Chapter 9 Power and Refrigeration System s With...

Fundamentals of Thermodynamics

Chapter 9Power and Refrigeration System

sWith Phase Change

Thermal Engineering Lab. 2

9.1 Introduction to power systems

Chapter 9. Power and Refrigeration Systems – With Phase Change

Shaft Work: Steady-state Process-

( )2 22 1 2

1 2 1 21 2 2 lV Vw vdp gz gz wæ ö

= - + - + - -ç ÷è ø

ò

l genw T s= ×

. , ,rev process negligible KE PED D

w vdpÞ = -ò

- Boundary-movement Work2

12 1w pdv= ò



-cycle 에대해서는

2 21 2

12 12 1 1 2 202 2

V Vq w h gz h gzæ ö æ ö

= - + + + - + +ç ÷ ç ÷è ø è ø

Tds dh vdp= -

( )2 2

2 1 121 1 lTds h h vdp q w= - - = +ò ò2 2

121 1gen genqds s Tds q T s

Td d d= + Þ = +ò ò

lw

òò =- pdvvdp


9.2 The Rankine cycle


1- 2 : Reversible adiabatic pumping process

2 - 3 : P = const, reversible heat transfer (addition)

3 - 4 : Reversible adiabatic expansion

4 -1 : P = const, reversible heat transfer (rejection)



,

,

1 1 1 L avgL Lth

H H avgH

Tds T Sqq T STds

hD

= - = - = -D

òò

,

,

1 L avg

H avg

TT

= -

area 1-2-2'-3-4-1area a-2-2'-3-b-a

net H Lth

H H

w q qq q

h -= = =



average temperature*

2 1avg

Tds QTS S S

= =- Dò

avgT h ® 고온측에서는

avgT h¯ ® 저온측에서는

avgT


Ex. 9.1 Determine the efficiency of a Rankine cycle using steam as the working fluid in which the condenser pressure is 10 kPa. The boiler pressure is 2 MPa. The steam leaves the boiler as saturated vapor.


( )

2

2 1 1

2 1 2 1

Assume isentropic process, Integration

For incompressible substance

Tds dh vdp

h h vdP

h h v P P

= -

- =

- = -

ò


9.3 Effect of pressure and temperature on the Rankine cycle


i) Exhaust pressure ↓

thh

:x Corrosion¯

, :H avgT slighly ¯

, : L avgT ¯

,

,

1 L avgth

H avg

TT

h = -



ii) Superheating

thh

x

, : H avgT

,L avgT const=

,

,

1 L avgth

H avg

TT

h = -



iii) Boiler pressure ↑

thh

x ¯

, : H avgT

,L avgT const=

,

,

1 L avgth

H avg

TT

h = -


Ex. 9.2 In a Rankine cycle, steam leaves the boiler and enters the turbine at 4 MPa and 400℃. The condenser pressure is 10 kPa. Determine the cycle efficiency.



9.4 The Reheat cycle


: thh 큰 변화 없음 :x


Ex. 9.3 Consider a reheat cycle utilizing steam. Steam leaves the boiler and enters the turbine at 4 MPa, 400℃. After expansion in the turbine to 400 kPa, the steam is reheated to 400℃ and then expanded in the low-pressure turbine to 10 kPa. Determine the cycle efficiency.



9.5 The Regenerative cycle and feedwater heaters


2-2' Rankine cycle

.

의평균 온도가 2'-3보다

낮기때문에 의효율은

Carnot cycle 보다 낮음



:not practicalheat transfer impossible

Ideal Regenerative cycle Carnot cycle Þ 과 동일한 열효율

• Ideal regenerative cycle – Carnot cycle과동일한열효율



• Regenerative cycle with open feedwater heater


Ex. 9.4 Consider a regenerative cycle using steam as the working fluid. Steam leaves the boiler and enters the turbine at 4 MPa, 400℃. After expansion to 400 kPa, some of the steam is extracted from the turbine to heat the feedwater in an open FWH. The pressure in the FWH is 400 kPa, and the water leaving it is saturated liquid at 400 kPa. The steam not extracted expands to 10 kPa. Determine the cycle efficiency.




• Regenerative cycle with closed feedwater heater



• Actual power plant utilizing regenerative feedwater heaters


9.6 Deviation of actual cycles from ideal cycles


Turbine Losses

Pump Losses

21 :actual21 :ideal

®® s

43 :actual43 :ideal

®® s



Piping Losses

a b : Pressure Loss®

b c : Heat Transfer®


Ex. 9.5 A steam power plant operates on a cycle with pressures and temperatures as designated in Fig. 9.17. The efficiency of the turbine is 86%, and the efficiency of the pump is 80%. Determine the thermal efficiency of this cycle.



9.7 Combined heat and power: other configurations


• Cogeneration system (열병합발전) : Electricity & Heat


9.8 Introduction to refrigeration systems



9.9 The vapor-compression refrigeration cycle


: refrigerator

: heat pump

L

c

H

c

qCOPw

qw

b

b

= =

¢ =

4 - 1 : P = const, evaporation3 - 4 : isenthalpic expansion2 - 3 : P = const, condensation1 - 2 : isentropic compression

COP: Coefficient of Performance

T

S

h

ln P

3

4

2

1

23

4 1S const=

2 1s s=4 3h h=

H L cq q w= +



1 : wet compression¢ 문제점

4 : isentropic expansion ¢ 문제점

구성의어려움


Ex. 9.6 Consider a refrigeration cycle that uses R-134a as the working fluid. The temperature of the refrigerant in the evaporator is -20℃, and in the condenser it is 40℃. The refrigerant is circulated at the rate of 0.03 kg/s. Determine the COP and the capacity of the plant in rate of refrigeration.



9.10 Working fluids for vapor-compression refrigeration systems


3R -12, R - 22, R -11, NH134 , 22 407 , 410R -12 R a R R c R a® - ®

2

3

2

Natural Working Fluid:CONHH OPropane+Butane


9.11 Deviation of the actual vapor-compression refrigeration cycle from the ideal cycle


h

ln P

24

7 8

5

6 1

3


Ex. 9.7 A refrigeration cycle utilizes R-134a as the working fluid. The following are the properties at various points of the cycle designated in Fig. 9.24:


P1 = 125 kPaP2 = 1.2 MPaP3 = 1.19 MPa,P4 = 1.16 MPa,P5 = 1.15 MPa,P6 = P7 = 140 kPa,P8 = 130 kPa

T1 = -10℃T2 = 100℃T3 = 80℃T4 = 45℃T5 = 40℃x6 = x7T8 = -20℃

The heat transfer from R-134a during the compression process is 4 kJ/kg. Determine the COP of this cycle.



9.12 Refrigeration cycle configurations



3 2 Cascade System(Netsle)NH CO-



9.13 The ammonia-absorption refrigeration cycle

L

H

qCOPq

b = =

Chapter 9 Power and Refrigeration System s With...

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