ME 200 L18:ME 200 L18: Conservation Laws: Heat ExchangersHW 7 Posted Due in One Week:

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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

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Rotating Machinery

A turbine is a steady-flow device used to produce mechanical work (W) by reducing the internal &/or kinetic &/or potential energy of the working fluid.

•For gas turbines, the fluid drives rotating blades while the υ increases from inlet to exit as the working fluid expands (or the p drops).

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4

Steam Turbine Example

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6 Kg/s of steam at an inlet velocity of 75 m/s enter a turbine stage at 3 MPa and 400 oC and exit at a velocity of 125 m/s at a pressure of 2 Mpa at 360 oC. Find: (a) the power developed by this steam turbine stage, (b) the percentage change in the steam density across the turbine and (c) the percentage change in the flow area. The heat loss through the casing is 33 kW.

kWcvQ 33

cvWskgm /61

skgm /61

2 21 2

1 2

2 2 2 1 12

2 2 1

2 2

2 1 1 1 2 2

2 2

75 12533 6 3230.9 3159.3

2000 2000

( ) 33 6 (71.6 5) 366.6

( ) ( ) / ( ) / (0.094 0.141) / 0

Turbine Turbine

e

Turbine

Turbine

V VQ W m h h

V A V AVm m m

v v v

W

a W kW

b v v v

2 2 2 1 1 1 2 1

2 1

.141 33%

( ) / / 125 / 0.141 75 / 0.094

/ (75 /125)*(0.141/ 0.094) 0.9

c A V v AV v A A

A A

Air Compressors

September 17th, 2010 ME 200 5

A pump is a steady-flow device that consumes shaft work from rotating blades that compress the fluid.•Compressors are used for gas systems, pumps for liquids so operating assumptions are similar.

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Example

At steady state, a well-insulated compressor takes in air at 60 ºF, 14.2 psi, with a volumetric flow rate of 1200 ft3/min, and compresses it to 500 ºF, 120 psi. Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in hp, and the volumetric flow rate at the exit, in ft3/min.

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Example

Find– Wcv = ? in hp– A2V2 = ? in ft3/min

System (air flowing through compressor)

Assumptions•The control volume is at steady state; the flow is steady•Q, Δke, and Δpe are negligible. •The air is an ideal gas.

Basic Equations

e

ee

eei

ii

iicvcv

cv gzV

hmgzV

hmWQdt

dE

22

22

RTP airP1 = 14.2 psiT1 = 60 ºFA1V1 = 1200 ft3/min

P2 = 120 psiT2 = 500 ºF

compressor1 2

7

e

eii

cv mmdt

dm

8

Example

Solution

e

ee

eei

ii

iicvcv

cv gzV

hmgzV

hmWQdt

dE

22

22

P

RT

AV

m

8

e

eii

cv mmdt

dm mmm 21

21 hhmWcv

1

111RT

PVAm

9

Example

9

3 2

2

14.2 1200 min 60min 1441540 1 152028.9

f

m

psi ft inm

ft lb h ftRlb R

hlbm m5310

From Table A-22E

mlbBtuh 1241 mlbBtuh 2312

21 hhmWcv

15310 124 231

2540

m

cvm

lb Btu hpW

h lb Btu h 223cvW hp

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Example

10

222 mVA

2

2 2 2

15405310 960

1 128.9120 144 60min

m

m

lb lbf ftR

ft hh lb RAV

psi in

min262 322 ftVA

2

222 P

RTmVA

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

min/11.51)300)(/287.0(

min)/340)(1.1(

1

11

1

1

kgKKkgkJ

mbars

RT

AVP

v

AVmAir

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

2Assumptions: 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

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SummarySummary

• Control volume energy and mass conservation equations that we learned are applicable to many practical energy devices and equipment.

• Important learning comes from application of appropriate assumptions, considering the appropriate working substances and their and their propertiesproperties in the proper range of operation to estimate different energy quantities.

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