19-200-0105B –

BASIC MECHANICAL
ENGINEERING
Module - 1

Asif S
Assistant Professor
Division of Mechanical Engineering
School of Engineering
CUSAT
Module I
 Module I
Thermodynamics: Thermodynamics systems – open, closed and isolated
systems, equilibrium state of a system, property and state, process, cycle,
Zeroth law of thermodynamics- concept of temperature, temperature scales.
First law – internal energy, enthalpy, work and heat, Different processes
(isobaric, isochoric, isothermal, adiabatic and polytropic processes). Second
law – Kelvin-plank and Claussius statements and their equivalence, Carnot
Cycle (Elementary problems only). Thermodynamic properties of Steam,
Steam Generator. Different types of boilers, boiler mountings and accessories.
Formation of steam at constant pressure, working of steam turbines,
compounding of turbines.
 Thermodynamics
 Thermodynamics is the science that deals with heat and work.

 The word "thermodynamics" was derived from the Greek words thermé
(heat) and dynamics (force).

 The science of energy, that concerned with the ways in which energy is
stored within the body

 Thermodynamics is that part of science which is concerned with the

conditions that material systems may assume and the changes in conditions
that may occur either spontaneously or as a result of interactions between
systems.
 Macroscopic and Microscopic Approach
 A thermodynamic analysis can be carried out either by considering the gross
behavior of matter (Macroscopic approach) or by considering the behavior
of individual molecules of the matter (Microscopic approach).

 Macroscopic Approach (Classical thermodynamics):- Effect of many

molecules together – effects can be perceived by human senses and can be
measured directly.

 Microscopic Approach (Statistical thermodynamics):- Matter is

considered to be composed of molecules and the analysis is carried out by
considering the position, velocity and energy of each molecule at a given
instant.
 Thermodynamic System
 A system is any prescribed and identifiable collection of matter.

 It can be any object, any quantity of matter, any region of space etc.

 Surroundings:- The matter or region outside the system.

 Boundary:- Real or imaginary envelope which encloses a system and

separate it from its surroundings.
 Thermodynamic System
 An open system can exchange both energy and matter with its
surroundings.

 A closed system, on the other hand, can exchange only energy with its
surroundings, not matter.

 An isolated system is one that cannot exchange either matter or energy with
its surroundings
Thermodynamic System
 Thermodynamic State of a System
 The condition of physical existence of a system at any instant is called its
state.

 The state of a thermodynamic system is described by specifying its

thermodynamic co-ordinates – pressure, temperature, volume etc.

 The state of a thermodynamic system is its condition or configuration which

can be well defined by the co-ordinates.

 The state of the system can be described by a number of state quantities that
do not depend on the process by which the system arrived at its state.
 Thermodynamic Equilibrium
 When a system is isolated from its surroundings all its properties such as
pressure, temperature, density etc may change with time but eventually
these changes will cease – Equilibrium.

 Equilibrium – No unbalanced potential which tends to promote a change of

state.
 Conditions - Thermodynamic Equilibrium

 Mechanical equilibrium :- No unbalanced forces within the system - there

is no pressure gradient within the system.

 Thermal equilibrium:- No variation of temperature within the system.

 Chemical equilibrium:- No chemical reaction within the system.

 Thermodynamic equilibrium:- A system which is in mechanical, thermal

and chemical equilibriums.

 Properties of a system
 Properties of a system is a measurable characteristic of a system that is in
equilibrium.

 Properties may be intensive or extensive

 Intensive – Are independent of the amount of mass: eg. Temperature,

Pressure, and Density,

 Extensive – varies directly with the mass. eg. mass, volume, energy,
enthalpy
 Properties of a system
 Specific properties – The ratio of any extensive property of a system to that
of the mass of the system is called an average specific value of that property
(also known as intensives property)
 Process
 If any one or more properties of the system undergo a change due to energy

or mass transfer we say that the system has undergone a change of state

 The successive change of state of the system due to energy or mass transfer

defined by definite path is called a process.

 The curve joining the successive state represents the process path

 If a system undergoes two or more processes and returns to its original state

after conclusion of processes, the system is said to have undergone a cycle

 Thermodynamic process
 Process – change from one equilibrium state to another.

Property held
Process
constant

Isobaric Pressure

Isothermal Temperature

Isochoric Volume

Isentropic Entropy
 Thermodynamic processes
 The prefix iso- is often used to designate a process for which a particular
property remains constant.

 Isobaric process: A process during which the pressure P remains constant.

 Pressure is Constant (ΔP = 0)

 Thermodynamic process
 Isothermal process: A process during which the temperature T remains
constant.
 Thermodynamic processes
 Isochoric process: A process during which the volume V remains constant
 Thermodynamic processes
 Isentropic process (Reversible adiabatic): A process during which the
entropy S remains constant. – No heat transfer
 Types of Thermodynamic processes
 Cyclic process:- When a system in a given initial state goes through various

processes and finally return to its initial state, the system has undergone a

cyclic process or cycle.

 Reversible process:- It is defined as a process that, once having take place it

can be reversed. In doing so, it leaves no change in the system or boundary.

 Irreversible process:- A process that cannot return both the system and

surrounding to their original conditions

 Types of Thermodynamic processes
 Adiabatic process:- A process that has no heat transfer into or out of the

system. It can be considered to be perfectly insulated.

 Isentropic process:- A process where the entropy of the fluid remains

constant.

 Polytropic process:- When a gas undergoes a reversible process in which

there is heat transfer, it is represented with a straight line, PVn = constant.

 Throttling process:- A process in which there is no change in enthalpy, no

work is done and the process is adiabatic.

 Point Function
 When two co-ordinates are located on the graph, They define a

point and the two properties on the graph define state. These

properties (p, T, v) are called point function.

 They depend on the state only, and not on how a system reaches

that state.
 Path Function
 If a thermodynamic system passes through a series of states, it is
said to describe a path.

 There are certain quantities like heat and work can not be located
on a graph by a point but there are represented by the area. It is not
a state or point function, rather it depends on the path of the
process. Such quantities are called path function and they are
inexact differentials

 If the value of thermodynamic variable depends upon the path

followed in going from one state to another, then the variable is a
path function.
 Quasi Static Process
 A quasi static process is defined as a process in which the
properties of the system depart infinitesimally (extremely small)
from the thermodynamic equilibrium state.

 If the properties of the system has finite departures from

thermodynamic equilibrium state the process is said to be non
quasi static

 Quasi static process is the succession of thermodynamic

equilibrium state while in case of non-quasi static process the end
states only represent the thermodynamic equilibrium.
 Condition for Reversible process

 No Friction

 Heat transfer is through infinitely temperature difference.

 There are no spontaneous changes in the system.

All processes in nature are irreversible.

Cyclic Process
 Zeroth Law of Thermodynamics

“If two systems are in thermal equilibrium with a third system,

separately, then they are also in thermal equilibrium with each other.’
 Concept of Temperature

 Temperature of a system is an intensive property that

determines whether or not a system is in thermal equilibrium

with another system.

 It can also be defined as the sense of hotness or coldness of a

body when we touch the body.

 It is the driving potential causing the flow of energy as heat.

 Heat
 Heat is the form of energy which transfer without transfer of
mass, from one body to another body (or between system and
surroundings) from higher temperature to lower temperature by
virtue of temperature difference between two bodies.

 Abbreviated as ‘Q’ and Unit is J (Joule)

 Heat Addition into system :- Positive ( +Q)

 Heat Rejection from system :- Negative (-Q)

 Total energy is Extensive Property and heat is a Path Function

(Inexact Differential)
 Work
 Analogous to heat, work is also a transient form of energy

which is observed when it crosses the boundaries of the system

without transfer of mass.

 It is a path function.

 Small work done due to displacement dS

 Work done in moving boundary of close system in
quasi-static process displacement work
 Force Exerted on piston, F = P A

 Small work done, δw = F dl

 δw = P A dl

 δw = P dV
Work done in moving boundary of close system in
quasi-static process displacement work
 Difference between heat and work

 Heat can only transfer when there is difference of

temperature between the system and surrounding, while work
transfer can take place even without the change in
temperature

 In constant volume process though work can not take place,

however heat can be transferred.

 In case of work transfer, its sole effect could be raising or

lowering a weight in the surrounding but in case of heat
transfer other effects are also observed.
 Similarities between heat and work

 Both heat and work exist in transit and these are never

possessed or contained in a system.

 Both heat and work refer to boundary phenomena.

 Both heat and work are path function and do not represent as

the properties of system (Inexact difference)

 Temperature

 Temperature is the degree of sensible heat or cold,

expressed in terms of a specific scale.

 Several scales and units exist for measuring

temperature, the most common being Celsius,

Fahrenheit, and especially in science Kelvin.

 Celsius Scale
 The Celsius scale is used for common temperature
measurements in most of the world.

 It is an empirical scale.

 Its zero point is zero degree Celsius being defined by the

freezing point of water, and 100 degree Celsius was the
boiling point of water, both at sea level atmospheric pressure.

 Because of the hundred degree interval, it is called a

centigrade scale.
 Fahrenheit Scale
 The United States commonly uses the Fahrenheit scale, on
which water freezes at 32 degree Fahrenheit and boils at 212
degree Fahrenheit at sea level atmospheric pressure.
 Kelvin Scale
 Many scientific measurements use the Kelvin temperature
scale.

 It is a thermodynamic or absolute temperature scale.

 Its zero point 0 K, is defined to coincide with coldest

physically-possible temperature.

 The temperature of absolute zero occurs at 0 K= -273.15

degree Celsius and the freezing point of water at sea level
atmospheric pressure occurs at 273.15K= Zero degree
Celsius.
Pressure
Pressure
Boyle’s Law

T = Constant

P1V1 = P2V2 = Constant

 Charles’ Law
P = Constant

T1/V1 = T2/V2 = Constant

First Law
 First Law

 Limitations of the first law of thermodynamics

 No restriction on the direction of the flow of heat: the first

law establishes definite relationship between the heat
absorbed and the work performed by a system. The first law
does not indicate whether heat can flow from a cold end to a
hot end or not. For example: we cannot extract heat from the
ice by cooling it to a low temperature. Some external work
has to be done.
 First Law

 Limitations of the first law of thermodynamics

 Does not specify the feasibility of the reaction: first law does
not specify that process is feasible or not for example: when a
rod is heated at one end then equilibrium has to be obtained
which is possible only by some expenditure of energy.

 Practically it is not possible to convert the heat energy into an

equivalent amount of work.
 Enthalpy

 It is total energy of the system

 Specific Enthalpy,

 U, p, V are point function, so H is point function and

property of system

 Unit of Enthalpy (H) is kJ

 Unit of Specific Enthalpy (h) is kJ/kg

 H = U + pV

 h = u + pv
 Internal Energy
 Internal energy is defined as the energy associated with the
random, disordered motion of molecules. It is separated in
scale from the macroscopic ordered energy associated with
moving objects; it refers to the invisible microscopic
energy on the atomic and molecular scale.

 For example, at room temperature glass of water sitting on a

table has no apparent energy, either potential or kinetic. But
on the microscopic scale it is a mass with high speed
molecules traveling at hundreds of meters per second.
Perpetual Motion Machine of first kind (PMM1)

 It is a device, which operates in a cycle without receiving

energy from any source, but delivers work to the
surroundings.

 Such a device would create work from nothing.

 This is clearly a violation of the first law of thermodynamics

and hence a perpetual motion machine of the first kind is
impossible.
 The Second Law of Thermodynamics

 Work can be completely converted into heat in a single

process but continuous conversion of heat in to work
requires a cyclic process ( a heat engine)

 All attempts to construct a heat engine that is 100% efficient

have failed

 The Kelvin- Planck statement of the second law of

thermodynamics is a qualitative statement of the
impossibility of certain types of processes
 Kelvin- Planck statement

It is impossible to construct an engine which operating in a

cycle will produce no effect other than the exchange of heat
from a single reservoir and produce an equal amount of work.

OR

It is impossible for an engine working in a cycle to transform

a given amount of heat from a reservoir completely into work.
In other words, no heat engine can have a thermal efficiency
of 100%
Kelvin- Planck statement
 Clausius statement

It is impossible for a self acting machine working in a cyclic

process, to transfer heat from a body at a lower temperature to
a body at a higher temperature, without the aid of an external
agency.

OR
Heat cannot, of itself, pass from a colder to a hotter body, i.e.,
heat cannot flow from itself from a system at low temperature
to a system at high temperature. The only alternative is that
some external work must be supplied to the machine
Clausius statement
 Perpetual Motion Machine of Second Kind
(PMM2)
 Without violating the first law, a machine can be imagined
which would continuously absorb heat from a single thermal
reservoir and would convert this heat completely into work.
 The efficiency of such a machine would be 100%.
 This machine is called the perpetual motion machine of the
second kind (PMM2)
 When the thermal energy is equivalent to the work done, this
does not violate the law of conservation of energy. However it
does violate the more subtle second law of thermodynamics
 Carnot Cycle
 The cycle was first suggested by Sadi Carnot, in 1824, which
works on reversible cycle
 Any fluid may be used to operate the Carnot cycle, which is
performed in an engine cylinder the head of which is supposed
alternatively to be perfect conductor or a perfect insulator of a
heat
 Heat is caused to flow into the cylinder by the application of
high temperature energy source to the cylinder head during
expansion, and to flow from the cylinder by the application of a
lower temperature energy source to the head during compression
Carnot Cycle
 Carnot Cycle
The assumptions made for describing the working of the Carnot
engine are as follows:
 The piston moving in a cylinder does not develop any friction during
motion
 The walls of piston and cylinder are considered as perfect insulators of
heat
 The cylinder head is so arranged that it can be a perfect heat conductor
or perfect heat insulator
 The transfer of heat does not affect the temperature of source or sink
 Working medium is a perfect gas and has constant specific heat
 Compression and expansion are reversible
Carnot Cycle
Carnot Cycle
Carnot Cycle
 Efficiency of a Reversible Heat Engine

 From the above expression, it may be noted that as T2 decreases

and T1 increases, efficiency of the reversible cycle increases
 Since 𝜂 is always less than unity, T2 is always greater than zero
and positive (+ ve)
 Carnot cycle cannot be performed in practice
because of the following reasons
 It is impossible to perform a frictionless process
 It is impossible to transfer the heat without temperature potential
 Isothermal process can be achieved only if the piston moves very slowly to
allow heat transfer so that the temperature remains constant
 Adiabatic process can be achieved only if the piston moves as fast as
possible so that the heat transfer is negligible due to very short time
available
 The isothermal and adiabatic processes take place during the same
stroke therefore the piston has to move very slowly for part of the
stroke and it has to move very fast during remaining stroke
 This variation of motion of the piston during the same stroke is not
possible
 Problem No. 1
5 kg of gas contained in a cylinder fitted with a piston. 160 kJ of heat is
transferred to the gas and simultaneously the piston is forced to compress the
gas with an expenditure of work equivalent to 120 kJ. Determine the change in
Specific Internal Energy of the gas.

Δu = ?
 Problem No. 1
5 kg of gas contained in a cylinder fitted with a piston. 160 kJ of heat is
transferred to the gas and simultaneously the piston is forced to compress the
gas with an expenditure of work equivalent to 120 kJ. Determine the change in
Specific Internal Energy of the gas.

Δu = ?
M = 5 kg
Q = 160 kJ
W = -120kJ
ΔU = Q – W = 160 – (-120) = 160 +120 = 280 kJ
Δu = ΔU/m
= 280/5 = 56 kJ/kg = 56000 J/kg
 Problem No. 2
A tank containing a fluid is stirred by a stirrer. The power input to the
stirrer is 3 kW. Heat is transferred from the tank at the rate of 6000 kJ/hour.
Considering the tank and the fluid as the system, determine the change in
internal energy of the system in one hour.
ΔU = ?
 Problem No. 2
A tank containing a fluid is stirred by a stirrer. The power input to the
stirrer is 3 kW. Heat is transferred from the tank at the rate of 6000 kJ/hour.
Considering the tank and the fluid as the system, determine the change in
internal energy of the system in one hour.
ΔU = ?
W = -3 kW = -3 kJ/s = -3 x 60 x 60 = -10800 kJ/hr.
Q = -6000 kJ/hr
ΔU = Q – W = -6000 – (-10800) = -6000 + 10800
= 4800 kJ/hr.
 Problem No. 3
One kg of an ideal gas is heated from 20 degree Celsius to 100 degree
Celsius. Assuming R = 264 J/kgK. and γ = 1.18 for the gas, find (i)
Specific heats (ii) Change in internal energy (iii) Change in enthalpy.
cp = ?
cv = ?
 Problem No. 3
One kg of an ideal gas is heated from 20 degree Celsius to 100 degree
Celsius. Assuming R = 264 J/kgK. and γ = 1.18 for the gas, find (i)
Specific heats (ii) Change in internal energy (iii) Change in enthalpy.
cp = ?
cv = ?
ΔU = m.cv. ΔT = ?
ΔH = m.cp. ΔT = ?
m = 1 kg
T1 = 20 + 273 = 293 K
T2 = 100 + 273 = 373 K
ΔT = 373 – 293 = 80 K
R = 264
γ = 1.18
cv = R/(γ -1) = 264/0.18 = 1467 J/kgK
cp = γR/(γ -1) = 1730.66 J
ΔU = 117328 J = 117.32 kJ
ΔH = 138448 J = 138.448 kJ
 Isothermal Process

pV = C

ΔU = m.cv. ΔT = 0

ΔH = m.cp. ΔT = 0

W = p1V1. ln(V2/V1) = p1V1. ln(p1/p2)

Q = p1V1. ln(V2/V1) = p1V1. ln(p1/p2)

 Isochoric Process

V=C

ΔU = m.cv. ΔT

ΔH = m.cp. ΔT

W=0

Q = m.cv. ΔT
 Isobaric Process

p=C

ΔU = m.cv. ΔT

ΔH = m.cp. ΔT

W = p(V2-V1)

Q = m.cp. ΔT
 Adiabatic Process

pVγ = C

ΔU = m.cv. ΔT

ΔH = m.cp. ΔT

W = (p1V1-p2V2)/(γ-1) = mR(T1-T2)/(γ-1)

Q=0
 Adiabatic Process

pVT relationship

p1V1γ = p2V2γ = C

p1/p2 = (V2/V1) γ = (T1/T2)(γ/γ-1)

 Polytropic Process

pVn = C

ΔU = m.cv. ΔT

ΔH = m.cp. ΔT

W = (p1V1-p2V2)/(n-1) = mR(T1-T2)/(n-1)

Q = W x (γ-n)/ (γ-1)
 Polytropic Process

pVT relationship

p1V1n = p2V2n = C

p1/p2 = (V2/V1) n = (T1/T2)(n/n-1)

 Problem No. 4
How much heat will be necessary to raise the temperature of 1.5 kg of a

gas from 90 degree Celsius to 225 degree Celsius, the volume remaining

constant during the heat supply. Specific heat of a gas at constant volume

may be taken as 0.718 kJ/kg.K

 Problem No. 4
m = 1.5 kg

T1 =90 deg C

T2 = 225 deg C

ΔT = 135 K

cv = 0.718 kJ/kg.K

W=0

Q – W = ΔU = Q

Q = mcv ΔT = 1.5 x 0.718 x 135 = 145.395 kJ

 Problem No. 5
An insulated cylinder of capacity 2.8 m^3 contains 15 kg of nitrogen.

Paddle work is done on the gas till the pressure inside the cylinder

increases from 5 bar to 10 bar. Determine ,

(i) Change in internal energy

(ii) Work done

(iii) Heat transfer

Assume cp = 1.04 kJ/kg.K and cv = 0.7432 kJ/kg.K

 Problem No. 5
V1 = 2.8 m^3
m = 15 kg
p1 = 5 bar = 5 x10^5 Pa
p2 = 10 bar = 10 x 10^5 Pa
cp = 1.04 kJ/kg.K and cv = 0.7432 kJ/kg.K
V1 = V2

R = 1.04-0.7432 = 0.2968 kJ/kg = 296.8 J/kg

p1V1 = mRT1
p2V2 = mRT2
T2 – T1 = (p2V2 – p1V1)/mR = (p2 – p1)V1/mR
= (10-5) x (10^5) x 2.8/(15 x 296.8)
ΔT = 314.4 K
ΔU = m.cv.ΔT = 15 x 0.7432 x 314.4 = 3505.9 kJ

Q – W = ΔU, Q = 0
-W = ΔU
W = -3505.9 kJ
 Problem No. 6
A stationary mass of gas is compressed at constant pressure from an initial

state of 2.5 m^3 and 2 bar to a final volume of 1.5 m^3. There is a transfer

of 400 kJ of heat from the gas during compression. Find the change in

internal energy of the gas

 Problem No. 6
p1 = 2 bar = 2 x 10^5 Pa = p2

V1 = 2.5 m^3

V2 = 1.5 m^3

Q = - 400 kJ

Q – W = ΔU

W = p x ΔV = p(V2-V1) = 2 x 10^5 x (1.5-2.5)

= -200 kJ

ΔU = Q-W = -400-(-200) = -400+200 = -200kJ

 Problem No. 7
5 cubic meter of air at 0 degree Celsius and a pressure of 3 bar is heated to

80 degree Celsius at constant pressure. Find,

(i) Change in Internal Energy

(ii) Heat Supplied

(iii) Mechanical work done

 Problem No. 7
V1 = 5 m^3 For air cv = 0.718 kJ/kg.K
p1 = p2 = 3 bar = 3 x 10^5 Pa cp = 1.005 kJ/kg.K
T1 = 0 deg C = 273 K R = 0.287 kJ/kg.K
T2 = 80 deg C = 80+273 = 353 K γ = 1.4
ΔT = T2-T1 = 353-273 = 80 K
p1V1 = mRT1
m = p1V1/RT1 = 19.14 kg
ΔU = m.cv.ΔT = 1099.40 kJ
Q = m.cp.ΔT = 1538.86 kJ
Q – W = ΔU
W = Q – ΔU = 1538.86 – 1099.40 = 439.46 kJ
 Problem No. 8
A piston cylinder device initially contains 0.4 m^3 of air at 100 kPa and 80

degree Celsius. The air is now compressed to 0.1 m^3 in such a way that

the temperature inside the cylinder remains constant. Determine the work

done during this process.

 Problem No. 8
V1 = 0.4 m^3

V2 = 0.1 m^3

p1 = 100 kPa = 100 x 10^3 Pa

T1 = T2 = 80 deg C = 80+273 = 353 K

W = p1V1. ln(V2/V1) = 10^5 x 0.4. ln(0.1/0.4) = -55.152 kJ

 Problem No. 9
Determine the volume of 2 kg of air at 30 deg. C and under a pressure of 2

bar. What would be its volume after isothermal compression to a pressure

of 4 bar. Calculate the work done.

 Problem No. 9
p1 = 2 bar = 2 x 10^5 Pa
p2 = 4 bar = 4 x 10^5 Pa
T1 = 30 deg. C = 30 + 273 = 303 K
m = 2 kg
pV = p1V1 = p2V2 = Constant
R = 0.287 kJ/kg.K
p1V1 = mRT1
V1 = mRT1/p1 = 0.8696 m^3
W = p1V1. ln(V2/V1) = p1V1. ln(p1/p2)
W = -120.55 kJ
 Problem No. 10
0.2 m^3 of gas at 1 bar and 100 degree Celsius is compressed adiabatically

to 0.05 m^3. Determine

(i) The mass of gas compressed

(ii) Final pressure and temperature

(iii) Increase in internal energy.

Take γ = 1.4 and R =295 J/kg.K

 Problem No. 10
V1 =0.2 m^3
V2 = 0.05 m^3
p1 = 1 bar = 10^5 Pa
T1 = 100 deg C = 373 K
γ = 1.4 and R =295 J/kg.K
p1V1 = mRT1
m = p1V1/RT1 = 0.1818 kg
p1/p2 = (V2/V1) γ = (T1/T2)(γ/γ-1)
p2 = 6.96 bar
p2/p1 = (T2/T1)(γ/γ-1)
T2/T1 = (p2/p1) (γ-1/γ)
T2 = T1 x (p2/p1) (γ-1/γ)
T2 =649.43 K
Q – W = ΔU, Q = 0
ΔU = -W = -(p1V1-p2V2)/(γ-1) = 37.06 kJ
 Problem No. 11
3.5 m^3 of hydrogen gas at a pressure of 100 kPa and 20 degree Celsius

are compressed adiabatically to 4.5 times its original pressure. It is then

expanded isothermally to its original volume. Determine the final pressure

of the gas and the heat transfer. Also determine the quantity of heat that is

to be exchanged to reduce the gas to its original pressure and volume. Take

the specific heat at constant pressure for hydrogen as 14.3 kJ/kg.K and γ =

1.4
Problem No. 11
 Problem No. 11
V1 = 3.5 m^3 =V3
p1 = 100 kPa = 10^5 Pa
T1 = 20 deg. C = 20+273 = 293 K
p2 = 4.5 x p1 = 4.5 x 10^5 Pa
Cp = 14.3 kJ/kgK
γ = 1.4
p1/p2 = (V2/V1) γ
V2 = 1.2 m^3
p3V3 = p2V2
p3 = p2V2/V3 =1.54 x 10^5 Pa
Q(1-2) = 0
Q (2-3) = W(2-3), ΔU = 0
Q(2-3) = p2V2.ln (V3/V2) = 578 kJ
(V2/V1) γ-1 = T1/T2
T2 = T1 x (V1/V2) γ-1
T2 = 452 K = T3
cv = cp/ γ = 10.2 kJ/kgK
m = p1V1/RT1= 0.287 kg
R= cp-cv = 4.1 kJ/kgK
Q(3-1) = mcv(T1-T3) = - 462.855 kJ
 Problem No. 12
1 kg of air at a pressure of 6 bar and a temperature of 40 deg. C expands to
a pressure of 1 bar according to the law pV 1.3 is constant. Determine
(i) Change in internal energy
(ii) Work done
(iii) Heat supplied
Assume R = 287 J/kgK and cv = 0.718 kJ/kgK
 Problem No. 12
P1 = 6 bar = 6 x 10^5 Pa
P2 = 1 bar = 1 x 10^5 Pa
m = 1 kg
T1 = 40 deg. C = 40+273 = 313 K
R = 287 J/kgK and cv = 0.718 kJ/kgK
n = 1.3
p1/p2 = (T1/T2)(n/n-1)
T2 = T1 x (p1/p2) (n-1/n)
T2 = 207 K
(i) ΔU = mcv(T2-T1) = -76.11 kJ
V1 = mRT1/p1 = 0.1497 m^3
V2 = V1 x (p1/p2) (1/n) = 0.594 m^3
(ii) W = (p1V1-p2V2)/(n-1) = mR(T1-T2)/(n-1) = 101.4 kJ
(iii) Q = ΔU + W = 25.29 kJ
 Problem No. 13
A gas mixture, which behaves as a perfect gas, has a molecular weight of
28. It is compressed according to the law pV 1.25 = Constant to 12 times its
original pressure. The initial conditions are 90 kPa and 60 deg. C
temperature. Find the work done and heat flow across the cylinder walls,
per kg of the gas mixture.
 Problem No. 13
n = 1.25 Ru = 8.314 J/mol.K
Assume γ = 1.4
M = 28 g
p1 = 90 kPa = 90 x 10^3 Pa
p2 = 12 x 90 x 10^3 Pa = 108 x 10^4 Pa
T1 = 60 deg. C = 333 K
m = 1 kg
R = Ru / M = 8.314/28 = 0.2969 J/gK = 296.9 J/kgK
V1 = mRT1/p1 = 1.1 m^3
V2 = V1 x (p1/p2) (1/n) = 0.15 m^3
T2 = T1 x (p1/p2) (n-1/n) = 547 K
W = (p1V1-p2V2)/(n-1) = mR(T1-T2)/(n-1) = -252 kJ
Q = W x (γ-n)/ (γ-1) = -94.5 kJ