CHAPTER

MEC 451

5 Thermodynamics
Lecture Notes:
MOHD HAFIZ MOHD NOH
HAZRAN HUSAIN & MOHD
SUHAIRIL
Faculty of Mechanical Engineering
Universiti Teknologi MARA, 40450
Air Standard
Shah Alam, Selangor Cycle

For students EM 220 and EM 221 only

 Review – Carnot Cycle

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

TL
η th , Carnot = 1−
TH

Upon derivation the performance of the real cycle is often

measured in terms of its thermal efficiency

Wnet
η th =
Qin

2
 Review – Ideal Gas Law

The ideal gas equation is defined as

Pv = RT or PV = mRT

where P = pressure in kPa

v = specific volume in m3/kg (or V = volume in
m3)
R = ideal gas constant in kJ/kg.K
m = mass in kg
T = temperature in K

3 3
The Δu and Δh of ideal gases can be expressed as

∆u = u 2 − u1 = C v (T2 − T1 )

∆h = h2 − h1 = C P (T2 − T1 )

Δu - constant volume process

Δh - constant pressure process

4
 Review – Thermodynamics Processes

 According to a law of P V n = constant

Process Description Result of IGL

P1 P2
isochoric constant volume (V1 = V2) =
T1 T2
constant pressure (P1 = V1 V2
isobaric
P 2)
=
T1 T2
constant temperature
isothermal
(T1 = T2) P1V1 = P2V2
n
polytropic -none- n
P1  V2   T1  n −1
=   =  
isentropic constant entropy (S1 = S2) P2  V1   T2 

5
 Review – Properties of Air

R = 0.2871 kJ/kg.K
Cp = 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
6
 Introduction

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

cylinder.

7
 Air-Standard Assumptions

 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.

8
 Air-Standard Assumptions

 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.

9
 Terminology for Reciprocating Devices

 Top dead center (TDC), bottom dead center (BDC), stroke,

bore, intake valve, exhaust valve, clearance volume,
displacement volume, compression ratio, and mean
effective pressure

10
 The compression ratio r of an
engine is defined as

V max VBDC
r= =
V min VTDC

 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 Wnet wnet
MEP = =
cycle. Vmax − Vmin vmax − vmin
11
 Otto Cycle
The 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

13
 Related formula based on basic thermodynamics:

n
n
P1  V2  T  n −1
=   =  1 
P2  V1   T2 

Qin = mCv ( T3 − T2 )

n
n
P1  V2   T1  n −1
=   =  
P2  V1   T2 

Qout = mCv ( T4 − T1 )

14
 Thermal efficiency of the Otto cycle:

Wnet Qnet Qin − Qout Qout

η th = = = = 1−
Qin Qin Qin Qin

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

Qnet ,23 − Wnet ,23 = ∆U 23

3
Wnet ,23 = Wother ,23 + Wb,23 = 0 + ∫ PdV = 0
2

 Thus, for constant specific heats

Qnet , 23 = ∆U 23
Qnet , 23 = Qin = mCv (T3 − T2 )
15
 Apply first law closed system to process 4-1, V = constant.
Qnet ,41 − Wnet ,41 = ∆U 41
1
Wnet ,41 = Wother ,41 + Wb ,41 = 0 + ∫ PdV = 0
4
 Thus, for constant specific heats,
Qnet , 41 = ∆U 41
Qnet , 41 = −Qout = mCv (T1 − T4 )
Qout = −mCv (T1 − T4 ) = mCv (T4 − T1 )

 The thermal efficiency becomes

Qout
η th , Otto = 1 −
Qin
mCv (T4 − T1 )
= 1−
mCv (T3 − T2 ) 16
 (T4 − T1 )
η th , Otto = 1−
(T3 − T2 )
T1 (T4 / T1 − 1)
= 1−
T2 (T3 / T2 − 1)
 Recall processes 1-2 and 3-4 are isentropic, so
k −1 k −1
T2  V1  T3  V4 
= ÷ and = ÷
T1  V2  T4  V3 

 Since V3 = V2 and V4 = V1,

T2 T3 T4 T3
= or =
T1 T4 T1 T2

17
 The Otto cycle efficiency becomes

T1
η th , Otto = 1 −
T2

 Since process 1-2 is isentropic,

k −1
T2  V1 
= ÷
T1  V2 
k −1 k −1
T1  V2  1
= ÷ = ÷
T2  V1  r

 where the compression ratio is

r = V1/V2 and

1
η th , Otto = 1 −
r k −1 18
 Example 5.1

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, ηth, the back work ratio and the
mean effective pressure.
Solution:
Data given:
T1 = 290 K
V1
=9
V2
Q23 = 7.5 kJ
P1 = 95 kPa
V1 = 3.8 Litres
19
Example 5.1
Pr ocess 1 − 2 ( isentropic compression )
k −1
T2  V1 
⇒ T2 = 290 ( 9 )
0.4
= ÷ = 698.4 K
T1  V2 
k −1
P2  V1 
⇒ P2 = 95 ( 9 )
1.4
= ÷ = 2059kPa
P1  V2 
Pr ocess 2 − 3 ( Const. volume heat addition )
0
1 law : Qnet − Wnet = ∆U
st

Q23 = mCv ( T3 − T2 )
0.2871( 290 )
1 1 = RT1 ⇒ v1 = = 0.875 mkg
3
IGL : Pv
95
Q23 v1
q23 = = Q23 = 1727 kJ
kg
m V1 20
Example 5.1

Back to IGL : But V3 = V2

P3 P2
q23 = Cv ( T3 − T2 ) =
T3 T2
= 0.718 ( T3 − 698.4 ) P3 = 9.15 MPa
T3 = 3103.7 K
Pr ocess 3 − 4 ( isentropic exp ansion )
k −1
T4  V3 
⇒ T4 = T3 ( 1 / 9 )
0.4
= ÷ = 1288.8 K
T3  V4 
k
P4  V3 
=  ÷ ⇒ P4 = P3 ( 1/ 9 ) = 422 kPa
1.4

P3  V4 

21
 Example 5.1

Pr ocess 4 − 1( Const. volume heat rejection )

Q41 = mCv ( T4 − T1 )
q41 = Cv ( T4 − T1 )
= 0.718 ( 1288.8 − 290 )
= 717.1 kJ
kg

Then:
Wnet = qin − qout
= q23 − q41
= 1009.6 kJ
kg

Wnet
ηth,Otto = = 0.585 ( 58.5% )
qin
22
 Example 5.1

What else?

Wnet wnet
MEP = =
Vmax − Vmin vmax − vmin
wnet wnet
= =
v1 − v2 v1 ( 1 − v2 / v1 )
wnet 1009.6
= = = 1298 kPa
v1 ( 1 − r ) 0.875 ( 1 − 1/ 9 )
1

wcompr ∆u12 Cv ( T2 − T1 )
rbw = = =
wexp ans −∆u34 Cv ( T3 − T4 )
= 0.225 ( 22.5% )

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
 Diesel Cycle

 The processes in the Diesel cycle are as per following:

25
 Diesel Cycle

v1 v3
= Compression ratio, rv and = Cut − off ratio, rc
v2 v2

26
 Related formula based on basic thermodynamics:

n
n
P1  V2  T  n −1
=   =  1 
P2  V1   T2 

Qin = mCP ( T3 − T2 )

n
n
P1  V2  T  n −1
=   =  1 
P2  V1   T2 

Qout = mCv ( T4 − T1 )

27
 Thermal efficiency of the Diesel cycle

Wnet Q
η th , Diesel = = 1 − out
Qin Qin
 Apply the first law closed system to process 2-3, P = constant.
Qnet ,23 − Wnet ,23 = ∆U 23
3
Wnet ,23 = Wother ,23 + Wb ,23 = 0 + ∫ PdV = 0
2

= P2 ( V3 − V2 )
 Thus, for constant specific heats
Qnet , 23 = ∆U 23 + P2 (V3 − V2 )
Qnet , 23 = Qin = mCv (T3 − T2 ) + mR (T3 − T2 )
Qin = mC p (T3 − T2 )
28
 Apply the first law closed system to process 4-1, V = constant
Qnet ,41 − Wnet ,41 = ∆U 41
1
Wnet ,41 = Wother ,41 + Wb ,41 = 0 + ∫ PdV = 0
4

 Thus, for constant specific heats

Qnet , 41 = ∆U 41
Qnet , 41 = −Qout = mCv (T1 − T4 )
Qout = −mCv ( T1 − T4 ) = mCv (T4 − T1 )

 The thermal efficiency becomes

Qout
η th , Diesel = 1 −
Qin
mCv (T4 − T1 )
= 1−
mC p (T3 − T2 ) 29
 PV PV
4 4
= 1 1 where V4 = V1
T4 T1
T4 P4
=
T1 P1

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

1 1 = PV 4 4 = PV
k k k k
PV 2 2 and PV 3 3

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

the first equation and obtain
k
P4  V3 
=  ÷ = rc k
T4  V2 
 Therefore,
1 rc k − 1
ηth , Diesel = 1 −
r k −1 k ( rc − 1)
30
 Example 5.2

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.

Solution:
Data given:

V1
= 18
V2
V3
= 2.5
V2
31
Example 5.2

Pr ocess 1 − 2 ( isentropic compression )

k −1
T2  V1 
⇒ T2 = 300 ( 18 )
0.4
= ÷ = 953.3K
T1  V2 
Pr ocess 2 − 3 ( Const. pressure heat addition )

V2 V3  V3 
P2 = P3 ⇒ = ⇒ T3 = T2  ÷ = 2383.3 K
T2 T3  V2 
Pr ocess 3 − 4 ( isentropic exp ansion )
V4 V1 V2
= . = 18 ( 1/ 2.5 ) = 7.2
V3 V2 V3
k −1
T4  V3 
⇒ T4 = 2383.3 ( 1/ 7.2 )
0.4
= ÷ = 1082 K
T3  V4 
32
 Example 5.2

Qin = Q23 = mCP ( T3 − T2 ) ⇒ qin = C p ( T3 − T2 ) = 1437.15 kJ

kg

Qout = Q41 = mCP ( T4 − T1 ) ⇒ qout = C p ( T4 − T1 ) = 561.48 kJ

kg

wnet = qin − qout = 875.67 kJ

kg

What we need?
wnet
( i ) ηth,diesel = = 0.6093 ( 60.93% )
qin
( ii ) Pmax = P2 = P3
k −1
 P2   T2  k
 ÷ =  ÷ ⇒ P2 = 5148 kPa ( Pmax )
 P1   T1 
wnet 875.67
( iii ) MEP = = = 969.1 kPa
V1 ( 1 − 1/ r ) 0.9566 ( 1 − 1/ 18 )
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 1700°C and the minimum cycle
temperature is 15°C. Calculate:
i. the specific heat transfer to the cycle
ii. the specific work of the cycle
iii. the thermal efficiency of the cycle

34
 Dual Cycle

 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.
35
 Dual Cycle

 The same procedure as to Otto and Diesel cycles can be applied to

Dual cycle. Upon substitutions, the thermal efficiency of Dual
cycle becomes

rp rck − 1
η th = 1 −
[( r p ]
− 1) + k c p ( rc − 1) rvk −1
36
 Example 5.3

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!!)
Solution:
Data given:

V1 P1
= 18 = 1.5
V2 P2
V4
T1 = 300 K = 1.2
V3
P1 = 1bar
37
Pr ocess 1 − 2 ( isentropic compression )
k −1
T2  V1 
⇒ T2 = 300 ( 18 )
0.4
= ÷ = 953.3K
T1  V2 
Pr ocess 2 − 3 ( Const. pressure heat addition )

V2 V3  V3 
P2 = P3 ⇒ = ⇒ T3 = T2  ÷ = 2383.3 K
T2 T3  V2 
Pr ocess 3 − 4 ( isentropic exp ansion )
V4 V1 V2
= . = 18 ( 1/ 2.5 ) = 7.2
V3 V2 V3
k −1
T4  V3 
⇒ T4 = 2383.3 ( 1/ 7.2 )
0.4
= ÷ = 1082 K
T3  V4 
38
 Pr ocess 4 − 5 ( isentropic exp ansion )
k −1 k −1 k −1
T5  V4   V4   V4   V3  
= ÷ ⇒ T5 = T4  ÷ = T4  ÷ ÷
T4  V5   V5   V3   V5  
= 1715.94 ( 1.2 ) ( ) 
0.4
1
18

= 584.85 K
Information needed?

Qout = Q51 = mCv ( T5 − T1 ) = 204.52 kJ

kg

Qin = Q23 + Q34 = mCv ( T3 − T2 ) + mC p ( T4 − T3 )

= 629.65 kJ
kg

39
 Answer the questions ?

Wnet Qin − Qout Qout 204.52

( ) th
a η = = = 1 − = 1 − = 0.675 ( 67.5% )
Qin Qin Qin 629.65
Wnet
( b ) MEP =
v1 ( 1 − 1r )
425.13
=
0.8613 ( 1 − 181 )
= 522.63 kPa

40
 Criteria of Performance

 Indicated power (IP)

 Brake power (bp)
 Friction power (fp) and mechanical efficiency, ηm
 Brake mean effective pressure (bmep), thermal
efficiency and fuel consumption
 Volumetric efficiency, ηv

41
 Indicated Power
 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.

ip = work done per cycle × cycle per minute

 For four-stroke engine,

pi LANn
IP =
2
 And for two-stroke engine, n is the number of cylinders.

IP = pi LANn

42
 Indicated Power

 Indicated mean effective pressure, pi given by:,

net area of diagram work done per cycle

pi = × constant pi =
length of diagram displacement volume

 For one engine cylinder

Work done per cycle = pi × A
×L
Where A = area of piston
L = length of stroke
 Power output = (work done per cycle) x (cycle per
minute) ip = P AL ×cycle per unit time
i

 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. 43
 Brake Power

 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.

τ = Wr

44
 Therefore

bp = 2πNτ
 Brake power is also can be expressed as

η m pi LANn Pb LANn
bp = η m IP = =
2 2
 Then the brake mean effective pressure (Pb) is

Pb = η m Pi

45
 Friction Power

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.
fp = IP − bp
46
 Mechanical Efficiency

 Power input to the shaft is usually bigger than the

indicated power due to frictional losses or the
mechanical efficiency.

brake power
η mech =
indicated power

47
ηm
Brake Mean Effective Pressure

 From the definition of Brake power BP = ηm IP

pi LANn for 4 stroke engine η m pi LANn Pb LANn

 Since IP = bp = =
2 and 2 2

 Since η m and Pi are difficult to obtain, they may be combined and

replaced by a brake mean effective pressure, Pb
Pb LANn
 Equating this equation to another definition of bp: = 2η NT
2
4η
 So: Pb = T
LAn
 Its observed that bmep is proportional to torque.

48
ηm
Brake Thermal Efficiency

 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,
brake power brake power bp bp
ηbp = = = =
power given fuel equivalent power Pfe m f Qnet ,v

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

49
 Specific Fuel Consumption

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

m f
sfc =
bp

50
 Volumetric Efficiency

 The parameter used to measure the effectiveness of an

engine’s induction process is the volumetric efficiency.

 Volumetric efficiency is only used with four-stroke cycle

engine, which have a distinct induction process.

V
ηV =
Vs

51
 Example 5.4

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.

Solution:

52
 Example 5.4

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
Solution:
and the volumetric efficiency of the engine.

53
Real Case

pi LANn
IP =
2

bp = 2πNτ

54