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Elementary Thermodynamic Cycle and Components

Elementary Thermodynamic

Cycles and Components




9.1. Power and Refrigeration Cycle

9.2. Rankine Cycle

9.3. Reheat and Regeneration

9.4. Brayton Cycle

9.5. Regeneration and Reheating in Brayton Cycle

9.6. Jet Propulsion Cycle

9.7. Otto, Diesel and Stirling Cycle

9.8. Vapor Compression Refrigeration Cycle

9.9. Ammonia Absorption Refrigeration Cycle

9.10. Combined Power and Refrigeration Cycle




Comparison of shaft work and

boundary-movement work

9.1 Power and Refrigeration Cycle

Power and Refrigeration Cycle




Four-process power cycle

Net work is the same for two different cycles, one

made of four SSSF devices and the other made of

four cylinder/piston devices.





Heat Engine (Power Cycle)

(1) Modes of Work

-SSSF devices involving shaft work,e.g. Rankine, Brayton cycles

-Cylinder/piston devices involving boundary

movement work,

e.g. Otto, Diesel, Stirling cycles

(2) Working Fluid

-Single phase : e.g. Brayton, Otto, Diesel, Stirling

-Two phase : e.g. Rankine

Refrigerator or Heat Pump (Refrigeration Cycle)

Reverse of a power cycle

w vdP

w Pdv





Closed Cycle Open Cycle





Rankine Cycle (Two-phase Power Cycle)

Simple steam power plant which

operates on the Rankine cycle

9.2 Rankine Cycle

Rankine Cycle




1-2: Reversible adiabatic pumping (pump)

2-3: Constant pressure heat addition (boiler)

3-4: Reversible adiabatic expansion (turbine)

4-1: Constant pressure heat rejection (condenser)

Heat and work may be represented by various areas

in the T-s diagram. PE and KE negligible.

Carnot Cycle;

Pumping of two-phase mixture – difficult !!

Superheating at dropping pressure – difficult !!

-> Rankine cycle is the ideal cycle that can be

approximated in practice

Rankine Cycle




depends on the avg T at which heat is added

and the avg T at which heat is rejected.

th

'

'

1 2 2 3 4 1

2 2 3

net th

H

areaW

qarea a b a

Rankine Cycle




Example 9.1

Determine the efficiency of a Rankine cycle using steam

as the working fluid in which the condenser pressure is

10kPa. The boiler pressure is 2MPa. The steam leavesthe boiler as saturated vapor.

In solving Rankine-cycle problems, we let wp denote

the work into the pump per kilogram of fluid flowing,

and qL the heat rejected from the working fluid per

kilogram of fluid flowing.

To solve this problem we consider, in succession, a

control surface around the pump, the boiler, the turbine,and the condenser. For each the thermodynamic model

is the steam tables, and the process is SSSF with

negligible changes in kinetic and potential energies.

Rankine Cycle




Control volume: Pump.

Inlet state: P1 known, saturated liquid; state fixed.

Exit state: P2 known.

2 1

2 1

2

2 11

2 1

2

2.0 /

193.8 /

p

p

w h h

s s

h h vdP

w v P P kJ kg

h kJ kg

Rankine Cycle




Control volume: Boiler.

Inlet state: P2, h2 known; state fixed.

Exit state: P3 known, saturated vapor; state fixed.

Control volume: Turbine.

Inlet state: state 3 known.Exit state: P4 known.

3 22605.7 / H q h h kJ kg

3 4

3 4 4

4

4

6.3409 0.6493 7.5009

0.7588

2007.5 /

792.0 /

t

t

w h h

s s x

x

h kJ kg

w kJ kg

Rankine Cycle




Control volume: Condenser.

Inlet state: State 4 known.

Exit state: State 1 known.

4 11815.7 /

30.3%

L

t pnet H Lth

H H H

q h h kJ kg

w ww q q

q q q

Rankine Cycle




Effect of P and T on the Rankine Cycle

Effect of exhaust pressure

on Rankine cycle efficiency

-> Increase in

Increase in moisture of turbine

th

Rankine Cycle




Effect of superheating on Rankine cycle efficiency

-> Increase in

Quality of steam leaving the turbine increases.

th

Rankine Cycle




Effect of boiler pressure on Rankine cycle efficiency

Max T, Exhaust P : constant

Max P increase -> Increase in

Quality of the steam leaving the

turbine decreases.

th

Rankine Cycle




Reheat

Ideal reheat cycle

9.3 Reheat and Regeneration

Reheat and Regeneration




To achieve a higher boiler pressure and a higher

quality at the turbine exit at the fixed max T

-> Little gain in

Decrease in the moisture content from turbineth





Example 9.3

Consider a reheat cycle utilizing steam. Steam leaves the

boiler and enters the turbine at 4MPa, 400C. After

expansion in the turbine to 400kPa, the steam is reheated

to 400C and then expanded in the low-pressure turbine

to 10kPa. Determine the cycle efficiency.For each control volume analyzed, the thermodynamic

model is steam tables, the process is SSSF, and changes

in kinetic and potential energies are negligible.





Control volume: High-pressure turbine.

Inlet state: P3, T3 known; state fixed.


3 4

3 4

3

3

4 4

4

4

3213.6 /

6.7690 /

1.7766 5.1193

0.9752

2685.6 /

h pw h h

s s

h kJ kgs kJ kg

s x

x

h kJ kg





Control volume: Low-pressure turbine.



5 6

5 6

5

5

6 6

6

6

3273.4 /

7.8985 /

0.6493 7.5009

0.9664

2504.3 /

1297.1 /

l p

t

w h h

s s

h kJ kg

s kJ kg

s x

x

h kJ kg

w kJ kg





Control volume: Pump.

Inlet state: P1 known,saturated liquid; state fixed.


2 1

2 1

2

2 1 2 11

2

4.0 /

195.8 /

p

p

w h h

s s

h h vdP v P P

w kJ kg

h kJ kg






Inlet state: State 2 and 4 both known.

Exit state: State 3 and 5 both known.

3 2 5 43605.6 /

1293.1 /

35.9%

H

net t p

net th

H

q h h h h kJ kg

w w w kJ kg

w

q





Regeneration

T-s diagram showing the relationship between

Carnot cycle efficiency and Rankine cycle efficiency

(Rankine) < (Carnot) due to the area, 11’2’2th th





Ideal regenerative cycle

(Rankine) = (Carnot) w/ reversible heat transfer

-> Impractical heat transfer from turbine

Moisture content from turbine

th th





Regenerative cycle with an open feedwater heater

Open Feedwater:

Less expensive

Requires a pump between each heater





Schematic arrangement for

a closed feedwater heater





Arrangement of regenerative feedwater

heaters in an actual power plant





Most powerplants combine one reheat stage with a

number of extraction stages, though rarely more

than 5.





Example 9.4

Consider a regenerative cycle using steam as the

working fluid. Steam leaves the boiler and enters theturbine at 4MPa, 400C. After expansion to 400kPa,

some of the steam is extracted from the turbine forthe purpose of heating the feedwater in an open

feedwater heater. The pressure in the feedwaterheater is 400kPa and the water leaving it is saturatedliquid at 400kPa. The steam not extracted expands

to 10kPa. Determine the cycle efficiency.

Process: SSSF.





Control volume: Low-pressure pump.

Inlet state: P1 known, saturated liquid; state fixed.


5

6

7

1

3213.6 /

2685.6 /

2144.1 /

191.8 /

h kJ kg

h kJ kg

h kJ kg

h kJ kg

1 2 1

2 1

2

2 1 2 11

pw h h

s s

h h vdP v P P







Exit state: P6 known; P7 known.

1 2 1

2 1

0.4 /

192.2 /

p

p

w v P P kJ kg

h h w kJ kg

5 6 1 6 7

5 6 7

1t w h h m h h

s s s





Control volume: Feedwater heater.

Inlet state: State 2 and 6 both known.

Exit state: P3 known, saturated liquid; state fixed.

Control volume: High-pressure pump.



1 6 1 2 3

1

5 6 1 6 7

1

0.1654

1 979.9 / t

m h m h h

m

w h h m h h kJ kg

2 4 3

4 3

3.9 / pw h h kJ kg

s s






Inlet state: P4, h4 known; state fixed.


4 3 2

1 1 2

608.6 /

1 975.7 /

p

net t p p

h h w kJ kg

w w m w w kJ kg

5 42605.0 /

37.5%

H

net th

H

q h h kJ kg

w

q





Deviation of Actual Cycle from Ideal Cycle

Piping Loss:

Effects of losses between boiler and turbine

Pressure drop(a-b), Heat loss(b-c): Both decrease

availability of the steam.





Turbine and Pump Losses:

Effects of turbine and pump losses





Turbine: flow loss, (heat loss)

Pump: flow loss, (heat loss)

Condenser loss : cooling below Tsat (minor)

2 1s p

p

h h

w

3 4

t t

s

w

h h





Cogeneration

Cogeneration system

(Process steam) + (Electricity)





Single-Phase Power Cycle

(Air-Standard Power Cycle)

Brayton cycle – Shaft work, gas turbine

Otto cycle – PdV work, gasoline engine

Diesel cycle –

PdV work, Diesel engine

IC engine with an open cycle

-> Approximation by a closed cycle

Combustion replaced by heat transferFixed mass of air as the working fluid





Both Rankine and Brayton Cycles

(Two isobaric processes)

+ (Two isentropic processes)

Two phase : Rankine cycle – Steam Power Plant

Single phase : Brayton cycle – Gas Turbine

9.4 Brayton Cycle

Brayton Cycle




Gas turbine operating on the Brayton cycle

(a) open cycle

(b)closed cycle

Brayton Cycle




Air-standard Brayton cycle

Brayton Cycle




1

1

2

2

1

11 1

th k

k

T

T P

P

Brayton Cycle




Cycle efficiency as a function of pressure ratio

for Brayton and regenerative Brayton cycles

Brayton Cycle




Example 9.6

In an air-standard Brayton cycle the air enters the

compressor at 0.1MPa, 15C. The pressure leaving the

compressor is 1.0MPa, and the maximum temperature in

the cycle is 1000C. Determine

1. The pressure and temperature at each point in the

cycle

2. The compressor work, turbine work, and cycle

efficiency

Brayton Cycle




For each of the control volumes analyzed, the model is

ideal gas with constant specific heat, value at 300K, and

each process is SSSF with no kinetic or potential energychanges.

Control volume: Compressor.

Inlet state: P1, T1 known; state fixed.Exit state: P2 known.

2 1

2 1

1

2 2

1 1

c

k

k

w h h

s s

T P

T P

Brayton Cycle







1

2

1

2

2 1 2 1

1.932

556.8

269.5 /

k

k

c p

P

P

T K

w h h C T T kJ kg

3 4

3 4

1

3 3

4 4

t

k

k

w h h

s s

T P

T P

Brayton Cycle




Control volume: High-temperature heat exchange.

Inlet state: state 2 fixed.

Exit state: State 3 fixed.

1

3

4

4

3 4 3 4

1.932

710.8

664.7 /

395.2 /

k

k

t p

net t c

P

P

T K

w h h C T T kJ kg

w w w kJ kg

3 2 3 2 819.3 / H pq h h C T T kJ kg

Brayton Cycle




Control volume: Low-temperature heat exchange.

Inlet state: state 4 fixed.

Exit state: State 1 fixed.

4 1 4 1424.1 /

48.2%

L p

net th

H

q h h C T T kJ kg

w

q

Brayton Cycle




Brayton Cycle

In an actual gas turbine the max. temperature of thegas enter ing the turbine is f ixed by mater ia l

considerations.

Large amount of compressor work vs that of turbinework (40~80%)

-> due to difference in specific volumes

cf. Rankine cycle (1~2% for pumping work)

Brayton Cycle




Gas Turbine Cycle with a Regenerator

Ideal regenerative cycle

9.5 Regeneration and Reheating in

Brayton Cycle

Regeneration and Reheating in Brayton Cycle




In general

For an ideal regenerator

,

In contrast to the Brayton cycle, decreases with

an increase in the pressure ratio for the cycle with aregenerator.

th

1

1 2

3 1

1

k

k

th T PT P

4 2T T

4 xT T 2 yT T





Efficiency of a Regenerator

T-s diagram to illustrate definition

of the regenerator efficiency

2

'

2

xreg

x

h h

h h





Ideal gas-turbine cycle using multistage

compression with intercooling, multistage

expansion with reheating, and a regenerator

Reversible Adiabatic Process (Compressor, Turbine)-> Reversible Isothermal Process (Ericsson Cycle)





Ideal gas-turbine cycle utilizing

intercooling, reheat and a regenerator





Multistage Compression with Intercooling

Multistage Expansion with Reheating

T-s diagram that shows how the gas-turbine cycle

with many stages approaches the Ericsson cycle





Air-Standard Cycle for Jet Propulsion

Ideal gas-turbine for a jet engine

(Brayton cycle)+(Reversible adiabatic nozzle)

9.6 Jet Propulsion Cycle

Jet Propulsion Cycle




Example 9.10

Consider an ideal jet propulsion cycle in which air enters

the compressor at 0.1MPa, 15C. The pressure leaving

the compressor is 1.0MPa, and the maximum temperature

is 1100C. The air expands in the turbine to a pressure at

which the turbine work is just equal to the compressor

work. On leaving the turbine, the air expands in a nozzleto 0.1MPa. The process is reversible and adiabatic.

Determine the velocity of the air leaving the nozzle.

The model used is ideal gas, constant specific heat, value

at 300K, and each process is SSSF with no potential

energy change. The only kinetic energy change occurs in

the nozzle.





1 1

2 2

3 3

3 4

3 4 4

1

3 3

4 4

3

4

4

0.1 288.2

1.0 556.8

269.5 /

1.0 1373.2

269.5 /

268.6 1104.6

1.2432

2.142 0.4668

c

c t p

k

k

P MPa T K

P MPa T K

w kJ kg

P MPa T K

w w C T T kJ kg

T T T K

T P

T P

PP MPa

P





Control volume: Nozzle.

Inlet state: State 4 fixed.


2

5

4 5

4 5

5 5

2

5 0 4 5

2

0.1 710.8

2 889 / p

V h h

s s

P MPa T K

V C T T m s





Otto Cycle

Air-standard Otto cycle

1 4

2 3

c

V V r

V V

1

11th k

cr

9.7 Otto, Diesel and Stirling Cycle

Otto, Diesel and Stirling Cycle




Example 9.11

The compression ratio in an air-standard Otto cycle is 8.

At the beginning of the compression stroke the pressure

is 0.1MPa and the temperature is 15C. The heat transfer

to the air per cycle is 1800kJ/kg air. Determine

1. The pressure and temperature at the end of each

process of the cycle.

2. The thermal efficiency.

3. The mea effective pressure.


Control mass: Air inside cylinder.




y

State information: P1=0.1MPa, T1= 288.2K.

Process information: Four process known. Also rv=8

and qH=1800kJ/kg.Model: Ideal gas, constant specific heat, value at

300K.

2 1

1

2 1

1 2

2 1

1 2

2 3 3 2 3 2

4 3

k

k

H v

s s

T V

T V

P V

P V q q u u C T T

s s


1

3 4

k

T V




3 4

4 3

3 4

4 3

1

1 2

3

1

1

2 1

2

1 2

2 1

2

1 2

3

2

2 3 3 2

11

0.827 /

2.3 662

18.38 1.838

0.1034 /

1800 /

k

net th k

v

k

k

v

T V

T V

P V

P V

wmep

r v v

v m kg

T V T K

T V

P V P MPa

P V

v m kg

q C T T kJ kg





3 2 3

3 3

3

2 2

1

3 4

4

4 3

3 44

4 3

1

4 1 1 4

1 2

2512 3174

4.795 8.813

2.3 1380

18.38 0.4795

11 0.565 56.5%

782.3 /

1017.7 /

1406

k

k

th k

v

v

net

T T T K

T PP MPa

T P

T V T K

T V

P V P MPa

P V

r

q C T T kJ kgw kJ kg v v mep

mep kPa





Diesel Cycle

Air-standard Diesel cycle

1

1 11

1th k

cr

3

2

V

V





Example 9.12

An air-standard Diesel cycle has a compression ratio of

18, and the heat transferred to the working fluid percycle is 1800kJ/kg. At the beginning of the compression

process the pressure is 0.1MPa and the temperature is

15C. Determine

1. The pressure and temperature at each point in the

cycle

2. The thermal efficiency3. The mean effective pressure


C t l : Ai i id li d




Control mass: Air inside cylinder.

State information: P1=0.1MPa, T1=288.2K.

Process information: Four process known. Also

rv=18 and qH=1800kJ/kg.

Model: Ideal gas, constant specific heat, value at

300K.

2 1

1

2 1

1 2

2 1

1 2

2 3 3 2

k

k

H p

s s

T V

T V

P V P V

q q C T T





4 3

1

3 4

4 3

1 2

3

1

32

1

2 1

2

1 2

2 1

2

1 2

2 3 3 2

0.827 /

0.04595 /

3.1777 915.8

57.2 5.72

1800 /

k

net net th

H

k

k

H p

s s

T V

T V w w

mepq v v

v m kg

v m kg

T V T K

T V

P V P MPaP V

q q C T T kJ kg





3 2 3

33 3

32 2

1

3 4

4

4 3

4 1 1 4

1 2

1794 2710

2.959 0.13598 /

2.0588 1316

736.6 /

1063.4 /

59.1%

1362

k

L v

net

net th

H

net

T T T K

V T

v m kgV T

T V T K

T V

q q C T T kJ kgw kJ kg

w

q

wmep kPav v





Comparison of Otto and Diesel Cycles

For the same and piston displacement,

For the same and ,

Otto Diesel

c cr r

Otto Diesel

th th

Otto Diesel

th th

cr

maxT maxP





Stirling Cycle

Air-standard Stirling cycle

-> External combustion engine with regeneration

Cylinder/piston application with a high mep


9.8 Vapor Compression




Vapor Compression Refrigeration Cycle

Ideal vapor-compression refrigeration cycle

Refrigeration Cycle





Turbine replaced by a throttling device

Compressor handling vapor only

reciprocating : low specific volume

centrifugal : high specific volume

Coefficient of performance

refrigerator

heat pump

Working fluids to protect the ozone layer

CFC(R-12, Freon, etc) -> HCFC, HFC

L

C

q

W

' H

C

q

W





Deviation from the Ideal Cycle

Actual vapor-compression refrigeration cycle





Example 9.13

Consider an ideal refrigeration cycle which uses R-12 as

the working fluid. The temperature of the refrigerant in

the evaporator is – 20C and in the condenser it is 40C.

The refrigerant is circulated at the rate of 0.03kg/s.

Determine the coefficient of performance and the

capacity of the plant in rate of refrigeration.

For each control volume analyzed, the thermodynamic

model is the R-12 tables. Each process is SSSF with no

change in kinetic or potential energy.






Control volume Compressor.

Inlet state: T1 known, saturated vapor; state fixed.

Exit state: P2 known(saturation pressure at T3).

At T3=40C

2 1

2 1

cw h h

s s

2

1

1 2

2

2

2 1

0.9607

178.61 /

0.7082

50.8

211.38 /

32.77 /

g

o

c

P P MPa

h kJ kg

s s

T C h kJ kg

w h h kJ kg


C t l l : E i l




Control volume: Expansion valve.

Inlet state: T3 known, saturated liquid; state fixed.

Exit state: T4

known.

Control volume: Evaporator.

Inlet state: State 4 known.Exit state: State 1 known.

3 474.53 / h h kJ kg

1 4104.08 /

3.18

3.12

L

L

c

q h h kJ kg

q

w

Capacity kW





Example 9.14

A refrigeration cycle utilizes R-12 as the working fluid.

Following are the properties at various points of the

Cycle.

1 1

2 2

3 3

4 4

5 5

6 7 6 7

8 8

125 10

1.2 100

1.19 80

1.16 45

1.15 40

140

130 20

P kPa T C

P MPa T C

P MPa T C

P MPa T C

P MPa T C

P P kPa x x

P kPa T C

°

°

°

°

°

°





The heat transfer from R-12 during the compression

process is 4kJ/kg. Determine the coefficient of

performance of this cycle.For each control volume, the model is the R-12

tables. Each process is SSSF with no changes in

kinetic or potential energy.



Exit state: P2, T2 known; state fixed.

1 2

2 1c

q h h w

w w h h q





Control volume: Throttling valve plus line.


Exit state: P7=P6 known, x7=x6.

1

2

185.16 /

245.52 /

64.36 / c

h kJ kg

h kJ kg

w kJ kg

5 6 774.53 / h h h kJ kg





Control volume: Evaporator.

Inlet state: P7, h7 known.

Exit state: P8, T8 known; state fixed.

8 7104.59 /

1.625

L

L

c

q h h kJ kg

q

w


9.9 Ammonia Absorption




Ammonia Absorption Refrigeration Cycle

Ammonia-absorption refrigeration cycle

Refrigeration Cycle





Air-Standard Refrigeration Cycle

Air-standard refrigeration cycle





Example 9.15

Consider the simple air-standard refrigeration cycle. Air

Enters the compressor at 0.1MPa, -20C, and leaves at0.5MPa. Air enters the expander at 15C. Determine

1. The coefficient of performance for this cycle

2. The rate at which air must enter the compressor toProvide 1kW of refrigeration

For each control volume in this example, the model is

ideal gas with constant specific heat, value at 300K, andEach process is SSSF with no kinetic or potential energy

Changes.








2 1

1 2

1

2 2

1 1

2

2 1 2 1

1.5845

401.2

148.5 /

c

k

k

c p

w h h

s s

T P

T P

T K

w h h C T T kJ kg





Control volume: Expander.

Inlet state: P3(=P2) known, T3 known; state fixed.

Exit state: P4(=P1) known.

3 4

3 4

1

3 3

4 4

4

3 4

1.5845

181.9

106.7 /

t

k

k

t

w h h

s s

T P

T P

T K

w h h kJ kg





Control volume: High-temperature heat exchanger.



Control volume: Low-temperature heat exchanger.



2 3 2 3113.4 / H pq h h C T T kJ kg

1 4 1 471.6 /

41.8 /

L p

net c t

q h h C T T kJ kg

w w w kJ kg





1.713

0.014 /

L

net

L

L

q

w

Qm kg s

q


9.10 Combined Power and




Combined Cycle

Mercury-water binary power system

Refrigeration Cycle

Combined Power and Refrigeration Cycle




Combined Brayton/Rankine Cycle System

Combined Power and Refrigeration Cycle



Combined-Cycle Cascade Refrigeration System

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