U24ME302- Engineering Thermodynamics Mechanical Engineering 2024-2025

U24ME302- Engineering Thermodynamics

UNIT I - FIRST LAW of thermodynamics

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U24ME302‑ Engineering Thermodynamics Mechanical Engineering 2024‑2025
Thermodynamics
It is the science of the relations between heat, work and the properties of the systems.
Approaches in Thermodynamics:
In macroscopic approach, certain quantity of matter is considered, without considering the
events occurring at the molecular level. These effects can be perceived by human senses or
measured by instruments.
eg: pressure, temperature
Thermodynamics based on macroscopic approach is called Classical thermodynamics.
In microscopic approach, the effect of molecular motion is considered.
eg: At microscopic level the pressure of a gas is not constant, the temperature of a gas is a
function of the velocity of molecules. Most microscopic properties cannot be measured with
common instruments nor can be perceived by human senses
Thermodynamics based on microscopic approach is called Statistical Thermodynamics
System
The system is a quantity of matter or a region in space on which we focus our attention (eg: the
water kettle or the aircraft engine).
Surroundings
The rest of the universe outside the system close enough to the system to have some
perceptible effect on the system is called the surroundings.

Boundaries
The surface which separates the system from the surroundings are called the boundaries as
shown in fig below (eg: walls of the kettle, the housing of the engine).

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Open System
A system in which, mass and energy (work or heat) can be transferred across the boundary.

Closed System
A system in which there is no mass transfer but only energy transfer across the boundary.

Isolated System
A system in which there is no mass transfer and energy transfer across the boundary.

Properties
It is some characteristic of the system by which the condition of system is described.
Extensive property:
Properties whose value depends on the size or extent of the system
If mass is increased, the value of extensive property also increases. eg: volume, mass
Intensive property:
Properties whose value is independent of the size or extent of the system. eg: pressure,
temperature (p, T).
State
A system is said to be in a state when it has definite values for properties. Any operation in
which one or more properties change is called a ‘change of state’.

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Equilibrium State
It is a state of balance. A system is said to be in equilibrium state if it is under mechanical,
chemical and thermal equilibrium.
Mechanical equilibrium: No unbalanced forces, ie no difference in pressure within the system
Chemical equilibrium: No chemical reaction within the system
Thermal equilibrium: No difference in temperature within the system
Process
The succession of states passed through during a change of state is called the path of the system.
A system is said to go through a process if it goes through a series of changes in state.
Consequently:
A system may undergo changes in some or all of its properties in a process.
Quasi-static Process
A process in which the intermediate states are equilibrium states is called quasistatic or quasi-
equilibrium process. Generally quasi-static processes are slow processes.
Cycle
A series of processes at the end of which the system comes back to initial state.
Open and closed cycle
In a closed cycle, the same working substance will be undergoing the cycle again and again.
In an open cycle, the working substance will be exhausted at the end of a series of processes
which will be repeated again and again.
Zeroth Law of Thermodynamics
If two systems (say A and B) are in thermal equilibrium with a third system (say C) separately
(that is A and C are in thermal equilibrium; B and C are in thermal equilibrium) then they are
in thermal equilibrium themselves (that is A and B will be in thermal equilibrium.

Two systems are said to be equal in temperature, when there is no change in their respective
observable properties when they are brought together. In other words, “when two systems are
at the same temperature they are in thermal equilibrium”(They will not exchange heat).
Note: They need not be in thermodynamic equilibrium.

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Point and path function
Path function: The function whose value is dependent on the path of the process. e.g. work
transfer and heat transfer
Point function: The function the change in whose value is independent on the path of the
process. e.g. pressure, temperature, etc.
Reversible and irreversible process
Reversible process: A reversible process is one which can be stopped at any stage and reversed
so that the system and surroundings are exactly restored to their initial states.
Irreversible process: An irreversible process is one which can be stopped at any stage and
reversed but the system and surroundings are not restored to their initial states.
Work and its Sign convention
Work is transient quantity which appears at the boundary when a system changes its state due
to the movement of a part of the boundary under the action of a force.
Sign convention:
If the work is done by the system, Work output of the system = + W
If the work is done on the system, Work input to system = – W
Heat and its Sign convention
Heat is transient quantity which appears at the boundary when a system changes its state due
to a difference in temperature between the system and its surroundings.
Heat received by the system = + Q, Heat rejected or given up by the system = – Q.
Flow energy
Energy required to introduce a quantity of fluid in a pipe section is flow energy. It is equal to pV.
Internal Energy
Internal energy of a gas is the energy stored in a gas due to its molecular interactions. It is denoted as U
Enthalpy of a system
It is the sum of internal energy and flow energy. i.e. H = U+ pV
Latent heat
Amount of heat required to cause a phase change in unit mass of a substance at constant pressure and
temperature.

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Units Constant Volume Constant Pressure Isothermal Adiabatic Process Polytropic Process
Process
(V=C) (p=C) (pV=C) (𝒑𝑽𝜸 = 𝑪) (𝒑𝑽𝒏 = 𝑪)

p-V Diagram

kJ V2 p1 V1 − p2 V2 p1 V1 − p2 V2
Work Done W=0 W = p1 (V2 − V1 ) W = p1 V1 ln ( ) W= W=
V1 γ−1 n−1
Change In Internal kJ
∆U = mCv (T2 − T1 ) ∆U = mCv (T2 − T1 ) ∆U = 0 ∆U = −W ∆U = Q − W
Energy
kJ V2 𝛾−𝑛
Heat Transfer Q = mCv (T2 − T1 ) Q = mCP (T2 − T1 ) Q = W = p1 V1 ln ( ) Q=0 Q=[ ]×W
V1 (𝛾 − 1)
kJ/K T2 T2 V2 T2
Change In Entropy ∆S = mCv ln ( ) ∆S = mCP ln ( ) ∆S = mR ln ( ) ∆S = 0 ∆S = mCn ln ( )
T1 T1 V1 T1
n
p2 V1
p2 V1 γ =( )
=( ) p1 V2
p1 V2
T2 p2 T2 V2 V2 p1 T2 V1 γ−1 T2 V1 n−1
p , V , T Relations = = = =( ) =( )
T1 p1 T1 V1 V1 p2 T1 V2 T1 V2
γ−1 n−1
T2 P2 γ
T2 P2 𝑛
=( ) =( )
T1 P1 T1 P1
 Where C=Constant, 1 and 2 indicates as initial  pv = RT Entropy for all process can also be calculated by:
𝑉 T V T P
and final state, Specific Volume ′𝑣′ = 𝑚 m3/kg ∆S = mCv ln (T2 ) + mR ln (V2 ), ∆S = mCP ln (T2 ) − mR ln (P2 )
1 1 1 1
 Equation for Ideal Gas : pV = mRT V2 P2
 Pressure ‘p’ – kN/m2 (1 bar =1 X 102 kN/ m2)  𝐩𝐕̇ = 𝐦̇𝐑𝐓 ∆S = mCP ln ( ) + mCv ln ( )
V1 P1
 Volume ‘ V ‘ - m3 Volume flow rate ′V̇ ′ − m3 /s For Ideal Gas (Air):
 Mass ‘m’ - kg Mass flow rate ‘ṁ’ – kg/s Specific Heat ‘Cp’ = 1.005 kJ/kg.K, ‘Cv’ = 0.718 kJ/kg.K
 Gas Constant ‘R’ - 0.287 kJ/kg.K For Water :
 𝐩𝐕̅=𝐧 ̅𝐑𝐓 Specific Heat ‘Cp’ = 4.186 kJ/kg.K,
 Temperature ‘T’ - K ( 0°C = 273K )
Molar volume V- m3 /kgmol p
 Enthalpy ‘h’ = u + Pv or h = CpT ln (p2 )
Number of moles ‘n’ Index of Expansion ′𝐧′ = 1
V1
(ln V )
2

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STEADY FLOW ENERGY EQUATION: (i) Rate of heat transfer in the heat exchanger
From steady flow energy Equation:
INLET OUTLET
C2 gz C2 gz
1
𝐦̇ (2000 1
+ 1000 + u1 + p1 v1 ) + Q̇ = 𝐦̇ (2000
2 2
+ 1000 + u2 + p2 v2 ) + Ẇ
z1 = z2 , C1 = 0, C2 = 0 , W=0, h1 = u1 + p1 v1 and h2 = u2 + p2 v2
Z1 Z2 𝐐𝟏𝟐 = 𝐦̇(𝐡𝟐 − 𝐡𝟏 ) (kJ/s)
(ii) The power output from the turbine assuming no heat loss
From steady flow energy Equation:
C2 gz C2 gz
Inlet: 1
𝐦̇ (2000 1
+ 1000 + u1 + p1 v1 ) + Q̇ = 𝐦̇ (2000
2 2
+ 1000 + u2 + p2 v2 ) + Ẇ
Kinetic energy + Potential energy+ Internal energy+ Flow energy + Heat
𝐂𝟏𝟐 Where, z1 = z2 , h1 = u1 + p1 v1 and h2 = u2 + p2 v2 , Q = 0
+ 𝐠𝐳𝟏 + 𝐮𝟏 + 𝐩𝟏 𝐯𝟏 + 𝐐 (J/kg)
𝟐
Exit 𝟏 𝐂 𝟐 −𝐂 𝟐
𝟐
𝐖𝟏𝟐 = 𝐦̇ [ 𝟐𝟎𝟎𝟎 + (𝐡𝟏 − 𝐡𝟐 )] (kJ/s)
Kinetic energy + Potential energy+ Internal energy+ Flow energy + Heat
𝐂𝟐𝟐
𝟐
+ 𝐠𝐳𝟐 + 𝐮𝟐 + 𝐩𝟐 𝐯𝟐 + 𝐖 (J/kg)
According to the Energy Conservation Priciple (iii) The power required to run the compressor assuming no heat loss
Inlet Energy = Exit Energy From steady flow energy Equation:
𝐂𝟏𝟐 𝐂𝟐𝟐 C1 2 gz1 C2 2 gz2
𝟐
+ 𝐠𝐳𝟏 + 𝐮𝟏 + 𝐩𝟏 𝐯𝟏 + 𝐐 =
𝟐
+ 𝐠𝐳𝟐 + 𝐮𝟐 + 𝐩𝟐 𝐯𝟐 + 𝐖 𝐦̇ (2000 + 1000 + u1 + p1 v1 ) + Q̇ = 𝐦̇ (2000 + 1000 + u2 + p2 v2 ) + Ẇ
Where, z1 = z2 , h1 = u1 + p1 v1 and h2 = u2 + p2 v2 , Q = 0
𝐂𝟐𝟐 − 𝐂𝟏𝟐
𝐐−𝐖= + 𝐠(𝐳𝟐 − 𝐳𝟏 ) + (𝐮𝟐 − 𝐮𝟏 ) + (𝐩𝟐 𝐯𝟐 − 𝐩𝟏 𝐯𝟏 ) (J/kg) 𝟏 𝐂 𝟐 −𝐂 𝟐
𝟐
𝟐 𝐖𝟏𝟐 = 𝐦̇ [ 𝟐𝟎𝟎𝟎 + (𝐡𝟏 − 𝐡𝟐 )] (kJ/s) (Negative work will attain)
𝐂𝟐𝟐 − 𝐂𝟏𝟐 𝐠(𝐳𝟐 − 𝐳𝟏 )
𝐐−𝐖= + + (𝐮𝟐 − 𝐮𝟏 ) + (𝐩𝟐 𝐯𝟐 − 𝐩𝟏 𝐯𝟏 ) (kJ/kg)
𝟐𝟎𝟎𝟎 𝟏𝟎𝟎𝟎 (iv) The velocity at exit of the nozzle
𝐂𝟐𝟐 − 𝐂𝟏𝟐 𝐠(𝐳𝟐 − 𝐳𝟏 ) From steady flow energy Equation:
𝐐̇ − 𝐖̇ = 𝐦̇ [ + + (𝐮𝟐 − 𝐮𝟏 ) + (𝐩𝟐 𝐯𝟐 − 𝐩𝟏 𝐯𝟏 )] (kJ/s) C1 2 gz1 C2 2 gz2
𝟐𝟎𝟎𝟎 𝟏𝟎𝟎𝟎 ṁ (2000 + 1000 + u1 + p1 v1 ) + Q̇ = m (2000 + 1000 + u2 + p2 v2 ) + Ẇ

𝐂𝟐𝟐 − 𝐂𝟏𝟐 𝐠(𝐳𝟐 − 𝐳𝟏 ) Where, z3 = z4 , Q=0, W=0, h3 = u3 + p3 v3 and h4 = u4 + p4 v4

𝐐̇ − 𝐖̇ = 𝐦̇ [ + + (𝐡𝟐 − 𝐡𝟏 )] (kJ/s)
𝟐𝟎𝟎𝟎 𝟏𝟎𝟎𝟎 C22 C2
1
= 2000 + (h1 − h2 ) C2 = √C12 + 2000 (h1 − h2 )
2
FOR OPEN SYSTEM: 𝐂𝟐 = 𝟒𝟒. 𝟕√(𝐡𝟏 − 𝐡𝟐 )) (m/s) If C1 is negligible
𝐐 − 𝐖 = ∆𝐄

FOR CLOSED SYSTEM:

𝐐 − 𝐖 = ∆𝐔

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Problems on First law of thermodynamics closed system:

1. A mass of 1.5 kg of air is compressed in a quasi-static process from 0.1 MPa to 0.7 MPa for
which pV = C. The initial density of air is 1.16 kg/m3. Find the work done by the piston to
compress the air.
Given: m=1.5 kg, P1=0.1MPa, P2=0.7MPa, ρ=1.16kg/m3

Find: The work done by the piston (W)

Solution:
m m 1.5
ρ= V
V1 = ρ
= 1.16 𝐕𝟏 = 𝟏. 𝟐𝟗𝟑 𝐦𝟑

For quasi-static process

V dV V
W = ∫ pdV W = p1 V1 ∫V 2 W = p1 V1 ln (V2 )
1 V 1

V2 p1
At pV = C p1 V1 = p2 V2 = C V1
=
p2
V p 0.1
W = p1 V1 ln (V2 ) W = p1 V1 ln (p1 ) W = 0.1 × 1.293 ln (0.7) 𝐖 = − 𝟐𝟓𝟏. 𝟔𝟑 𝐤𝐉
1 2

2. A mass of gas is compressed in a quasi-static process from 80 kPa, 0.1 m3 to 0.4 MPa, 0.03
m3. Assuming that the pressure and volume are related by pVn = constant, find the work
done by the gas system.
Given: P1=80 kPa, P2=0.4MPa, V1=0.1 m3, V2=0.03 m3
Find: (i) n, (ii) The work done (W)
Solution:
At pV n = C p1 V1n = p2 V2n
Taking Log on
V p
ln p1 + n ln V1 = ln p2 + n n(ln V1 − ln V2 ) = ln p2 − ln p1 n (ln V1 ) = ln (p2 )
2 1
p 400
ln( 2 ) ln( )
p1 80
n= V n= 0.1 𝐧 = 𝟏. 𝟑𝟑𝟔𝟕 ≈ 𝟏. 𝟑𝟒
(ln 1 ) (ln
0.03
)
V2

p1 V1 −p2 V2 80×0.1−400×0.03
W = ∫ pdV W= n−1
W= 1.34−1
𝐖 = −𝟏𝟏. 𝟕𝟔𝟒𝐤𝐉

3. At the beginning of the compression stroke of a two-cylinder internal combustion engine

the air is at a pressure of 101.325 kPa. Compression reduces the volume to 1/5 of its
original volume, and the law of compression is given by pV1.2 = constant. If the bore and
stroke of each cylinder is 0.15 m and 0.25 m, respectively, determine the power absorbed in
kW by compression strokes when the engine speed is such that each cylinder undergoes 500
compression strokes per minute.
Given: P1=101.325 kPa, V2=1/5 V1 , D=0.15 m, L=0.25m, N=500rpm and pV1.2 = C
Find: The power absorbed in kW (P)

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Solution:
V 1.2
p1 V11.2 = p2 V21.2 p2 = (V1 ) × p1 = (5)1.2 × 101.325 𝐩𝟐 = 𝟔𝟗𝟗. 𝟒𝟏𝐤𝐏𝐚
2

πd2 π×0.152
Initial Volume (𝐕𝟏 ) = 4
×L V1 = 4
× 0.25 𝐕𝟏 = 𝟎. 𝟎𝟎𝟒𝟒𝟐 𝐦𝟑
V1 0.00442
Final Volume (𝐕𝟐 ) = 5
V2 = 5
𝐕𝟐 = 𝟎. 𝟎𝟎𝟎𝟖𝟖𝟒 𝐦𝟑
n 1.2
Work done 𝐖 = [p V − p1 V1 ] W= [(699.41 × 0.000884) − (101.325 × 0.00442)]
n−1 2 2 1.2−1

𝐖 = 𝟑. 𝟓𝟗𝐤𝐉
4. Air at 1.02 bar, 22°C, 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 :(i) The final
temperature ;(ii) The final volume ;(iii) The work done.
Given: P1=1.02 bar, T1=22°C+273= 295K, V1= 0.015 m , P2= 6.8 bar, pV n = C
3

Find : (i) T2 , (ii) V2, (iii) W

Solution:
(i) Final temperature
From the relation
γ−1 γ−1 1.4−1
T2 P2 γ P2 γ 6.8 1.4
T1
= (P ) T2 = (P ) × T1 T2 = (1.02) × 295 𝐓𝟐 = 𝟓𝟎𝟕. 𝟐𝟒𝐊
1 1

(ii) Final Volume

I I
γ γ p1 V2 γ P γ 1.02 1.4
p1 V1 = p2 V2 p2
= (V ) V2 = (P1 ) × V1 = ( 6.8 ) × 0.015 𝐕𝟐 = 𝟎. 𝟎𝟎𝟑𝟖𝟕 𝐦𝟑
1 2

(iii) Work
p1 V1 −p2 V2 102×0.015−680×0.00387
W = ∫ pdV W= γ−1
W= 1.4−1
𝐖 = −𝟐. 𝟕𝟓𝟒𝐤𝐉

5. 0.15 m3 of an ideal gas at a pressure of 15 bar and 550 K is expanded isothermally to 4

times the initial volume. It is then cooled to 290 K at constant volume and then compressed
back polytropically to its initial state. Calculate the net work done and heat transferred
during the cycle.
3 V2
Given: V1= 0.15 m , P1=15 bar, T1= T2 =550K, =4
V1

T3=290K, V2 = 4 × V1 = 0.6 m3
Find : (i) W , (ii) Q

Solution:
Process 1-2: Isothermal Process

V 0.15
At pV = C p1 V1 = p2 V2 p2 = V1 × p1 = 4×0.15 × 15 𝐩𝟐 = 𝟑. 𝟕𝟓𝐛𝐚𝐫
2

V
Work done 𝐖𝟏𝟐 = p1 V1 ln V2 = (15 × 105 ) × 0.15 ln(4) 𝐖𝟏𝟐 = 𝟑𝟏𝟏. 𝟗𝐤𝐉
1

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Process 2-3: Constant Volume Process, Work done 𝐖𝟐𝟑 = 0

p T 290
At =C p2 T3 = p3 T2 p3 = T3 × p2 = 550 × 3.75 𝐩𝟑 = 𝟏. 𝟗𝟖 𝐛𝐚𝐫
T 2

Process 3-1: Polytropic Process

At pV n = C p3 V3n = p1 V1n
Taking Log on
V3 p
ln p3 + n ln V3 = ln p1 + n ln V1 n(ln V3 − ln V1 ) = ln p3 − ln p1 n (ln ) = ln (p1 )
1 3
p 15
ln( 1 ) ln( )
p3 1.98
n= V n= 𝐧 = 𝟏. 𝟒𝟔
(ln 3 ) (ln(4))
V1

Work done:
p3 V3 −p1 V1 1.98×105 ×0.6−680×105 ×0.15
𝐖𝟑𝟏 = 𝐖𝟑𝟏 = 𝐖𝟑𝟏 = −𝟐𝟑𝟎. 𝟖𝟕𝐤𝐉
n−1 1.46−1

Total Work done

𝐖 = 𝐖𝟏𝟐 + 𝐖𝟐𝟑 + 𝐖𝟑𝟏 𝐖 = 𝟑𝟏𝟏. 𝟗 + 𝟎 + (−𝟐𝟑𝟎. 𝟖𝟕) 𝐖 = 𝟖𝟏. 𝟎𝟑𝐤𝐉
6. 0.2 m3 of air at 4 bar and 130°C is contained in a system. A reversible adiabatic expansion
takes place till the pressure falls to 1.02 bar. The gas is then heated at constant pressure till
enthalpy increases by 72.5 kJ. Calculate (i) The work done (ii) The index of expansion, if
the above processes are replaced by a single reversible polytropic. Process giving the same
work between the same initial and final states. Take Cp = 1 kJ/kg K, Cv = 0.714 kJ/kg K.
Given: V1= 0.2 m3 , P1=4 bar, T1= 130°C, P2= 1.02
bar, ∆h = 72.5kJ, Cp = 1 kJ/kgK, Cv = 0.714
kJ/kgK.
Find : (i) W , (ii) n, (iii)

Solution:
From the characteristic gas equation:
𝐤𝐉
R = CP − CV R = 1 − 0.714 𝐑 = 𝟎. 𝟐𝟖𝟔 .𝐊
𝐤𝐠
C 1
γ = CP γ = 0.714 𝛄 = 𝟏. 𝟒
V

Process 1-2: Reversible adiabatic process:

I I
γ γ P 1.4 4 1.402
p1 V1 = p2 V2 V2 = (P1 ) × V1 = (1.02) × 0.2 𝐕𝟐 = 𝟎. 𝟓𝟑 𝐦𝟑
2
γ−1 γ−1 1.4−1
T2 P2 γ P2 γ 1.02 1.4
From the relation, = (P ) T2 = (P ) × T1 T2 = ( ) × 403 𝐓𝟐 = 𝟐𝟕𝟐. 𝟕𝐊
T1 1 1 4

Work
p1 V1 −p2 V2 (4×105 ×0.2)−(1.02××105 ×0.53)
W12 = W12 = 𝐖𝟏𝟐 = 𝟔𝟒. 𝟖𝟓𝟎𝐤𝐉
γ−1 1.4−1

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Process 2-3: Constant pressure process

Q23 = mCp (T3 − T2 ) 72.5 = 0.694 × 1 × (T3 − 272.7) P1 V1 (4 × 105 ) × 0.2
m= =
T3 = 𝟑𝟕𝟕𝐊 RT1 286 × 403
𝐦 = 𝟎. 𝟔𝟗𝟒𝐤𝐠
V T 377
At =C V2 T3 = V3 T2 V3 = T3 × V2 = 272.7 × 0.53 𝐕𝟑 = 𝟎. 𝟕𝟑𝟐𝐦𝟐
T 2

W23 = p2 (V3 − V2 ) W23 = 1.02 × 105 (0.732 − 0.53) 𝐖𝟐𝟑 = 𝟐𝟎. 𝟔𝟎𝟒𝐤𝐉

Total Work done:

𝐖 = 𝐖𝟏𝟐 + 𝐖𝟐𝟑 𝐖 = 𝟔𝟒. 𝟖𝟓 + 𝟐𝟎. 𝟔𝟎𝟒 𝐖 = 𝟖𝟓. 𝟒𝟓𝟒𝐤𝐉

Index of expansion, n: (If the work by the polytropic process is the same)

p1 V1 −p3 V3 (4×102 ×0.2)−(1.02××102 ×0.53)

W13 = 85.454 = 𝐧 = 𝟏. 𝟎𝟔𝟐
n−1 n−1

7. A cylinder contains 0.45 m3 of a gas at 1 × 105 N/m2 and 80°C. The gas is compressed to a
volume of 0.13 m3, the final pressure being 5 × 105 N/m2. Determine :(i) The mass of gas (ii)
The value of index ‘n’ for compression (iii) The increase in internal energy of the gas (iv)
The heat received or rejected by the gas during compression.. Take  = 1.4, R = 294.2
J/kg°C.
Given: V1= 0.45 m3 , P1=1x 105 N/m2 , T1= T2 =80°C, V2= 0.13 m3 , P2=5x 105 N/m2
Find : (i) m , (ii) n , (iii) ΔU , (iv) Q

Solution:
(i) The mass of gas :
P1 V1 (1×105 )×0.45
m= m= 𝐦 = 𝟎. 𝟒𝟑𝟑𝐤𝐠
RT1 294.2×353

(ii) The value of index ‘n’ for compression

At pV n = C p1 V1n = p2 V2n
Taking Log on
V p
ln p1 + n ln V1 = ln p2 + n ln V2 n(ln V1 − ln V2 ) = ln p2 − ln p1 n (ln V1 ) = ln (p2 )
2 1
p 5
ln( 2 ) ln( )
p1 1
n= V n= 0.45 𝐧 = 𝟏. 𝟐𝟗𝟔 ≈ 𝟏. 𝟑
(ln 1 ) (ln
0.13
)
V2

(iii) The increase in internal energy of the gas (polytropic process)

T2 V n−1 V n−1 0.45 1.3−1
= (V1 ) T2 = (V1 ) × T1 T2 = (0.13) × 353 𝐓𝟐 = 𝟓𝟎𝟗. 𝟕𝐊
T1 2 2

0.2942
∆U = mCv (T2 − T1 ) ∆U = 0.433 × (509.7 − 353) ∆𝐔 = 𝟒𝟗. 𝟗𝐤𝐉
1.4−1

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(iv) The heat received or rejected by the gas during compression

p1 V1 −p2 V2 1×105 ×0.45−5×105 ×0.13
W = ∫ pdV W= W= 𝐖 = −𝟔𝟕. 𝟒𝟒𝐤𝐉
γ−1 1.3−1

Q = ∆U + W Q = 49.9 − 67.44 𝐐 = −𝟏𝟕. 𝟒𝟒𝐤𝐉

8. Three grams of nitrogen gas at 6 atm and 160°C in a frictionless piston cylinder device is
expanded adiabatically to double its initial volume, then compressed at constant pressure
to its initial volume and then compressed again at constant volume to its initial state.
Calculate the network done on the gas. Draw the P-V diagram for the processes.
Given:
m = 3g = 3 × 10−3 kg
P1 = 6atm = 6.08 × 102 kN/m2
T1 = 160℃ + 273 = 433K
V1 = V3
V2 = 2V1
Find:Network done?
Assume γ = 1.4
P1 V1 mRT1 3×10−3 ×0.287×433
m= V1 = V1 = 𝐕𝟏 = 𝟎. 𝟎𝟎𝟎𝟔𝟏𝒎𝟑
RT1 P1 6.08×102

V2 = 2 × 0.00061 𝐕𝟐 = 𝟏. 𝟐𝟐 × 𝟏𝟎−𝟑 𝒎𝟑
Process 1-2: Revesible adiabatic expansion:
p2 𝑉 𝛾 0.00061 1.4
γ γ
p1 V1 = p2 V2 = (𝑉1 ) p2 = 6.08 × 102 × (𝟏.𝟐𝟐×𝟏𝟎−𝟑 ) 𝐩𝟐 = 𝟐. 𝟓 × 𝟏𝟎𝟐 𝒌𝑵/𝒎𝟐
p1 2

p1 V1 −p2 V2 6.08×102 ×0.00061−2.5×102 ×𝟏.𝟐𝟐×𝟏𝟎−𝟑

W= W= 𝐖 = 𝟐𝟐𝟓. 𝟔𝟑𝐤𝐉
γ−1 1.4−1

Process 2-3: Constant pressure cooling

W23 = P(V3 − V2 ) W23 = 2.5 × 102 (0.00061 − 1.22 × 10−3 ) 𝐖𝟐𝟑 = −𝟎. 𝟏𝟓𝟐𝟓 𝐤𝐉

Process 2-3: Constant volume compression

𝐖𝟑𝟏 = 𝟎
Network done:
Wnet = 225.63 − 0.1525 + 0 𝐖𝐧𝐞𝐭 = 𝟐𝟐𝟓. 𝟒𝟖𝐤𝐉
9. A gas undergoes a thermodynamic cycle consisting of three processes beginning at an
initial state where p1 = 1 bar, V1 = 1.5 m3 and U1 = 512 kJ. The processes are as follows:
(i) Process 1–2: Compression with pV = constant to p2 = 2bar, U2 = 690 kJ,
(ii) Process 2–3: W23 = 0, Q23 = –150 kJ, and
(iii) Process 3–1: W31 = +50 kJ. Neglecting KE and PE changes,
Determine the heat interactions Q12 and Q31.

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3
Given: P1=1 bar , V1= 1.5 m , U1= 512kJ, P2=2 bar , U2= 690kJ , W23 = 0,
Q23 = –150 kJ, W31 = +50 kJ
Find : (i) Q12 , (ii) Q31

Solution:
Process 1–2: Isothermal Process
V 1
𝑄12 = ∆𝑈 + ∫ pdV 𝑄12 = 𝑈2 − 𝑈1 + p1 V1 ln (V2 ) 𝑄12 = (690 − 512) + 100 × 1.5 ln (2)
1

𝑄12 = 178 − 103.972 𝐐𝟏𝟐 = 𝟕𝟒. 𝟎𝟑𝐤𝐉

Process 2–3: Constant Volume Process
W23 = 0, Q23 = –150 kJ
Process 3–1: Adiabatic Expansion Process
Q23 = (U2 − U3 ) + W −150 = (690 − U3 ) + 0 𝐔𝟑 = 𝟓𝟒𝟎𝐤𝐉
𝑄31 = (U3 − U1 ) + W 𝑄31 = (540 − 512) + 50 𝐐𝟑𝟏 = 𝟐𝟐𝐤𝐉
10. A gas undergoes a thermodynamic cycle consisting of the following processes:
i. Process 1–2: Constant pressure p = 1.4 bar, V1 = 0.028 m3, W12 = 10.5 kJ
ii. Process 2–3: Compression with pV = constant, U3 = U2
iii. Process 3–1: Constant volume, U1 – U3 = – 26.4 kJ.
There are no significant changes in KE and PE. (a) Sketch the cycle on a p–V diagram, (b)
Calculate the network for the cycle in kJ (c) Calculate the heat transfer for process 1–2, (d)
Show that cycle ΣQ = ΣW.
Given: P1=1.4 bar , V1= 0.028 m3 , W12= 10.5kJ,
U3= U2 U1 – U3 = – 26.4 kJ,
Find: (i) Sketch the cycle, (ii) The net work for the
cycle (iii) 𝑄12 (iv) Show that cycle ΣQ = ΣW
Solution:
Process 1–2: Constant Pressure Process
𝑊12 = 𝑃(V2 − V1 ) 10.5 = 1.4 × 100(V2 − 0.028) 𝐕𝟐 = 𝟎. 𝟏𝟎𝟑𝐦𝟑
U1 − U3 = −26.4 U1 − U2 = −26.4 𝐔𝟐 = 𝐔𝟏 + 𝟐𝟔. 𝟒
𝑄12 = ∆𝐸 + ∫ pdV 𝑄12 = 𝑈2 − 𝑈1 + 𝑊12 𝑄12 = (26.4) + 10.5 𝐐𝟏𝟐 = 𝟑𝟔. 𝟗𝐤𝐉
W12 = 10.5kJ
Process 2–3: Isothermal compression Process
V 0.028
W23 = p2 V2 ln (V3 ) W23 = 1.4 × 100 × 0.103 ln (0.103) 𝐖𝟐𝟑 = −𝟏𝟖. 𝟕𝟖𝟑 𝐤𝐉
2

Q23 = (U2 − U3 ) + W23 Q23 = (0) + (−18.783) 𝐐𝟐𝟑 = −𝟏𝟖. 𝟕𝟖𝟑𝐤𝐉

Process 3–1: Constant Volume Process W31 = 0
Q31 = (U3 − U1 ) + W Q31 = (−26.4) + 50 𝐐𝟑𝟏 = 𝟐𝟔. 𝟒𝐤𝐉

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(b) The net work for the cycle

W = W12 + W23 + W31 W = 10.5 − 18.783 + 0 𝐖 = −𝟖. 𝟐𝟖𝟑𝐤𝐉
(d) Show that cycle ΣQ = ΣW
W12 + W23 + W31 = 𝑄12 + 𝑄23 + 𝑄31 −8.283 = 36.9 − 18.783 − 26.4
ΣQ = ΣW is proved
11. 90 kJ of heat are supplied to a system at a constant volume. The system rejects 95 kJ of
heat at constant pressure and 18 kJ of work is done on it. The system is brought to original
state by adiabatic process. Determine :
(i) The adiabatic work ;
(ii) The values of internal energy at all end states if initial value is 105 kJ.
Given:
Heat supplied at constant volume =90kJ
Heat rejected at constant pressure=-95kJ
Work done on the system = -18kJ
Initial value of internal energy = 105kJ
Solution:
Process l-m: Constant volume heat supplied
W𝑙−m = 0
Q𝑙−m = W𝑙−m + (Um − U𝑙 ) Q𝑙−m = (Um − U𝑙 ) 90 = (Um − 105) 𝐔𝐦 = 𝟏𝟗𝟓𝐤𝐉
Process m-n: Constant pressure heat rejected
Q𝑚−n = W𝑚−n + (Un − U𝑚 ) −95 = −18 + (Un − U𝑚 ) Un − U𝑚 = −77kJ 𝐔𝐧 = 𝟏𝟏𝟖𝐤𝐉
Process n-l: Adiabatic Expansion Q𝑛−𝑙 = 0, for Adiabatic process

∫ 𝛿𝑄 = 90 − 95 = −5kJ

∮ δQ = ∫ 𝛿𝑤 ∫ 𝛿𝑊 = −18 + 𝑊𝑛−𝑙 −5 = −18 + 𝑊𝑛−𝑙 𝐖𝐧−𝐥 = 𝟏𝟑𝐤𝐉

12. A fluid system, contained in a piston and cylinder machine, passes through a complete cycle of four
processes. The sum of all heat transferred during a cycle is -340 kJ. The system completes 200
cycles per min.
Process Q (kJ/min) W (kJ/min) E (kJ/min)
1-2 0 4340 -
2-3 42,000 0 -
3-4 - 4,200 0 -73,200
4-1 - - -
Complete the above table showing the method for each item, and compute the net rate of
work output in kW.

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Solution:
Sum of all heat transferred during the cycle = -340 kJ
Number of cycles completed by the system = 200 cycle/min
Process 1-2:
Q12 = W12 + ∆E12 0 = 4340 + ∆E12 ∆𝐄𝟏𝟐 = −𝟒𝟑𝟒𝟎 𝐤𝐉/𝐦𝐢𝐧
Process 2-3:
Q23 = W23 + ∆E23 42000 = 0 + ∆E23 ∆𝐄𝟐𝟑 = 𝟒𝟐𝟎𝟎𝟎 𝐤𝐉/𝐦𝐢𝐧
Process 3-4:
Q34 = W34 + ∆E34 −4200 = W34 − 73200 𝐖𝟑𝟒 = 𝟔𝟗𝟎𝟎𝟎 𝐤𝐉/𝐦𝐢𝐧
Process 4-1:
∑Cycle Q = −340 kJ/min
The system completes 200 cycles / min
Q12 + Q23 + Q34 + Q41 = 200 × (−340) 0 + 42000 − 4200 + Q41 = 200 × (−340)
𝐐𝟒𝟏 = −𝟏𝟎𝟓𝟖𝟎𝟎 𝐤𝐉/𝐦𝐢𝐧
Now, ∫ 𝑑𝐸 = 0 , since cyclic integral of any property is zero
∆E12 + ∆E23 + ∆E34 + ∆E41 = 0 −4340 + 42000 + (−73200) + ∆E41 = 0
∆𝐄𝟒𝟏 = 𝟑𝟓𝟓𝟒𝟎 𝐤𝐉/𝐦𝐢𝐧
Q41 = W41 + ∆E41 −105800 = W41 − 35540 𝐖𝟒𝟏 = −𝟏𝟒𝟏𝟑𝟒𝟎 𝐤𝐉/𝐦𝐢𝐧
kJ 68000
Rate od work output = −68000 min = − = 𝟏𝟏𝟑𝟑. 𝟑𝟑 𝐤𝐖
60

13. 680kg of fish at 5°C are to be frozen and stored at – 12°C. The specific heat of fish above
freezing point is 3.182, and below freezing point is 1.717 kJ/kg K. The freezing point is –
2°C, and the latent heat of fusion is 234.5 kJ/kg. How much heat must be removed to cool
the fish, and what per cent of this is latent heat?
kJ
Given: m=680 kg, T1=5°C, T2= – 2°C, T3= – 12°C, Cafp=3.182 ,
kg K
kJ
Cbfp=1.717 kg K, hfg=234.5 kJ/kg

Find : (i) Heat removed to cool the fish, (ii) percentage of this is latent heat.
Solution:
Heat to be removed above freezing point:
Qaf = mCafp (T1 − T2 ) Qaf = 680 × 3.182 × (5 − (−2)) 𝐐𝐚𝐟 = 𝟏𝟓. 𝟏𝟒𝟔𝐌𝐉
Latent heat to be removed:
QLH = m × hfg QLH = 680 × 234.5 𝐐𝐋𝐇 = 𝟏𝟓𝟗. 𝟒𝟔𝐌𝐉
Heat to be removed below freezing point:
Qaf = mCbfp (T2 − T3 ) Qaf = 680 × 1.717 × (−1 − (−12)) 𝐐𝐚𝐟 = 𝟏𝟏. 𝟔𝟕𝟔𝐌𝐉
Total Heat:
Q = Qaf + QLH + Qaf Q = 15.146 + 159.46 + 11.676 𝐐 = 𝟏𝟖𝟔. 𝟐𝟖𝟏𝟔𝐌𝐉

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14. A gas of mass 1.5 kg undergoes a quasi static expansion, which follows a relationship P=a+bV,
where ‘a’ and ‘b’ are constants. The initial and final pressures are 1000 kPa and 200 kPa
respectively and the corresponding volumes are 0.2 m3 and 1.2 m3. The specific internal energy of
the gas is given by the relation u=(1.5 pv-85) kJ/kg, where p is in kPa and V is in m3/kg. Calculate
the net heat transfer and the maximum internal energy of the gas attained during expansion.
Given: m=1.5 kg, P1=1000kPa , P2=200kPa , V1=0.2 m3, V2=1.2 m3,
Find : (i) The net heat transfer, (ii) The maximum internal energy

Solution:
p = a + bV, Where a and b are constants
Substitute P1 V1 and P2, V2, in the above quation to find the a and b are constants
1000 = a + 0.2b 1
200 = a + 1.2b 2
b = - 800
Substitute value b in the first equation
1000 = a + 0.2b 1000 = a + (0.2x(-800)) a=1160
The equation becomes
p = 1160 - 800V
Work done:
V V
W = ∫ pdV W = ∫V 2 [1160 − 800V] 𝑑𝑉 W = [1160V − 400V 2 ]V21
1

𝑊 = [1160(V2 − V1 ) − 400[𝑉22 − 𝑉12 ]]

𝑊 = [1160(1.2 − 0.2) − 400[1.22 − 0. 22 ]]

𝐖 = 𝟔𝟎𝟎 𝐤𝐉
Change in internal energy:
0.2
u1 = 1.5p1 v1 − 85 u1 = 1.5 × 1000 × 1.5 − 85 𝐮𝟏 = 𝟐𝟏𝟓 𝐤𝐉/𝐤𝐠
1.2
u2 = 1.5p2 v2 − 85 u2 = 1.5 × 200 × 1.5 − 85 𝐮𝟐 = 𝟏𝟓𝟓 𝐤𝐉/𝐤𝐠

∆U = m(u1 − u2 ) ∆U = 1.5(215 − 155) ∆𝐔 = 𝟔𝟎𝐤𝐉

(i) The net heat transfer
Q = ∆U + W Q = 60 + 600 𝐐 = 𝟔𝟔𝟎𝐤𝐉
(ii) The maximum internal energy
u = 1.5pv − 85 u = 1.5(1160 − 800v)v − 85 u = 1.5(1160v − 800v 2 ) − 85
∂u ∂u 1160
= 1.5(1160v − 1600v) − 85, For maximum u, ∂v = 0, v = 1600 = 0.725
∂v

umax = 1.5(1160 × 0.725 − 800 × 0.7252 ) − 85 𝐮𝐦𝐚𝐱 = 𝟑𝟑𝟓. 𝟓 𝐤𝐉/𝐤𝐠

Umax = m × umax Umax = 1.5 × 335.5 𝐔𝐦𝐚𝐱 = 𝟓𝟎𝟑. 𝟐𝟓𝐤𝐉

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15. A system of volume V contains a mass m of gas at pressure p and temperature T. The
macroscopic properties of the system obey the following relationship:
𝒂
[𝑷 + 𝟐 ] (𝑽 − 𝒃) = 𝒎𝑹𝑻
𝑽
Where a, b, and R are constants. Obtain an expression for the displacement work done by
the system during a constant-temperature expansion from volume V1 to volume
V2.Calculate the work done by a system which contains 10 kg of this gas expanding from 1
m3 to 10 m3 at a temperature of 293 K. Use the values a = 15.7 ×10 Nm4 , b = 1.07 ×10−2m3 ,
and R = 0.278 kJ/kg-K.
Given: m=10kg, V1=1 m3,V2=10 m3, a = 15.7 ×10 Nm4, b = 1.07 ×10−2m3, R = 0.278 kJ/kg-K
Find : The work done (W)

Solution:
Displacement work done by the system during a constant-temperature expansion.
a a
[P1 + 2]
(V1 − b) = [P2 + 2 ] (V2 − b) = mRT = K
V1 V2
From the equation
a K a
[P + V2 ] (V − b) = K P = (V−b) − V2
Work done
2 K a
W = ∫ pdV W = ∫1 [(V−b) − V2 ] dV
2 1 1
W = ∫1 [K ((V−b)) − a (V2 )] dV
1 2
W = [K ln(V − b) + a (V)]
1
(V2 −b) 1 1
W = K ln (V + a (V − V )
1 −b) 2 1
(10−1.07 ×10−2) 1 1
W = 10 × 0.278 × 293 ln (1−1.07 ×10−2)
+ 15.7 × 10 (10 − 1)
𝐖 = 𝟏𝟕𝟒𝟐. 𝟏𝟒𝐤𝐉
16. The properties of a certain fluid are related as follows:
u = 196 + 0.718t,
pv = 0.287 (t + 273)
Where u is the specific internal energy (kJ/kg), t is in °C, p is pressure (kN/m2), and v is
specific volume (m3/kg). For this fluid, find Cv and Cp.
Specific Heat at Constant Pressure:
∂h ∂(u+pv) ∂(196+0.718t+0.287(t+273)
CP = (∂T) CP = ( ) CP = ( ) T=t+273, 𝛛𝐓 = 𝛛𝐭
P ∂T P ∂T P
0.718 ∂t+0.287 ∂t
CP = ( ) 𝐂𝐏 = 𝟏. 𝟎𝟎𝟓 𝐤𝐉/𝐤𝐠. 𝐊
∂t P

Specific Heat at Constant Volume:

∂u ∂(196+0.718t
Cv = (∂T) Cv = ( ) 𝐂𝐯 = 𝟎. 𝟕𝟏𝟖 𝐤𝐉/𝐤𝐠. 𝐊
P ∂T P

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Problems on First law of thermodynamics Open system:

1. In a gas turbine unit, the gases flow through the turbine is 15 kg/s and the power developed
by the turbine is 12000 kW. The enthalpies of gases at the inlet and outlet are 1260 kJ/kg
and 400 kJ/kg respectively, and the velocity of gases at the inlet and outlet are50 m/s and
110 m/s respectively. Calculate :
(i) The rate at which heat is rejected to the turbine, and (ii) The area of the inlet pipe given
that the specific volume of the gases at the inlet is 0.45 m3/kg.

Given: m=15kg/s, W=12000kW,

h1=1260kJ/kg, h2=400 kJ/kg,
C1=50 m/s, C2= 110 m/s
v=0.45 m3/kg
Find : (i) Q (ii) AREA

Solution:
From steady flow energy Equation: (per unit mass)
𝐂𝟏𝟐 𝟐 𝐂𝟐
+ 𝐠𝐳𝟏 + 𝐮𝟏 + 𝐩𝟏 𝐯𝟏 + 𝐪𝟏−𝟐 = 𝟐𝟎𝟎𝟎 + 𝐠𝐳𝟐 + 𝐮𝟐 + 𝐩𝟐 𝐯𝟐 + 𝐰𝟏−𝟐
𝟐𝟎𝟎𝟎

Where,
h1 = u1 + p1 v1 and h2 = u2 + p2 v2 , z1 = z2
The equation becomes
C21 C22 502 1102 12000
+ h1 + q1−2 = + h2 + W + 1260 + q1−2 = 2000 + 400 + 𝐪𝟏−𝟐 = −𝟓𝟓 𝐤𝐉/𝐤𝐠
2 2 2000 15

Q = ṁ × −55 kJ/kg Q = 15 kg/s × 55 kJ/kg 𝐐 = 𝟖𝟐𝟖 𝐤𝐖

Inlet Area
From continutity equation
A1 C1 A×50
ṁ = 15 = 𝐀 = 𝟎. 𝟏𝟑𝟓𝐦𝟐
v1 0.45

2. In an air compressor air flows steadily at the rate of 0.5 kg/s through an air compressor. It
enters the compressor at 6 m/s with a pressure of 1 bar and a specific volume of 0.85 m3/kg
and leaves at 5 m/s with a pressure of 7 bar and a specific volume of 0.16 m3/kg. The
internal energy of the air leaving is 90 kJ/kg greater than that of the air entering. Cooling
water in a jacket surrounding the cylinder absorbs heat from the air at the rate of 60 kJ/s.
Calculate : (i) The power required to drive the compressor (ii) The inlet and output pipe
cross-sectional areas.

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Given: m=0.5 kg/s, W=12000kW,

C1=6 m/s, C2= 5 m/s
P1=1 bar, v1=0.85 m3/kg
P2=7 bar, v2=0.16 m3/kg
Δu= - 90 kJ/kg, Q̇= - 60 kJ/s
Find : (i) W (ii) A1 and A2

Solution:
From steady flow energy Equation: (per unit mass)
𝐂𝟏𝟐 𝟐 𝐂𝟐
+ 𝐠𝐳𝟏 + 𝐮𝟏 + 𝐩𝟏 𝐯𝟏 + 𝐪𝟏−𝟐 = 𝟐𝟎𝟎𝟎 + 𝐠𝐳𝟐 + 𝐮𝟐 + 𝐩𝟐 𝐯𝟐 + 𝐰𝟏−𝟐
𝟐𝟎𝟎𝟎

Where, z1 = z2
C12 − C22 Q̇
w= + (u1 − u2 ) + (p1 v1 − p2 v2 ) +
2000 m
62 − 52 (1 × 105 × 0.85) − (7 × 105 × 0.16) −60
w= + (−90) + +( )
2000 1000 0.5
𝐰 = −𝟐𝟑𝟕 𝐤𝐉/𝐤𝐠
Therefore,
Ẇ = ṁ × 237 kJ/kg W = 0.5 kg/s × 237 kJ/kg 𝐖 = 𝟏𝟏𝟖. 𝟓 𝐤𝐖
Inlet Area and Exit Area
From continutity equation
A1 C1 A1 ×6
ṁ = 0.5 = 𝐀 𝟏 = 𝟎. 𝟎𝟕𝟎𝟖𝐦𝟐
v1 0.85
A2 C2 A1 ×5
̇ =
m 0.5 = 𝐀 𝟐 = 𝟎. 𝟎𝟏𝟔 𝐦𝟐
v2 0.16

3. 12 kg of air per minute is delivered by a centrifugal air compressor. The inlet and outlet
conditions of air are C1 = 12 m/s, p1 = 1 bar, v1 = 0.5 m3/kg and C2 = 90 m/s,p2 = 8 bar, v2 =
0.14 m3/kg. The increase in enthalpy of air passing through the compressor is150 kJ/kg and
heat loss to the surroundings is 700 kJ/min. Find : (i) Motor power required to drive the
compressor ; (ii) Ratio of inlet to outlet pipe diameter. Assume that inlet and discharge
lines are at the same level.
Given: m=0.2 kg/s, C1= 12 m/s, C2= 90 m/s
P1=1 bar, v1=0.5 m3/kg
P2= 8 bar, v2=0.14 m3/kg
Δh= 150 kJ/kg, Q= - 11.67 kJ/s
d
Find : (i) W, (ii) d1
2

Solution:

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020

(i) Motor power required to drive the compressor

From steady flow energy Equation: (per unit time)
𝟐 𝟐
𝐂𝟏 𝐂𝟐
𝐦̇ (𝟐𝟎𝟎𝟎 + 𝐠𝐳𝟏 + 𝐮𝟏 + 𝐩𝟏 𝐯𝟏 ) + 𝐐̇ = 𝐦̇ (𝟐𝟎𝟎𝟎 + 𝐠𝐳𝟐 + 𝐮𝟐 + 𝐩𝟐 𝐯𝟐 ) + 𝐖̇

Where, z1 = z2 , h1 = u1 + p1 v1 and h2 = u2 + p2 v2
C1 −C2 2 2 12 −90 2 2
Ẇ = ṁ [ 2000 + (h1 − h2 )] + Q̇ Ẇ = 0.2 [ 2000 + (−150)] + (−11.67)

𝐖̇ = −𝟒𝟐. 𝟒𝟔 𝐤𝐉/𝐬 𝐖 = −𝟒𝟐. 𝟒𝟔 𝐤𝐖

(ii) Ratio of Inlet to outlet pipe diameter
From continutity equation
A1 C1 A2 C2 A1 C v A1 90 0.5 𝐀𝟏 𝐝𝟏
ṁ = = = v2 × C1 = 0.14 × 12 = 𝟐𝟔. 𝟕𝟖 = 𝟓. 𝟏𝟕𝟓
v1 v2 A2 2 1 A2 𝐀𝟐 𝐝𝟐

4. At the inlet to a certain nozzle the enthalpy of fluid passing is 2800 kJ/kg, and the velocity
is 50 m/s. At the discharge end the enthalpy is 2600 kJ/kg. The nozzle is horizontal and
there is negligible heat loss from it.
(i) Find the velocity at exit of the nozzle.
(ii) If the inlet area is 900 cm2 and the specific volume at inlet is 0.187 m3/kg, find
the mass flow rate.
(iii) If the specific volume at the nozzle exit is 0.498 m3/kg, find the exit area of
nozzle
Given: h1= 2800 kJ/kg, C1= 50 m/s,
h2= 2600 kJ/kg
A1=900 cm2, v1=0.187 m3/kg
v2=0.498 m3/kg
Find : (i) C2, (ii) ṁ , (iii) A2
Solution:
(i) The velocity at exit of the nozzle
From steady flow energy Equation: (per unit mass)
𝐂𝟏𝟐 𝟐 𝐂𝟐
+ 𝐠𝐳𝟏 + 𝐮𝟏 + 𝐩𝟏 𝐯𝟏 + 𝐪𝟏−𝟐 = 𝟐𝟎𝟎𝟎 + 𝐠𝐳𝟐 + 𝐮𝟐 + 𝐩𝟐 𝐯𝟐 + 𝐰𝟏−𝟐
𝟐𝟎𝟎𝟎

Where, z1 = z2 , Q=0, W=0, h1 = u1 + p1 v1 and h2 = u2 + p2 v2

C22 C2
1
= 2000 + (h1 − h2 ) C22 = 502 + 2000(2800 − 2600) 𝐂𝟐 = 𝟔𝟑𝟒. 𝟒 𝐦/𝐬
2000

(ii) Mass flow rate

From continutity equation
𝐴1 𝐶1 900×10−4 ×50
𝑚̇ = 𝑚̇ = 𝒎̇ = 𝟐𝟒. 𝟎𝟔 𝒎/𝒔
𝑣1 0.187

(iii) Area at the exit

𝐴2 𝐶2 𝐴2 ×634.4
𝑚̇ = 24.06 = 𝑨𝟐 = 𝟎. 𝟎𝟏𝟖𝟖𝟖𝟕 𝒄𝒎𝟐
𝑣2 0.498

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5. Air at a temperature of 20°C passes through a heat exchanger at a velocity of 40 m/s where
its temperature is raised to 820°C. It then enters a turbine with same velocity of 40 m/s and
expands till the temperature falls to 620°C. On leaving the turbine, the air is taken at a
velocity of 55 m/s to a nozzle where it expands until the temperature has fallen to510°C. If
the air flow rate is 2.5 kg/s, calculate :
(i) Rate of heat transfer to the air in the heat exchanger
(ii) The power output from the turbine assuming no heat loss
(iii) The velocity at exit from the nozzle, assuming no heat loss

Given: t1= 20°C, C1= 40 m/s,

t2= 820°C, C2= 40 m/s
t3=620°C, C3=55 m/s
t4=510°C, 𝑚̇ = 2.5 𝑘𝑔/𝑠
Find : (i) Q (ii) W , (iii) C4

Solution:
(i) Rate of heat transfer to the air in the heat exchanger
From steady flow energy Equation:
𝟐 𝟐
𝑪𝟏 𝒈𝒛𝟏 𝑪𝟐 𝒈𝒛𝟐
𝒎̇ (𝟐𝟎𝟎𝟎 + 𝟏𝟎𝟎𝟎 + 𝒖𝟏 + 𝒑𝟏 𝒗𝟏 ) + 𝑸̇𝟏𝟐 = 𝒎̇ (𝟐𝟎𝟎𝟎 + 𝟏𝟎𝟎𝟎 + 𝒖𝟐 + 𝒑𝟐 𝒗𝟐 ) + 𝑾̇𝟏−𝟐

Where, 𝑧1 = 𝑧2, 𝐶1 = 0, 𝐶2 = 0 , W=0, ℎ1 = 𝑢1 + 𝑝1 𝑣1 and ℎ2 = 𝑢2 + 𝑝2 𝑣2

𝑄̇12 = 𝑚̇(ℎ2 − ℎ1 ) 𝑄̇12 = 𝑚̇𝐶𝑃 (𝑡2 − 𝑡1 ) = 2.5 × 1.005(820 − 20) 𝑸̇𝟏𝟐 = 𝟐𝟎𝟏𝟎𝒌𝑱/𝒔

(ii) The power output from the turbine assuming no heat loss
𝟐 𝟐
𝑪𝟐 𝒈𝒛𝟐 𝑪𝟑 𝒈𝒛𝟑
𝒎̇ (𝟐𝟎𝟎𝟎 + 𝟏𝟎𝟎𝟎 + 𝒖𝟐 + 𝒑𝟐 𝒗𝟐 ) + 𝑸̇𝟐𝟑 = 𝒎̇ (𝟐𝟎𝟎𝟎 + 𝟏𝟎𝟎𝟎 + 𝒖𝟑 + 𝒑𝟑 𝒗𝟑 ) + 𝑾̇𝟐𝟑

Where, z2 = z3 , h2 = u2 + p2 v2 and h3 = u3 + p3 v3 , Q = 0

𝐶2 −𝐶3 2 2 2
𝐶2 −𝐶32
𝑊̇23 = 𝑚̇ [ 2000 + (ℎ2 − ℎ3 )] 𝑊̇23 = 𝑚̇ [ 2000 + 𝐶𝑃 (𝑡2 − 𝑡3 )]

40 −55 2 2
Ẇ23 = 2.5 [ 2000 + 1.005(820 − 620)] 𝐖̇ = 𝟓𝟎𝟒. 𝟑 𝐤𝐉/𝐬 𝐖̇ = 𝟓𝟎𝟒. 𝟑 𝐤𝐖

(iii) The velocity at exit of the nozzle

𝐂𝟐 𝐠𝐳 𝐂𝟐 𝐠𝐳
𝟑
𝐦̇ (𝟐𝟎𝟎𝟎 𝟑
+ 𝟏𝟎𝟎𝟎 + 𝐮𝟑 + 𝐩𝟑 𝐯𝟑 ) + 𝐐̇𝟑𝟒 = 𝐦̇ (𝟐𝟎𝟎𝟎
𝟒 𝟒
+ 𝟏𝟎𝟎𝟎 + 𝐮𝟒 + 𝐩𝟒 𝐯𝟒 ) + 𝐖̇𝟑𝟒

Where, z3 = z4 , Q=0, W=0, h3 = u3 + p3 v3 and h4 = u4 + p4 v4

C24 C23 C24 552

= + CP (t 2 − t 3 ) = 2000 + 1.005(620 − 510) 𝐂𝟒 = 𝟒𝟕𝟑. 𝟒 𝐦/𝐬
2 2 2

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020

ADDITIONAL SOLVED PROBLEMS

1. A piston-cylinder assembly contains air (ideal gas with 𝛄=1.4) at 200 kPa and occupies a volume of
0.01 m3 . The piston is attached to one end of a spring and the other end of the spring is fixed to a
wall. The force exerted by the spring on the piston is proportional to the decrease in the length of
the spring from its natural length. The ambient atmospheric pressure is 100 kPa. Now, the air in
the cylinder is heated till the volume is doubled and at this instant it is found that the pressure of
the air in the cylinder is 500 kPa. Calculate the work done by the gas. [NOV/DEC 2015]
Solution:

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020

2. An insulated rigid tank having 5 kg of air at 3 atm and 30°C is connected to an air supply line at 8
atm and 50°C through a valve. The valve is now slowly opened to allow the air from the supply line
to flow into the tank until the tank pressure reaches 8 atm, and then the valve is closed. Determine
the final temperature of the air in the tank. Also, find the amount of air added to the tank.
Solution: [NOV/DEC 2015]

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020

3. A vessel of constant volume 0.3 m3 contains air at 1.5 bar and is connected via a valve, to a large
main carrying air at a temperature of 38°C and high pressure. The valve is opened allowing air to
enter the vessel and raising the pressure therein to 7.5 bar. Assuming the vessel and valve to be
thermally insulated, find the mass of air entering the vessel. [APR/MAY 2015]

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020

4. Prove that energy is a property of a system.

Property is characteristic nature of substance (system) expressed in terms of some numerical
values followed by some unit. During process, properties get changed.
Energy can be classified as stored energy and transit energy. Stored energy (internal energy)
is a property. Heat and work are transit energies
We can prove that energy is a property - by first law of TD. As per law, on completion of a cycle the
net change in property value should be zero. (∑ 𝑄)𝑐𝑦𝑐𝑙𝑒 = (∑ 𝑊)𝑐𝑦𝑐𝑙𝑒 or (∑ ∆U)Cycle = 0
by applying 1st law of TD to the processes A, B and C which are carried out between the
states 1&2

Internal energy is not a path function. It is a point function. So, it is proved that energy is
property

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ME 8391- Engineering Thermodynamics Mechanical Engineering 2019-2020
5. A rigid tank containing 0.4 m3 of air at 400 kPa and 30°C is connected by a valve to a piston
cylinder device with zero clearance. The mass of the piston is such that a pressure of 200kPa is
required to raise the piston. The valve is opened slightly and air is allowed to flow into the cylinder
until the pressure of the tank drops to 200 kPa. During this process, heat is exchanged with the
surrounding such that the entire air remains at 30°C at all times. Determine the heat transfer for
this process. [NOV/DEC 2010]

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