Notes 15
Chapter 5
Components of
Thermodynamic Cycles
Components of Thermodynamic
Cycles
• What can be the components of
thermodynamic cycles?
• Turbines, valves, compressors, pumps,
heat exchangers (evaporators,
condensers), mixers,
Open Systems Control Volume
Where Exactly
Are They
Used?
Vapor-cycle
Power Plants
Inside The Household
Refrigerator
Review of 1st Law
V V
q  w  ( h2  h1 ) 
 g ( z 2  z1 )
2
2
2
2
1
Applications to some
steady state systems
• Start simple
– nozzles
– diffusers
– valves
• Includes systems with power in/out
– turbines
– compressors/pumps
• Finish with multiple inlet/outlet devices
– heat exchangers
– mixers
You will need to use all
that you’ve learned to date
•
•
•
•
•
•
Conservation of mass
Conservation of energy
Property relationships
Ideal gas equation of state
Property tables
Systematic analysis approach
Nozzles and Diffusers
• Nozzle--a device which accelerates a fluid
as the pressure is decreased.
V1, P1
V2, P2
This configuration is for sub-sonic flow.
Nozzles and Diffusers
• Diffuser--a device which decelerates
a fluid and increases the pressure.
V1, P1
V2, P2
For supersonic flow, the shape
of the nozzle is reversed.
Nozzles
General shapes of
nozzles and diffusers
nozzle
diffuser
Subsonic flow
nozzle
diffuser
Supersonic flow
Common assumptions we’ll
make for nozzles and diffusers
• Steady state, steady flow
• No work - device produces or uses
shaft work
• Heat loss/gain is often neglected
• Potential energy change is usually
small
We start our analysis of
diffusers and nozzles with
the conservation of mass
If we have steady state, steady flow, then:
dm cv
0
dt
And
1  m
2 m

m
We continue with
conservation of energy
2
2
dEcv
V

V
e
 [( hi  he )  i
 Q  W shaft  m
 g ( zi  ze )]  0
dt
2
We can simplify by dividing by mass flow:
0
V22  V12
q  w  ( h2  h1 ) 
 g ( z 2  z1 )
2
0
0
q = 0 (adiabatic)
w = 0 (these are not work producing devices;
neither is work done on them)
We can rearrange to get a
much simpler expression:
V V
( h2  h1 ) 
2
2
1
2
2
With a nozzle or diffuser, we are
converting flow and internal energy,
represented by Dh =D(u+Pv) into
kinetic energy, or vice-versa.
System Approach to
Problem Solving
• State assumptions for the device
• Write basic form of conservation of
mass and energy equations
• Apply assumptions to get to previous
equation
V V
( h2  h1 ) 
2
2
1
2
2
Sample Problem
An adiabatic diffuser is employed to
reduce the velocity of a stream of air
from 250 m/s to 35 m/s. The inlet
pressure is 100 kPa and the inlet
temperature is 300°C. Determine
the required outlet area in cm2 if the
mass flow rate is 7 kg/s and the final
pressure is 167 kPa.
Sample problem: diagram
and basic information
INLET
T1 = 300C
P1 = 100 kPa
V1 = 250 m/s
m = 7 kg/s
OUTLET
Diffuser
P2 = 167 kPa
V2 = 35 m/s
Sample Problem:
Assumptions
• SSSF (Steady state, steady flow) no time dependent terms
• adiabatic
• no work
• potential energy is zero
• air is ideal gas
Sample Problem:
apply basic equations
Conservation of Mass
1  m
2 m

m
 
m
Solve for A2
V1 A1
1

V2 A2
m  2
A2 
V2
2
How do we get
specific volumes?
Remember ideal gas equation of state?
P  RT
or
RT1
1 
P1
and
RT2
2 
P2
We know T1 and P1, so v1 is simple. We
know P2, but what about T2?
NEED ENERGY EQUATION!!!!
Sample Problem:Energy Eqn
V V
q  w  ( h2  h1 ) 
 g ( z 2  z1 )
2
2
2
V V
( h2  h1 ) 
2
2
1
2
1
2
2
V1 and V2 are given. We need h2 to get T2 and v2.
If we assumed constant specific heats, we could
get T2 directly
V V
c p (T2  T1 ) 
2
2
1
2
2
Sample problem - con’t
However, I’m going to use variable specific heats..
we get h1 from air tables at T1 = 300 + 273 = 573K.
kJ
h1  578.73
kg
From energy equation:
kJ  ( 250 )2  ( 35 )2  m 2  3 kJ s 2 
 2  10

h2  578.73
 
2 
kg 
2
kg m 
 s 
kJ
h2  609.4
kg
This corresponds to an exit
temperature of 602.2 K.
Now we can get solution.
3
RT2
m
2 
 1.0352
P2
kg
and
m  2
A2 
V2
kg 
m3 

 1.0352
s 
kg 

7


2

 m  4 m 

 35  10
2 
s 
cm 

A2  2070 cm
2
Throttling Devices (Valves)
Typical assumptions
for throttling devices
• No work
• Potential energy changes are zero
• Kinetic energy changes are usually small
• Heat transfer is usually small
• Two port device
Look at energy equation:
Apply assumptions from previous page:
0
0
0
0
V V
q  w  ( h2  h1 ) 
 g ( z 2  z1 )
2
We obtain:
or
2
2
2
1
( h2  h1 )  0
h2  h1
Discussion Question
Does the fluid temperature
increase,
decrease, or
remain constant
as it goes through an adiabatic valve?
Look at implications:
If the fluid is liquid/vapor in equilibrium:
During throattling
process:
h const.
T
P const.
• The pressure
drops,
• The temperature
drops,
h const.
s
• Enthalpy is
constant
Look at implications:
If fluid is an ideal gas:
( h2  h1 )  C p ( T2  T1 )  0
Cp is always a positive number, thus:
T2  T1
TEAMPLAY
Refrigerant 12 enters a valve as
a saturated liquid at 0.9607 Mpa
and leaves at 0.1826 MPa.
What is the quality and the
temperature of the refrigerant at
the exit of the valve?
TEAMPLAY
Liquid R12
State (1)
Liquid saturated, x=0
Psat = 0.9607 MPa
Tsat = ?
Hliq = ?
Liquid + Vapor
R12
State (2)
Liquid+vapor, x=?
Psat = 0.1826 MPa
Tsat = ?
Hliq = ?
Turbine
• A turbine is device in which work is produced
by a gas passing over and through a set of
blades fixed to a shaft which is free to rotate.

m in

m out

W CV
TURBINES
We’ll assume steady state,
V22  V12
q  w  ( h2  h1 ) 
 g ( z 2  z1 )
2
Sometimes Almost
neglected
always
neglected
q  w  ( h2  h1 )
Turbines
• We will draw turbines like this:
inlet
w
maybe q
outlet
Compressors,
pumps, and fans
• Machines developed by engineers to
make life easier, decrease world
anxiety, and provide exciting
engineering problems from the
industrial revolution for students.
• Machines which do work on a fluid to
raise its pressure, potential, or speed.
• Mathematical analysis proceeds the
same as for turbines, although the signs
may differ.
Primary differences
• Compressor - used to raise the pressure
of a compressible fluid
• Pump - used to raise pressure or
potential of an incompressible fluid
• Fan - primary purpose is to move large
amounts of gas, but usually has a small
pressure increase
Compressors,
pumps, and fans
Compressor
Side view
of pump
End view
of pump
Sample Problem
Air initially at 15 psia and 60°F is
compressed to 75 psia and 400°F.
The power input to the air is 5 hp
and a heat loss of 4 Btu/lbm occurs
during the process. Determine the
mass flow in lbm/min.
Draw Diagram
15 psia
60oF
Compressor
Wshaft = 5 hp
75 psia
400oF
Q = 4 Btu/lb
Assumptions
•
•
•
•
Steady state steady flow (SSSF)
Neglect potential energy changes
Neglect kinetic energy changes
Air is an ideal gas
What do we know?
INLET
T1 = 60F
P1 = 15 psia
OUTLET
T2 = 400F
P2 = 75 psia
Apply First Law:
0
0
0
0
2
2




V
V
1
2


  h1 
  h2 
Q - Wshaft  m
 gz1   m
 gz2   0
2
2




Simplify and rearrange:
 q  W shaft  m
 (h2  h1 
m
m 
W shaft
q  (h2  h1 
Continuing with the solution..
Get h1 and h2 from air tables
Btu
h1  124.27
lbm
Btu
h2  206.46
lbm
Follow through with solution

ft  lbf  60s 
( 5hp 550


hp  s  min 

m 
Btu
Btu 
ft  lbf 

 (206.46  124.27 
 4
 778

lbm
lbm 
Btu 

lbm
m  2.46
min
Heat Exchangers
and mixing devices
• Heat exchangers are devices which
transfer heat between different fluids
• Mixing devices (also called open heat
exchangers) combine two or more
fluids to achieve a desired output,
such as fluid temperature or quality
Heat exchangers are used
in a variety of industries
•
•
•
•
•
Automotive - radiator
Refrigeration - evaporators/condensers
Power production - boilers/condensers
Power electronics - heat sinks
Chemical/petroleum industry- mixing
processes
Heat Exchangers
Condenser/evaporator
for heat pump
Something a little closer to home..
Heat Exchangers
• Now, we must deal with multiple inlets
and outlets:
m 4
m 2
m 1
m 3
Question…...
• At steady flow, what is the relationship
between
1, m
2, m
 3 , and m
4?
m
Conservation of Mass
What assumptions to make?
• See any devices producing/using shaft
work?
• What about potential energy effects?
• What about kinetic energy changes?
• Can we neglect heat transfer?
Apply conservation of
mass on both streams...
If we have steady
flow, then:
m 4
dm cv
0
dt
m 2
m 1
And
1  m
2m
A
m
3 m
4  m
B
m
Fluid A
m 3
Fluid B
Conservation of energy can
be a little more complicated...
I’ve drawn the control
volume around the
whole heat exchanger.
Implications:
No heat
transfer from
the control
volume.
m 4
m 2
m 1
Fluid A
m 3
Fluid B
Heat Exchangers
• Now if we want the energy lost or gained
by either fluid we must let that fluid be
the control volume, indicated by the red.
m 2
1  m
A
m
Conservation of energy
looks pretty complicated:
Q  WCV
2
2




V1
V3
 1  h1 
 3  h3 
m
 gz1   m
 gz3 
2
2








V22
V42
 2  h2 
 4  h4 
m
 gz2   m
 gz4   0
2
2




We know from conservation of mass:
1  m
 2 m
A
m
 3 m
4 m
B
m
Conservation of energy
for the heat exchanger
Apply what we know about the
mass flow relationships:
Q  WCV


V12 V22
 A  (h1  h2  
m

 g (z1  z2 
2
2

A


V32 V42
 B  (h3  h4  
m

 g (z3  z4   0
2
2

B
Heat Exchangers
• Generally, there is no heat transfer from
or to the overall heat exchanger, except
for that leaving or entering through the
inlets and exits.
• So, Q  0

• And, the device does no work, W
CV  0
• Also, potential and sometimes kinetic
energy changes are negligible.
Heat Exchangers
0
0
Q  W CV


V12 V22
 A  (h1  h2  
m

 g (z1  z2 
2
2

A
0, (sometimes 0, (usually
negligible)
negligible)


V32 V42
 B  (h3  h4  
m

 g (z3  z4   0
2
2

B
0, (sometimes 0, (usually
negligible)
negligible)
Heat Exchangers
• And we are left with
 A ( h1  h2 )  m
 B ( h4  h3 )
m
The energy change of fluid A is equal to the
negative of the energy change in fluid B.
QUESTION…..
How would the energy equation
differ if we drew the boundary
of the control volume around
each of the fluids?
Heat Exchangers
The energy equation for one side:

 A (h1  h2   0
QCV , A  m
Or dividing through by the mass flow:
q A  h2  h1
Example Problem
Refrigerant 134a with a mass flow rate of 5
kg/min enters a heat exchanger at 1.2 MPa
and 50C and leaves at 1.2 MPa and 44C.
Air enters the other side of the heat exchanger
at 34 C and 101 kPa and leaves at 42 C and
101 kPa. Calculate:
a) the heat transfer from the refrigerant
b) the mass flow rate of the air
Draw diagram
AIR
OUTLET
T3=42C
P3 = 101 kPa
 Ref  5 kg / min
m
R-134a
INLET
R-134a
OUTLET
T2=44C
P2 = 1.2 MPa
AIR
INLET
T3=34C
P3 = 101 kPa
T1=50C
P1 = 1.2 MPa
 air  ?
m
State assumptions
•
•
•
•
•
Steady state, steady flow
No work
Air is ideal gas
Kinetic energy change is zero
Potential energy change is zero
Start analysis with R-134a
0
Q Ref  WCV
0
0
V22  V12
 Ref [( h2  h1 ) 
m
 g ( z2  z1 )]
2
Apply assumptions
 Ref ( h2  h1 )
Q Ref  m
We can get h1 and h2 from tables. The
refrigerant mass flow is given.
From R-134a tables
h1 = 275.52 kJ/kg
h2 = 112.22 kJ/kg
Plugging back into energy equation:
kg
kJ

QRef  5
( 112.22  275.52 )
min
kg
Q Ref
kJ
 816.5
min
On to part (b) of the problem
We want to get the mass flow of the air...
Start by writing the energy equation for the air side:
0
Q air  WCV
0
0
V42  V32
 air[( h4  h3 ) 
m
 g ( z4  z3 )]
2
Simplify

 air ( h4  h3 )
Qair  m
Sample Problem, Con’t
If air is an ideal gas, then we can rewrite the
enthalpy difference as:
 airC p (T4  T3 )
Q air  m
Rearrange to solve for mass flow:
Q air
m air 
C p (T4  T3 )
How do we get the heat transfer rate to/from the air?
Almost there!!!!!
We can write:
Q air   Q ref  Q air   816.5 kJ / min
Get specific heat from table: Cp = 1.006 kJ/kg.K
kJ
816.5
kg
min
m air 
 101.4
kJ
min
(1.006
)(42  34 )K
kgK