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ME 2101
Basic Thermodynamics
Dr
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mpe@aust
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J   Q   W

  Cyclic integral

Joule’s Experiment
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

In SI units, 1st Law of TD for closed system undergoing cycle:

  Q   W

Dr
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Mahbubul Muttakin, Assoc
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, MPE, AUST

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5

1st Law of TD for a Closed System Undergoing a non-cyclic process


If a system undergoes a change of state during which both heat transfer and
work transfer are involved, the net energy transfer will be stored or
accumulated within the system
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Mahbubul Muttakin, Assoc
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, MPE, AUST

Energy of a System


The physical significance of the property E is that it represents all the energy of
a system at a given state
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Such as:

Kinetic Energy (KE) : energy of a system associated with motion
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

Internal Energy (U): sum of all molecules energy of a system
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These are associated with co-ordinate frame that
we select and can be specified by macroscopic parameters of mass, velocity and
elevation
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Mahbubul Muttakin, Assoc
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, MPE, AUST

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Energy of a System

 The

KE of an object is an organized
form of energy associated with the
orderly motion of all molecules in one
direction in a straight path or around an
axis
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Dr
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Prof
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Mahbubul Muttakin, Assoc
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, MPE, AUST

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Energy of a System
 Most

closed systems remain stationary during a process and thus
experience no change in their kinetic and potential energies
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Hence,  Q  dU   W
or in intergral form, Q12  (U 2 - U1 )  W12
This equation is known as Non-flow Energy Equation
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Mahbubul Muttakin, Assoc
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, MPE, AUST

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Corollaries of

1st

Law of TD

From First Law of TD, Q12  (U 2 - U1 )  W12
Now, in an Isolated system, Q12  0 & W12  0
 U 2 - U1  0
“The internal Energy of a closed system remains unchanged if the system is
isolated from its surrounding
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“It is impossible to construct a perpetual motion machine of first kind
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 If

δQ is the amount of heat transferred to raise the temperature of m kg of
substance by dT, then, specific heat C = ( 1/m)*(δQ/dT)

For Constant
Volume
Process

For
Constant
Pressure
Process

Dr
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Prof
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Mahbubul Muttakin, Assoc
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, MPE, AUST

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Specific Heat at Constant Pressure, CP

  Q  dH
Then, CP = ( 1/m)*[dH/dT]P = [dh/dT]P

dh = CPdT
And, dH = mCPdT

Hence, in isobaric Process, Heat Transfer, Q = H2-H1= mCP (T2-T1)
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005 kJkg-1k-1
Cv = 0
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314 J K -1 mol-1 );
M
M  Molecular weight of the substance
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e
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P1V1g = P2V2g
P2/P1 = (V1/V2)g

T2/T1 = (V1/V2)g-1
T2/T1= (P2/P1)(g-1)/g
PV = mRT

P = rRT

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But in many
engineering applications we come across open systems where in both mass and energy transfer takes
place
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1
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2
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E, V 2 / 2, thus causing the energy of the system to change by this amount for
every kg of matter entering the system boundary
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Potential energy: P
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is measured with reference to some base
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E
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4
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But it is associated with the fact that there must be some pumping process which is
responsible for the movement of the matter across the system boundary
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Flow energy is associated with pressure and volume of the
process i
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PV
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Mahbubul Muttakin, Assoc
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, MPE, AUST

1st

law of TD: Open system

Control Volume: The first and most important step in the analysis of an open system is to
imagine a certain region enclosing the system
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A C
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is any volume of fixed shape, and of fixed position and orientation relative to the
observer
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Flow process:
1
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Unsteady flow process

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e
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Assumptions of steady flow process—
 The mass flow rate at the inlet is constant with respect to time and equal to the mass flow rate at
the outlet
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 Properties are constant over the cross section of the flow at the inlet and outlet
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For a steady flow system,
Energy entering the system = Energy leaving the system
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Mahbubul Muttakin, Assoc
...
, MPE, AUST

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1st

20

law of TD: Open system

Min
U1

P1V1

Mout

PE1

U2

KE1

P2V2

Z1

PE2
KE2
Z2
So,
PE1 + KE1 + P1V1 + U1 + Q = PE2 + KE2 + P2V2 + U2 + W
Q = DPE + DKE + [(P2V2 + U2) – (P1V1 + U1)] + W
= DPE + DKE + DH + W
[as, H = U + PV]
= g(Z2-Z1) + ½ (C22-C12) + (h2-h1) + W [considering unit mass ,i
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m = 1kg]
This is called the Steady Flow Energy Equation (SFEE)
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Mahbubul Muttakin, Assoc
...
, MPE, AUST

Applications of

1st

law of TD: Open system

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Nozzle and Diffuser: Nozzle is a duct of varying cross sectional area in which the velocity increases
with a corresponding drop in pressure
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And also there is no work interaction during the process, i
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, W = 0, Q = 0, Z1 = Z2
We have from SFEE,
Q –W = Δh + ΔPE + ΔKE
0 = (h2-h1) + ½ (C22-C12) [for unit mass]
(h1-h2) = ½ (C22-C12)
i
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, the gain in KE during the process is equal to the
decrease in enthalpy of the fluid
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Thus its function is
reverse to that of a nozzle
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The fluid is first accelerated in a set of nozzle and then directed
through curved moving blades which are fixed on the rotor shaft
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Since the velocity of flow of the fluid
through the turbine is very high, the flow process is generally assumed to be adiabatic, hence heat
transfer Q = 0
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SFEE is, 0 = ½ (C22-C12) + (h2-h1) + W
If we consider the system frictionless then (C2 ~ C1)
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A rotary compressor can be regarded as a
reversed turbine
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, the pressure of the fluid drops substantially and the
process is called throttling
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Hence the
process is assumed to occur adiabatically
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SFEE is Q1-2 – W1-2 = Δh + ΔKE + ΔPE
We have,
Q = 0;
W = 0;
Z1 = Z2,
C1 ~ C2
0 – 0 = h2 – h1 + 0 + 0 i
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, h1 = h2

In a throttling process, the enthalpy remains constant
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It is used to add or reduce heat energy of the fluid flowing through the device
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There will be no work interaction during the
flow of the fluid through any heat exchanger
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Mahbubul Muttakin, Assoc
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, MPE, AUST

24

Math Problem-01
Consider a cyclic process in a closed system which includes three heat

interactions, namely Q1 = 20 kJ, Q2 = -6 kJ, and Q3 = -4 kJ and two
work interactions for which W1 = 4500 N-m
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(a) How much heat flows into the system along the path 1-4-3 if
work done by the system is 10 kJ (b) when the state of the system is returned
from state 3 to state 1 along the curved path, the work done on the system is 20
kJ
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(c) If U1 = 0
and U4 = 40kJ, find the heat absorbed in the process 1-4 and 4-3 respectively
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Mahbubul Muttakin, Assoc
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, MPE, AUST

Math Problem-03
A closed system undergoes an isochoric process in which 85 kJ of heat

is supplied to it
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Finally the system is brought back to its original state by an
adiabatic process
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The system then undergoes an isochoric process in which 125
kJ of heat is rejected by the system
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Determine and comment
on the work involved in the adiabatic process
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Mahbubul Muttakin, Assoc
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, MPE, AUST

28

Math Problem-05
A frictionless cylinder-piston device contains Helium gas initially at
150 kPa, 20 °C and 0
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The gas is now compressed in polytropic
process to 400 kPa and 140 °C
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iv
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The polytropic exponent ‘n’
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The change in internal energy
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Take CV = 3
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1 m3
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Water is heated to
produce saturated vapor
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At the discharge end, the
enthalpy is 2762 kJ/kg
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(i) Determine the velocity at exit from the nozzle
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1 m2 and the specific volume at inlet is 0
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(iii) If the specific volume at the nozzle exit is 0
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Prof

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