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ME 200 L19:ME 200 L19: Conservation Laws: CyclesHW 7 Due Wednesday before 4 pm
HW 8 Posted Start earlyKim See’s Office ME Gatewood Wing Room 2172
https://engineering.purdue.edu/ME200/ThermoMentor© Program Launched
Spring 2014 MWF 1030-1120 AMJ. P. Gore
gore@purdue.eduGatewood Wing 3166, 765 494 0061
Office Hours: MWF 1130-1230
TAs: Robert Kapaku rkapaku@purdue.edu Dong Han han193@purdue.edu
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Common Steady-flow Energy DevicesCommon Steady-flow Energy Devices
NozzlesNozzles
CompressorsCompressorsHeat Exchangers and MixersHeat Exchangers and Mixers
ThrottlesThrottles
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Water, Steam, Gas TurbinesWater, Steam, Gas Turbines
PumpPump
DiffusersDiffusers
Heat Exchangers
►Direct contact: A mixing chamber in which hot and cold streams are mixed directly.
►Tube-within-a-tube counterflow: A gas or liquid stream is separated from another gas or liquid by a wall through which energy is conducted. Heat transfer occurs from the hot stream to the cold stream as the streams flow in opposite directions.
► if there is no stirring shaft or moving boundary.
► ΔKE = (Vi2/2-Ve
2/2) negligible unless specified.
► ΔPE = negligible unless specified.
► If Heat transfer with surroundings is negligible.
►Control Volume includes both hot and cold flows. The “heat exchange,” between them is internal!
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Heat Exchanger Modeling
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Example Problem: Heat Exchanger
Given: Air and Refrigerant R-22 pass through separate streams through an insulated heat exchanger. Inlet and exit states of each are defined.
Find: (a) Mass flow rates, (b) Energy transfer from air to the refrigerant.
R-22
Air
3
4
1
2
Assumptions: Flow work only, insulated casing, steady state, steady flow, no leaks.
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2122
242213220
hh
hhmm
hmhmhmhm
AirR
AirRAirR
Data: Av1= 40m3/min,1=27 C=300K, P1= 1.1 barsT2=15 C= 288 K, P2= 1barP4= 7 bars, T4=15 C, P3=7 bars, x3=0.16
31 1 1
1 1
(1.1 )(100 / )((40 / min) / (60 / min))
(0.287 / )(300 )
0.8517( / )(60 / min) 51.11 / min
Air
AV P AV bar kPa bar m sm
v RT kJ kg K K
kg s s kg
R22 properties: Table A-9, A-8. P=7 bars, Tsat=10.91 C. Therefore, 4 is superheated and h4=256.86 kJ/kg. Table A-8 h3=hf3+x3hfg3= 58.04+(.16)(195.6) =89.34 kJ/kg.
Example Problem: Heat Exchanger
Given: Air and Refrigerant R-22 pass through separate streams through an insulated heat exchanger. Inlet and exit states of each are defined.
Find: (a) Mass flow rates, (b) Energy transfer from air to the refrigerant.
R-22
Air
3
4
1
2
Assumptions: Flow work only, insulated casing, steady state, steady flow, no leaks.
22 3 1 22 4 2
1 222
4 3
( ) ( )
0 0 0
cvR Air R Air
R Air
dEQ W m h m h m h m h
dt
h hm m
h h
Data: Av1= 40m3/min,1=27 C=300K, P1= 1.1 barsT2=15 C= 288 K, P2= 1barP4= 7 bars, T4=15 C, P3=7 bars, x3=0.16
R22 properties: Table A-9, A-8. P=7 bars, Tsat=10.91 C. Therefore, 4 is superheated and h4=256.86 kJ/kg. Table A-8 h3=hf3+x3hfg3= 58.04+(.16)(195.6) =89.34 kJ/kg.
Example Problem: Heat Exchanger
Given: Air and Refrigerant R-22 pass through separate streams through an insulated heat exchanger. Inlet and exit states of each are defined.
Find: (a) Mass flow rates, (b) Energy transfer from air to the refrigerant.
R-22
Air
3
4
1
2
Assumptions: Flow work only, insulated casing, steady state, steady flow, no leaks.
min/673.334.8986.256
15.28819.30011.51
34
2122
kghh
hhmm AirR
Data: Av1= 40m3/min,1=27 C=300K, P1= 1.1 barsT2=15 C= 288 K, P2= 1barP4= 7 bars, T4=15 C, P3=7 bars, x3=0.16
min/3.61519.30015.28811.51)(
min/3.615)34.8986.256(673.3)(
12
342222
kJhhmQ
kJhhmQ
AirAir
RR
►Engineers creatively combine components to achieve some overall objective, subject to constraints such as minimum total cost. This engineering activity is called system integration.
System Integration
►The simple vapor compression refrigeration cycle provides an illustration.
►An integrated system
that transfers heat from a low T reservoir to a high T reservoir using work is called a Heat Pump or a Refrigerator
Heat Pumps and Refrigerators
Heat Pumps and Refrigerators are thermodynamic devices that take heat from a low temperature reservoir and pump it into a high temperature reservoir using cyclic external work input.•The only difference between a refrigerator and a heat pump is in the “desired energy result”:– it’s QC the heat removed from a low temperature
reservoir for a refrigerator.– it’s QH the heat input into a high temperature reservoir
for a heat pump.•The required energy input is the compressor work in each case.
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Heat Pumps and Refrigerators
The figure of merit is the same, desired energy output/required energy input.•It’s termed the coefficient of performance
•Carnot or Max performance
H
Cthermal Q
Q1
CH
C
cycle
CQQ
Q
W
Q
CH
H
cycle
HQQ
Q
W
Q
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C C
H c H C
H H
H c H C
Q T
Q Q T T
Q T
Q Q T T
Heat Pumps and Refrigerators
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Vapor-compression refrigeration cycle
120 kPa, -22 oC, Sat. Vapor
120 kPa, -22 oC, Sat. L+V
800 kPa, 60 oC, SHV800 kPa,
31oC, SCL
Refrigerator and Heat Pump Example
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(273 22) 2518.37
30 30
C CARNOT C CARNOT CCARNOT
cycle CARNOT H CARNOT C CARNOT H C
Q Q T
W Q Q T T
(273 60) 33311.1
30 30
H CARNOT H CARNOT HCARNOT
cycle CARNOT H CARNOT C CARNOT H C
Q Q T
W Q Q T T
Heat Engines
A generic heat engine can be represented as: •Features include:– receiving heat from a
high T source.– producing net
mechanical work.– rejecting heat to a low
T sink.– operating in a cyclic
manner.
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Heat Engines• Application of the First Law to our (cyclic) heat
engine gives• Cyclic operation implies no net change in
(h+V2/2+gZ) over the cycle hence:
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2 2
, ,
2 2
, ,
2 2
2 2
0 ( ) ( ) 0 0
cv i ecv cv i i i e e e
i o i e i e
i ecv cv cv i i i e e e
i o i o i e
i o o i
dE V VQ W m h gz m h gz
dt
V VdE Q dt W dt m h gz m h gz
Q Q W W
inoutoutin WWQQ
Heat Engines• Application of the First Law to our (cyclic) heat
engine gives• Cyclic operation implies no net change in
(h+V2/2+gZ) over the cycle hence:
• This is usually written in terms of the net work.
inoutoutin WWQQ
outinnet QQW
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Heat Engines
Thermal efficiency is the “figure of merit” for heat engines.•It’s utility comes from indicating what fraction of the energy added to the system is converted to mechanical work.
H
C
H
CH
H
cycle
Q
Q
Q
Q
W
1
Thermal efficiency = Energy you get / Energy you pay for
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Thermal efficiency = Desired energy result / Required energy input
2nd Law Corollaries for Power Cycles
No power cycle can have a thermal efficiency of 100%.
•Carnot Corollaries– The thermal efficiency of an irreversible power cycle is
always less that the thermal efficiency of a reversible power cycle when each operates between the same two thermal reservoirs.
– All reversible power cycles operating between the same two thermal reservoirs have the same thermal efficiency.
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