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

---

CLB 20703
Chemical Engineering
Thermodynamics
Chapter 3:
The First Law of Thermodynamics

Objective of Chapter 3
To discuss ideas about energy for
engineering analysis and develop
equations for applying the principle of
the First Law of Thermodynamics on
conservation of energy in open and
closed systems

Outline


Introduction to First Law of
Thermodynamics
...
1 INTRODUCTION
 First

Law of Thermodynamics (also known as conservation of
energy principle) states that energy can be neither created
nor destroyed during a process but can only change forms
...


Energy Balance for any system
Ein  Eout
 
 
Energy tra nsfer of a system

• Energy
transfer
are
recognized at the system
boundary as they cross it 
represent the energy gained
or lost by a system during a
process
...

• Only two forms of energy
transfer associated with a
closed system are heat
transfer and work
...

• The change in the total energy of a system
during a process is the sum of the changes
in its internal, kinetic, and potential
energies:
E system= U + KE + PE

3
...

• By using the sign convention of
heat and work, heat to be
transferred into the system (heat
input) in the amount of Q and
work to be done by the system
(work output) in the amount of W:



Esystem

Change in kinetic, potetial,
internal, etc
...

 Then the energy balance for a cycle simplifies to Ein - Eout = 0 or
Ein = Eout
...
1
A rigid tank contains a hot fluid that is cooled while being stirred by a
paddle wheel
...
During
the cooling process, the fluid loses 500J of heat, and the paddle wheel
does 100kJ of work on the fluid
...
Neglect the energy stored in the paddle wheel
...
2
A 0
...
Heat is now transfer to the refrigerant until the final
pressure reaches 700kPa
...
3
A piston cylinder device initially contains steam at 200 kPa, 200 OC and
0
...
at this state, a linear spring (F α x) is touching the piston but
exerts no force on it
...
6 m3,
respectively
...
3 MASS AND ENERGY BALANCE FOR
OPEN SYSTEM


Open systems are characterized by flowing
streams, there are 4 common measures of flow:
 Velocity, u
 Volumetric flowrate, q = uA

 Molar flowrate, n  uA


 Mass flowrate, m  Mn  uA
where M = molar mass/molecular weight
A = cross-sectional area
ρ = specific or molar density

Energy Analysis Of Open System






An open system or a control volume (CV) = a selected region in space
and usually encloses a device that involves mass flow in and out of the
system such as a compressor, turbine or nozzle
...


To simplify the energy analysis of CV:
 The system should be assumed undergoing steady-flow process, and
 Conservation of Mass Principle for CV should be firstly defined
before the 1st Law of Thermodynamics can be applied to CV
...

Steady-flow process = a process during which a fluid flows through a
control volume steadily  the fluid properties within the control volume
may change with position but not with time
...
The mass and volume flow
rates are related by:


 V

m  V
v

or



V A
  Vav A  av
m
v

kg s 

Conservation of Energy Principle
conservation of energy principle (1st Law of Thermodynamics)
for control volumes has the similar definition with that of closed
systems:
For steady-flow

 The



E in  Eout

 


Esystem




Rate of net energy tra nsfer across
CV by heat, work and mass

(kW)

process,Ė=0

Rate of change in internal, kinetic,
potential etc energies of CV



E in




Rate of net energy tra nsfer in
by heat, work and mass


Eout

Rate of net energy tra nsfer out
by heat, work and mass




Qin  Win   m Qout  Wout   m
in

out









V2
V2
  Pv 
  Pv 
Qin  Win   m   u 
 gz  Qout  Wout   m   u 
 gz 




2
2
in
out
 h

 h








V2
V2


Qin  Win   m h 
 gz  Qout  Wout   m h 
 gz 


in
out
 2  
 2  






for each inlet

for each outlet

 = energy per
unit mass
flowing in and
out of CV

Energy Balance for control volumes
2
2




V
V
 Qout  Wout   m h 
h 

Qin  Win   m
 gz 
 gz 

in
out
 2  
 2  






for each inlet

for each outlet



2
2

 

  W  m h  Vi  gz   Q  W  m h  Ve  gz 


Qin in    i
  e 2
i
out
out
e
2


 

  





Qnet   Qin   Qout



Qnet  Wnet

for each inlet

for each outlet



2
2




V
V


  m he  e  gze    m hi  i  gzi 




2
2
  



 


for each inlet




Wnet  Wout  Win

for each outlet


 



Q  W  H  KE  PE

Overall 1st Law of
thermodynamics for
CV undergoing
steady-flow process

Steady-flow Engineering Devices

Nozzles and Diffusers
 Nozzle and diffuser are commonly utilized in jet engines, rockets, spacecraft and
even garden hoses
...







Mass balance : m in  m out m1 m 2  m


Energy balance : E in  E out
2
2



V
V 


m h1  1   m h2  2 


2 
2 





2
V2  2h1  h2   V1

Turbines
Control
Surface

1

W
2

In steam, gas, or hydroelectric power plants, the device that drives
the electric generator  TURBINE
...

 By ignoring the change in KE and PE energies (ke=pe=0) through
an adiabatic turbine (Q=0) with a single stream (one inlet-one outlet)
that undergoes a steady flow process:




 

Mass balance : m in  m out m1 m 2  m
 
Energy balance : E  E
in

out




mh1  mh2  W
 
W  mh  h
1

2



Compressors and Fans
 Compressors,

as well as pumps and fans, are devices used to increase the
pressure of a fluid
...

 The differences between the three devices:
 A fan increases the pressure of a gas slightly and is mainly used to
mobilize a gas at low pressure
...

 Pumps work very much like compressors except that they handle liquids
instead of gases
...

 Unlike turbines, they produce a pressure drop without involving
any work but often accompanied by a large drop in temperature
 devices are commonly used in refrigeration and airconditioning applications
...
The section where the mixing process
takes place  mixing chamber
...
” An ordinary T-elbow or a Y-elbow in a shower =
mixing chamber for the cold and hot water streams
...

 Mixing chambers are usually well insulated (q=0), usually do
not involve any kind of work (w=0), the kinetic and potential
energies of the fluid streams are usually negligible (ke=0,
pe=0):


1
2

1

MIXER

2
????


 

  m out m1  m 2  m3
 
Energy balance : E in  E out



 m1h1  m2 h2  m3h3
Mass balance :

3

3


m

in

Heat Exchangers
 Heat

exchangers are devices where two moving fluid streams exchange
heat without mixing
...

The simplest form of a heat exchanger is a double-tube (also called
tube and-shell) heat exchanger
...

 Heat exchangers typically involve no work interactions (w=0) and
negligible kinetic and potential energy changes (ke=0, pe=0) for each
fluid stream
...

3

If only one of the fluids is selected as
the control volume, then heat will
cross this boundary as it flows from
one fluid to the other and will not be
zero  the rate of heat transfer
between the two fluids
...

 Flow through a pipe or a duct usually satisfies the steady-flow
conditions
...
Water flow through the pipes in the furnace of a power
plant, the flow of refrigerant in a freezer, and the flow in heat
exchangers are some examples of this case
...
Heat transfer in this case is
negligible
...

 For compressed liquid, the density and specific volumes are
constant (v2=v1=v) and the process of pumping compressed liquid
is isothermal (u=cvdT=0)
...
4
Steam enter a nozzle at 400 OC and 800 kPa with a velocity
of 10m/s, and leaves at 300 OC and 200 kPa while losing
heat at a rate of 25 kW
...


Example 4
...
it is to be cooled by water
from a nearby river by circulating the
water through the tubes within the
condenser
...
If the steam is to leave
the condenser as saturated liquid at 20
kPa, determine the mass flow rate of
the cooling water required

---