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!

e

ee

eei

ii

ii gzhmgzhmWQ )2

()2

(022 VV

cvcv

Heat Exchanger Modeling

ee

eiii hmhm 0

)( iiee gzmgzm

0cvW

0cvQ

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.

34

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

10

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

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Steam Power Plant

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

QQ

Q

W

1

Thermal efficiency = Energy you get / Energy you pay for

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Thermal efficiency = Desired energy result / Required energy input

Heat Engines

max

1

1

C

H

C

H

Q

Q

T

T

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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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