Download - Lec 13: Machines (except heat exchangers)
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Lec 13: Machines (except heat exchangers)
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• 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
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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
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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
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Nozzles and Diffusers
• Nozzle--a device which accelerates a fluid as the pressure is decreased.
V1, p1V2, p2
This configuration is for subsonic flow.
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Nozzles and Diffusers
• Diffuser--a device which decelerates a fluid and increases the pressure.
V1, p1V2, p2
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Nozzles
For supersonic flow, the shape of the nozzle is reversed.
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General shapes of nozzles and diffusers
Subsonic Flow
Nozzle Diffuser
Nozzle Diffuser
Supersonic Flow
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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.
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TEAMPLAYTEAMPLAY
• 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?
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We start our analysis of diffusers and nozzles with the conservation
of mass
dt
dmCV 0
mmm 21
If we have steady state, steady flow, then:
And
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We continue with conservation of energy
dt
dECV 0)]zz(g2
)hh[(mWQ ei
2e
2i
eiCV
VV
)zz(g2
)hh(wq 12
21
22
12
VV
We can simplify by dividing by mass flow:
Applying the definition that w=0 and using some other assumptions...
0 0 0
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We can rearrange to get a much simpler expression:
With a nozzle or diffuser, we are converting flow energy and internal energy, represented by Dh into kinetic energy, or vice-versa.
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Sample Problem
An adiabatic diffuser is employed to reduce thevelocity of a stream of air from 250 m/s to 35 m/s.The inlet pressure is 100 kPa and the inlettemperature is 300°C. Determine the requredoutlet area in cm2 if the mass flow rate is 7 kg/sand the final pressure is 167 kPa.
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Sample Problem:Assumptions
• SSSF (Steady state, steady flow) - no time dependent terms
• adiabatic• no work• potential energy change is zero• air is ideal gas
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Sample problem:diagram and basic information
INLET
T1=300C
P1=100 kPa
V1=250 m/s
m = 7 kg/s
OUTLET
P2=167 kPa
V2=35 m/s
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Sample Problem: apply basic equations
Conservation of Mass
mmm 21
2
22
1
11
ν
AV
ν
AVm
Solve for A2
2
22
V
νmA
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How do we get specific volumes?
Remember ideal gas equation of state?
RTP or
1
11 P
RT
2
22 P
RTand
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’tEnergy
V1 and V2 are given. We need h2 to get T2 and v2. If we assumed constant specific heats, we could get T2 directly
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Sample problem - con’t
However, use variable specific heats...get h1 from air tables at T1 = 300+273 = 573 K.
2
23
2
222
2 m
s
kg
kJ10
s
m
2
)35()250(
kg
kJ578.73h
kg
kJ578.73h1
From energy equation:
kg
kJ4.609h2 This corresponds to an exit
temperature of 602.2 K.
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Now we can get solution.
kg
m0352.1
P
RT 3
2
22
2
24
3
cmm
10sm
35
kgm
0352.1s
kg 7
and
22 cm 2070A
2
22
V
νmA
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TEAMPLAYTEAMPLAY
Work problem 5-65
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Throttling Devices (Valves)
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Short tube orifice for 2.5 ton air conditioner
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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.
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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
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Look at energy equation:
Apply assumptions from previous page:
0 0 00
We obtain:
or
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Look at implications:
If fluid is an ideal gas:
cp is always a positive number, thus:
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Discussion QuestionDoes the fluid temperature
increase, decrease, or
as an ideal gas goes through an adiabatic valve?
remain constant
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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?
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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.
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Turbines
)zg(zVV
)h(hwq 12
21
22
12 2
Sometimes neglected
)( 12 hhwq
Almost always neglected
We’ll assume steady state,
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Turbines
• We will draw turbines like this:
inlet
outlet
w
maybe q
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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.
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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
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Compressors, pumps, and fans
Axial flow Compressor
Side view End view
Centrifugal pump
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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.
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Draw Diagram
15 psia 60 F
W sh = 5 hp
75 psia 400 F q = 4 Btu/lb
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Assumptions
• Steady state steady flow (SSSF)• Neglect potential energy changes• Neglect kinetic energy changes• Air is ideal gas
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What do we know?
INLET
T1 = 60F
P1 = 15 psia
OUTLET
T2 = 400F
P2 = 75 psia
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Apply First Law:
022 2
22
21
21
1
gz
Vhmgz
VhmWQ sh
0 00 0
Simplify and rearrange:
12 hhmWqm sh
12 hhq
Wm sh
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Continuing with the solution..
Get h1 and h2 from air tables
m2
m1 lb
Btu206.46h
lb
Btu124.27h
Follow through with solution
Btulbft
778lbBtu
124.27206.46lbBtu
4
min60s
shplbft
5505hp
mf
mm
f
min
lb2.46m m
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TEAMPLAYTEAMPLAY
Work problem 5-73E
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TEAMPLAYTEAMPLAY
• 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
out12out W)h(hmQ
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TEAMPLAYTEAMPLAY
• 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.
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TEAMPLAYTEAMPLAY
• 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”.
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TEAMPLAYTEAMPLAY
• 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.”
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TEAMPLAYTEAMPLAY
• Under “Plot” select “New Plot Window” and “X-Y Plot”.