Thermo 2 - Ch. 12

Thermodynamics Research LaboratoryMapúa Institute of Technology

Chung Yuan Christian UniversitySchool of Chemical Engineering and Chemistry

POWER AND REFRIGERATION SYSTEMS: GASEOUS WORKING FLUIDS

Lecturer: Prof. Allan N. Soriano, Ph.D. Ch.E.Email: [email protected]



Overview• Previously, we have studied power and refrigeration systems that utilize condensing working fluids, in particular those involving steady-state flow processes with shaft work.

• In this part, we continue the power and refrigeration study involving steady-state flow processes, but those with gaseous working fluids throughout, recognizing that the difference in expansion and compression work terms is considerably smaller.

- It was noted that condensing working fluids have the maximum difference in the - vdP work terms between the expansion and compression processes.



Air-Standard Power Cycles• The basic model used with gaseous power systems.• A closed cycle devised to closely approximate the open cycles such as the internal combustion engines.

- In internal combustion engines there is a change in the composition of the working fluid, because during combustion it changes from air and fuel to combustion products.

- In contrast, the steam power plant may be called an external combustion engines because heat is transferred from the products of the combustion to the working fluid.



Air-Standard Power Cycles

• Model Assumptions:

1. A fixed mass of air is the working fluid throughout the entire cycle, and the air is always an ideal gas. Thus, there is no inlet process or exhaust process.

2. The combustion process is replaced by a process transferring heat from an external source.

3. The cycle is completed by heat transfer to the surroundings (in contrast to the exhaust and intake process of an actual engine).

4. All processes are internally reversible. 5. An additional assumption is often made that air has a constant

specific heat, evaluated at 300 K, called cold air properties, recognizing that this is not the most accurate model.



Air-Standard Power Cycles

• The principal value of the air-standard cycle is to enable us to examine qualitatively the influence of a number of variables on performance.

- The qualitative results obtained from the air-standard cycle, such as efficiency and mean effective pressure, will differ from those of the actual engine. - Mean effective pressure is defined as the pressure that, if it acted in the piston during the entire power stroke, would do an amount of work equal to that actually done on the piston. - The work for one cycle is found by multiplying this mean effective pressure by the area of the piston (minus the area of the rod on the crank end of a double-acting engine) and by the stroke.



Air-Standard Brayton Cycle• Ideal cycle for the simple gas turbine.

Comp Turb

CombChamber

2 3

1 4Wnet

Fuel

Comp Turb

CombChamber

2 3

1 4Wnet

HeatExchanger

QH

QL

Open Cycle Closed Cycle



Air-Standard Brayton Cycle

1/

1/111232

141

23

14th

TTTTTT

TTCTTC

qq

p

p

H

L

• The efficiency of the air-standard Brayton cycle is found as follows:

kk

kkkk

PPTT

TT

TT

TT

TT

TT

TT

TT

TT

PP

PP

/1122

1th

1

4

2

3

1

4

2

3

1

2

4

3

1/

4

3

1/

1

2

1

2

4

3

/111

11 and

thathowever, note, We

The efficiency is a function of the isentropic

pressure ratio.




SAMPLE PROBLEM: Ex. 12.1, p. 425:

In an air-standard Brayton cycle the air enters the compressor at 0.1 MPa and 15oC. The pressure leaving the compresor is 1.0 MPa, and the maximum temperature in the cycle is 1100oC. Determine 1. The pressure and temperature at each point in the cycle. 2. The compressor work, turbine work, and cycle efficiency.

Consider the compressor, the turbine, and the high-temperature and low-temperature heat exchangers in turn.





Comp Turb

CombChamber

2 3

0.1 MPa, 15oC 4

Wnet

HeatExchanger

QH

QL

1

1.0 MPa 1100oC





Consider a gas turbine with air entering the compressor under the same conditions as in Ex. 12.1 and leaving at a pressure of 1.0 MPa. The maximum temperature is 1100oC. Assume a compressor efficiency of 80%, a turbine efficiency of 85%, and a pressure drop between the compressor and turbine of 15 kPa. Determine the compressor work, turbine work, and cycle efficiency.




12

12

12

12

TTTTη

hhhhη

sc

sc


Comp Turb

CombChamber

2 3

0.1 MPa, 15oC 4

Wnet

HeatExchanger

QH

QL

1

1.0 MPa 1100oC, (P2-drop)

st

st

TTTTη

hhhhη

43

43

43

43



Simple Gas-Turbine Cycle with a Regenerator• The efficiency of the gas-turbine cycle may be improved by introducing a regenerator.

Comp Turb

CombChamber2 31

4

Wnet

Regenerator

y

x



Simple Gas-Turbine with Regenerator

433

th

and TTCwTTCqq

wwqw

ptxpH

H

ct

H

net

• The efficiency of this cycle with regeneration is found as follows:

But for an ideal regenerator, T4 = Tx, and therefore qH = wt. Consequently,

3

2

/1

1

2

3

1/1

213

/1121

th

343

121

43

12th

11/1

1/1

/11/111

TT

PP

TT

PPTPPT

TTTTTT

TTCTTC

ww

kk

kk

kk

p

p

t

c




• Thus, for the ideal cycle with regeneration, the thermal efficiency depends not only on the pressure ratio but also on the ratio of the minimum and maximum temperature.

• A higher efficiency can be achieved by using a regenerator with a greater heat-transfer area, however, this also increases the pressure drop, which represents a loss, and both the pressure drop and the regenerator efficiency must be considered in determining which regenerator gives maximum thermal efficiency for the cycle.

- From an economic point of view, the cost of the regenerator must be weighed against the savings that can be effected by its use.





If an ideal regenerator is incorporated into the cycle of Ex. 12.1, determine the thermal efficiency of the cycle.

Comp Turb

CombChamber2 31

4

Wnet

Regenerator

y

x

QH



Ericsson Cycle

• It is found that the isothermal process would be preferable to the adiabatic process on both the compressor and turbine.

• It consists of two reversible, constant-pressure processes and two-reversible constant-temperature processes.

- The Brayton cycle, being the idealized model for the gas-turbine power plant, has a reversible, adiabatic compressor and a reversible, adiabatic turbine.



Ericsson Cycle


An air-standard cycle has the same states given in Ex. 12.1. In this cycle, however, the compressor and turbine are both reversible, isothermal processes. Calculate the compressor work and the turbine work, and compare with those of Ex. 12.1.



Ericsson Cycle


Comp Turb

CombChamber

2 3

0.1 MPa, 15oC 4

Wnet

HeatExchanger

QH

QL

1

1.0 MPa 1100oC



Air-Standard Cycle for Jet Propulsion• In this cycle, the work done by the turbine is just sufficient to drive the compressor.

- The gases are expanded in the turbine to a pressure for which the turbine work is just equal to the compressor work.

• The exhaust pressure of the turbine will then be greater than that of the surroundings, and the gas can be expanded in a nozzle to the pressure of the surroundings.

- Since the gases leave at a high velocity, the change in momentum that the gases undergo gives a thrust to the aircraft in which the engine is installed.



Air-Standard Cycle for Jet Propulsion

• The principles governing this cycle follow from the analysis of the Brayton cycle plus that for a reversible, adiabatic nozzle.

Comp Turb

Burner

2 3

1 4 5Nozzle





Consider an ideal jet propulsion cycle in which air enters the compressor at 0.1 MPa and 15oC. The pressure leaving the compressor is 1.0 MPa, and the maximum temperature is 1100oC. 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 nozzle to 0.1 MPa. The process is reversible and adiabatic. Determine the velocity of the air leaving the nozzle.





Comp Turb

Burner

2 3

1 4 5Nozzle

0.1 MPa, 15oC

1.0 MPa 1100oC

0.1 MPa



Air-Standard Refrigeration Cycle• The reverse of the Brayton cycle, and it is used in practice in the liquefaction of air and other gases and also in certain special situations that require refrigeration, such as aircraft cooling systems.

Expander Compressor

3 2

4 1

Wnet

QH

QL



Air-Standard Refrigeration Cycle

4312

41

4312

41

net TTCTTCTTC

hhhhhh

wwq

wq

PP

P

EC

LL

• The COP of the air-standard refrigeration cycle involves the net work between the compressor and expander work terms:

12/1

1

2

14

23

1

24312

41

1/

4

3

4

3

1/

1

2

1

2

/ ration, pressure the where;1

1

1

1

1/1/1

1and

:processes isentropic twofor the relationspower theWriting

PPrr

TT

TTTT

TTTTTT

TT

TT

PP

TT

PP

pkkp

kkkk





Consider the simple air-standard refrigeration cycle where air enters the compressor at 0.1 Mpa and -20oC and leaves at 0.5 Mpa. Air enters the expander at 15oC. Determine

1. The COP of this cycle.2. The rate at which air must enter the compressor to

provide 1 kW of refrigeration.





Expander Compressor

3 2

4 1

Wnet

QH

QL

0.1 MPa, 15oC

0.5 MPa15oC



Piston Cylinder Power Cycles• Otto Cycle• Diesel Cycle• Stirling Cycle• Atkinson Cycle• Miller Cycle



Power Cylinder Power Cycles

The Otto Cycle • Approximates a spark-ignition internal combustion engine.

P

v1

4

3

2

T

s1

4

3

2

s = const

s = const

v = const

v = const

The air-standard Otto cycle.



The Otto Cycle

1

4

2

1

11

2

1th

1

4

2

3

4

3

1

1

4

1

2

1

1

2

232

141th

23

14th

ration compressio where

1111 and

atfurther th note We1/1/1

11

VV

VVr

rr

TT

TT

TT

TT

VV

VV

TT

TTTTTT

TTmCTTmC

QQ

QQQ

v

kv

kv

kk

v

v

H

L

H

LH

• The efficiency of this cycle is found as follows, assuming constant specific heat of air:




The Diesel Cycle • The ideal cycle for the diesel engine, which is called the compression ignition engine.

P

v1

4

32

T

s1

4

3

2

1/

1/1111232

141

23

14th

TTkTTTT

TTCTTC

QQ

p

v

H

L

The air-standard Diesel cycle.




The Stirling Cycle • Stirling cycles engines have been developed in recent years as external combustion engines with regeneration

The air-standard Stirling cycle.

P

v1

4

3

2

T = const

T = const

T

s1

43

2

v = const

v = const




The Atkinson Cycle • A cycle slightly different from the Otto cycle that has a higher expansion ratio than the compression ratio allowing more work to be extracted.

P

v1

4

3

2

T

s

1

4

3

2s

s

v

P = const




The Miller Cycle • Modification of Atkinson cycle using a supercharger, which approximates the Ford Escape and the Toyota Prius hybrid car engines.

P

v1

4

3

2

T

s

15

3

2

s = const

s

v

v = const

5

4

P



See you in TAIWAN!

END



LEARNING TASK No. 5

Solve Homework ProblemsNo. 12.15 (p. 455) and 12.41 (p. 457)

Documents

Thermo 2 - Ch. 12