CHAPTER

5MEC 451Thermodynamics

Air Standard Cycle

Lecture Notes:MOHD HAFIZ MOHD NOHHAZRAN HUSAIN & MOHD SUHAIRIL Faculty of Mechanical EngineeringUniversiti Teknologi MARA, 40450 Shah Alam, Selangor

For students EM 220 and EM 221 only

thnet

in

W

Q

th CarnotL

H

T

T, 1

Upon derivation the performance of the real cycle is often measured in terms of its thermal efficiency

The Carnot cycle was introduced as the most efficient heat

engine that operate between two fixed temperatures TH

and TL. The thermal efficiency of Carnot cycle is given by

Review – Carnot Cycle

2

The ideal gas equation is defined as

mRTPVorRTPv

where P = pressure in kPav = specific volume in m3/kg (or V = volume in

m3)R = ideal gas constant in kJ/kg.Km = mass in kgT = temperature in K

Review – Ideal Gas Law

3 3

The Δu and Δh of ideal gases can be expressed as

)( 1212 TTCuuu v

)( 1212 TTChhh P

Δu - constant volume processΔh - constant pressure process

4

Process Description Result of IGL

isochoricconstant volume (V1 =

V2)

isobaricconstant pressure (P1 =

P2)

isothermalconstant temperature

(T1 = T2)

polytropic -none-

isentropicconstant entropy (S1 =

S2)

According to a law of constantnVP

2

2

1

1

T

P

T

P

2

2

1

1

T

V

T

V

2211 VPVP

1

2

1

1

2

2

1

n

nn

T

T

V

V

P

P

Review – Thermodynamics Processes

5

R = 0.2871 kJ/kg.KCp = 1.005 kJ/kg.K

Cv = 0.718 kJ/kg.K

k = 1.4

where R = ideal gas constant

Cp = specific heat at constant pressure

Cv = specific heat at constant volume

k = specific heat ratio

Review – Properties of Air

6

IC Engine – combustion of fuel takes place inside an engine’s cylinder.

Introduction

7

Air continuously circulates in a closed loop.

Always behaves as an ideal gas.

All the processes that make up the cycle are

internally reversible.

The combustion process is replaced by a heat-

addition process from an external source.

Air-Standard Assumptions

8

A heat rejection process that restores the working

fluid to its initial state replaces the exhaust process.

The cold-air-standard assumptions apply when the

working fluid is air and has constant specific heat

evaluated at room temperature (25oC or 77oF).

No chemical reaction takes place in the engine.

Air-Standard Assumptions

9

Top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure

Terminology for Reciprocating Devices

10

The compression ratio r of an engine is defined as

rV

V

V

VBDC

TDC

max

min

The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle.

MEPW

V V

w

v vnet net

max min max min11

Otto CycleThe Ideal Cycle for Spark-Ignition Engines

12

The processes in the Otto cycle are as per following:

Process

Description

1-2 Isentropic compression

2-3 Constant volume heat addition

3-4 Isentropic expansion

4-1 Constant volume heat rejection

1

2

3

4Qout

Qin

PvgConstant

v1v2 v

P

s

T

Qout

Qin

1

2

3

4

(a) P-v diagram (b) T-s diagram 13

Related formula based on basic thermodynamics:

1

2

1

1

2

2

1

n

nn

T

T

V

V

P

P

1

2

1

1

2

2

1

n

nn

T

T

V

V

P

P

3 2in vQ mC T T

4 1out vQ mC T T

14

Thermal efficiency of the Otto cycle:

thnet

in

net

in

in out

in

out

in

W

Q

Q

Q

Q Q

Q

Q

Q

1

Apply first law closed system to process 2-3, V = constant.

Thus, for constant specific heats

Q U

Q Q mC T T

net

net in v

,

, ( )

23 23

23 3 2

,23 ,23 23

3

,23 ,23 ,23

2

0 0

net net

net other b

Q W U

W W W PdV

15

Apply first law closed system to process 4-1, V = constant.

Thus, for constant specific heats,

Q U

Q Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )

( ) ( )

41 41

41 1 4

1 4 4 1

The thermal efficiency becomes

th Ottoout

in

v

v

Q

Q

mC T T

mC T T

,

( )

( )

1

1 4 1

3 2

,41 ,41 41

1

,41 ,41 ,41

4

0 0

net net

net other b

Q W U

W W W PdV

16

th Otto

T T

T T

T T T

T T T

,

( )

( )

( / )

( / )

1

11

1

4 1

3 2

1 4 1

2 3 2

Recall processes 1-2 and 3-4 are isentropic, so

Since V3 = V2 and V4 = V1,

3 32 4

1 4 1 2

T TT Tor

T T T T

11

32 1 4

1 2 4 3

kkTT V V

andT V T V

17

The Otto cycle efficiency becomes

th Otto

T

T, 1 1

2

Since process 1-2 is isentropic,

where the compression ratio is r = V1/V2 and

th Otto kr, 11

1

1

2 1

1 2

1 1

1 2

2 1

1

k

k k

T V

T V

T V

T V r

18

An Otto cycle having a compression ratio of 9:1 uses air as

the working fluid. Initially P1 = 95 kPa, T1 = 17°C, and V1 =

3.8 liters. During the heat addition process, 7.5 kJ of heat

are added. Determine all T's, P's, th, the back work ratio

and the mean effective pressure.

Example 5.1

Solution:Data given:

1

1

2

23

1

1

290

9

7.5

95

3.8

T K

VV

Q kJ

P kPa

V Litres

19

Example 5.1

10.42 1

21 2

11.42 1

21 2

290 9 698.4

95 9

Pr 1 2

Pr 2 3 .

2059

1 :

k

k

stnet net

T VT K

T V

P VP kPa

P V

law Q W

ocess isentropic compression

ocess Const volume heat addition

3

0

23 3 2

1 1 1 1

23 123 23

1

0.2871 290: 0.875

95

1727

v

mkg

kJkg

U

Q mC T T

IGL Pv RT v

Q vq Q

m V

20

Example 5.1

3 2

3 223 3 2

3 2

3 3

3

10.434

4 33 4

1.4344 3

3 4

:

0.718 698.4 9.15

3103.7

1/ 9 1288.8

1/ 9 422

Pr 3 4 exp

v

k

k

ocess isentr

Back to IGL ButV V

P Pq C T T

T T

T P MPa

T K

VTT T

opic ansi

KT V

VPP P kPa

P

on

V

21

Example 5.1

41 4 1

41 4 1

0.718 1288.8 290

717.

Pr 4 1 .

1

v

v

kJkg

Q mC T

ocess Const volume heat rejec

T

q C T

tion

T

Then:

23 41

,

1009.6

0.585 58.5%

net in out

kJkg

netth Otto

in

W q q

q q

W

q

22

Example 5.1

What else?

max min max min

1 2 1 2 1

11

12

exp 34

1 /

1009.61298

1 0.875 1 1/ 9

net net

net net

net

r

vcomprbw

ans

W wMEP

V V v v

w w

v v v v v

wkPa

v

Cw ur

w u

2 1

v

T T

C

3 4

0.225 22.5%

T T

23

Supplementary Problems 5.1

1. An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 750 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Taking into account the variation of specific heats with temperature, determine (a) the pressure and temperature at the end of the heat addition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle.

[(a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 percent,(d ) 495 kPa]2. The compression ratio of an air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is at 100 kPa, 35°C, and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Using specific heat values at room temperature, determine (a) the highest temperature and pressure in the cycle; (b) the amount of heat transferred in, in kJ; (c) the thermal efficiency; and (d) the mean effective pressure.

[(a) 1969 K, 6072 kPa,(b) 0.59 kJ, (c) 59.4 percent, (d) 652 kPa] 24

The processes in the Diesel cycle are as per following:

Diesel Cycle

25

cv rratiooffCutv

vandrrationCompressio

v

v,,

2

3

2

1

Diesel Cycle

26

Related formula based on basic thermodynamics:

1

2

1

1

2

2

1

n

nn

T

T

V

V

P

P

1

2

1

1

2

2

1

n

nn

T

T

V

V

P

P

3 2in PQ mC T T

4 1out vQ mC T T

27

Thermal efficiency of the Diesel cycle

th Dieselnet

in

out

in

W

Q

Q

Q, 1

Apply the first law closed system to process 2-3, P = constant.

Thus, for constant specific heats

Q U P V V

Q Q mC T T mR T T

Q mC T T

net

net in v

in p

,

,

( )

( ) ( )

( )

23 23 2 3 2

23 3 2 3 2

3 2

,23 ,23 23

3

,23 ,23 ,23

2

2 3 2

0 0

net net

net other b

Q W U

W W W PdV

P V V

28

Apply the first law closed system to process 4-1, V = constant

Q U

Q Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )

( ) ( )

41 41

41 1 4

1 4 4 1

Thus, for constant specific heats

The thermal efficiency becomes

th Dieselout

in

v

p

Q

Q

mC T T

mC T T

,

( )

( )

1

1 4 1

3 2

,41 ,41 41

1

,41 ,41 ,41

4

0 0

net net

net other b

Q W U

W W W PdV

29

PV

T

PV

TV V

T

T

P

P

4 4

4

1 1

14 1

4

1

4

1

where

Recall processes 1-2 and 3-4 are isentropic, so

PV PV PV PVk k k k1 1 2 2 4 4 3 3 and

Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain

Therefore,

34

4 2

k

kc

VPr

T V

, 1

111

1

kc

th Diesel kc

r

r k r

30

An air-standard Diesel cycle has a compression ratio of 18 and a cut-off ratio of 2.5. The state at the beginning of compression is fixed by P = 0.9 bar ant T = 300K. Calculate:

i. the thermal efficiency of the cycle,

ii. the maximum pressure, Pmax, and

iii. The mean effective pressure.

Example 5.2

Solution:

Data given:

1

2

3

2

18

2.5

V

V

V

V

31

Example 5.2

10.42 1

21 2

3 322 3 3 2

2 3 2

4 1 2

3 2 3

4

Pr 1 2

Pr 2 3

300 18 953.3

2383.

.

Pr 3 4 exp

3

. 18 1/ 2.5 7.2

k

ocess isentropic compression

ocess Const pressure heat addition

ocess isentropic ansio

T VT K

T V

V VVP P T T K

T T V

V V V

V V V

T

n

1

0.434

3 4

2383.3 1/ 7.2 1082k

VT K

T V

32

Example 5.2

23 3 2 3 2

41 4 1 4 1

1437.15

561.48

875.67

kJin P in p kg

kJout P out p kg

kJnet in out kg

Q Q mC T T q C T T

Q Q mC T T q C T T

w q q

What we need?

,

max 2 3

1

2 22 max

1 1

1

0.6093 60.93%

5148

875.67969.1

1 1/ 0.9566 1 1/18

netth diesel

in

k

k

net

wi

q

ii P P P

P TP kPa P

P T

wiii MEP kPa

V r

33

Supplementary Problems 5.2

1. An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kPa and 20°C. If the maximum temperature in the cycle is not to exceed 2200 K, determine (a) the thermal efficiency and (b) the mean effective pressure. Assume constant specific heats for air at room temperature.

[ (a) 63.5 percent, (b) 933 kPa]

2. An ideal diesel cycle has a compression ratio of 16 to 1. The maximum cycle temperature is 1700C and the minimum cycle temperature is 15C. Calculate:i. the specific heat transfer to the cycleii. the specific work of the cycleiii. the thermal efficiency of the cycle

34

Dual cycle gives a better approximation to a real engine. The heat addition process is partly done at a constant volume and partly at constant pressure. From the P-v diagram, it looks like the heat addition process is a combination of both Otto and Diesel cycles.

Dual Cycle

35

The same procedure as to Otto and Diesel cycles can be applied to Dual cycle. Upon substitutions, the thermal efficiency of Dual cycle becomes

111

11

kvcpp

kcp

th rrckr

rr

Dual Cycle

36

At the beginning of the compression process of an air-standard dual cycle with a compression ratio of 18, the temperature is 300 K and the pressure is 1 bar. The pressure ratio for the constant volume part of the heating process is 1.5 to 1. The volume ratio for the constant pressure part of the heating process is 1.2 to 1. Determine (a) the thermal efficiency and (b) the mean effective pressure. (WRONG SOLUTION!!)

Example 5.3

Solution:

1 1

2 2

41

3

1

18 1.5

300 1.2

1

V P

V P

VT K

V

P bar

Data given:

37

10.42 1

21 2

3 322 3 3 2

2 3 2

4 1 2

3 2 3

4

Pr 1 2

Pr 2 3

300 18 953.3

2383.

.

Pr 3 4 exp

3

. 18 1/ 2.5 7.2

k

ocess isentropic compression

ocess Const pressure heat addition

ocess isentropic ansio

T VT K

T V

V VVP P T T K

T T V

V V V

V V V

T

n

1

0.434

3 4

2383.3 1/ 7.2 1082k

VT K

T V

38

11 1

5 34 4 45 4 4

4 5 5 3 5

0.41

181715.94 1.2

5

Pr 4 5

84.85

exp

kk k

ocess isentropic ansio

T VV V VT T T

V

n

T V V V

K

Information needed?

51 5 1

23 34 3 2 4 3

204.52

629.65

kJout v kg

in v p

kJkg

Q Q C T T

Q Q Q C T T C T T

m

m m

39

Answer the questions ?

11

118

204.521 1 0.675 67.5%

629.65

1

425.13

0.8613 1

522.63

net in out outth

in in in

net

r

W Q Q Qa

Q Q Q

Wb MEP

v

kPa

40

Indicated power (IP) Brake power (bp) Friction power (fp) and mechanical efficiency,

m

Brake mean effective pressure (bmep), thermal efficiency and fuel consumption

Volumetric efficiency, v

Criteria of Performance

41

Defined as the rate of work done by the gas on the piston as evaluated from an indicator diagram obtained from the engine using the electronic engine indicator.

2

LANnpIP i

For four-stroke engine,

And for two-stroke engine,

LANnpIP i

Indicated Power

42

ip = work done per cycle × cycle per minute

n is the number of cylinders.

Indicated Power

43

constantdiagramoflength

diagramofareanetip

Indicated mean effective pressure, pi given by:,

For one engine cylinder Work done per cycle = pi

A L Where A = area of piston

L = length of stroke

timeunitpercycleALPip i Power output = (work done per cycle) x (cycle

per minute)

For four-stoke engines, the number of cycles per unit time is N/2 and for two-stroke engines the number of cycles per unit time is N, where N is the engine speed.

volumentdisplaceme

cycleperdoneworkip

Brake power is a way to measure the engine power output.

The engine is connected to a brake (or dynamometer) which can be loaded so that the torque exerted by the engine can be measured.

The torque is obtained by reading off a net load, w at known radius, r.

Wr

Brake Power

44

Therefore

Nbp 2

22

LANnPLANnpIPbp bim

m

Brake power is also can be expressed as

Then the brake mean effective pressure (Pb) is

imb PP

45

Friction Power

46

The difference between the Ip and bp is the friction power (fp). It is the power that overcome the frictional resistance of the engine parts.

bpIPfp

Power input to the shaft is usually bigger than the indicated power due to frictional losses or the mechanical efficiency.

powerindicated

powerbrakemech

Mechanical Efficiency

47

Brake Mean Effective Pressure

48

From the definition of Brake power IPBP m

Since 2

LANnpIP i for 4 stroke engine

and 22

LANnPLANnpbp bim

Since and Pi are difficult to obtain, they may be combined and replaced by a brake mean effective pressure, Pb

m

m

Equating this equation to another definition of bp: NTLANnPb 22

TLAn

Pb

4 So:

Its observed that bmep is proportional to torque.

Brake Thermal Efficiency

49

The power output of the engine is obtained from the chemical energy of the fuel supplied. The overall engine efficiency is given by the brake thermal efficiency,

m

vnetf

p

fe

pbp Qm

b

P

b

,power equivalent fuel

power brake

givenpower

power brake

mf = mass flow fuel , Qnet,v = net calarofic value of the fuel.

sfc is the mass flow rate of fuel consumed per unit power output and is a criterion of economical power production.

bp

msfc f

Specific Fuel Consumption

50

Volumetric efficiency is only used with four-stroke cycle engine, which have a distinct induction process.

The parameter used to measure the effectiveness of an engine’s induction process is the volumetric efficiency.

sV V

V

Volumetric Efficiency

51

An engine operating at 2400 rpm consumes 12 ml of fuel (s.g. 0.85) in 60 second. The engine indicates a load of 30 N on the pony brake system and the brake’s torque arm is 20 cm. Determine (a) the brake power, (b) the mass flow rate of fuel, and (c) the specific fuel consumption.

Example 5.4

Solution:

52

A four-cylinder petrol engine has a bore of 57 mm and a stroke of 90 mm. Its rated speed is 2800 rpm and it is tested at this speed against a brake which has a torque arm of 0.356 m. The net brake load is 155 N and the fuel consumption is 6.741 l/h. The specific gravity of the petrol used is 0.735 and it has a net calorific value of 44,200 kJ/kg. The engine is tested in an atmospheric condition at 101.325 kPa and 15 oC at air-fuel ratio of 14.5/1. Calculate for this speed, the engine torque, the bmep, the brake thermal efficiency, the specific fuel consumption and the volumetric efficiency of the engine.

Example 5.4

Solution:

53

2

LANnpIP i

Nbp 2

Real Case

54