Theory:

6.1: Heat is the energy associated with the random motion of particles, while work is the
energy of ordered motion in one direction.
6.2: A cyclic heat engine is a type of heat engine that operates on a cyclic process,
meaning it undergoes a series of thermodynamic processes that return it to its initial state.
This allows the engine to perform continuous work as it cycles through these processes.
6.3: A heat engine cycle in a closed system involves cyclic energy conversion where heat
is taken in, converted to work, and rejected as waste heat. It is fundamental to many
power generation and propulsion systems, with different cycle types optimized for
specific applications.
6.4: A heat engine cycle in a steady-flow system continuously processes a working fluid
through compressors, turbines, heat exchangers, and other components to convert heat
energy into work. The Brayton cycle (gas turbines) and Rankine cycle (steam power
plants) are common examples. These cycles are widely used in power generation and
propulsion systems.
𝑊
6.5: The thermal efficiency, η of a heat engine is defined W as 𝜂 = , where W is the
𝑄𝐻
work done (output) by QH the working substance and QH is the amount of heat absorbed
(input) by it.
No, a heat engine cannot have 100% thermal efficiency unless the cold reservoir is at
absolute zero (0 K), which is impossible in practice. Therefore, no real heat engine can
convert all the input heat into useful work.
6.7: A thermal energy source supplies heat to a system, whereas a thermal energy sink
absorbs heat from a system.
reservoir that supplies energy in the form of heat is called a source and one that absorbs
energy in the form of heat is called a sink
6.8: Mechanical energy reservoir is a large body enclosed by an adiabatic impermeable
wall capable of storing work as Potential or Kinetic energy.
6.10: It states that “It is impossible to construct a device which operates on a cycle and
produces no other effect than the transfer of heat from a single body to produce work.”
6.11: A heat engine produces work from heat, operating in cycles. By Kelvin's statement
of the 2nd Law, that cannot be achieved with a single thermal reservoir: at least two
reservoirs at different temperatures are needed.
6.13: It states that “It is impossible to construct a device that operates in a cycle and
produces no effect other than the transfer of heat from a lower-temperature body to a
higher-temperature body.”
Exercise:
6-1: Q1 = 105 MJ, T1 = 400K, Q2 = 42 MJ, T2 = 200K, W = 15kWh
maximum thermal efficiency of his engine possible
η = 1 – 200/400 = 50 % 𝜂 = 1 − 200/400 = 50 %
That engine and deliver output = η × input = 0.5 × 105 MJ = 52.5 MJ = 14.58 kWh
As he claims that his engine can deliver more work than ideally possible so I would not
advise to investing money.
6-2: COP = 5, 𝜂𝑚𝑜𝑡𝑜𝑟 = 90% = 0.9, Q1 = 1 kW
COP = desired effect / input
(COP)ref = (COP)H.P – 1
 6 = H/W => (COP)H.P = 6
So input (W) = H/6
But motor efficiency 90% so Electrical energy require (E)
𝑊 𝐻
(E) = = = 0.1852 H = 18.52% of Heat (direct heating)
0.9 0.9×6
100
H= = 5.4kW
18.52

6-3: (COP)ref = 5, η = 30% = 0.3, Q1 = ?

(COP)ref = 5
So for each MJ removed from the cold body we need work = 1 MJ / 5 = 200 kJ
For 200 kJ work output of heat engine hair η = 30%
We have to supply heat = 200 kJ / 0.3 = 666.67 kJ
Now COP of H.P. = COP of Ref. + 1 = 5 + 1 = 6
Heat input to the H.E. = 1 MJ
Work output (W) = 1 × 0.3 MJ = 300 kJ
That will be the input to H.P.
 𝑄1
(COP)H.P =
𝑊

And Q1 = (COP)H.P × W = 6 × 300 kJ = 1.8 MJ

6.4: process 1-2: W1-2 = 2.8 kWh = 10080 kW, Q1-2 = 732 kJ, process 2-1: W2-1 = 2.4
kWh = 8640 kW
From the first Law of thermodynamics
(a) For process 1–2
Q1-2 = E2 - E1 + W1–2 <=> - 732 = (E2 - E1) – 10080 [2.8 kWh = 2.8 x 3600 kJ]
=> E2 – E1 = 9348 kJ
For process 2–1
Q2-1 = E1 – E2 + W2-1 = - 9348 + 8640 = – 708 kJ i.e. Heat flow out to the atmosphere.
(b) Yes Second Law limits the maximum possible work. As Electric energy stored in a
battery is High grade energy so it can be completely converted to the work.
Then, W = 9348 kJ
(c) Q21 = –9348 + 9348 = 0 kJ
6-5:

275
Ideal COP of Ref. = = 9.82143
30−2

Actual COP = 0.15 × COPideal = 1.4732

Heat to be removed in a day Q2 = 420 × 20 kJ = 8400 kJ
Work required = 5701.873 kJ/day = 1.58385 kWh/day
Electric bill per month = 1.58385 × 0.32 × 30 Rupees = Rs. 15.20
6-6:

𝑇𝐻 333
(COP)H.P = = = 6.05454
𝑇𝐻 − 𝑇3 333− 278

𝑊𝐻.𝑃 + 17
Q3 = WH.P + 17 => = 6.05454 => WH.P = 3.36 kW
𝑊𝐻.𝑃

WH.E = 30 + 3.36 = 33.36 kW

𝑇𝐻 333
𝜂𝐻.𝐸 = 1 − =1− = 0.7
𝑇1 1113
𝑊 𝑊 33.36
a/ => = 0.7  Q1 = = = 47.66 kW
𝑄1 0.7 0.7

b/ From H.E = Q1 – W = 47.66 – 33.36 = 14.3 kW

For H.P = 17 + 3.36 = 20.26 kW
=> total = 34.56 kW
6-7: T1 = - 5˚C = 268K, T2 = 25˚C = 298K, QC = 5 kW.
𝑇1 268
COPcarnot = = = 8.93
𝑇2 − 𝑇1 298− 268

COPact = 0.4 × COPcarnot = 0.4 × 8.93 = 3.572

𝑄𝐶 𝑄𝐶 5
COP = => 𝑊𝑖𝑛𝑝𝑢𝑡 = = ≈ 1.4 kW
𝑊𝑖𝑛𝑝𝑢𝑡 𝐶𝑂𝑃 3.572

QH = QC + Winput = 5 + 1.4 = 6.4 kW

6-8: COPH.P = 4, ηH.E = 27% = 0.27
 𝑄2 𝑄2 𝑄2
𝜂𝐻.𝐸 = 1 − <=> 0.27 = 1 − <=> 0.73 =
𝑄1 𝑄1 𝑄1

<=> 0.73Q1 = Q2
𝑄4 𝑄4
COPH.P = <=> 4 = <=> 4W = Q4
𝑊 𝑊

W = Q1 – Q2 = Q1 – 0.73Q1 = 0.27Q1
=> 4 × 0.27Q1 = Q4 <=> 1.08Q1 = Q4
𝑄2 + 𝑄4 0.73𝑄1 + 1.08𝑄1
=> = = 1.81
𝑄1 𝑄1

6-9: Q1 = 20 kJ, T1 = 100˚C, Q2 = 14.6 kJ, T2 = 0˚C

Let 𝑓 (𝑡 ) = 𝑎𝑡 + 𝑏
𝑄1 𝑓(𝑡1 ) 20 100𝑎+𝑏 100𝑎 𝑎
=> = )
=> = = + 1 => = 3.6986 × 10−3
𝑄2 𝑓(𝑡2 14.6 𝑏 𝑏 𝑏

For absolute zero => Q2 = 0

𝑄1 100𝑎+𝑏 −𝑏 −1
=> = or at + b = 0 <=> t = = = - 270.37˚C
0 𝑎𝑡 + 𝑏 𝑎 3.6986×10−3

6-10: TH = 421˚C = 694 K, Q1 = 200 kJ, TC = 4.4˚C = 277.4 K

 𝑄1 𝑄2 𝑄1 − 𝑄2 𝑄3 𝑄2 − 𝑄3
= = = =
694 𝑇 694 − 𝑇 277.4 𝑇 − 277.4
With Q1 – Q2 = 2W2 and Q2 – Q3 = W2
2 1
=> = <=> 2𝑇 − 554.8 = 694 − 𝑇
694 − 𝑇 𝑇 − 277.4
=> 𝑇 = 416.27 K = 143.27˚C
𝜂1 = 40%
277.4
𝜂2 = 1 − = 33.36%
416.27
6-11:

333
Carnot efficiency η = 1 – = 0.647
944
 0.647 𝑄1′ 𝑄1′
=> Actual efficiency η = = 0.3236 = 1 - => = 0.6764
2 𝑄1 𝑄1

305.2
Ideal COP = =7.866
305.2 -266.4
7.866 𝑄3
Actual COP = = 3.933 =
2 𝑊

If Q3 = 1 kJ
𝑄3 1
𝑤= = = 0.254
3.933 3.933
W = Q1 – Q1’ = Q1 – 0.6764
0.254
𝑄1 = = 0.78 kJ/kJ heat input
1 – 0.6764

6-12:

273
Maximum (COP) = = 15.167
291 – 273

𝑄 𝑄 1000 × 333.5
= 15.167 <=> 𝑊𝑚𝑖𝑛 = = = 21.988 𝑀𝐽 = 6,108 𝑘𝑊ℎ
𝑊𝑚𝑖𝑛 15.167 15.167
6-13:
η of H.E. between A and C
𝑇𝐶
𝜂𝑒𝑛𝑔𝑖𝑛𝑒 = 𝛼 (1 − )
𝑇𝐴
𝑄1 × 𝑇𝐶 𝑄1 × 𝑇𝐶
𝑄2 = = 𝑄3 =
𝑇𝐴 𝑇𝐵
Total heat rejection:
 1 1
𝑄2 + 𝑄3 = 𝑄1 × 𝑇𝐶 × ( + )
𝑇𝐴 𝑇𝐵

Total heat input = 2Q1

1 1
𝑄1 × 𝑇𝐶 × ( + )
𝑇𝐴 𝑇𝐵
𝜂𝑒𝑛𝑔𝑖𝑛𝑒 = (1 − )
2𝑄1

𝛼𝑇𝐶 𝑇𝐶 𝑇𝐶
=> 𝛼 − =1 − −
𝑇𝐴 2𝑇𝐴 2𝑇𝐵
Multiply both side by TA and divide by TC
𝑇𝐴 𝑇𝐴 1 𝑇𝐴 𝑇𝐴 𝑇𝐴
=> 𝛼−𝛼 = − − => − 𝛼 = (2𝛼 − 1) + 2(1 − 𝛼)
𝑇𝐶 𝑇𝐶 2 2𝑇𝐵 𝑇𝐵 𝑇𝐶
6-14:
a/ If the engines produce the same amount of work output
𝑄1 𝑄2
So W1 = W2 <=> η1Q1 = η2Q2 With =
𝑇1 𝑇

𝑇1 𝑄2 𝑇 𝑇1 𝑇2 𝑇 𝑇2
=> 𝑄1 = => (1 − ) ( ) 𝑄2 = (1 − ) 𝑄2 => 1 − =1 −
𝑇 𝑇1 𝑇 𝑇 𝑇1 𝑇
𝑇1 + 𝑇2 𝑇1 + 𝑇2
=> 2 = => 𝑇 =
𝑇 2
b/ f the engines have the same cycle efficiencies
𝑇 𝑇2
1 − = 1 − => 𝑇 = √𝑇1 𝑇2
𝑇1 𝑇
(as T is + ve so – ve sign neglected)
6-15:

a.b/ As their efficiency is same so ηA = ηB

𝑇 100
1 − =1 − => T = 316.23 K
1000 𝑇
T𝑄1 1680 × 316.23
𝑄2 = = = 531.27 kJ
1000 1000
𝑄2 ×100 531.27×100
𝑄3 = = = 168 kJ
316.23 316.23

c/ WA = Q1 – Q2 = 1880 – 531.27 = 1348.73 kJ

WB = Q2 – Q3 = 531.27 – 168 = 363.27 kJ
d/ If the equal work
100 + 1000
=> T = = 550 K
2
𝑇𝑄1 1680 × 550
=> 𝑄2 = = = 924 kJ
1000 1000
e/
550
𝜂2 = 1 − = 0.45
1000
100
𝜂2 = 1 − = 0.82
550
6-16:

a/ Estimated Heat rate = 0.525 × (20 – 5) = 7.875 kJ/s

293
COP = = 19.53
293 - 278
Q̇ 7.875
Wmin = = = 0.403 kW = 403 W
(COP)max 19.53
b/ Ẇ = 403 W
Q1 = 0.525 (T – 293) kW = 525 (T – 293) W
525 (T – 293) 293
COP = = => T = 308 K = 35℃
403 (T – 293)
Tmax = 35˚C
6-17:

Heat have to radiate = Q2

=> Q2 = 𝜎AT24 (1)
Engine side:
𝑄1 𝑄2 W W𝑄2
= = ⇒ 𝑇2 = (2)
𝑇1 𝑇2 𝑇1 - 𝑇2 𝑇1 - 𝑇2
From (1) and (2):
W𝑄2 W 𝑇2 W 1
= σAT24 => A = 4 ( ) = 4( )
𝑇1 - 𝑇2 σT2 𝑇1 - 𝑇2 σT2 𝑇1 T23 - T24
For minimum area:
𝜕A 𝜕
= 0 <=> (𝑇1 T23 - T24 ) = 0
𝜕𝑇2 𝜕𝑇2
𝑇2 3
=> 𝑇1 × 3T22 - 4T23 = 0 <=> 3𝑇1 = 4𝑇2 <=> =
𝑇1 4
6-18:
 10 293
(COP)H.P = = => W = 0.6826 kW
W 293 - 273
6-19:

Suppose A is any refrigerator and B is reversible refrigerator and also assume

(COP)A > (COP)B and Q1A = Q1B = Q
𝑄1A 𝑄1B Q Q
> <=> > <=> WB > WA
WA WB WA WB
Then we reversed the reversible refrigerator ‘B’ and then work output of refrigerator ‘B’
is WB and heat rejection is Q1B = Q (same)
So we can directly use Q to feed for refrigerator and Reservoir ‘T2’ is eliminated then
also a net work output (WB – WA) will be available. But it violates the Kelvin Plank
statement ( Second Law of thermodynamic) so our assumption is wrong.
So (COP)R ≥ (COP)A
6-20:
Q̇ H = Q̇ leak = 25 kW
Q̇ H 𝑇H 293.2
𝛽' = = = = 9.773
Ẇin 𝑇H - 𝑇L 30
25
=> Ẇin = = 2.56 kW
9.773