MPE 371 Ch 1 - Air standard cycles-1 PDF

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CHAPTER 1 HEAT ENGINE CYCLES 1.1

Carnot cycle

This cycle consists of two isothermal processes joined by two adiabatic processes. It is most conveniently represented on a T-s and p-v diagrams as follows: Process 1-2 = isentropic expansion from T1 to T2 Process 2-3 = isothermal heat rejection Process 3-4 = isentropic compression from T2 to T1 Process 4-1 = isothermal heat supply.

The cycle is completely independent of the working substance used. T

p p4

T1

4

4 1

P1

1

P3 p2

T2 3

3 2

2 B

A

C s

The cycle efficiency is given by   

net work output heat supplied

heat supplied  heat rejected heat supplied

D v

2



T1  s B  s A   T2  s B  s A  T1  s B  s A 



T1  T2  s B  s A  T1  sB  sA 

T 1   2  T1

  

(1.1) There is no attempt to use the Carnot cycle with gas as working substance in practice because of two reasons: 1. The pressure of the gas changes continuously from p 4 to p1 during the isothermal heat supply and from p2 to p3 during the isothermal heat rejection. But in practice it is much more convenient to heat a gas at approximately constant pressure or at constant volume. 2. The Carnot cycle, despite its high thermal efficiency, has a small work ratio. [Work ratio is the ratio of the net work output (area 12341) to the gross work output of the system (area 412DC4); the work done on the gas is given by 234CD2.]

Example 1.1 What is the highest possible theoretical efficiency of a heat engine operating with a hot reservoir of furnace gases at 2000 oC when the cooling water is available at 10oC? Solution From Eq. (1.1)  Carnot 1 

T2 T1

3

So

 C 1

10  273 283 1  0 .8754 2000  273 2273

Example 1.2 A hot reservoir at 800oC and a cold reservoir at 15oC are available. Calculate the thermal efficiency and the work ratio of a Carnot cycle using air as the working fluid, if the maximum and minimum pressures in the cycle are 210 bar and 1 bar. Solution The cycle is shown below on T-s and p-v diagrams.

Using Eq. (1.1)

 Carnot 1 

T2 288 1   0. 732 T1 1073

In order to find the work output and the work ratio it is necessary to find the entropy change (s1 – s4). For an isothermal process from 4 to A,

p   210  s A  s 4  R ln 4   0.287 ln   = 1.535 kJ/kg K  1   p2 

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At constant pressure from A to 2,

T   1073  s A  s2  c p ln 1   1. 005ln  = 1.321 kJ/kg K  288   T2  Therefore s1 – s4 = 1.535 – 1.321 = 0.214 kJ/kg K

Then Net work output = area 12341 = (T1 – T2)(s1 – s4) = (1073 – 288) x 0.214 = 168 kJ/kg

Gross work output is Work output 4 to 1 + work output 1 to 2 Now, for an isothermal process, Q + W = 0 i.e. –W4-1 = Q4-1 = area under the line 4-1 on Fig (a) = (s1 – s4) x T1 = 0.214 x 1073 = 229.6 kJ/kg For an isentropic process from 1 to 2, W = (u2 – u1), therefore for a perfect gas -W2-1 = cv(T1 – T2) = 0.718(1073 – 288) = 563.6 kJ/kg

Therefore Gross work output = 229.6 + 563.6 = 793.2 kJ/kg

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i.e. Work ratio 

1.2

net work output 168  0 .212 gross work output 793 .2

The air standard cycle

- Cycles in which the fuel is burned directly in the working fluid are not heat engines in the true meaning of the term since the system is not reduced to its initial state. - The working fluid undergoes a chemical change by combustion and the resulting products are exhausted to the atmosphere. - In practice such cycles are used frequently and are called internalcombustion cycles; the fuel is burned directly in the working fluid, which is normally air. - By supplying fuel inside the cylinder, higher temperatures for the working fluid can be attained; the maximum temperature of all cycles is limited by the metallurgical limit of the material used and the efficiency of the cooling system. - Examples of internal combustion cycles are the open cycle gas turbine unit, the petrol engine, the diesel or oil engine, and the gas engine. o In the open cycle gas turbine the working fluid flows at a steady rate from one component to another round the cycle. o In the petrol engine a mixture of air and petrol is drawn into the cylinder, compressed by the piston, then ignited by

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an electric spark. The hot gases expand, pushing the piston back, and are then swept out to exhaust, and the cycle recommences with the introduction of a fresh charge of petrol and air. o In the diesel or oil engine, the oil is sprayed under pressure into the compressed air at the end of the compression stroke, and the combustion is spontaneous due to the high temperature of the air after compression. o In a gas engine a mixture of gas and air is induced into the cylinder, compressed and then ignited as in the petrol engine by an electric spark. - To give a basis of comparison for the actual internal-combustion engine the air standard cycle is defined. - In an air standard cycle the working substance is assumed to be air throughout, all processes are assumed to be reversible, and the source of heat supply and the sink for heat rejection are assumed to be external to the air. - The cycle can be represented on any diagram of properties, and is usually drawn on the p-v diagram, since this allows a more direct comparison to be made with the actual engine machine cycle. - Note that an air standard cycle on a p-v diagram is a true thermodynamic cycle, whereas a record of the pressure variations in an engine cylinder against piston displacement is a machine cycle.

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1.3

Joule Cycle

This is also known as Brayton or Constant Pressure cycle and forms the basis for the closed cycle gas turbine unit. In this cycle the heat supply and heat rejection processes occur reversibly at constant pressure. The expansion and compression processes are isentropic. T

p p2

T3 T2

3

p2

2

3

2 p1

T4 T1

4

p1

1

1

A

B

4

s

v

Heat supplied

3 2

Heater Net work output

Compressor Turbine 4 1

Cooler Heat rejected

Neglecting velocity changes and applying the steady-flow energy equation to each part of the cycle gives: Work input to compressor = (h2 – h1) = cp(T2 – T1)

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Work output from turbine = (h3 – h4) = cp(T3 – T4) Heat supplied in heater = (h3 – h2) = cp(T3 – T2) Heat rejected in cooler = (h4 – h1) = cp(T4 – T1) Thus



c p  T3  T 2   c p  T 4  T1  c p  T3  T 2 

1 

T4  T1 T3  T2

(1.2)

Since process 1 to 2 and 3 to 4 are isentropic between the same pressures p2 and p1, then T2 T1



 p   2  T4  p1  T3

  1 / 

 rp 1  / 

where rp is the pressure ratio, p2/p1.   1 /   1 / i.e. T3 T4 r p  and T 2 T 1r p 

   1 /   T  T  so T3  T2 r p 4 1

Hence substituting in the expression for the efficiency gives

 1 

T4  T1

1

1     1 /  rp T4  T1 rp  1 / 

(1.3)

Thus for the Joule cycle the cycle efficiency depends only on the pressure ratio. The work ratio (rw) is: rw 

net work output gross work output

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

c p T 3  T 4   c p T 2  T 1  c p T 3  T 4 

1 

T1  T2  T3  T4 

(1.4)

Now, as previously

T2 T 3   rp 1 /  T1 T 4 Therefore T

3   T 2 T 1r p 1 /  and T4    1  / 

rp

Hence substituting

rw 1 





T3 1 

1 / r p 1 / 

T1 rp   1 /   1

 



1 

T1  1 /  rp T3

(1.5)

Thus the work ratio depends not only on the pressure ratio but also on the ratio of the minimum and maximum temperatures. For a given inlet temperature, T1, the maximum temperature, T3, must be made as high as possible for a high work ratio. For an open-cycle gas turbine unit the actual cycle is not such a good approximation to the ideal Joule cycle, since fuel is burned with the air, and a fresh charge is continuously induced into the compressor. The ideal cycle nevertheless provides a good basis for comparison, and in many calculations for the ideal open-cycle gas turbine the effects of the mass of fuel and the charge in the working fluid are neglected.

Example 1.3

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In a gas turbine unit, air is drawn at 1.02 bar and 15 oC, and is compressed to 6.12 bar. Calculate the thermal efficiency and the work ratio of the ideal cycle, when the maximum cycle temperature is limited to 800oC. Solution The ideal cycle is shown below on a T-s diagram.

From Eq. (1.3)  1 

1 r p(   1 ) / 

 1 .02  1     6 .12 

(1 .4  1 ) / 1 .4

= 0.401

The net work output of the cycle is given by the work output of the turbine minus the work input in the compressor, i.e., Net work output = cp(T3 – T4) – cp(T2 – T1) Now, T2  p 2  T1  p1

(   1) / 

  



T3  6.12    T4  1.02 

( 1.4  1 ) / 1. 4

Therefore T2 = 1.669 x T1 = 1.669 x 288 = 480.5 K

= 1.669

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and T 1073 T4  3  = 642.9 K 1.669 1. 669 Thus, Net work output = 1.005(1073 – 642.9) – 1.005(480.5 – 288) = 238.8 kJ/kg Gross work output = work output of the turbine = cp(T3 – T4) = 1.005(1073 – 642.9) = 432.3 kJ/kg Then Work ratio 



1.4

net work output gross work output

238.8 0 .553 432.3

The Otto cycle

This is the ideal air standard cycle for the petrol engine, the gas engine and the high speed oil engine. It consists of the following processes: Process 1-2 = isentropic compression Process 2-3 = reversible constant volume heating Process 3-4 = isentropic expansion Process 4-1 = reversible constant volume heating 3

p

12 2 4 1 v2

v1

v

To give a direct comparison with an actual engine the ratio of the specific volumes, v1/v2, is taken to be the same as the compression ratio of the actual engine, i.e. Compression ratio,

rv  

v1 v2 swept volume  clearance volume clearance volume

The heat supplied at constant volume between T2 and T3 is given by: Q1 = cv(T3 – T2) Similarly the heat rejected per unit mass at constant volume between T4 and T1 is given by Q2 = cv(T4 – T1) Processes 1 to 2 and 3 to 4 are isentropic and therefore there is no heat flow. Thus



T  T1 c v T3  T2   c v  T4  T1  1  4 T3  T 2 c v T3  T 2 

Now for processes 1 to 2 and 3 to 4, which are isentropic,

(1.6)

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T2 T3  v1     T1 T4  v2 

 1

v   4   v3 

 1

 rv 1

  Then T3 T4 rv 1 and T2 T1 rv 1

Hence substituting gives

 1

T 4  T1

 T4





T1 rv 1

1 

1 rv 1

(1.7)

Thus the thermal efficiency of the Otto cycle depends only on the compression ratio, rv.

Example 1.4 Calculate the ideal air standard cycle efficiency based on the Otto cycle for a petrol engine with a cylinder bore of 50 mm, a stroke of 75 mm and a clearance volume of 21.3 cm3.

Solution Swept volume =

 50 2 75 = 147200 mm3 = 147.2 cm3 4

Therefore, Total cylinder volume = 147.2 + 21.3 = 168.5 cm3 i.e., 168 .5 7 .914 21 .3

Compression ratio, rv 

14

Then  1 1.5

1 rv  1

1 

1 7.914 0. 4

 0 .563

The Diesel Cycle

This is ideal air standard cycle for the original diesel engine and consists of the following processes: Process 1-2 = isentropic compression Process 2-3 = reversible constant pressure heating Process 3-4 = isentropic expansion Process 4-1 = reversible constant volume cooling

p p3 = p2

2

3

4 1 v2

v1

v

Heat supplied, Q1 = cp(T3 – T2) Heat rejected, Q2 = cv(T4 – T1) There is no heat flow in processes 1-2 and 3-4 since they are isentropic.

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By substituting in the equation of thermal efficiency, i.e.,



Q1  Q 2 Q1

and expressing each temperature in terms of T 1 and compression ratio ( v1/v2) or cut-off ratio ( v3/v2) the following equation may be derived:  1

  1    1rv  1

(1.8)

where

 = v3/v2 = cut-off ratio rv = v1/v2 = compression ratio Thus the thermal efficiency of a Diesel cycle depends not only on the compression ratio but also on the heat supplied between 2 and 3, which fixes the ratio v3/v2.

Example 1.5 A diesel engine has an inlet temperature and pressure of 15 oC and 1 bar respectively. The compression ratio is 12/1 and the maximum cycle temperature is 1100oC. Calculate the air standard thermal efficiency based on the diesel cycle. Solution

Referring to the figure below,

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T1 = 15 + 273 = 288 K and T3 = 1100 + 273 = 1373 K

Now, v   1  T1  v 2 

T2

1

 rv  1 12 0. 4 = 2.7

i.e., T2 = 2.7 x 288 = 778 K At constant pressure from 2 to 3, since pv = RT for a perfect gas, then

T3



T2 i.e.,

v3 v2

v3 v2

1373  1. 765 778

Therefore

v4 v3



v4 v2 v2 v3



v 1 v2 v 2 v3

12 

1 = 6.8 1. 765

Also, T3  v 4    T4  v3 

 1

6 .8 0. 4 = 2.153

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i.e., T4 

1373 = 638 K 2.153

Heat input, per kg of air, is Q1 = cp(T3 – T2) = 1.005(1373 – 778) = 598 kJ/kg

Heat rejected, per kg of air, is Q2 = cv(T4 – T1) = 0.718(638 – 288) = 251 kJ/kg

Therefore,



1.6

Q1  Q 2 Q1



598  251 = 0.58 598

The dual-combustion cycle

This is also known as the limited-pressure or mixed cycle and is the ideal air standard cycle of modern diesel and oil engines. It consists of the following processes: Process 1-2 = isentropic compression Process 2-3 is reversible constant volume heating Process 3-4 = reversible constant pressure heating Process 4-5 = isentropic expansion Process 5-1 = reversible constant volume cooling

3 p3 = p4 2

4

p

18

5 1 v1

v2 = v3

v

The heat is supplied in two parts, the first part at constant volume and the remainder at constant pressure, hence the name ‘dualcombustion’. In order to get the thermal efficiency, three factors are necessary. These are: -

The compression ratio, rv = v1/v2,

-

The ratio of pressure, rp = p3/p2, and

-

The ratio of volumes,  = v4/v3.

Then it can be shown that

  1

r p   1

 rp  1  r p   1rv  1

(1.9)

Thus the thermal efficiency of a dual-combustion cycle depends not only on the compression ratio but also on the relative amounts of heat supplied at constant volume and at constant pressure. Equation (8.9) is much too cumbersome to use, and the best method of calculating thermal efficiency is to evaluate each temperature round the cycle and then get the total heat supplied ( Q1) and the total heat rejected (Q2) as: Q1 = cv(T3 – T2) + cp(T4 – T3) Q2 = cv(T5 – T1) Note that when rp = 1 (i.e. p3 = p2), then Eq (1.9) reduces to the thermal efficiency of the diesel cycle.

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Example 1.6 An oil engine takes in air at 1.01 bar, 20oC and the maximum cycle pressure is 69 bar. The compressor ratio is 18/1. Calculate the air standard thermal efficiency and the mean effective pressure based on the dual-combustion cycle. Assume that the heat added at constant volume is equal to the heat added at constant pressure. [Mean effective pressure is the height of a rectangle having the same length and area as the cycle plotted on a p-v diagram.]

Solution The cycle is shown below on a p-v diagram.

T2  v1  T1  v2

   

1

18 0.4 = 3.18

i.e., T2 = 3.18 x T1 = 3.18 x (20 + 273) = 931 K From 2 to 3 the process is at constant volume, hence

p 3 T3 p3 v3 p v   2 2 and v3 = v2 since p 2 T2 T3 T2

20

p3

i.e., T3 

T2 

p2

69 931 p2

To find p2, use the equation: 

v   1  181.4 = 57.2  v2 

p2 p1

i.e., p2 = 57.2 x 1.01 = 57.8 bar Then substituting, T3 

69 931 = 1112 K 57. 8

Now the heat added at constant volume is equal to the heat added at constant pressure in this example, therefore cv(T3 – T2) = cp(T4 – T3) i.e., 0.718(1112 – 931) = 1.005(T4 – 1112) 0.718 181 1112 = 1241.4 K 1 .005

Therefore, T4 

To find T5 it is necessary to know the value of the volume ratio, v5/v4.

At constant pressure from 3 to 4

v4 v3



T4 T3

v5

Therefore,

And



v4

1241. 4 = 1.116 1112



v    5  T5  v 4  T4

v1 v4



v1 v3

1 18  = 16.14 v2 v4 1.116

 1

16.14 0.4 = 3.04

1241 .4 = 408 K 3.04

i.e., T5 

21

Now the heat supplied, Q1, is given by Q1 = cv(T3 – T2) + cp(T4 – T3) Or Q1 = 2cv(T3 – T2) since in this example the heat added at constant volume is equal to the heat added at constant pressure. Therefore Q1 = 2 x 0.718 x (1112 – 931) = 260 kJ/kg The heat rejected is given by Q2 = cv(T5 – T1) = 0.718(408 – 293) = 82.6 kJ/kg

Then



260  82.6 Q1  Q 2  = 0.682 Q1 260

The net work done, - W = Q1 – Q2 = Q1 = 0.682 x 260 = 177 kJ/kg From the definition of mean effective pressure, - W = pm(v1 – v2) Using pv = RT and rv = v1/v2 = 18, then

v  17 17 RT1  v 1  v 2  v 1  1   v1  18 p1 18  18  

17 287 293 = 0.786 m3/kg 181. 0110 5

Then substituting, - W = pm x 0.786

22

or pm = - W/0.786 kJ/m3 i.e. Mean effective pressure 

177 10 3

10 5 0.786

1.7

= 2.25 bar

The Stirling cycle

This has an efficiency equal to that of the Carnot cycle but has a higher work ratio. It consists of the following processes Process 1-2 = reversible constant volume heating Process ...