THERMODYNAMICS Practice Workbook

First Law – Non Flow Process /50

Example 4
The pressure of a perfect gas is 4.6 bar when 0.04 m3 at 31 oC. Determine the mass of gas if R is
valued at 0.287 kJ/kgK. The gas pressure increased to 11.3 bar while the volume is constant.
Calculate the final temperature and heat supplied to the gas.

V1 = 0.04m3
p1 = 4.6bar
p2=11.3bar
T1=31oC
V1=V2
T2=?
V2 = 0.32 m3
T2=?

Example 5
A rigid tank contain air at 500kPa and 150 oC. As a result of heat transfer to the surroundings,
the temperature and pressure inside the tank drop to 65 oC and 400kPa, respectively. Determine
work done during this process,

3.4.3 Isothermal Process, First Law – Non Flow Process /51

j if during the process no temperature change
or approaching 0 it can be considered
isothermal process .

The properties relationship between the
beginning and end of the perfect gas can be
written as when, T1 = T2 and the above

relation becomes p1V1  p2V2

Constant temperature (Isothermal) process
From the equation

an increase volume, decrease pressure.
an isothermal process, the pressure is
inversely proportional to the volume.

Work transfer: Heat transfer:
U2  U1  0
W = p1V1 ln V2 = p1V1 ln V2
V1 V1 Q W

Thus, for a perfect gas, all the heat added during a constant temperature process is converted into
work and the internal energy of the system remains constant.

Example Questions

Example 6
A quantity of a certain perfect gas is heated at a constant temperature from an initial state of 0.22
m3 and 325 kN/m2 to a final state of 170 kN/m2. Calculate the final pressure of the gas.

From the question,
V1 = 0.22 m3
V2 = ?
P1 = 325 kN/m2
P2 = 170 kN/m2
T1 = T2

First Law – Non Flow Process /52

Example 7
In an industrial process, 0.4 kg of oxygen is compressed isothermally from 1.01 bar and 22o C to
5.5 bar. Determine the work done and the heat transfer during the process. Assume that oxygen is
a perfect gas and take the molecular weight of oxygen to be M = 32 kg/kmole.

[Ans :Q = W = 52kJ]

Example 8
berdasarkan rajah dibawah, kirakan kerja terlaku ( W12 ) untuk bendalir 3kg

First Law – Non Flow Process /53

3.4.4 Polytrophic Process (pVn = C)

If the process will be isothermal. In this case the index n = 1
If the process will be adiabatic. In this case the index n = .

By the manipulation of equations the

following relationship can be determined:

n1 n1

T2  p2  n VV21 
T1  p1  
    

By examining equations the following
conclusions for a polytropic process on a
perfect gas can be drawn as:

An increase in volume , decrease in pressure.
An increase in volume, decrease temperature.
Formulae:

Work transfer: Heat transfer:

2 = p1V1  p2V2 Q  U 2  U1  mCv(T2  T1 )
n 1 , Q    n W
W  pdV
 1
1

W  p1v1  p2v2  mR(T2  T1)
n 1 n 1

Example questions

Example 9
0.9kg gas sempurna pada tekanan bar 15 dan 250 ° C berkembang secara polytropic
kepada 1.5 bar. Hitung suhu akhir, kerja dan haba yang dipindahkan apabila indeks
pengembangan adalah 1.25 [Ans :330K, 199kJ, 74 kJ ]

First Law – Non Flow Process /54

3.4.5 Adiabatic Process
The system is thermally well insulated Q ≈ 0.

For example, the outer casing of steam engine, steam turbines and gas turbines are well insulated
to minimize heat loss. The fluid expansion process in such machines may be assumed to be
adiabatic.

For a perfect gas the equation for an
adiabatic process is pV = C

Cp
ratio of specific heat,  = C v

The above equation is applied to states 1

and 2 as: p1V1  p2V2

By manipulating equations, the following
relationship can be determined

 1  1

T2  p2   VV21 
T1  p1 
   

An increase in volume , decrease pressure.
An increase volume, decrease temperature.
An increase pressure, increase temperature.

Work transfer:

2 p1V1  p2V2

W   pdV =  1
1

W  p1v1  p2v2  mR(T2  T1)
 1  1

Heat transfer: In an adiabatic process, Q = 0.

Referring to the process represented on the p-V diagram it is noted that during adiabatic

process the volume increases and the pressure decreases.

First Law – Non Flow Process /55

Example question
Example 11

Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015 m3, is compressed reversibly
and adiabatically by a piston to a pressure of 6.8 bar. Calculate the final temperature, the final
volume, and the work done on the mass of air in the cylinder.

Data: p1=1.02 bar; T1=22 + 273 = 295 K;
v1= 0.015 m3; p2= 6.8 bar T2= ?, v2= ?, W = ?

From equation

 1 For an adiabatic process,
W = u1 – u2
T2  p2   for a perfect gas,
T1  p1  W = cv(T1- T2)= 0.718(295-507.5)
  
= - 152.8 kJ/kg
T2 = 295 x  6.8 (1.41) /1.4 = 507.5 K i.e. work input per kg = 152.8 kJ
1.02 
The mass of air can be found using equation
(where  for air = 1.4) pV = mRT
m  p1v1  1.02x105 x0.015  0.018kg
Final temperature
= 507.5 – 273 RT1 0.287x103195

= 234.5oC Total work done
= 0.0181 x 152.8 = 2.76kJ
p2  VV12  or v1   p2 1/ 
p1  v2  p1 
  

0.015   6.8 1/1.4 v2 = 0.0038 m3
v2 1.02 

Practice 3.2

1. 0.046 m3 of perfect gas are contained in a sealed cylinder at a pressure of 300kN/m2 and a
temperature of 45 oC. The gas is compressed to 1270kN/m2 and the temperature is 83oC.
, determine:
a) the mass of gas (kg)
b) the final volume of gas (m3)
Given: R = 0.29 kJ/kg K

First Law – Non Flow Process /56

2. A gas initially at a pressure of 1.4 bar, volume 0.1m3 and a temperature of 250oC is
compressed until the pressure is 7 bar and 600oC. Calculate the new volume of the gas

3. 0.5kg of air and 0.02m3 volume expanded reversibly at constant pressure of 5.0bar to

170oC and 0.05m3 volume. Sketch the pv diagram and calculate the amount of heat

transfer. [ Ans :Q = 9.16kJ]

4. Sketch the p-v diagram and calculate the work done and the heat transferred when 1kg
of air is compressed at constant pressure from 10bar and temperature 115oC to 250 oC.

[Ans : W =38.7kJ, Q = 136kJ]

First Law – Non Flow Process /57

5. 5kg of air is heated in a rigid container of 385 oC and a 4bar pressure to a 14bar. Sketch
the pv diagram and get the amount of heat transferred.

6. 2kg of gas receive 200 kJ as heat at constant volume process. If the temperature of the
gas increases by 100 oC, determine the Cv of the process.
[Ans :0.268kJ/kJK]

7. A perfect gas is contained in a rigid vessel at 3 bar and 315 oC. The gas is then cooled
until the pressure falls to 1.5 bar. Calculate the heat rejected per kg of gas. Given:M = 26
kg/kmol and  = 1.26.
[Ans : Q =361.6 kJ/kg]

First Law – Non Flow Process /58

8. A quantity gas has an initial pressure of 1.4 bar and a volume of 0.14m3 the gas is
compressed to 7 bar while the temperature is constant. Calculate the final volume of gas.
[Ans : 0.028m3]

10’ 1 kg of nitrogen (M = 28) is compressible and isothermally is compressed from 1.01 bar,

; 200 oC to 4.2 bar. Calculate the work done and the heat flow during the process. Assume

nitrogen as perfect gas. [Ans :Q = W = -124 kJ/kg]

11 1kg gas sempurna pada tekanan 5.0bar dan isipadu 0.03m3 menjalani proses
isothermally reversible sehingga tekanan 1.0bar. Lakarkan gambarajah pv dan kirakan
haba mengalir dalam proses.
[Ans : Q = 24.1kJ]

First Law – Non Flow Process /59

1 1 kg of air at 1 bar, 15oC is compressed reversible and adiabatic to 4 bars pressure.

2 Calculate the final temperature and work done on the air if Cp = 1.005 kJ / kgK, Cv =

0.718 kJ / kgK. [Ans :155oC. -100.5 kJ/kg]

13 The air at 117kPa pressure and 240 oC temperature and 0.018m3 volume is adiabatically
compressed and reversed by a piston until the pressure is 7 bar. Determine the
temperature and final volume and work done on the mass of the air in the cylinder.

14 Volume of air 0.5 m3 fulled at 2 bar and temperature 27oC. It is compressed to a pressure
of 0.2m3 according to the law pV1.25 = constant. Determine the amount of heat energy

and work energy. Given Cv=0.7kJ/kgK.

[Ans :W= -102.9 kJ, Q = -33.57 kJ]

First Law – Non Flow Process /60

15 0.112 m3 of gas has a pressure of 138 kN/m2. It is compressed to 690 kN/m2 according
to the law pV1.4 = C. Determine the new volume of the gas. [Ans : V2 = 0.0348 m3]

16 Oxygen (M = 32) is heated at constant volume from 1.05bar, 15oC and 0.03m3 to 5bar
pressure. By assuming oxygen as a perfect gas, calculate:
a. gas mass
b. heat flowing
c. work
d. entropy change
[Ans : 0.042kg, 29.53kJ, 0, 0.04254kJ/K]

First Law – Non Flow Process /61

17 Nitrogen (M = 28) is in one cylinder of 0.03m3 at 1.05bar pressure and 15oC
temperature. This gas is compressed isothermally reversible until the pressure of 4.2bar.
calculate
a. Gas mass
b. Entropy change
c. Heat flow
d. Work done
Sketch the pv and Ts diagrams for the above process by assuming N2 as the perfect gas.
[Ans :- ( 0.0368kg, -0.01515kJ/K, -4.36kJ, -4.36kJ.
]

18 A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is compressed
adiabatically to 5 bar. Determine the following:
a. final temperature
b. final volume
c. work transfer
d. heat transfer
e. change in internal energy
[Ans : 222.7oC, 14230 cm3, 6.56 kJ input, 0 kJ, 6.56 kJ increase]

First Law – Flow Process /62

3.5 FLOW PROCESSES

In heat engine it is the steady flow processes which are generally of most interest. The
conditions which must be satisfied by all of these processes are :
i. The mass of fluid flowing past any section in the system must be constant with

respect to time.
ii. The properties of the fluid at any particular section in the system must be constant

with respect to time.
iii. All transfer of work energy and heat which takes place must be done at a uniform

rate.

section within the system must be constant with respect to time.

3.6 STEADY FLOW ENERGY EQUATION

This equation is a mathematical statement on the principle of Conservation of Energy as
applied to the flow of a fluid through a thermodynamic system.

The various forms of energy which the fluid can have are as follows:

gZ1  u1  p1v1  C12 Q  gZ 2  u2  p2v2  C 2 W
2 2

2

C2
g Z = Potential energy = Kinetic energy, U = Internal energy

2

p2v2 = Flow or displacement energy
Q = Heat received or rejected

W = External work done

First Law – Flow Process /63

The application of the principle of energy conservation to the system is,

Total energy entering the system = Total energy leaving the system, for unit mass of

substance,

gZ1  u1  P1v1  C12 Q  gZ 2  u2  P2 v 2  C 2 W
2 2

2

This is called the steady flow energy equation.

In equation was stated that the particular combination of properties of the form, h = u +
Pv. Thus, the steady flow energy equation is written as

gZ1  h1  C12 Q  gZ 2  h2  C 2 W
2 2

2

     h2  h1    C22  C12   Z 2  Z1  g  (kJ/s)
2 
Q W m 



qw h2  h1    C 2  C12   Z 2  Z  g 
2 2 
 (kJ/kg)
1 

 

qQ (kJ/kg) w  W (kJ/kg)

m 

m

3.7 APPLICATION OF STEADY FLOW EQUATION
Boilers
Condensers
Turbine
Nozzle
Throttle
Pump

Since h2 > h1, W will be found to be negative.

First Law – Flow Process /64

Equation of Continuity
Mass flowing through a system

̇ = ̇ i.e. A1C1  A2C2
v1 v2

Practice 3.3 First Law – Flow Process /65

Soalan 1
fluid enters the condenser at 35kg/s with the specific enthalpy 2200kJ/kg, and leaves with
specific enthalpy 255 kJ/kg. Determine the rate of heat loss from the system

[Jaw- 68 075 kJ/s]

Soalan 2
Bendalir mengalir melalui turbin pada kadar 0.75kg / s. dan enthalpi telah berkurang
580kJ/ kg dan kehilangan haba 2100 kJ / min . Tentukan kuasa yang dihasilkan oleh
turbin, dengan menganggap bahawa tenaga kinetik dan perubahan tenaga potensi boleh
diabaikan. [Jaw: 400 kW]
Solution

Soalan 3
A pump delivers fluid at the rate of 45 kg/min. At the inlet the specific enthalpy is 46 kJ/kg,
and at the outlet the specific enthalpy of the fluid is 175kJ/kg. If 105kJ/min of heat energy
are lost to the surroundings by the pump, determine the power required to drive the pump.

[Jaw:W = - 98.5 kW]

First Law – Flow Process /66
Soalan 4
Fluid with specific enthalpy 2800kJ/kg enters a horizontal nozzle with negligible velocity at
14 kg/s. At outlet the specific enthalpy nozzle and the specific volume of the fluid is
2250kJ/kg and 1.25m3/kg. Assuming the flow is adiabatic, determine the area of the outlet.
[Jaw:0.01668 m2]

Soalan 5
In a turbo jet engine the momentum of the gases leaving the nozzle produces the
propulsive force. The enthalpy and velocity of the gases at the nozzle entrance are 1200
kJ/kg and 200 m/s respectively. The enthalpy of the gas at exit is 900 kJ/kg. If the heat loss
from the nozzle is negligible, determine the velocity of the gas jet at exit from the nozzle.

[Jaw;800 m/s]

First Law – Flow Process /67

Soalan 6
One nozzle is used to increase the velocity of a fixed fluid flow. At the entrance has
enthalpy 3025kJ/kg, and 60m/s. At exit specific enthalpy 2790kJ / kg. The nozzle is
horizontal. And no heat loss happens, calculate
a. The velocity at the exit
b. Fluid flow if the cross section of the cross section is 0.1m2 and the volume of course is

0.19m3 / kg.

Question 7
A fluid flowing along a pipeline undergoes a throttling process from 10 bar to 1 bar in
passing through a partially open valve. Before throttling, the specific volume of the fluid is
0.3 m3/kg and after throttling is 1.8 m3/kg. Determine the change in specific internal energy
during the throttling process.
Solution
For a throttling process, the steady flow energy equation becomes
0 = m (h2 - h1)
or h2 = h1
But h2 = u2 + P2v2
and h1= u1 + P1v1
Therefore the change in specific internal energy
∆u = u2 – u1

= ( h2 – P2v2 ) - (h1 – P1v1)
= ( h2 - h1 ) – ( P2v2 - P1v1 )
= 0 – ( 100 x 1.8 – 1000 x 0.3 )
= 120kJ/kg

First Law – Flow Process /68
Question 8
Steam enters a turbine with specific enthalpy 2990 kJ/kg and leaves the turbine with a
specific enthalpy 2530 kJ/kg. The heat lost to the surroundings as the steam passes through
the turbine is 25 kJ/kg. The steam flow rate is 324000 kg/h. Determine the work output from
the turbine in kilowatts.

Question 9
1500kg of steam is generated within 1 hour at 30bar pressure and 300oC. The water supply
into the boiler is at 40oC. Determine the heat rate to be supplied.

First Law – Flow Process /69
Question 10
Steam enters a throttle at 0.1 bar, 150oC and 180 m/s. It comes out in saturated steam at
2.7bar and 50 m/s. The cross sectional area of the diffuser is 0.08 m2. Determine: steam mass
flow rate, heat transfer rate, cross sectional area of diffuser.

Question 11
Steam enters an adiabatic turbine steadily at 10 Mpa and 400oC and exits at 20 kPa with a
90% quality. By ignoring the change in kinetic energy and potential, determine the mass
flow required to produce 1MW output power. [Jaw: 1.38 kg/ s]

First Law – Flow Process /70
Question 12
Steam at 1 bar pressure leaves the nozzle at dryness 0.9 and velocity 800m/s. The cross-
sectional area of the outlet is 1.0x103 mm2. calculate the steam mass flow rate.[Jaw: 0.52
kg/s ]

Question 13
Steam enters a nozzle in steady state with a pressure of 40 bar and a temperature of 400oC.
Heat transfer and potential energy changes can be ignored. At the exit pressure of 15 bar,
and mass flow is 2kg/s. Get an outlet area. Determine the area at the exit.

[jaw; 4.89 x 10-4 m2]

First Law – Flow Process /71
Question 14
Steam enters an overhead nozzle with a velocity of 60 m / s at p = 8 bar and a temperature
of 250 oC. Steam leaves the nozzle at 1.6 bar pressure and 0.96 dryness fraction. The cross-
sectional area of the outlet is 1.2 x 103 mm2. The stream is adiabatic. Calculate the steam
velocity of the steam outlet and steam mass flow rate.[Jaw:832 m/s,0.95kg/s]

Question 15
Steam flows into a condenser with a steady stream. The steam enthalpy of the input is
3523kJ/kg and the steam flow velocity is 42m/s. Enthalpy outlet is 530 kJ/kg and the exit
velocity is 8m/s. Calculate the heat transferred to the cooling water.

[Jaw: -2995.85 kJ / kg]

First Law – Flow Process /72
Question 16
A steam nozzle supplied with a specific enthalpy 2780kJ/kg at a rate of 9.1kg/min. Output
steam velocity of 1070m/s. Assuming that the steam inlet velocity is negligible and that
this process is adiabatic, determine:
a. The enthalpy is of course steam at the outlet of the nozzle
b. The required area of the outlet if the final volume of the steam is 18.75m3/kg.

[Jaw: 2208 kJ/kg; 2660 mm2]

Question 17
A gas turbine generates 7000kW of power with a mass flow rate of 25kg/s. The gas flows
steadily into the turbine at a pressure of 400kN/m2 with a velocity of 150m/s. Gas out of
turbine at 105kN/m2, temperature 500oC and 280m/s speed. The processes in the turbine
are considered as adiabatic with Cp = 1.15kJ/kgK and R = 0.28kJ/kgK. Find: gas
temperature when entering the turbine, the cross-sectional area of the entrance.

First Law – Flow Process /73

Question 18
Steam at P = 20bar and T = 250 oC flows steadily at a turbine with a velocity that can be
neglected. Steam exit at P = 0.20bar with velocity 200 m/s. The heat transfer rate from the
turbine to the atmosphere is 160 kW. The power output of the turbine is 3440 kW. The
mass flow rate is 6.1 kg/s. Calculate the dryness fraction of the steam out of the turbine
and the cross section of the cross section. [Ans: 0.867, 0.2m3]

Question 19
Steam in steady flow conditions enters a fully insulated turbine at 12.5Mpa, 500oC and 80m/s
velocity. Steam exit at 10kPa, velocity 40m/s and dryness fraction 0.92, mass flow rate is
25kg/s. Determine
a. output power,
b. the cross sectional area of the turbine outlet

[Ans: 23.8 MW, 8.44 m2]
Question 20
The air in an open system flows in steady-state conditions at 10bar, 400K and 20m/s at
ducts with a cross-sectional area of 20cm2, the state of the output section is 6bar, 345.7K
and velocity 330m/s. Assuming the air acts as a perfect gas, determine the mass flow rate
in kg/s and the cross section of the outlet. [Ans: 0.3484 kg/s , 1.75cm2 ]

Question 21
Steam at 30bar pressure and 300oC temperature flows steadily into the turbine and exits at
8 bar and 200°. Flow is adiabatic, kinetic energy change and potential can be neglected.
Calculate the work done by a unit of steam mass that flows.[ Ans: 155 kJ/kg]

Question 22
Refrigerant 134a at 200 kPa, 40% quality, flows through a 1.1 cm inside diameter, d, tube
with a velocity of 50 m/s. Find the mass flow rate of the refrigerant. .
[ Ans: 0.0134 kg/s]

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4.0 SECOND LAW OF THERMODYNAMICS

4.1 Concept of the second law of thermodynamics

First Law of Thermodynamics stated,

` when a system undergoes a complete cycle, then the net heat supplied is equal to the net
work done.

 dQ =  dW

The Second Law of Thermodynamics;

Although the net heat supplied in a cycle is equal to the net work done, the gross heat
supplied must be greater than the work done; some heat must always be rejected by the
system.

In symbols, Q1 = W – Q2

4.2 The heat engine

The devices to convert heat to work
A heat engine is a system operating in a complete cycle and developing net work from a supply
of heat.
The second law implies that a source of heat supply (or hot reservoir) and a sink (or cold
reservoir) for the rejection of heat are both necessary, since some heat must always be rejected
by the system.

Heat engines differ considerably from one another, but all can be characterised by the following:
 They receive heat from a high-temperature source (for example solar energy, oil furnace,

nuclear reactor, steam boiler, etc.)
 They convert part of this heat to work (usually in the form of a rotating shaft, for example

gas turbine, steam turbine, etc.)
 They reject the remaining waste heat to a low-temperature sink (for example the

atmosphere, rivers, condenser, etc.)
 They operate on a cycle.
A diagrammatic representation of a heat engine is shown in Figure 4.1.

The thermal efficiency of a heat engine is defined as the ratio of the net work done in the cycle

to the gross heat supplied in the cycle. It is usually expressed as a percentage.

 Thermal efficiency,  
  W  Q1  Q2  1   Q2 
Q1 Q1
 Q1 

It can be seen that the second law implies that the thermal efficiency of a heat engine must always

be less than 100%

(Q1 > W ).

S e c o n d L a w (Heat Engine and Heat Pump)| 75
Figure 4.1 Part of the heat received by the heat engine is converted to work,while the rest is

rejected to cold reservoir.

Figure 4.2 Reverse heat engine (Heat Pump)

S e c o n d L a w (Heat Engine and Heat Pump)| 76

Example question

Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat
rejection to a nearby river is 45 MW, determine the net work done and the thermal efficiency
for this heat engine

Solution
Q1 = 80 MW
Q2 = 45 MW

the net work done is;
W = QH – QL

= (80 – 45) MW
= 35 MW
thermal efficiency
  W  35 MW  0.4375 (or 43.75%)

Q1 80 MW
That is, the heat engine converts 43.75 percent of
the heat it receives to work.

Figure 4.3 Heat Engine

Practice 4.1
Question 1
Heat is transferred to a heat engine from a hot reservoir at a rate of 120 MW. If the net work
done is 45 MW, determine the rate of waste heat rejection to a cold reservoir and the thermal
efficiency of this heat engine. [Jaw: Q2 = 75 MW,  = 37.5 %]

S e c o n d L a w (Heat Engine and Heat Pump)| 77

Question 2
The work done by heat engine is 20 kW. If the rate of heat that
enters into the hot reservoir is 3000 kJ/min, determine the
thermal efficiency and the rate of heat rejection to the cold
reservoir.

W = 20 kW
Q1 = 3000 kJ/min

= 50 kJ/s or kW

  W  20  40%
Q1 50

i.e. Thermal efficiency of heat engine is 40 %.
W = Q1 – Q2
Q2 = Q1 – W = 50 – 20 = 30 kW

i.e. The rate of heat rejection to the cold reservoir is 30 kW.

Question 3
A HE receives heat from a heat source at 1200 oC and has a thermal efficiency of 40%. The HE
does maximum work equal to 500kJ. Determine the
a. heat supplied to heat engine by the heat source,
b. the rejected to the heat sink,
c. the temperature of the heat sink.

S e c o n d L a w (Heat Engine and Heat Pump)| 78
Question 4
A HE is operating on a Carnot cycle and has a measured work output of 900kJ and heat
rejection of 150kJ to heat reservoir at 27C. Determine the
a. heat supplied to the heat engine by the heat source and
the temperature of the heat source.

Question 5
A HE is operating on a Carnot cycle and has a thermal efficiency of 75%. The waste heat from
this engine is rejected to a nearby lake at 15C at a rate of 14kW. Determine the power output of
the engine and the temperature of the source.

[42kW, 879oC]

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Example 6
A refrigerator with a temperature of 12oC in a range of 30o C. The total heat exited is estimated
at 3300kJ/h and the heat rejection in the condenser is 4800kJ/h. Determine the power required
by the compressor, in kW and COP of the refrigerator..

Example 7
A household refrigerator with COP 1.2 removes heat from the fridge space at 60kJ / min.
Determine the electric power used by the refrigerator and heat rate into the kitchen air.

S e c o n d L a w (Heat Engine and Heat Pump)| 80
Example 8
The refrigerator operates by releasing heat to the atmosphere of 750W and has a COP 3.5. A
Carnot operates in a room where the temperature is 15oC. Determine the rate of heat transfer
from the cold space and the temperature of the cold space..

Example 9
A household refrigerator with COP of 1.2 removes heat from the refrigerated space at rate of
60kJ/min. Determined the electric power consumed by the refrigerator and the rate of heat to
the kitchen air.
[0.83kW, 110kJ/min]

S e c o n d L a w (Heat Engine and Heat Pump)| 81
Example 10
A refrigerator used for cooling food in grocery store is to produce a 25000kJ/h, cooling effect,
and it has a COP 1.6. how many kilowatts of power wills this refrigerator require to operate?.

[Jaw: 4.34kW]

Example 11
A heat pump is used to meet the heating needs of the home and maintain it at 20 oC. When the
outdoor temperature drops to -2C, it is estimated to heat loss at 80000 kJ / h. If the heat pump
under this condition has COP 2.5, determine the power used by the heat pump and the rate at
which the heat is absorbed from the cold air outside.

S e c o n d L a w (Heat Engine and Heat Pump)| 82
Example 12
HP is used to heat the house during the winter and to maintain the
temperature of 21C at all times. The house is estimated to lose heat
at 135000kJ / h when the outside temperature drops to -5C.
Determine the minimum required power to drive this heat pump.

Example 13
A heat pump is use to maintain a house at a constant temperature
of 23C. The house is losing heat to the outside air through the walls
at a rate of 60,000kJ/h while the energy generated within the house
from people, lights and appliance amounts to 4000kJ/h. For a COP
of 2.5, determine the required power input to the heat
pump.[6.22kW]

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4.3 Entropy and Isentropic

Entropy is a thermodynamic property that is the measure of a system’s thermal energy per unit
temperature. kJ/kgK

Closed system entropy is constantly increasing. In the case of heat transfer, heat energy is
transferred from a higher temperature component to a lower temperature component.

The value of entropy will increase with temperature.

the isentropic process means the process without the increase or decrease of entropy in the
system is also called adiabatic.
∆s = 0, s1 = s2

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Figure 4.4 P-V And T-S Diagrams
Example question
If the water temperature is 350K. In the heat of 700kJ process the heat is transferred to the
water, so some water will be evaporated. So what is the change in water entropy?.
∆S = 700kJ/350K = 2kJ/K
Practice 4.2
1. Sketch Ts diagrame from pv diagram below

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2. Determine value of s1, s2, h1,and h2 from below diagrame

(1 2 isentropic process, s 1 = s2)

3. 1 kg of steam compressed from 4 bars, saturated liquid to 40bar with isentropic process.
Calculate the specific enthalpy change.

4. 1 kg of steam condensed from 40 bars, saturated vapour to saturated liquid isothermally
process. Calculate the specific enthalpy change.

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4.4 The Carnot cycle

Figure 4.5 The Carnot cycle

A brief summary of the essential features is as follows:
4 to 1: The heat energy is supplied to the boiler resulting in evaporation of the water,

therefore the temperature remains constant.
1 to 2: Isentropic expansion takes place in the turbine or engine.
2 to 3: In the condenser, condensation takes place, therefore the temperature remains

constant.
3 to 4: Isentropic compression of the wet steam in a compressor returns the steam to its

initial state.

Boiler : W = 0, Q41 = h1 – h4

Turbine : Q = 0 W12 = h1 – h2

Condenser : W = 0 Q23 = h3 – h2 = - ( h2 – h3 )
 Heat rejected in condenser = h2 – h3

Compressor : H3 + Q34 = h4 + W34
 W34 = ( h3 – h4 ) = -( h4 – h3 )

i.e. Work input to pump = ( h4 – h3 )

Thermal efficiency of Carnot cycle

Hence we have carnot 1 T2
T1

or  ca rn ot  Net work output  (h1  h2 )  (h4  h3 )
Heat supplied in the boiler h1  h4

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The work ratio for Carnot cycle

The ratio of the net work output to the gross work output of the system is called the work ratio.

The Carnot cycle, despite its high thermal efficiency, has a low work ratio.

Work ratio  net work = (h1  h2 )  (h4  h3 )
gross work (h1  h2 )

This cycle is never used in practice owing to:

1. The difficulty in stopping the condensation at 3, so that subsequent compression would
bring the state point to 4.

2. A very large compressor would be required.
3. Compression of wet steam in a rotary compressor is difficult as the water tends to separate

out.
4. Friction associated with the expansion and compression processes would cause the net

work done to be very small as compared to the work done in the turbine itself.

The Carnot cycle is modified to overcome the above difficulties and this modified cycle, known
as the Rankine cycle, is widely used in practice.

Example question

What is the highest possible theoretical efficiency of a Carnot cycle with a hot reservoir of
steam at 200oC when the cooling water available from condenser is at 10oC?

Solution to Example

carnot 1 T2 = 1 10  273 = 1 283
T1 200  273 473

i.e. Highest possible efficiency = 1 – 0.598
= 0.402 or 40.2 %

S e c o n d L a w ( C a r n o t C y c l e ) | 88

Example question

A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of
0.035 bar. Calculate for these limits the cycle efficiency and the work ratio for a Carnot cycle
using wet steam.

Solution to Example

T1 saturation temperature at 42 bar = 253.2 + 273 = 526.2 K

T2 saturation temperature at 0.035 bar= 26.7 + 273 = 299.7 K

carnot  T1  T2 526.2  299.7 = 0.432 or 43.2 %
T1 = 526.2

Didapati Wratio  net work, Wnet
gross work

 ca r n o t  Wnet Wnet  Q  carnot
Q
Dan Wnet diperolehi dari atau

Dan Q = h1-h4 = hfg at 42 bar = 1698 kJ/kg
Then,Wnet = 0.432 x 1698 = 734 kJ/kg

To find the gross work , W12 = h1 – h2, dan h1 = 2800kJ/kg

h2 =?, use formula h2 = hf2 + x2hfg2 ,
i.e h2 =112+ x2(2438 )=112 + 0.696 x 2438 = 1808kJ/kg

and x2 using s1 = s2.

From the tables, s1 = s2 = 6.049 kJ/kg K

From the equation s2 = 6.049 = sf2 + x2sfg2 = 0.391 + x2 8.13

 x2 = 6.049  0.391  0.696
8.13

Then, h2 = hf2 + x2hfg2 = 112 + 0.696 x 2438 = 1808kJ/kg

W12 = (h1 – h2) = (2800 – 1808) = 992 kJ/kg

Therefore, Work ratio  net work 734

= = 0.739
gross work 992

S e c o n d L a w ( C a r n o t C y c l e ) | 89

Example qeustion

A steam power plant operates between a boiler pressure of 40 bar and a condenser pressure of
0.045 bar. Calculate for these limits the cycle efficiency and the work ratio for a Carnot cycle
using wet steam.

Solution to Example
A Carnot cycle is shown in the figure below.
T1 = 250.3 + 273 = 523.3 K
T2 = 31.0 + 273 = 304.0 K

carnot  T1  T2 = 523.3  304
T1 =

523.3

=0.419 or 41.9 %

Wratio  net work, Wnet
gross work

Wnet  Q41  carnot

,
Q41 = h1 – h4 = hfg at 42 bar = 1714 kJ/kg

 Wnet = 0.419 x 1714 = 718.2 kJ/kg

To find the gross work , W12 = h1 – h2, dan h1 = 2801 kJ/kg

h2 =?, use formula h2 = hf2 + x2hfg2 ,

i.e h2 = 130 + x2 (2428 )

dan x2 using s1 = s2.

From the tables, s1 = s2 = 6.070 kJ/kg K

From the equation s2 = 6.070 = sf2 + x2sfg2 = 0.451 + x2 7.98

 x2 = 6.070  0.451  0.704
7.98

Then, h2 = hf2 + x2hfg2 = 130 + 0.704 x 2428 = 1839.3 kJ/kg

Gross work, W12 = (h1 – h2) = (2801 – 1839.3) = 961.7 kJ/kg

Work ratio  net work = 718.2
gross work = 0.747

961.7

S e c o n d L a w ( R a n k i n e C y c l e ) | 90

4.5 Rankine Cycle

Many of the impracticalities associated with the Carnot cycle can be eliminated by condensing
it completely in the condenser, as shown schematically on a T-s diagram. The cycle that results
is the Rankine cycle, is the ideal cycle for vapour power plants. The ideal Rankine cycle does
not involve any internal irreversibility and consists of the following processes:

4,5 to 1: Constant pressure heat addition in a boiler
1 to 2: Isentropic expansion taking place in the turbine or engine
2 to 3: Constant pressure heat rejection in the condenser
3 to 4: Isentropic compression of water in the feed pump

From a comparison made between Fig. 11.1 and Fig. 11.2, the similarities between the Carnot
and the Rankine cycles can be clearly seen. In the Rankine cycle, the exhaust steam is completely
condensed into water in the condenser. It actually follows the isentropic expansion in the turbine.
This water is then pumped into the boiler by a boiler feed pump. After the feed pump, since the
water is not at the saturation temperature corresponding to the pressure, some of the heat energy
supplied in the boiler is taken up by the water as sensible heat before evaporation can begin. This
results in the boiler process being no longer completely isothermal; the process is, therefore,
irreversible, causing the Rankine cycle to be an irreversible cycle and to have a lower efficiency
than the Carnot cycle.

Figure 4.6 Steam Powerplant schematic

S e c o n d L a w ( R a n k i n e C y c l e ) | 91

Figure 4.7 The Rankine cycle
.

i.e. h1 + Q = h2 + W

In this statement of the equation the subscripts 1 and 2 refer to the initial and final state points of
the process; each process in the cycle can be considered in turn as follows:

Boiler: W = 0, Q451 = h1 – h4
h4 + Q451 = h1 + W 

Turbine: is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2 ),

h1 + Q12 = h2 + W12  W12 = ( h1 – h2)

Condenser: W = 0
h2 + Q23 = h3 + W23 i.e. Q23 = - ( h2 – h3 )
 Heat rejected in condenser = h2 – h3

Pump: Q = 0
h3 + Q34 = h4 + W34  W34 = ( h3 – h4 ) = -( h4 – h3 )

i.e. Work input to pump = ( h4 – h3 )

This is the feed pump term, and as the quantity is small as compared to the turbine work, W12.
The feed pump is usually neglected, especially when the boiler pressures are low.
The net work done in the cycle, W = W12 + W34

i.e. W = ( h1 – h2) – ( h4 – h3 )
Or, if the feed pump work is neglected,

W = ( h1 – h2 )

S e c o n d L a w ( R a n k i n e C y c l e ) | 92

Thermal efficiency of Rankine cycle

Rankine efficiency,

R  Net work output , R (h1  h2 )  (h4  h3 )
Heat supplied in the boiler h1  h4

or  R (h1  h2 )  (h4  h3 ) If the feed pump term, (h4 – h3) is neglected, equation
(h1  h3 )  (h4  h3 )

becomes R (h1  h2 )
(h1  h3 )

When the feed pump term is included, it is necessary to evaluate the quantity, W34.
Pump work = -W34 = f (p4 – p3)

The work ratio for Rankine cycle

Work ratio  net work
gross work

i.e. Work ratio = (h1  h2 )  (h4  h3 ) = (h1  h2 )  vf ( p4  p3 ) )
(h1  h2 ) (h1  h2 )

Specific steam consumption

i.e. s.s.c. = 3600 kg/kwh = 3600 kg/kwh
Wnet (h1  h2 )  (h4 h3 )

S e c o n d L a w ( R a n k i n e C y c l e ) | 93

Example question

A steam power plant operates between a boiler pressure of 42 bar and a condenser pressure of
0.035 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam
consumption for a Rankine cycle with dry saturated steam at entry to the turbine.

Solution to Example

h1= 2800 kJ/kg and h2 = 1808 kJ/kg
h3 = hf at 0.035 bar = 112 kJ/kg
with v = vf at 0.035 bar

Pump work = -W34 = f (p4 – p3)
= 0.001 x ( 42 – 0.035) x 102 = 4.2 kJ/kg

W12 = h1 – h2 = 2800 – 1808 = 992 kJ/kg

cycle efficiency

 R (h1  h2 )  (h4  h3 ) (992)  (4.2)
(h1  h3 )  (h4  h3 ) = (2800  112)  (4.2) = 0.368 or 36.8 %

Work ratio  net work = 992 - 4.2
gross work 992 = 0.996

3600 3600

s.s.c. = (h1  h2 )  (h4 h3 ) = (992)  (4.2) = 3.64 kg/kW h

S e c o n d L a w ( R a n k i n e C y c l e ) | 94

Example question

A steam power plant operates between a boiler pressure of 40 bar and a condenser pressure of
0.045 bar. Calculate for these limits the cycle efficiency, the work ratio, and the specific steam
consumption for a Rankine cycle with dry saturated steam at entry to the turbine.

Solution to Example

h1= 2801 kJ/kg and h2 = 1839.3 kJ/kg
h3 = hf at 0.045 bar = 130 kJ/kg

Pump work = -W34 = f (p4 – p3)
= 0.001 x ( 40 – 0.045) x 102
= 4.0 kJ/kg

W12 = h1 – h2 = 2801 – 1839.3 = 961.7 kJ/kg

Cycle efficiency,

 R (h1  h2 )  (h4  h3 ) (961.7)  (4.0) = 0.359 or 35.9 %
(h1  h3 )  (h4  h3 ) = (2801 130)  (4.0)

Work ratio  net work 961.7 - 4.0
gross work = 961.7 = 0.996

s.s.c. = 3600 = 3600 = 3.76 kg/kW h
(h1  h2 )  (h4 h3 ) (961.7)  (4.0)

S e c o n d L a w ( R a n k i n e C y c l e ) | 95
Practice 4.2
1. What is the highest thermal efficiency possible for a Carnot cycle operating between

210o C and 15o C. [ 40.37 %]

2. Sketch Ts diagrame for carnot cycle and determine the position of boiler, turbine,
condenser and pump for that cycle.

3. A steam power plant operates between a boiler pressure of 30 bar and a condenser
pressure of 0.04 bar. Calculate for these limits the cycle efficiency, the work ratio and
the specific steam consumption for a Carnot cycle using wet steam. [40.4 %, 0.771,
4.97 kg/kWh]

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4. A steam power plant operates between a boiler pressure of 30 bar and a condenser

pressure of 0.04 bar. Calculate for these limits the cycle efficiency, the work ratio, and
the specific steam consumption for a Rankine cycle with dry saturated steam at entry
to the turbine.
[34.6 %, 0.997, 3.88 kg/kWh]

5. In a steam power plant, dry saturated steam enters the turbine at 47 bar and is expanded
isentropically to the condenser pressure of 0.13 bar. Determine the Rankine cycle
efficiency when
a. the feed pump work is neglected
b. the feed pump work is taken into account
[33.67 %, 33.55 %]

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6. A steam power plant is working between boiler pressure 50bar and condenser pressure
0.05bar. calculate for carnot cycle:
a. Work tubine
b. Work input to pump
c. Heat rejected in condenser
d. Heat supplied
e. Cycle efficiency
f. Work ratio
g. SSC

Draw TS diagrame cycle

S t e a m T a b l e | 98

4.6 Steam Properties

Phase Remarks Stea
m
Compress  Pressure is higher than saturated pressure (p > ps at temperature T)
liquid  temperature lower than saturated temperature (T > Ts at pressure p) table
 energy in lower than energy in saturated liquid (u > uf at p or T)
 lower enthalpy of saturated liquid enthalpy (h > hf at p or T) Saturated Water & Steam

Liquid Dyness fraction = 0 dan nilai u = uf, h = hf, s = sf
saturated

Wet Steam  pressure equal to saturated pressure ( p = ps at temperature T)
 temperature equal to saturated temperature (T = Ts at pressure p)
The values for u, s and h are between subscript / suffix 'f' and subscript
/ suffix 'g'. In this case the dose fraction value (x) is between 0 <x <1
(and in this case the formula is used)

Steam When the dense fraction value, x = 1 and if the value v, u, h, s is the
saturated same value with vg, ug, hg and sg

Superheated  Pressure is lower than saturated pressure (p < ps at temperature T) Superheated Steam
 temperature higher than saturated temperature (T > Ts at pressure

p)
 volume higher than saturated volume (   g at p or T )

1. internal energy in higher than saturated internal energy ( u  ug at
p or T )

2. The enthalpy is higher than the saturated enthalpy ( h  hg pada p
atau T )

S t e a m T a b l e | 99

Flow Process

P1 Inlet SISTEM Outlet P2
v1 Z1 Z2 v2
T1 T2
C1 C2

A1 A2

Datum
Input (subscript 1) dan Output (subscript 2)

P Tekanan ( kN/m2 @ kPa ) o Kadaralir Jisim ( kg/s )

v Isipadu Tentu ( m3/kg ) m
C Halaju Aliran ( m/s2 )
A Luas keratan rentas salur ( m2 )

Z Ketinggian dari datum ( m )

o AC  o o
Kadaralir Jisim m  atau  AC
dan dlm aliran mantap/Sekata m1  m2
v

Steady Flow Energy Equation

Q12  W12  (P2v2  P1v1 )  (U 2  U1)  ( C2 2  C12 )  g(Z2  Z1 )
2

Q12  W12  (h2  h1 )  ( C2 2  C12 ) g (Z 2  Z1 ) bagi bendalir kerja stim
2

Tukarkam rumus (h2 – h1) kepada Cp(T2 – T1) jika bendalir kerja udara atau gas

REMINDER :

Semua unit dalam persamaan di atas dalam unit J/kg (m2/s2) tetapi dalam semua

pengiraan/penyelesaian rumus tenaga kinetik dan tenaga keupayaan hendaklah dibahagikan
dengan 103 sebab kerja (W12), Haba (Q12), Entalpi (h) biasanya dalam unit kJ/kg , contohnya

Q12  W12  (h2  h1 )   C2 2  C12    g(Z2  Z1  Unit in kJ/kg
 2 x 103   103 

------------------------------------------------------------------------------------------------------------

oo o   C 2  C12  
(h2  2   Z1 )
Q12 W 12  m   h1 )  2  g(Z 2 Unit in kW


Q12  W 12  m   h1 )   C2 2  C12   g (Z 2   Unit in kJ
(h2 2 Z1 )

