The document describes the working principles of a heat engine. It begins by explaining that a heat engine takes in energy by heat from a high temperature reservoir and expels a fraction of that energy as work during a cyclic process. It then provides more details on the specific processes involved, including absorbing heat, rejecting heat to a lower temperature reservoir, and doing work. The thermal efficiency of a heat engine is also defined.

1. HEAT ENGINE
It results in the motion of the
automobile
A fuel is burned and the high-
temperature gases produced are
used to vapourize water to steam.
This steam is directed at the
pistons to perform mechanical
work.
A heat engine is a device that takes in energy by heat and, operating
in a cyclic process, expels a fraction of that energy by means of
work.

2. The working substance absorbs
energy by heat from a high
temperature energy reservoir (Qh)
HEAT ENGINE
Working Principle
 Energy is expelled as heat to a
lower temperature reservoir
(Qc)
 Work done is by the engine
(Weng)

3.  The work done by the engine
equals the net heat energy
absorbed by the engine
HEAT ENGINE
Working Principle
Since it is a cyclical process, 0
in
E
 
 Its initial and final internal
energies are the same
eng net h c
W Q Q Q
  
From the first law 

4. Thermal Efficiency of a Heat Engine
eng
1
h c c
h h h
W Q Q Q
e
Q Q Q

   
HEAT ENGINE
Thermal efficiency is defined as the ratio of the net work done by
the engine during one cycle to the energy input at the higher
temperature
We can think of the efficiency as the ratio of what you gain to what
you give

5. Thermal Efficiency of a Heat Engine
HEAT ENGINE
In practice, all heat engines expel only a fraction of the input energy
by mechanical work. Therefore, their efficiency is always less than
100%
 To have e = 100%, QC must be zero  impossible!
Example: An engine transfers 2 x103 J of energy from a hot
reservoir during a cycle and transfers 1.50 x 103 J as exhaust to a
cold reservoir.
3
eng
3
1.50 x 10
1 0.25 25%
2.00 x 10
h
W
e
Q
    

6. Second Law  Kelvin-Planck Statement
AN IMPOSIBLE HEAT ENGINE
It is impossible to construct a heat
engine that, operating in a cycle,
produces no effect other than the
input of energy by heat from a
reservoir and the performance of an
equal amount of work
 Weng can never be equal to Qh
 Means that Qc cannot equal zero
 Some Qc must be expelled to the
environment
 Means that efficiency (e) cannot
equal 100%

7. HEAT PUMPS AND REFRIGERATORS
In a heat engine, the direction of energy transfer is from the hot
reservoir to the cold reservoir, which is the natural direction.
What if we wanted to transfer energy from the cold reservoir to the
hot reservoir?
Since it is not the natural direction of energy transfer, we must put
some energy (work) into a device to be successful.
Examples
A refrigerator is a common type of heat pump
An air conditioner is another example of a heat pump
 Devices that do this are called heat pumps or refrigerators
Therefore, a heat pump or a refrigerator is the reverse of a heat
engine.

8. Working
Principle
HEAT PUMP REFRIGERATOR
or a

9. It is impossible to construct a
cyclical machine whose sole effect
is to transfer energy continuously by
heat from one object to another
object at a higher temperature
without the input of energy by
work
OR
The energy does not transfer
spontaneously by heat from a cold
object to a hot object
Second Law  Clausius Statement
AN IMPOSIBLE HEAT PUMP

10. Coefficient of Performance
The effectiveness of a heat pump is described by a number called the
coefficient of performance (COP)
HEAT PUMP
For a heat pump operating in the cooling mode, “what you gain”
is energy removed from the cold reservoir to work done on the
pump.
energy transferred at low temp
COP
work done on the heat pump
c
Q
W
 
A good refrigerator should have a high COP (Cooling mode)
 Typical values are 5 or 6
COP, Cooling Mode

11. energy transferred at high temp
COP =
work done by heat pump
h
Q
W

Qh is typically higher than W  values of COP (heating mode) are
greater than unity
The use of heat pumps that extract energy from the air are most
satisfactory in moderate climates
In heating mode, the COP is the ratio of the energy transferred to
the hot reservoir to the work required to transfer that energy.
COP, Heating Mode
Coefficient of Performance
HEAT PUMP

12. Example:
A certain refrigerator has a COP of 5.00. When the refrigerator is
running, its power input is P = 500 W.
A sample of water of mass 500 g and temperature 20.0°C is placed
in the freezer compartment.
How long does it take to freeze the water to ice at 0°C?
Specific Latent heat of fusion of ice Lf = 3.33  105 J kg1
Assumptions:
All other parts of the refrigerator stay at the same temperature and
there is no leakage of energy from the exterior.
Coefficient of Performance

13. Solution:
The time interval t required for the freezing process
to occur:
W
t
P
 
P COP
f
mc T L m
t
  
  
 
 
1 0 1 0 5 1
0.5 kg 4186 J kg C x 20 C 3.33 x 10 J kg
500 W (5)
t
  
 
 
 
  
83.3 s
t
 

14. Efficiency of a Carnot Engine
1 c
h
Q
e
Q
 
We defined the thermal efficiency of a heat
engine earlier as
Temperatures must be in Kelvins
All Carnot engines operating between the same two temperatures
will have the same efficiency
c c
h h
Q T
Q T
 
For a Carnot cycle we showed that
1 c
c
h
T
e
T
 
The thermal efficiency of a Carnot engine
The equation can be applied to any working substance operating in
a Carnot cycle between two heat reservoirs.

15. Sadi Carnot
(1796 – 1832)

16. If Th = Tc  Efficiency is zero
If Tc = 0 K  Efficiency is 100%. Such reservoirs are not
available. Therefore, the efficiency is always less than 100%
1 c
c
h
T
e
T
 
The thermal efficiency of a Carnot engine
The efficiency increases as Tc is lowered and as Th is raised
In most practical cases, Tc is near room temperature, 300 K, so
generally Th is raised to increase efficiency
This experience has lead to the formulation of the THIRD LAW OF
THERMODYNAMICS, which can be stated as follows:
“It is impossible by any procedure, no matter how idealized, to
reduce any system to absolute zero of temperature.”

17.  Theoretically, a Carnot-cycle heat engine can run in reverse
 This would constitute the most effective heat pump available
 This would determine the maximum possible COPs for a given
combination of hot and cold reservoirs
c c
C
h c
Q T
COP
W T T
 

In cooling mode:
C
h h
h c
Q T
COP
W T T
 

In heating mode:
 As the difference between the temperatures of the two reservoirs
approaches zero in this expression, the theoretical COP
approaches infinity.
 In practice, the low temperature of the cooling coils and the high
temperature at the compressor limit the COP to values below 10.
Carnot Heat Pump and Coefficient of Performances
(COPs)

18. Carnot’s theorem
 No real heat engine operating between two energy reservoirs
can be more efficient than a Carnot engine operating
between the same two reservoirs
 All real engines are less efficient than a Carnot engine because
they do not operate through a reversible cycle
 The efficiency of a real engine is further reduced by friction,
energy losses through conduction, etc.

19. THE IDEAL OTTO CYCLE
The Otto cycle is the ideal cycle for
spark-ignition engines, in honor of
Nikolaus Otto, who invented it in
1867.
The ideal Otto cycle consists of four
internal reversible processes:
A  B Adiabatic compression
B  C Constant volume heat addition
C  D Adiabatic expansion
D  E Constant volume heat rejection

20. – Intake Stroke
The intake valve
opens, (exhaust
valve is closed) and
the air-fuel mixture
enters the cylinder
at atmospheric
pressure.
O → A
Energy enters the system by matter transfer
 potential energy stored in the fuel

21. – Compression Stroke
The work done on the gas is negative
A → B
The inlet valve closes,
piston moves upward
The temperature TA  TB (increases)
1 1
A A B B
T V T V
 
 

The air-fuel mixture is
compressed adiabatically
from volume V1  V2

22. Spark
B → C
Combustion occurs when
the spark plug fires
It occurs very quickly
while the piston is at its
highest position
Potential energy of the fuel  internal energy
The temperature increases rapidly TB  TC
Volume remains approximately constant because of the short time
interval  no work is done
 
h V C B
Q nC T T
 
Energy enters the system as heat

23. Power Stroke
C → D
The gas expands adiabatically
from volume V2  V1
1 1
C C D D
T V T V
 
 

The temperature TC  TD (decreases)
The work done by the gas is used to push the piston downward.

24. Exhaust Valve Opens
D→ A
An exhaust valve opens as the piston
reaches its bottom position
The pressure drops suddenly for a short
period
 
c V D A
Q nC T T
 
Energy leaves the system as heat
The temperature decreases rapidly TD  TA
Volume remains approximately constant because of the short time
interval  no work is done

25. Exhaust Stroke
A→ O
The piston moves
upward while the
exhaust valve remains
open.
Residual gases are expelled to the atmosphere.
Volume decreases
V1  V2
The cycle repeats again.

26. The Gasoline Engine –Complete Cycle

27. In the constant volume heat addition and heat rejection process, no
work is done. The energy balances for these two processes are:
 
h V C B
Q nC T T
 
Energy enters the system as heat
B→ C 
 
c V D A
Q nC T T
 
Energy leaves the system as heat
D→ A 
h c
net
otto
in h
Q Q
W
e
Q Q

 
The thermal efficiency for an ideal Otto cycle
1 c
otto
h
Q
e
Q
 
1 1
c D A
otto
h C B
Q T T
e
Q T T
 

    

 
The thermal efficiency for an ideal Otto cycle

28. 1
A D
V V V
  2
B C
V V V
 
and
1
B
A B
A
V
T T
V
 
 
  
 
1
C
D C
D
V
T T
V
 
 
  
 
1
2
1
A B
V
T T
V
 
 
  
 
1
2
1
D C
V
T T
V
 
 
  
 
No heat transfer is involved in the two
adiabatic processes.
1 1
C C D D
T V T V
 
 

C→ D 
1 1
A A B B
T V T V
 
 

A→ B 
1
2
1
D A
C B
T T V
T T V
 
 

  
  

29. The Otto cycle efficiency is only a function of the compression
ratio.
1
2
1
1
otto
V
e
V
 
 
  
 
Otto Cycle Efficiency
We assume that the air-fuel mixture behave as an ideal gas
 is the ratio of the molar specific heats and r = V1 / V2 is called the
compression ratio

30.  is the ratio of the molar specific heats
and r = V1 / V2 is called the
compression ratio
Typical values:
Compression ratio r = 8 and
 = 1.4
 eotto = 56% (ideal)
Efficiencies of real engines are
15% to 20%
Low efficiencies are mainly due to friction, energy transfer by
conduction, incomplete combustion of the air-fuel mixture

31. Theoretically, a higher r can generate a higher thermal efficiency.
It will cause the temperature of the air-fuel mixture to rise
above the auto ignition temperature of the fuel during the
compression process, and will cause an early and rapid burn
before the spark ignition.
This early and rapid burn produces an audible noise, which is
called “Engine knock”.
Engine knock in spark-ignition engine can cause engine damage.
Thus there is an upper limit of compression ratio for spark-
ignition engines
The Gasoline Engine –Practical situation

32. The Diesel cycle
In the 1892, Rudolf Diesel invented an efficient, compression
ignition, internal combustion engine that bears his name
Operate on a cycle similar to the Otto cycle without a spark plug
The compression ratio is much greater (r = 20) and so the cylinder
temperature at the end of the compression stroke is much higher
Fuel is injected and the
temperature is high enough for
the mixture to ignite without the
spark plug
Diesel engines are more
efficient than gasoline engines
Theoretical efficiency:
1
1 D A
diesel
C B
T T
e
T T

 

   

 