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Lec 21: Isentropic efficiencies, air standard cycle, Carnot cycle, Otto cycle

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• For next time:– Read: § 8-6 to 8-7– HW11 due Wednesday, November 12, 2003

• Outline:– Isentropic efficiency – Air standard cycle– Otto cycle

• Important points:– Realize that we already know how to analyze all

these new cycles, we just need to define what the cycle steps are

– Know the difference between the air standard cycle and the cold air approximations

– Know how to solve cycles using variable specific heats and constant specific heats

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Isentropic Efficiencies

act

s

act

sc w

w

W

W

We can use the isentropic process as an ideal by which to compare real processes in different engineering devices.

Actual compressors take more work than isentropic compressors. The efficiency will vary between zero and one.

COMPRESSORS

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Compressor Isentropic Efficiency

PEKEhhwq 12

For a steady-state, adiabatic compressor

If q, KE, and PE are all zero, then:

General expression

Isentropic compressorActual compressor

21 hhw

s21s hhw

a21a hhw

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Compressor Isentropic Efficiency

a21

s21

a

sc hh

hh

w

w

The compressor efficiency is then:

For an ideal gas with constant specific heats,

Thus:

2aTT

TT

1

2s1

c

)TT(chhw s21ps21s

)T(Tchhw 2a1p2a1a

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Compressor Isentropic Efficiency

Note that the work is directly proportional to T with constant specific heats. Real gases will also have dependence on P.

The work can then be represented by a change on the T-axis of a Ts diagram.

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Compressor Isentropic Efficiencies

T

S

1

2s

2a

T1–T2s

T1–T2a

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Isentropic Efficiencies

s

act

s

act

w

w

W

WT

TURBINES

With turbines, we’re interested in the work/power output not input. An isentropic turbine will produce the maximum output. Efficiency is given by:

2s1

2a1

hh

hh

T

Again, isentropic efficiency will vary between zero and one.

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Isentropic EfficienciesT

S

1

2s 2a2a

2a

Actual path will vary depending on amount of irreversibilities...

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Isentropic Efficiencies

s

2exit

a

2exit

N

2V

2V

For nozzles, the isentropic efficiency is given by

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TEAMPLAYTEAMPLAY

• Work problem 7-89

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Chapter 8, Gas Cycles

• Carnot cycle is the most efficient cycle that can be executed between a heat source and a heat sink.

• However, isothermal heat transfer is difficult to obtain in reality--requires large heat exchangers and a lot of time.

H

L

T

T-1

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Gas Cycles

• Therefore, the very important (reversible) Carnot cycle, composed of two reversible isothermal processes and two reversible adiabatic processes, is never realized as a practical matter.

• Its real value is as a standard of comparison for all other cycles.

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Gas Cycles

• Assumptions of air standard cycle• Analyze two cycles in detail

– Otto– Brayton

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Assumptions of air standard cycle

• Working fluid is air• Air is ideal gas• Combustion process is replaced by heat

addition process• Heat rejection is used to restore the fluid

to its initial state and complete the cycle• All processes are internally reversible• Constant or variable specific heats can be

used

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Gas cycles have many engineering applications

• Internal combustion engine– Otto cycle– Diesel cycle

• Gas turbines – Brayton cycle

• Refrigeration– Reversed Brayton cycle

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Some nomenclature before starting internal combustion

engine cycles

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More terminology

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Terminology

• Bore = d• Stroke = s• Displacement volume =DV =• Clearance volume = CV • Compression ratio = r

4

ds

2

CV

CVDVr

TDC

BDC

V

V

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Mean Effective Pressure

Mean Effective Pressure (MEP) is a fictitious pressure, such that if it acted on the piston during the entire power stroke, it would produce the same amount of net work.

minmax VV

WMEP net

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The net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volume

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Real Otto cycle

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Real and Idealized Cycle

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Idealized Otto cycle

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Idealized Otto cycle

• 1-2 - ADIABATIC COMPRESSION (ISENTROPIC)

• 2-3 - CONSTANT VOLUME HEAT ADDITION

• 3-4 - ADIABATIC EXPANSION (ISENTROPIC)

• 4-1 - CONSTANT VOLUME HEAT REJECTION

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Performance of cycle

in

net

q

w

Efficiency:

Let’s start by getting heat input:

23in uuq

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Cycle Performance

outinnet qqw Get net work from energy balance of cycle:

Efficiency is then:

Substituting for qin and qout:

netw )uu()uu( 1423

in

net

q

w

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Substituting for net work and heat input:

We can simplify the above expression:

)u-(u

)u-(u-)u-(u

23

1423

Cycle Performance

)u-(u

)u-(u1

23

14

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Teamplay

Problem 8-36

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Cold air standard cycle

cp, cv, and k are constant at ambient temperature ( 70 °F) values.

Assumption will allow us to get a quick “first cut”approximation of performance of cycle.

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Cycle performance with cold air cycle assumptions

If we assume constant specific heats:

)T-(Tc

)T-(Tc1

)u-(u

)u-(u1

23v

14v

23

14

)T-(T

)T-(T1

23

14

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Cycle performance with cold air cycle assumptions

Because we’ve got two isentropic processes in the cycle, T1 can be related to T2, and T3 can be related to T4 with our ideal gas isentropic relationships….

Details are in the book!

1k

1k

2

1

1

2 rV

V

T

T

1k

1k

3

4

3

4

r

1

V

V

T

T

2

3

1

4

T

T

T

TThus

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1k2

1

r

11

T

T1

This looks like the Carnot efficiency, but it is not! T1 and T2 are not constant.

Cycle performance with cold air cycle assumptions

What are the limitations for this expression?

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Differences between Otto and Carnot cycles

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Effect of compression ratio on Otto cycle efficiency

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Sample Problem

The air at the beginning of the compression stroke of an air-standard Otto cycle is at 95 kPa and 22C and the cylinder volume is 5600 cm3. The compression ratio is 9 and 8.6 kJ are added during the heat addition process. Calculate:

(a) the temperature and pressure after the compression and heat addition process(b) the thermal efficiency of the cycle

Use cold air cycle assumptions.

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Draw cycle and label points

T1 = 299 K

P1 = 95 kPa

r = V1 /V2 = V4 /V3 = 9

Q23 = 8.6 kJ

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Major assumptions

• Kinetic and potential energies are zero• Closed system • 1 is start of compression• Ideal cycle: 1-2 isentropic compression, 2-

3 const. volume heat addition, etc. • Cold cycle const. properties

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Carry through with solution

kg 10 x 29.6RT

VPm 3-

1

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Calculate mass of air:

Compression occurs from 1 to 2:

ncompressio isentropic V

VTT

1

2

112

k

11.42 9K 27322T

K 705.6T2 But we need T3!

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Get T3 with first law:

23v23 TTmcpe)keu(mWQ

Solve for T3:

2v

3 Tc

qT K705.6

kgkJ0.855

kg6.29x10kJ8.6 3

K2304.7T3

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Thermal Efficiency

11.41k 9

11

r

11

585.0

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Let’s take a look at the Diesel cycle.

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Idealized Diesel cycle

1-2 - ADIABATIC COMPRESSION (ISENTROPIC)

2-3 - CONSTANT PRESSURE HEAT ADDITION

3-4 - ADIABATIC EXPANSION (ISENTROPIC)

4-1 - CONSTANT VOLUME HEAT REJECTION

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Performance of cycle

in

net

q

wEfficiency:

Heat input occurs from 2 to 3 in constant pressure process:

23in hhq Why enthalpies?

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TEAMPLAY

• Work problem 8-16