FME 321 Lecture 5
The Carnot Cycle
This hypothetical heat engine cycle consists of two isothermal processes joined together by two
adiabatic processes. All its processes are reversible; hence the adiabatic processes are also isentropic.
The cycle is most conveniently represented on a TS diagram.
1 - 2 Isentropic expansion
2 - 3 Isothermal heat rejection
3 - 4 Isentropic compression
4 - 1 Isothermal heat supply.
Thermal efficiency of a heat engine is defined by
𝜂 =1−
𝑄2
𝑄1
With reference to the TS diagram, the
Heat supplied = Q1 = area 41BA4
= T1 (SB –SA)
Heat rejected
= Q2 = area 23AB2
=T2 (SB – SA)
Therefore
𝜂𝑐𝑎𝑟𝑛𝑜𝑡 = 1 −
=1 −
T2 (S𝐵 – S𝐴 )
T1 (S𝐵 – S𝐴 )
𝑇2
𝑇1
The Reversed Carnot cycle
The figure above shows the heat engine cycle in reverse, quantities Q1, Q2 and W having opposite
directions. The effect of the reversed heat engine is to transfer a quantity of heat Q1 from a cold source
at temperature T1 and reject a quantity Q2 at temperature T2.
The reversed heat engine fulfills the requirements of a refrigerator and a heat pump. In the case of a
refrigerator the important quantity is Q1 and for a heat pump it is the quantity Q2.
4 - 1 = Isothermal heat transfer at low temperature
𝑄41 = 𝑇1 (𝑆1 − 𝑆4 )
1 - 2= Isentropic compression from T1 to T2
Q12 = 0
2 - 3= Isothermal heat transfer at high temperature
𝑄23 = 𝑇2 (𝑆1 − 𝑆4 )
3 - 4 = Isentropic expansion from T3 to T4
Q34 = 0
For the cycle;
Heat received at low temperature
= Refrigeration effect
= 𝑇1 ( 𝑆1 − 𝑆 4 )
And for the cycle
∮𝑊=∮𝑄
Or Net work = Heat received – Heat rejected.
Therefore Net work = 𝑇1 (𝑆1 − 𝑆4 ) − 𝑇2 (𝑆1 − 𝑆4 )
=− (𝑇1 − 𝑇2 )(𝑆1 − 𝑆4 )
-ve sign shows that work must be supplied in order to effect the processes of the cycle.
Therefore
C. O. P =
𝑅𝑒𝑓𝑟𝑖𝑔𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑒𝑓𝑓𝑒𝑐𝑡
𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
=(
=
𝑇1 (𝑆1 −𝑆4 )
𝑇2 −𝑇1 )(𝑆1 −𝑆4 )
𝑇1
𝑇2 −𝑇1
C.O.P = Coefficient of Performance
Refrigerator and heat pump performances are defined by means of C.O.P. as follows:
C.O.P. refrigerator =
C.O.P. heat pump =
𝑄1
𝑊
𝑄2
𝑊
=
=
𝑄1
𝑄2−𝑄1
𝑄2
𝑄2−𝑄1
From the T-S diagram the areas are proportional to the heat quantities, therefore:
Q1 = T1 (S1-S2)
Q2 = T2 (S2-S3)
= T2 (S1-S4)
Therefore C.O.P.r =
𝑇1(𝑆1−𝑆4)
(𝑇2−𝑇1)(𝑆1−𝑆4)
And
C.O.P.hp =
And
C.O.P.r =
𝑇2(𝑆1−𝑆4)
(𝑇2−𝑇1)(𝑆1−𝑆4)
𝑇1
𝑇2−𝑇1
C.O.P.hp =
𝑇2
𝑇2−𝑇1
These equations give the maximum possible values of C.O.P. r and C.O.P. hp between given values of T1
and T2, the temperatures of the refrigerant in the evaporator and condenser coils respectively.
Example 4.1
2000 kJ of heat is transferred from a reservoir at 3000C to an engine that operates on the Carnot cycle.
The engine rejects heat to a reservoir at 230C. Determine the thermal efficiency of the cycle and the
work done by the engine.
Example 4.2
The following table consists of energy transfers in three hypothetical heat engines cycles. In each case
the temperature of the source and sink are 527oC and 27oC respectively. Using Clausius inequality
classify each cycle as either reversible, impossible or irreversible.
Rate of heat supply kW
Rate of Heat Rejection
kW
a
291
70
221
75.9
b
291
198
93
32
c
291
109.125
181.9
62.5
Cycle
Solution:
Rate of Work
W
Efficiency
%
Clausius inequality is ∮
𝛿𝑄
𝑇
≤0
Therefore the cyclic integral is performed for each cycle.
𝛿𝑄
𝑇
∮
i.e.
=∫
=
𝛿𝑄1
𝑇1
𝑄1
𝑇1
𝛿𝑄2
𝑇2
+∫
+
𝑄2
𝑇2
For case (a)
𝛿𝑄
𝑇
∫
=
291
800
−
70
300
= + 0.13
Case (b)
𝛿𝑄
𝑇
∮
=
291
800
−
198
300
= - 0.296
Case (c)
∮
𝛿𝑄
𝑇
=
291
109.125
−
800
300
= 0
According to Clausius inequality
∮
∮
∮
𝛿𝑄
𝑇
𝛿𝑄
𝑇
𝛿𝑄
𝑇
= 0 Cycle is reversible
< 0 Cycle is irreversible
> 0 Cycle is impossible
Example 4.3
A refrigerator that operates on a Carnot cycle is required to transfer 33 kW from a reservoir at 00C to the
atmosphere at 270C. What is the work input?
Example 4.4
An inventor claims to have developed a refrigeration unit which maintains the refrigerated space at 230C while operating in a room where the temperature is 270C and has a COP of 5.5
a) How do you evaluate his claim?
b) How would you evaluate his claim of a COP of 5.0?
Example 4.5
An office block is heated by means of a heat pump. The air temperature within the building is 200C and
the outside air temperature is -40C. The rate of heat transfer to the heat pump is 30 kW and the power
to drive the pump is 10 kW
Determine:
i)
ii)
iii)
iv)
The rate of heat transfer to the building
The coefficient of performance
The maximum possible COP of the heat pump
Minimum power input to the heat pump.