving the nozzle
...
285 kPa, 760 m / s)
p2
=6
∴ p2 = 600 kPa
p1

Page 251 of 265

Gas Power Cycles

By: S K Mondal

Chapter 13

3

p2

pi

5
T

2s

2

4s

p1

6

1
290 K, 100 kPa

S
γ −1

1
...
4
T1
⎝ p1 ⎠
T2s = 483
...
005 (518 – 290) = 229
...
9 kJ/kg = CP (T3 – T4s)
ηT
∴ T3 – T4s = 268
...
8 K
1
...
8 ⎞1
...
4

∴

pi = p2

⎛ 1131
...
4 − 1
× ⎜
= 285 kPa
⎟
⎝ 1400 ⎠

Δh = h5 – h6 = CP (T5 – T6)
T3 − T5
= ηT
T3 − T4 s
T5
⎛p ⎞
= ⎜ 5⎟
T6
⎝ p6 ⎠
∴

∴ T3 – T5 = 227
...
4 − 1
1
...
9 K

Δh = CP (1172 – 868
...
6 kJ/kg
Page 252 of 265

Gas Power Cycles

By: S K Mondal
∴

Q13
...
95 × 304
...
8 m/s

A stationary gas turbine power plant operates on the Brayton cycle and
delivers 20 MW to an electric generator
...
The minimum pressure is
95 kPa and the maximum pressure is 380 kPa
...
85 and 0
...

If a regenerator of 75% effectiveness is added to the plant, what would be
the changes in the cycle efficiency and the net work output?
(Ans
...
37 kg/s, (b) 110
...
528,
(d) 0
...
148 ΔWnet = 0)
T2
⎛p ⎞
= ⎜ 2⎟
T1
⎝ p1 ⎠

Solution:

T4 ⎛ p4 ⎞
=
T3 ⎜ p3 ⎟
⎝ ⎠

γ− 1
γ

γ −1
γ

∴ T2 = 431K
⎛p ⎞
=⎜ 1 ⎟
⎝ p2 ⎠

γ −1
γ

; T4 = 807
...
76 kJ/kg
20000
•
= 79
...
76

3

p2
380 kPa
1200 K

431 K
T

p1

2
4

807
...
234 MW
(b) η =

WC
T − T1
= 0
...
13 kg/s
(d) v1 =

RT1
= 0
...
33 m3/s

Page 254 of 265

Refrigeration Cycles

By: S K Mondal

Chapter 14

14
...
A steam power plant is an example of
a heat engine
...
Domestic
refrigerators and room air conditioners are the examples
...


•

A heat pump is similar to a refrigerator, however, here the required output is the heat
rejected to the high temperature body
...
(a) Heat Engine (b) Refrigeration and heat pump cycles

Page 255 of 265

Refrigeration Cycles

By: S K Mondal

Chapter 14

Fig
...

temperature of the cold reservoir
...
1

Solution:
Q14
...
12 and 0
...
The mass flow of refrigerant is 0
...

Determine
(a) The rate of heat removal from the refrigerated space
(b) The power input to the compressor
(c) The heat rejection to the environment
(d) The COP
(Ans
...
35 kW, (b) 1
...
20 kW, (d) 3
...

A Refrigerant-12 vapour compression cycle has a refrigeration load of 3
tonnes
...
Find
(a) The refrigerant flow rate in kg/s
(b) The volume flow rate handled by the compressor in m3/s
(c) The work input to the compressor in kW
(d) The heat rejected in the condenser in kW
(e) The isentropic discharge temperature
...

Solution: As 50°C temperature difference in evaporate so evaporate temperature = – 20°C and
Condenser temperature is 30°C
...
589 bar
4 3

p2 = 7
...
7 kJ/kg, h3 = 64
...
7 +
(190
...
7)
20
1
7
5 6
Δh = 3
...
7088 +
(0
...
7088)
h
20
= 0
...
025
∴ h4 = h3 – Δh = 61
...
e
...
7
30°C hg = 64
...
98/vc
∴ Degree of sub cooling = 3
...
7203 – 0
...
7321 – 0
...
3 − 199
...
63 kJ/kg
∴ h2 = 199
...
63 – 181
...
9 kJ/kg
Refrigerating effect (Q0) = h7 – h5 = h7 – h4 = (178
...
6) kJ/kg = 117
...
1
=
= 4
...
9
•

v1 = 0
...
014361 m3/s

•
π D2
N
×L×
× n × ηvol = V1
4
60

L
= 1
...
2 D

π × D2
900
× 1
...
95 = 0
...
1023 m = 10
...
1227 m = 12
...
4

A vapour compression refrigeration system uses R-12 and operates
between pressure limits of 0
...
15 MPa
...
A refrigerating load of 2 kW is required
...

(Ans
...
15, 243 cm3)

Solution:

p1 = 150 kPa: Constant saturated temperature (– 20°C)
p2 = 745 kPa: Constant saturated temperature (30°C)

ding
ubco
2°C s
4 3

2

p
5 6

7

1
10°C superheated

h7 = 178
...
6 kJ/kg
h4 = h4-5 = 59
...
6 – 59
...
64 kJ/kg = h5
5

10
(190
...
8 kJ/kg
20
10
(0
...
7088) = 0
...
7088 +
20

h1 = h7 +

s1

Page 258 of 265

Refrigeration Cycles

By: S K Mondal

Chapter 14

⎛ 0
...
6854 ⎞
h2 = 199
...
3 – 199
...
2 kJ/kg
⎝ 0
...
6854 ⎠
∴ Compressor work (W) = h2 – h1 = 29
...
8 – 62
...
16 kJ/kg
122
...
16
∴
COP =
29
...
1166m3 /kg
•

Mass flow ratio m × 122
...
016372 kg/s
•

•

∴ V1 = mv1 = 1
...
76 ×

600
60

∴ Vs = 251
...
6

Solution:

A R-12 vapour compression refrigeration system is operating at a
condenser pressure of 9
...
19 bar
...
The values of enthalpy at the
inlet and outlet of the evaporator are 64
...
7 kJ/kg
...
082 m3/kg
...
13
Determine:
(a) The power input in kW required for the compressor
(b) The COP
...
517 kW
...
(a) 11
...
56)
T1 = – 10°C
T3 = 40°C
h4 = 646 kJ/kg
h1 = 1057 kJ/kg
n = 1
...
082 m3 / kg
3

2

2′

p
4

1

1′

h

Refrigeration effect (195
...
6) kJ/kg = 131
...
4024 kg/s

m Qo = 15 × 3
...
19 ⎞1
...
022173 m3/kg
v1
p2 ⎠
⎝ 9
...
93 kJ/kg
n −1
(a) Wcompressor = 11
...
12

Solution :

Chapter 14

15 × 3
...
64

Determine the ideal COP of an absorption refrigerating system in
which the heating, cooling, and refrigeration take place at 197°C, 17°C,
and –3°C respectively
...
5
...
E
...
E = ηCarnot = ⎜1 − a ⎟
Th ⎠
⎝
To
T ⎞
⎛
× ⎜1 − a ⎟
Ta − To ⎝
Th ⎠
T [T − Ta ]
= o × h
Th [Ta − To ]

∴ (COP) ideal =

Given To = 270 K, Ta = 290 K, Th = 470 K
∴ (COP) ideal =
Q14
...
17
470 [290 − 270]

Derive an expression for the COP of an ideal gas refrigeration cycle with
a regenerative heat exchanger
...

⎜ Ans
...
23

= Th

(rP γ − 1)
γ −1

rP γ

=

Th
γ −1

T1 rP γ − Th

or COP =

T

Th r

1
( γ −1)/ γ
p

− T1

Large quantities of electrical power can be transmitted with relatively
little loss when the transmission cable is cooled to a superconducting
temperature
...
If the pressure
ratio is 10 and heat is rejected directly to the atmosphere at 300 K,
determine the COP and the performance ratio with respect to the
Carnot cycle
...
0
...
38)

Page 261 of 265

Refrigeration Cycles

By: S K Mondal
T2
⎛p ⎞
= ⎜ 2⎟
T1
⎝ p1 ⎠

Solution:

∴

Chapter 14
γ− 1
γ

= 10

γ −1
γ

1
...
1

T2 = 300 × 10 1
...
4

2
Q1

T

300 K

15 K

5

3

1

4

6

Q2
S

∴
T5 = 5
...
0284 CP
Work input (W) =CP [(T2 – T1) – (T4 – T5)] = 444
...
0284 CP
= 0
...
97 CP
T6
15
=
= 0
...
0203
= 0
...
05263
COPcarnot
Q14
...
The evaporator temperature is 10°C and the
condenser pressure is 0
...
Using an ideal vapour compression cycle,
estimate the power required to drive the compressor if steam/water
mixture is used as the working fluid, the COP and the mass flow rate of
the fluid
...

(Ans
...
0 kW, 3
...
001012 kg/s)

Page 262 of 265

Refrigeration Cycles

By: S K Mondal

Chapter 14

h1 = 2519
...
9008 kJ/kg-K
v1 = 106
...
33°C

hf = 340
...
5 bar

p
1

4

10°C
1
...
88642,
500°C 50 kPa s = 9
...
9
h = 3488
...
9008 − 8
...
7 – 3278
...
9 = 3305
...
1546 − 8
...
3 – 2519
...
5 kJ/kg

∴
∴

Heating (Q) = h2 – hf3 = (3305
...
5) kJ/kg = 2964
...
0101187 kg/s
2964
...
8
COP =
= 3
...
5
•

= m =

∴

•

Compressor power = mW = 7
...
26

A 100 tonne low temperature R-12 system is to operate on a 2-stage
vapour compression refrigeration cycle with a flash chamber, with the
refrigerant evaporating at – 40°C, an intermediate pressure of 2
...
Saturated vapour enters both the compressors
and saturated liquid enters each expansion valve
...
Determine:
(a) The flow rate of refrigerant handled by each compressor
(b) The total power required to drive the compressor
(c) The piston displacement of each compressor, if the clearance is 2
...

(Ans
...
464, 3
...
6274, 0
...
86,
(e) 3
...
54 kW, 1
...
433)

Page 263 of 265

Refrigeration Cycles

By: S K Mondal

Chapter 14

h1 = 169 kJ/kg
h3 = 183
...
6 kJ/kg = h6
h7 = h8 = 26
...

m1

7

2
...

m2

p

8
9
–90°C

1

7
...
6417 bar

h
S1 = S2 = 0
...
7020 kJ/kg – K
∴
From P
...
9
=
= 1
...
2 − 64
...
7367 kg/s
•

m1 = m2 × 1
...
7635 kg/s
•

•

(b) Power of compressor (P) = m2 (h 2 − h1 ) + m1 (h 4 − h3 )
= 14328 kW
(d) COP =
(e)

Refrigeration efficiency
100 × 14000
= 2
...
28

For single storage
From R12 chart ha′ = 2154 kJ/kg, hg = hs = 64
...
725 kg/s
3600
•

Compressor power (P) = m (h4′ – h1) = 3
...
35 kW
Page 264 of 265

Refrigeration Cycles

By: S K Mondal

Chapter 14
100 × 14000
3600
COP =
= 2
...
35

Page 265 of 265



---