Lec 13: Machines (except heat
exchangers)
1
• For next time:
– Read: § 5-4
– HW7 due Oct. 15, 2003
• Outline:
– Diffusers and nozzles
– Turbines
– Pumps and compressors
• Important points:
– Know the standard assumptions that go
with each device
– Know how to simplify the governing
equations using these assumptions
– Consider what each device would be
used for in real-world applications
2
Applications to some steady
state systems
• Start simple
– nozzles
– diffusers
– valves
• Include systems with power in/out
– turbines
– compressors/pumps
• Finish with multiple inlet/outlet devices
– heat exchangers
– mixers
3
We will need everything we have
covered
•
•
•
•
•
•
Conservation of mass
Conservation of energy
Property relationships
Ideal gas equation of state
Property tables
Systematic analysis approach
4
Nozzles and Diffusers
• Nozzle--a device which accelerates a fluid
as the pressure is decreased.
V1, p1
V2, p2
This configuration is for subsonic flow.
5
Nozzles and Diffusers
• Diffuser--a device which decelerates a
fluid and increases the pressure.
V1, p1
V2, p2
6
For supersonic flow, the shape of
the nozzle is reversed.
Nozzles
7
General shapes of nozzles and
diffusers
Subsonic Flow
Nozzle
Diffuser
Supersonic Flow
Nozzle
Diffuser
8
Common assumptions for
nozzles and diffusers
• Steady state, steady flow.
• Nozzles and diffusers do no work and
use no work.
• Potential energy changes are usually
small.
• Sometimes adiabatic.
9
TEAMPLAY
• For nozzles, diffusers and other
machines--just how important is PE?
• The energy in the head of a kitchen
match is reportedly about 1 Btu.
• How far does 1 lbm have to fall in a
standard earth gravity field to “match”
this much energy?
• Example 5-12 on p. 175 has an
enthalpy change h1 - h2 less than 20
Btu. What does your result mean
physically for a nozzle or diffuser?
10
We start our analysis of diffusers
and nozzles with the conservation
of mass
If we have steady state, steady flow,
then:
dm CV
dt
0
And
m 1  m 2  m
11
We continue with conservation of
energy



dE CV
Vi 2  Ve2
 Q WCV  m[( h i  h e ) 
 g( z i  z e )]  0
dt
2
We can simplify by dividing by mass flow:
0 0
0
V V
q  w  ( h 2  h1 ) 
 g( z 2  z 1 )
2
2
2
2
1
Applying the definition that
w=0 and using some other
assumptions...
12
We can rearrange to get a much
simpler expression:
V V
( h 2  h1 ) 
2
2
1
2
2
With a nozzle or diffuser, we are converting
flow energy and internal energy, represented
by Dh into kinetic energy, or vice-versa.
13
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 requred
outlet area in cm2 if the mass flow rate is 7 kg/s
and the final pressure is 167 kPa.
14
Sample Problem:Assumptions
• SSSF (Steady state, steady flow) - no time
dependent terms
• adiabatic
• no work
• potential energy change is zero
• air is ideal gas
15
Sample problem:diagram and
basic information
INLET
V1
Diffuser
OUTLET
V
2
T1=300C
P1=100 kPa
P2=167 kPa
V1=250 m/s
V2=35 m/s
m = 7 kg/s
16
Sample Problem: apply basic
equations
Conservation of Mass
1  m
2  m

m
V1 A1 V2 A2
m 

ν1
ν2
Solve for A2
m ν2
A2 
V2
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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!!!!
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Sample problem - con’t
Energy
V22  V12
q  w  (h2  h1 ) 
 g(z 2  z1 )
2
V12  V22
(h2  h1 ) 
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
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Sample problem - con’t
However, use variable specific heats...get h1
from air tables at
T1 = 300+273 = 573 K.
kJ
h1  578.73
kg
From energy equation:
kJ  (250) 2  (35) 2  m 2  3 kJ s 2 
 2 10

h 2  578.73  
2 
kg 
2
kg m 
 s 
kJ
h 2  609.4
kg
This corresponds to an exit
temperature of 602.2 K.
20
Now we can get solution.
3
RT2
m
2 
 1.0352
P2
kg
and
m ν2
A2 
V2
3

m 
 kg 

 7 1.0352
kg 
 s 

2

 m  4 m 

 35 10
2 
cm 
 s 
A 2  2070 cm
2
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TEAMPLAY
Work problem 5-65
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Throttling Devices (Valves)
23
Short tube orifice for 2.5 ton air
conditioner
24
Throttles (throttling devices)
• A major purpose of a throttling device is
to restrict flow or cause a pressure drop.
• A major category of throttling devices is
valves.
25
Typical assumptions for throttling
devices
• Do no work, have no work done on them
• Potential energy changes are zero
• Kinetic energy changes are usually small
• Heat transfer is usually small
26
Look at energy equation:
Apply assumptions from previous page:
0 2
0
0 0
V  V1
q  w  (h2  h1 ) 
 g(z 2  z1 )
2
2
2
We obtain:
or
(h2  h1 )  0
h2  h1
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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
28
Discussion Question
Does the fluid
temperature
increase,
decrease, or
remain constant
as an ideal gas goes through an adiabatic valve?
29
TEAMPLAY
Refrigerant 134a enters a valve as a
saturated liquid at 200 psia and leaves at
50 psia. What is the quality of the
refrigerant at the exit of the valve?
30
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.
31
32
Turbines
We’ll assume steady state,
V V
q  w  (h2  h1 ) 
 g(z 2  z1 )
2
2
2
2
1
Sometimes
neglected
Almost always
neglected
q  w  (h2  h1 )
33
Turbines
• We will draw turbines like this:
inlet
w
maybe q
outlet
34
Compressors, pumps, and fans
• Machines developed to make life easier,
decrease world anxiety, and provide
challenging problems for engineering
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.
35
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
36
Compressors, pumps, and fans
Axial flow
Compressor
Side view
End view
Centrifugal pump
37
38
39
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/lb occurs during the process.
Determine the mass flow in lbm/min.
40
Draw Diagram
15 psia
60 F
W sh
= 5 hp
75 psia q = 4 Btu/lb
400 F
41
Assumptions
•
•
•
•
Steady state steady flow (SSSF)
Neglect potential energy changes
Neglect kinetic energy changes
Air is ideal gas
42
What do we know?
INLET
T1 = 60F
P1 = 15 psia
OUTLET
T2 = 400F
P2 = 75 psia
43
Apply First Law:
20
1
0
20
2
0




V
V

Q  W sh  m  h1 
 gz1   m  h2 
 gz 2   0
2
2




Simplify and rearrange:
m q  W sh  m h2  h1 
W sh
m 
q  h2  h1 
44
Continuing with the solution..
Get h1 and h2 from air tables
Btu
h1  124.27
lb m
Btu
h 2  206.46
lb m
Follow through with solution

ft  lb f  60s 
 5hp  550


hp  s  min 

 
m

Btu
Btu 
ft  lb f 
  4
 778
 206.46  124.27 

lb m
lb m 
Btu 

lb m
  2.46
m
min
45
TEAMPLAY
Work problem 5-73E
46
TEAMPLAY
• Use EES and vary the exit pressure from 5
psia to 0.5 psia in increments of 1.0 psia.
Show the results as a table and a plot.
• Open EES and put in the basic equation
  m


Q
(h

h
)

W
out
2
1
out
47
TEAMPLAY
• You will have to use some new features of
EES
– 1. Under options always check and set
unit system, if necessary.
– 2. Under options, find function info,
and select fluid properties.
– 3. For steam, use Steam_NBS.
48
TEAMPLAY
• Parametric studies
• Under “Tables”, select “New Parametric
Table”
• Click and drag the variables you want to
see to the right--P2, Qdot, and h2.
• See that P2 is not specified in the
problem statement in the “Equations
Window”.
49
TEAMPLAY
• Enter P2 via “Alter Values” under
“Tables”
• Click on the column headings to be able
to enter units.
• You must solve the table before you can
plot it.
• Under “Calculate” select “Solve Table.”
50
TEAMPLAY
• Under “Plot” select “New Plot Window” and
“X-Y Plot”.
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