thermodynamics theory + questions.0001
TRANSCRIPT
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Basic ConceptsS K Mondal’s Chapter 1
1
1. Basic Concepts
Theory at a Glance (For GATE, IES & PSUs)
Intensive and Extensive PropertiesIntensive property: Whose value is independent of the size or extent i.e. mass of the system.
These are, e.g., pressure p and temperature T.
Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case
letters as the symbols). e.g., Volume, Mass (V, M). If mass is increased, the value of extensive
property also increases. e.g., volume V, internal energy U, enthalpy H, entropy S, etc.
Specific property: It is a special case of an intensive property. It is the value of an extensive
property per unit mass of system. (Lower case letters as symbols) eg: specific volume, density (v, ρ).
Thermodynamic System and Control Volume• In our study of thermodynamics, we will choose a small part of the universe to which we will
apply the laws of thermodynamics.
We call this subset a SYSTEM.
• The thermodynamic system is analogous to the free body diagram to which we apply the laws ofmechanics, (i.e. Newton’s Laws of Motion).
• The system is a macroscopically identifiable collection of matter on which we focusour attention (e.g., the water kettle or the aircraft engine).
SystemDefinition
• System: A quantity of matter in space which is analyzed during a problem.
• Surroundings: Everything external to the system.
• System Boundary: A separation present between system and surrounding.
Classification of the system boundary:-• Real solid boundary
• Imaginary boundary
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S
Th
as:
Th
be
T
C
K Mon
e system bo
• Contr
• Contre choice of bng analyzed
pes o
losed1. It’s
fixed i
2. Thi
referre
3. Thethe sys
4. Ene
into or
dal’s
ndary may
ol Mass Sys
ol Volume Soundary dep
.
Syste
ystemsystem of
dentity.
type of s
to as “clos
e is no matem bounda
rgy transfe
out of the sy
e further cl
em.
stem.ends on the
m
(Contrfixed mas
ystem is u
ed system”.
ss transfery.
may take
stem.
asic
2
assified
roblem
l Mass with
sually
cross
place
once
Syste
Fig. A C
or
pts
)
ntrol Mas
losed Sys
Ch
System
em
pter 1
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S
E
•
•
K Mon
pen S1. Its
2. Thi
ref
a "
3. Ma
acr
4. En
int
5. A c
fix
an
6. Co
bouacr
ma
7. Th
8. Wh
sys
9. Th
ma
10. Mo
ample:
Heat exch
across the s
Pump - A
energy from
dal’s
stema system of
s type of s
rred to as "
ontrol vol
ss transfer
oss a control
rgy transfe
or out of th
ontrol volum
d region a
energy tra
trol Surf
ndary ofoss which t
ss and energ
mass of a c
en the net i
tem is fixed
identity of
ss system (cl
st of the eng
anger - Flu
stem boun
ontinuous fl
the surroun
(Contixed volu
ystem is u
open syste
me"
can take
volume.
r may also
e system.
e can be see
ross which
sfers are st
ce – Its
control ve transfer o
y takes plac
ntrol volum
flux of mas
and vice-ver
mass in a
osed system
ineering dev
id enters an
ary.
ow of fluid t
dings to the
asic
3
ol Voe.
sually
” or
place
occur
n as a
mass
died.
the
olumef both
.
e (open syst
s across the
a.
ontrol volu
.
ices, in gene
d leaves th
akes place t
system.
once
ume S
Fig. A Cor
m) may or
control surf
e always c
al, represen
system co
rough the s
pts
ystem
ntrol Volu Open Syst
ay not be fi
ce equals z
anges unli
t an open sy
tinuously w
ystem with
Ch
)
e Systemem
ed.
ro then the
e the case f
tem or cont
ith the tran
transfer of
pter 1
mass of the
or a control
ol volume.
sfer of heat
mechanical
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S
I1.
2.
3.
Th
un
W
A
K Mon
olateIt is a sy
identity
No inter
place be
surround
In more
system i
busy mar
uasi-Se processe
restrained
need restra
uasi – stati
• The deequilibri
• All stateare equil
• If we reone the
displace
quasist
• On the opressure
because
• In both c
• AnotherOn the o
dal’s
Systtem of fixe
and fixed
ction of ma
tween the
ings.
informal
like a clos
ket.
atic Ps can be
ined process
process is o
iation fro
um is infinit
s of the syst
ibrium state
ove the wei
pressure
the piston
tic.
ther hand if
. (This is u
t is not in a
ases the sys
e.g., if a per
ther hand if
mmass with
nergy.
s or energy
system an
ords an is
ed shop am
ocess restraine
es in practic
ne in which
thermody
esimal.
m passes th
s.
ghts slowly
of the gas
gradually.
we remove
restrained
sustained m
ems have u
son climbs d
he jumps th
asic
4
same
takes
the
olated
idst a
d or
e.
namic
rough
ne by
will
It is
ll the weigh
expansion)
anner.
dergone a c
own a ladde
n it is not a
once
Fig. A
Fig. A q
ts at once th
ut we don’
ange of sta
r from roof
quasistatic
pts
Isolated
uasi – static
e piston will
consider t
e.
o ground, it
rocess.
Ch
ystem
process
be kicked u
at the wor
is a quasist
pter 1
p by the gas
k is done –
tic process.
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S
L
•
•
•
•
S•
•
•
Yo
Yo
K Mon
ws of
The Zeroth
temperatur
The First
energy.
The Second
matter into
The Thirdcrystalline
ummatFirstly, the
conversion
work outpu
Secondly, m
K is unattai
Because, w
scale!! Thatu can’t get
u can’t get
dal’s
Therm
Law deal
s.
aw deals
Law of ther
work. It als
aw of therubstance at
ion ofre isn’t a
f heat to w
.
ore interesti
nable. This i
don’t know
is why all te something
To get work
something
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be given.
odyna
s with the
ith the con
odynamics
introduces
odynamicsabsolute zer
Lawseaningful t
rk. Only at
ngly, there i
s precisely t
what 0 K lo
mperature s for nothin
output you
for very lit
work outpu
asic
5
ics
mal equili
ervation of
provides wit
he concept o
defines theo temperatu
emperature
infinite tem
sn’t enough
he Third law
oks like, we
cales are at:
must give so
tle:
t there is a
once
rium and
energy and
h the guidel
f entropy.
absolute zere is zero.
of the sour
perature on
work availa
.
haven’t got
est empiric
me thermal
inimum am
pts
provides a
introduces
nes on the c
o of entropy
e from whi
can dream
le to produ
a starting p
l.
nergy.
ount of ther
Ch
means for
the concept
onversion he
. The entro
ch we can
of getting t
e 0 K. In ot
int for the
al energy t
pter 1
measuring
of internal
at energy of
y of a pure
get the full
e full 1 kW
er words, 0
emperature
hat needs to
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S
Yo
Vi
Z•
ITo
pr
ha
Te
co
of
te
te
in
gifro
In
th
ra
K Mon
u can’t get
lation of a
rothIf two syste
(that is A a
thermal eq
• All te
ternati provide a
ctical consi
s been refi
mperature S
forms with
accuracy of
perature o
peratures
truments a
ing the temm 3.0 to 24.
the range fr
rmometers.
iation and
dal’s
every thin
However m
ll 3 laws:
Try to get e
aw ofms (say A a
d C are in t
ilibrium the
perature
onal Ttandard fo
erations, th
ed and ext
cale of 1990
the thermod
measureme
a number
is accomplis
d values of
perature as5561 K is b
om 13.8033
Above 123
easuremen
:
ch work yo
erything fo
hermnd B) are in
hermal equil
mselves (th
easureme
mper temperatu
e Internatio
ended in se
(ITS-90) is
ynamic tem
nt obtainabl
f reproducib
hed by for
the ITS. In
functions ofsed on meas
to 1234.93
.9 K the te
ts of the int
asic
6
are willing
nothing.
dyna thermal eq
ibrium; B a
t is A and B
ts are bas
ture Sre measure
al Tempera
veral revisi
defined in s
erature, th
e in 1990.
le fixed poin
ulas that g
the range fr
the vapor prurements u
, ITS-90 is
perature i
nsity of visi
once
to give 0 K c
icsilibrium wi
d C are in t
will be in th
d on Zerot
aleent taking
ture Scale (I
ns, most r
ch a way t
unit of whi
he ITS–90
ts (Table). I
ive the rela
m 0.65 to 5
essures of ping a heliu
defined by
defined us
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an’t be reac
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law of th
into accou
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cently in 1
at the tem
ch is the kel
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tion betwee
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rticular hel constant-v
eans of cer
ing Planck’s
radiation t
Ch
ed.
stem (say C
ibrium) the
rium).
rmodyna
t both the
pted in 192
990. The I
erature me
vin, to withi
the assign
between th
n readings
is defined b
ium isotopelume gas t
ain platinu
equation fo
e absolute
pter 1
) separately
they are in
ics
retical and
. This scale
ternational
sured on it
n the limits
d values of
fixed-point
of standard
y equations
. The rangeermometer.
resistance
r blackbody
emperature
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S
sc
Ke
a
w
Ti
63
m
ju
th
In
for
th
K Mon
le. The abs
lvin temper
arnot engin
ter arbitrari
e Consta
.2% of step
asured time
ction from t
thermocou
general, tim
measureme
velocity of
dal’s
lute tempe
ture scale a
e operating
ly assigned
ts: The tim
hange in te
constant ar
he welded c
le probe gre
e constants
nt of liquid.
he media.
ature scale
lso, the tripl
between res
he value of
e constant i
mperature
e sheath wa
n on an un
atly influen
for measure
The time co
asic
7
is also kno
e point of w
ervoirs at te
273.16 K.
the amoun
f a surroun
ll thickness,
rounded the
es the time
ent of gas
stant also v
once
n as Kelvin
ter is taken
perature
of time req
ing media.
degree of in
rmocouple. I
onstant me
an be estim
aries invers
pts
temperatu
as the stan
and tp, tp
uired for a t
Some of th
sulation com
n addition,
surement.
ated to be te
ly proportio
Ch
e scale. In
ard referen
being the tr
hermocoupl
factors infl
paction, an
he velocity
n times as l
nal to the s
pter 1
efining the
e point. For
iple point of
to indicate
uencing the
distance of
f a gas past
ng as those
uare root of
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S
W
no
th
st
do
un
FFr
an
ev
bec
ir
pLe
voco
m
pr
p
K Mon
ork ark is one of
t a point fun
at the wor
red in the s
e. Therefo
dergoes a
ree Exee Expansi
B by a thi
cuated. If th
ome equal.
eversible.
V-wot the gas in
ume V 1. Thrdinates 1
ve out to a
ssure 2 p a
nd the volu
dal’s
ath futhe basic mo
ction. Ther
is posses
ystem is en
re, work is
rocess.
ansioon Let us co
diaphragm.
diaphragm
his is know
Also work
k or Dihe cylinder
e system is, V 1. The pis
new final p
d volume V 2
me V. This
nctiondes of energ
fore, work
sed by the
rgy, but no
energy in
withsider an insu
Compartmen
is punctured,
as free or
done is zer
splace(Figure sho
in thermodton is the on
osition 2, w
. At any int
ust also b
asic
8
transfer. T
is not a pr
system. It
work. A de
transit an
ero Wlated contain
t A contains
the gas in A
unrestrained
during fre
Free Exp
mentn in below)
namic equily boundary
ich is also
rmediate po
an equilibr
once
he work don
operty of t
is an inter
crease in en
it can be
ork Tr r (Figure) w
a mass of g
will expand
expansion.
e expansio
ansion
orkbe a system
librium, thewhich move
thermody
int in the tr
ium state, s
pts
e by a syste
e system,
ction across
ergy of the
dentified
nsferich is divide
s, while com
into B until
he process.
having initi
state of ws due to gas
amic equili
vel of the pi
nce macros
Ch
is a path f
and it can
the bound
ystem appe
nly when
into two co
partment B
the pressures
of free ex
lly the pres
ich is descr pressure. L
rium state
ston, let the
opic proper
pter 1
nction, and
ot be said
ry. What is
ars as work
he system
partments A
s completely
in A and B
pansion is
sure 1 p and
ibed by thet the piston
specified by
pressure be
ies p and V
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S
si
for
th
ga
w
dr
W
a
K Mon
nificant onl
equilibrium
piston, the
on the pist
d
ere dV = a
wn at the t
en the pist
ount of wor
dal’s
y.
states. Wh
force F acti
n.
=l = infinite
p of it will b
n moves out
W done by
n the pisto
g on the pis
⋅ simal displ
e explained
from positi
the system
−1 2
asic
9
moves an i
on F = p.a.
l =cement volu
ater.
n 1 to positi
ill be
= ∫
once
nfinitesimal
nd the infin
pame. The di
n 2 with th
2
1
V
p
pts
distance dl,
itesimal am
l =ferential si
volume ch
V
Ch
and if ‘a' be
unt of work
pdn in dW w
nging from
pter 1
the area of
done by the
ith the line
V 1 to V2, the
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S
Th
th
Si
cosy
m
m
in
th
Th
on
HHe
fro
pa
Th
PEx
In
ch
p
So
K Mon
e magnitud
area unde
ce p is a
rdinate, alltem as the
st be equili
st be qu
initely slow
ough is an e
e integratio
a quasi-sta
eat Traat transfer
m state 1 to
th. Therefor
e displacem
OBLEample 1
a closed sys
nge in inter
2 10V V
⎛ ⎞+⎜ ⎟
⎝ ⎠
lution:
dal’s
of the wo
the path 1
t all times
the states p volume cha
brium state
si-static.
ly so that
quilibrium s
∫ pdV catic path.
nsfer-s a path f
state 2 dep
dQ is an in
nt work is g
S & S
em, volume
nal energy g
here p is i
k done is g
2, as shown
a thermo
ssed througnges from
s, and the
he piston
every state
tate.
be perform
Pathnction, tha
nds on the
exact differe
2
1∫ d
iven by
1 2−W
LUTIO
changes fro
iven the pre
kPa and V
asic
10
iven by
in Fig.
ynamic
by the1 to V 2
ath 1-2
moves
passed
ed only
Functit is, the am
intermediat
ntial, and w
1 2−= Q or
2 2
1 1= =∫ ∫dW
S
1.5m3 to 4.
ssure volum
is in m3.
once
Fig.
nunt of heat
states thro
write
1 2 2≠ −Q Q
2≠ − pdV W
5 m3 and he
relation as
pts
Quasi-Stat
transferred
gh which t
1
1
t addition is
Ch
ic pdV Wor
when a syst
e system p
2000 kJ. C
pter 1
em changes
sses, i.e. its
lculate the
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Basic ConceptsS K Mondal’s Chapter 1
11
( )
( )
( )
[ ]
2 2
1 1
2
3 3 2
2 11
3 3
10.
110
3
1 4.54.5 1.5 10
3 1.5
191.125 3.375 10 3
3
29.250 10.986 40.236
⎛ ⎞= = +⎜ ⎟⎝ ⎠
⎡ ⎤= − +
⎢ ⎥⎣ ⎦⎡ ⎤= − +⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥⎣ ⎦= + =
∫ ∫V V
V V
Work done p dV V dV V
V V V ln
V
ln
ln
kJ
First Law of Thermodynamics:-
Q = W + ΔU2000 = 40.236 + ΔU∴ ΔU = 2000 – 40.236 = 1959.764 kJ
Example 2.
A fluid is contained in a cylinder piston arrangement that has a paddle that imparts work to the
fluid. The atmospheric pressure is 760 mm of Hg. The paddle makes 10,000 revolutions during
which the piston moves out 0.8m. The fluid exerts a torque of 1.275 N-m one the paddle. What is net
work transfer, if the diameter of the piston is 0.6m?
Solution:
Work done by the stirring device upon the system
W1 = 2πTN= 2π × 1.275 × 10000 N-m = 80kJThis is negative work for the system.
(Fig.)
Work done by the system upon the surroundings.
W2 = p.dV = p.(A × L)
= 101.325 × 4
π (0.6)2 × 0.80 = 22.9kJ
This is positive work for the system. Hence the net work transfer for the system.W = W1 + W2 = - 80 + 22.9 = - 57.l kJ .
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Basic ConceptsS K Mondal’s Chapter 1
12
ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS)
Previous 20-Years GATE Questions
GATE-1. List-I List II [GATE-1998]
A. Heat to work 1. Nozzle
B. Heat to lift weight 2. Endothermic chemical reaction
C. Heat to strain energy 3. Heat engine
D. Heat to electromagnetic energy 4. Hot air balloon/evaporation
5. Thermal radiation
6. Bimetallic strips
Codes: A B C D A B C D
(a) 3 4 6 5 (b) 3 4 5 6
(c) 3 6 4 2 (d) 1 2 3 4
Open and Closed systemsGATE-2. An isolated thermodynamic system executes a process, choose the correct
statement(s) form the following [GATE-1999]
(a) No heat is transferred
(b) No work is done
(c) No mass flows across the boundary of the system
(d) No chemical reaction takes place within the system
GATE-2a. Heat and work are [GATE-2011]
(a) intensive properties (b) extensive properties
(c) point functions (d) path functions
Quasi-Static ProcessGATE-3. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and
0.015 m3. It expands quasi-statically at constant temperature to a final volume
of 0.030 m3. The work output (in kJ/kg) during this process will be: [GATE-2009]
(a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00
Free Expansion with Zero Work TransferGATE-4. A balloon containing an ideal gas is initially kept in an evacuated and
insulated room. The balloon ruptures and the gas fills up the entire room.
Which one of the following statements is TRUE at the end of above process?(a) The internal energy of the gas decreases from its initial value, but the enthalpy
remains constant [GATE-2008]
(b) The internal energy of the gas increases from its initial value, but the enthalpy
remains constant
(c) Both internal energy and enthalpy of the gas remain constant
(d) Both internal energy and enthalpy of the gas increase
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Basic ConceptsS K Mondal’s Chapter 1
13
GATE-5. Air is compressed adiabatically in a steady flow process with negligible change
in potential and kinetic energy. The Work done in the process is given by:
[GATE-1996, IAS-2000]
(a) – ∫Pdv (b) +∫Pdv (c) – ∫vdp (d) +∫vdp
pdV-work or Displacement WorkGATE-6. In a steady state steady flow process taking place in a device with a single inlet
and a single outlet, the work done per unit mass flow rate is given byoutlet
inlet
vdp=∫
, where v is the specific volume and p is the pressure. The
expression for w given above: [GATE-2008]
(a) Is valid only if the process is both reversible and adiabatic
(b) Is valid only if the process is both reversible and isothermal
(c) Is valid for any reversible process
(d) Is incorrect; it must be
outlet
inletvdpω = − ∫
GATE-7. A gas expands in a frictionless piston-cylinder arrangement. The expansion
process is very slow, and is resisted by an ambient pressure of 100 kPa. During
the expansion process, the pressure of the system (gas) remains constant at 300
kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work
that could be utilized from the above process is: [GATE-2008]
(a) 0kJ (b) 1kJ (c) 2kJ (d) 3kJ
GATE-8. For reversible adiabatic compression in a steady flow process, the work
transfer per unit mass is: [GATE-1996]
( ) ( ) ( ) ( )a pdv b vdp c Tds d sdT ∫ ∫ ∫ ∫
Previous 20-Years IES Questions
IES-1. Which of the following are intensive properties? [IES-2005]
1. Kinetic Energy 2. Specific Enthalpy
3. Pressure 4. Entropy
Select the correct answer using the code given below:
(a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4
IES-2. Consider the following properties: [IES-2009]
1. Temperature 2. Viscosity
3. Specific entropy 4. Thermal conductivity
Which of the above properties of a system is/are intensive?
(a) 1 only (b) 2 and 3 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4
IES-2a. Consider the following: [IES-2007, 2010]
1. Kinetic energy 2. Entropy
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Basic ConceptsS K Mondal’s Chapter 1
15
(a) Open system [IES-2010]
(b) Steady flow diabatic system
(c) Closed system with a movable boundary
(d) Closed system with fixed boundary
IES-8. Which of the following is/are reversible process(es)? [IES-2005]
1. Isentropic expansion
2. Slow heating of water from a hot source
3. Constant pressure heating of an ideal gas from a constant temperature
source
4. Evaporation of a liquid at constant temperature
Select the correct answer using the code given below:
(a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4
IES-9. Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a
reversible process is the most efficient process. [IES-2001]
Reason (R): The energy transfer as heat and work during the forward process
as always identically equal to the energy transfer is heat and work during thereversal or the process.
(a) Both A and R are individually true and R is the correct explanation of A
(b) Both A and R are individually true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
IES-9a Which one of the following represents open thermodynamic system?
(a) Manual ice cream freezer (b) Centrifugal pump [IES-2011]
(c) Pressure cooker (d) Bomb calorimeter
IES-10. Ice kept in a well insulated thermo flask is an example of which system?
(a) Closed system (b) Isolated systems [IES-2009] (c) Open system (d) Non-flow adiabatic system
IES-10a Hot coffee stored in a well insulated thermos flask is an example of
(a) Isolated system (b) Closed system [IES-2010]
(c) Open system (d) Non-flow diabatic system
IES10b A thermodynamic system is considered to be an isolated one if [IES-2011]
(a) Mass transfer and entropy change are zero
(b) Entropy change and energy transfer are zero
(c) Energy transfer and mass transfer are zero
(d) Mass transfer and volume change are zero
IES-10c. Match List I with List II and select the correct answer using the code given
below the lists: [IES-2011]
List I
A. Interchange of matter is not possible in a
B. Any processes in which the system returns to
its original condition or state is called
C. Interchange of matter is possible in a
List II
1. Open system
2. System
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D. The quantity of matter under consideration in
thermodynamics is called
3. Closed system
4. Cycle
Code: A B C D A B C D
(a) 2 1 4 3 (b) 3 1 4 2(c) 2 4 1 3 (d) 3 4 1 2
Zeroth Law of ThermodynamicsIES-11. Measurement of temperature is based on which law of thermodynamics?
[IES-2009]
(a) Zeroth law of thermodynamics (b) First law of thermodynamics
(c) Second law of thermodynamics (d) Third law of thermodynamics
IES-12. Consider the following statements: [IES-2003]
1. Zeroth law of thermodynamics is related to temperature
2. Entropy is related to first law of thermodynamics3. Internal energy of an ideal gas is a function of temperature and pressure
4. Van der Waals' equation is related to an ideal gas
Which of the above statements is/are correct?
(a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4
IES-13. Zeroth Law of thermodynamics states that [IES-1996, 2010]
(a) Two thermodynamic systems are always in thermal equilibrium with each other.
(b) If two systems are in thermal equilibrium, then the third system will also be in
thermal equilibrium with each other.
(c) Two systems not in thermal equilibrium with a third system are also not in thermal
equilibrium with each other.
(d) When two systems are in thermal equilibrium with a third system, they are in
thermal equilibrium with each other.
International Temperature ScaleIES-14. Which one of the following correctly defines 1 K, as per the internationally
accepted definition of temperature scale? [IES-2004]
(a) 1/100th of the difference between normal boiling point and normal freezing point of
water
(b) 1/273.15th of the normal freezing point of water
(c) 100 times the difference between the triple point of water and the normal freezing
point of water
(d) 1/273.15th of the triple point of water
IES-15. In a new temperature scale say °ρ, the boiling and freezing points of water at
one atmosphere are 100°ρ and 300°ρ respectively. Correlate this scale with the
Centigrade scale. The reading of 0°ρ on the Centigrade scale is: [IES-2001]
(a) 0°C (b) 50°C (c) 100°C (d) 150°C
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18
Free Expansion with Zero Work TransferIES-21. Match items in List-I (Process) with those in List-II (Characteristic) and select
the correct answer using the codes given below the lists:
List-I List-II [IES-2001]
A. Throttling process 1. No work done
B. Isentropic process 2. No change in entropy
C. Free expansion 3. Constant internal energy
D. Isothermal process 4. Constant enthalpy
Codes: A B C D A B C D
(a) 4 2 1 3 (b) 1 2 4 3
(c) 4 3 1 2 (d) 1 3 4 2
IES-22. The heat transfer, Q, the work done W and the change in internal energy U are
all zero in the case of [IES-1996](a) A rigid vessel containing steam at 150°C left in the atmosphere which is at 25°C.
(b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly
outwards.
(c) A rigid vessel containing ammonia gas connected through a valve to an evacuated
rigid vessel, the vessel, the valve and the connecting pipes being well insulated and
the valve being opened and after a time, conditions through the two vessels becoming
uniform.
(d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated
bottle.
pdV-work or Displacement WorkIES-23. One kg of ice at 0°C is completely melted into water at 0°C at 1 bar pressure.
The latent heat of fusion of water is 333 kJ/kg and the densities of water and
ice at 0°C are 999.0 kg/m3 and 916.0 kg/m3, respectively. What are the
approximate values of the work done and energy transferred as heat for the
process, respectively? [IES-2007]
(a) –9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ
(c) –333.0 kJ and –9.4 J (d) None of the above
IES-24. Which one of the following is the
correct sequence of the threeprocesses A, B and C in the
increasing order of the amount of
work done by a gas following ideal-
gas expansions by these processes?
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Basic ConceptsS K Mondal’s Chapter 1
19
[IES-2006](a) A – B – C (b) B – A – C (c) A – C – B (d) C – A – B
IES-25. An ideal gas undergoes an
isothermal expansion from
state R to state S in a turbineas shown in the diagram given
below:
The area of shaded region is
1000 Nm. What is the amount is
turbine work done during the
process?
(a) 14,000 Nm (b) 12,000 Nm
(c) 11,000 Nm (d) 10,000 Nm [IES-2004]
IES-26. Assertion (A): The area 'under' curve on pv plane, pdv∫
represents the work of
reversible non-flow process. [IES-1992]
Reason (R): The area 'under' the curve T –s plane Tds∫
represents heat of any
reversible process.
(a) Both A and R are individually true and R is the correct explanation of A
(b) Both A and R are individually true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
IES-27. If pdv∫
and vdp∫
for a thermodynamic system of an Ideal gas on valuation
give same quantity (positive/negative) during a process, then the process
undergone by the system is: [IES-2003]
(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal
IES-28. Which one of the following expresses the reversible work done by the system
(steady flow) between states 1 and 2? [IES-2008]2 2 2 2
1 1 1 1
(a) (b) (c) (d) pdv vdp pdv vdp− −∫ ∫ ∫ ∫
Heat Transfer-A Path FunctionIES-29. Assertion (A): The change in heat and work cannot be expressed as difference
between the end states. [IES-1999]
Reason (R): Heat and work are both exact differentials.(a) Both A and R are individually true and R is the correct explanation of A
(b) Both A and R are individually true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
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Previous 20-Years IAS Questions
Thermodynamic System and Control VolumeIAS-1. The following are examples of some intensive and extensive properties:
1. Pressure 2. Temperature [IAS-1995]
3. Volume 4. Velocity
5. Electric charge 6. Magnetisation
7. Viscosity 8. Potential energy
Which one of the following sets gives the correct combination of intensive and
extensive properties?
Intensive Extensive
(a) 1, 2, 3, 4 5, 6, 7, 8
(b) 1, 3, 5, 7 2, 4, 6, 8
(c) 1, 2, 4, 7 3, 5, 6, 8(d) 2, 3, 6, 8 1, 4, 5, 7
Zeroth Law of ThermodynamicsIAS-2. Match List-I with List-II and select the correct answer using the codes given
below the lists: [IAS-2004]
List-I List-II
A. Reversible cycle 1. Measurement of temperature
B. Mechanical work 2. Clapeyron equation
C. Zeroth Law 3. Clausius Theorem
D. Heat 4. High grade energy
5. 3rd
law of thermodynamics6. Inexact differential
Codes: A B C D A B C D
(a) 3 4 1 6 (b) 2 6 1 3
(c) 3 1 5 6 (d) 1 4 5 2
IAS-3. Match List-I with List-II and select the correct answer: [IAS-2000]
List-I List-II
A. The entropy of a pure crystalline 1. First law of thermodynamics
substance is zero at absolute zero
temperature
B. Spontaneous processes occur 2. Second law of thermodynamics
in a certain direction
C. If two bodies are in thermal 3. Third law of thermodynamics
equilibrium with a third body,
then they are also in thermal
equilibrium with each other
D. The law of conservation of energy 4. Zeroth law of thermodynamics.
Codes: A B C D A B C D
(a) 2 3 4 1 (b) 3 2 1 4
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Basic ConceptsS K Mondal’s Chapter 1
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(c) 3 2 4 1 (d) 2 3 1 4
International Temperature Scale
IAS-4. A new temperature scale in degrees N is to be defined. The boiling andfreezing on this scale are 400°N and 100°N respectively. What will be the
reading on new scale corresponding to 60°C? [IAS-1995]
(a) 120°N (b) 180°N (c) 220°N (d) 280°N
Free Expansion with Zero Work TransferIAS-5. In free expansion of a gas between two equilibrium states, the work transfer
involved [IAS-2001]
(a) Can be calculated by joining the two states on p-v coordinates by any path and
estimating the area below
(b) Can be calculated by joining the two states by a quasi-static path and then finding
the area below(c) Is zero
(d) Is equal to heat generated by friction during expansion.
IAS-6. Work done in a free expansion process is: [IAS-2002]
(a) Positive (b) Negative (c) Zero (d) Maximum
IAS-7. In the temperature-entropy diagram
of a vapour shown in the given figure,
the thermodynamic process shown by
the dotted line AB represents
(a) Hyperbolic expansion
(b) Free expansion
(c) Constant volume expansion(d) Polytropic expansion
[IAS-1995]
IAS-8. If pdv∫
and vdp∫
for a thermodynamic system of an Ideal gas on valuation
give same quantity (positive/negative) during a process, then the process
undergone by the system is: [IAS-1997, IES-2003]
(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal
IAS-9. For the expression pdv∫
to represent the work, which of the following
conditions should apply? [IAS-2002]
(a) The system is closed one and process takes place in non-flow system(b) The process is non-quasi static
(c) The boundary of the system should not move in order that work may be transferred
(d) If the system is open one, it should be non-reversible
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Answers with Explanation (Objective)
Previous 20-Years GATE Answers
GATE-1. Ans. (a)
GATE-2. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system.
0, 0, 0 or ConstantdQ dW dE E = = ∴ = =
GATE-2a. Ans. (d)
GATE-3. Ans. (a) Iso-thermal work done (W) = 211
lnV
RT V
⎛ ⎞⎜ ⎟⎝ ⎠
21 1
1ln
0.030800 0.015 ln
0.015
8.32kJ/kg
V P V V
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞= × × ⎜ ⎟
⎝ ⎠=
GATE-4. Ans. (c) It is free expansion. Since vacuum does not offer any resistance, there is no work
transfer involved in free expansion.
Here,
2
1
0δω =∫ and Q1-2=0 therefore Q1-2 = U Δ + W1-2 so, U Δ = 0
GATE-5. Ans. (c) For closed system W = pdv+∫ , for steady flow W = vdp−∫ GATE-6. (c)
GATE-7. Ans. (b) W = Resistance pressure.Δ V = 1 × Δ V = 100 × 0.1 kJ = 1kJGATE-8. Ans. (b) W vdp= −∫
Previous 20-Years IES Answers
IES-1. Ans. (b)
IES-2. Ans. (d) Intensive property: Whose value is independent of the size or extent i.e. mass of
the system.
Specific property: It is a special case of an intensive property. It is the value of anextensive property per unit mass of system (Lower case letters as symbols) e.g., specific
volume, density (v, ρ).
IES-2a. Ans. (c) Kinetic energy21 mv
2 depends on mass, Entropy kJ/k depends on mass so
Entropy is extensive property but specific entropy kJ/kg K is an intensive property.
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Basic ConceptsS K Mondal’s Chapter 1
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IES-3. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive
property.
IES-4. Ans. (a) Extensive property: Whose value depends on the size or extent i.e. mass of the
system (upper case letters as the symbols) e.g., Volume, Mass (V, M). If mass is
increased, the value of extensive property also increases.
IES-4a Ans. (d) The properties like temperature, viscosity which are Independent of the MASS of
the system are called Intensive property
IES-5. Ans. (d)
• But remember 100% heat can’t be convertible to work but 100% work can beconverted to heat. It depends on second law of thermodynamics.
• A thermodynamic system is defined as a definite quantity of matter or a region inspace upon which attention is focused in the analysis of a problem.
• The system is a macroscopically identifiable collection of matter on which we focusour attention
IES-5a Ans. (d)
IES-6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transferexists.
IES-7. Ans. (c)
IES-8. Ans. (d) Isentropic means reversible adiabatic. Heat transfer in any finite temp difference is
irreversible.
IES-9. Ans. (a) The energy transfer as heat and work during the forward process as always
identically equal to the energy transfer is heat and work during the reversal or the
process is the correct reason for maximum efficiency because it is conservative system.
IES-9a. Ans. (b)
IES-10. Ans. (b) Isolated System - in which there is no interaction between system and the
surroundings. It is of fixed mass and energy, and hence there is no mass and energy
transfer across the system boundary.
IES-10a Ans. (a)IES-10b. Ans. (c)
IES-10c. Ans. (d)
IES-11. Ans. (a) All temperature measurements are based on Zeroth law of thermodynamics.
IES-12. Ans. (a) Entropy - related to second law of thermodynamics.
Internal Energy (u) = f (T) only (for an ideal gas)
Van der Wall's equation related to => real gas.
IES-13. Ans. (d)
IES-14. Ans. (d)
IES-15.Ans. (d)0 300 0
150 C100 300 100 0
C C
− −= ⇒ = °
− −
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IES-16. Ans. (b) Both A and R are correct but R is not correct explanation for A. Temperature is
independent of thermometric property of fluid.
IES-17. Ans. (d)
IES-18. Ans. (d) But it will occur at absolute zero temperature.
IES-19. Ans. (c) Time Constants: The time constant is the amount of time required for athermocouple to indicated 63.2% of step change in temperature of a surrounding media.
Some of the factors influencing the measured time constant are sheath wall thickness,
degree of insulation compaction, and distance of junction from the welded cap on an
ungrounded thermocouple. In addition, the velocity of a gas past the thermocouple probe
greatly influences the time constant measurement. In general, time constants for
measurement of gas can be estimated to be ten times as long as those for measurement
of liquid. The time constant also varies inversely proportional to the square root of the
velocity of the media.
IES-20. Ans. (c)
IES-20a Ans. (d)
IES-21. Ans. (a)
IES-22. Ans. (c) In example of (c), it is a case of free expansion heat transfer, work done, and
changes in internal energy are all zero.
IES-23. Ans. (a) Work done (W) = P Δ V = 100× (V2 – V1) = 100×2 1
m m⎛ ⎞−⎜ ⎟ρ ρ⎝ ⎠
= 100 kPa× 1 1999 916
⎛ ⎞−⎜ ⎟
⎝ ⎠ = –9.1 J
IES-24. Ans. (d) 4 (2 1) 4kJ= = × − =∫ AW pdV 1
3 (7 4) 4.5kJ2
1 (12 9) 3kJ
= = × × − =
= = × − =
∫
∫
B
C
W pdV
W pdV
IES-25. Ans. (c) Turbine work = area under curve R–S
( )
( )
3
5
1 bar 0.2 0.1 m 1000 Nm
10 0.2 0.1 Nm 1000Nm 11000Nm
pdv= = × − +
= × − + =
∫
IES-26. Ans. (b)
IES-27. Ans. (d) Isothermal work is minimum of any process.
0[ is onstant]
pv mRT
pdv vdp T c
pdv vdp
=
+ =
= −∫ ∫∵
IES-28. Ans. (b) For steady flow process, reversible work given by
2
1 vdp−∫ .IES-29. Ans. (c) A is true because change in heat and work are path functions and thus can't be
expressed simply as difference between the end states. R is false because both work and
heat are inexact differentials.
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Previous 20-Years IAS Answers
IAS-1. Ans. (c) Intensive properties, i.e. independent of mass are pressure, temperature, velocity
and viscosity. Extensive properties, i.e. dependent on mass of system are volume, electriccharge, magnetisation, and potential energy. Thus correct choice is (c).
IAS-2. Ans. (a)
IAS-3. Ans. (c)
IAS-4. Ans. (d) The boiling and freezing points on new scale are 400° N and 100°N i.e. range is
300°N corresponding to 100°C. Thus conversion equation is
°N = 100 + 3 × °C = 100+ 3 × 60 = 100 + 180 = 280 °N
IAS-5. Ans. (c)
IAS-6. Ans. (c) Since vacuum does not offer any resistance, there is no work transfer involved in
free expansion.
IAS-7. Ans. (b)
IAS-8. Ans. (d) Isothermal work is minimum of any process.
IAS-9. Ans. (a)IAS-10. Ans. (c) For closed system W = pdv+∫ , for steady flow W = vdp−∫ IAS-12. Ans. (b) Start with D. ∫PdV only valid for quasi-static path so choice (c) & (d) out.
Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2.
IAS-13. Ans. (c) W = pdv∫ where pressure (p) is an intensive property and volume (v) is anextensive property
IAS-14. Ans. (a) Pressure is intensive property but such option is not there.
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ecific entha
st La
n Maeral principl
nsformed fr
which woul
simultaneo
of the first
e statement
without so
PMM1
H is define
y of a syste
y of a syste
equal to the
It is possib
no work ot
re substance
du + pdv
d(pv)
( )du d pv+
d(u+pv)
dh
enthalpy, a
, kJ/kg and
lpy, kJ/kg
of T
31
hine oe of the cons
m one form
continuous
usly (Fig. sh
kind, or in b
is also tru
e other for
d as
=and its uni
=and its uni
heat transf
e to derive
er than pd
.
property of s
also h = u +
herm
the Fiervation of
to another.
ly supply m
own in belo
rief, PMM1.
e, i.e. ther
of energy a
The
+t is kJ.
+t is kJ/kg.
rred in a c
n expressio
work. In s
ystem.
pdv
odyn
rst Kinnergy. Ener
echanical w
). Such a fic
A PMM1 is
can be no
ppearing si
Converse
V
vnstant volu
for the he
ch a proces
micsCh
d-PMMgy is neither
ork without
titious mach
hus imposs
machine
ultaneousl
f PMM1
me process i
t transfer i
s in a close
pter 2
1created nor
some other
ine is called
ible.
hich would
(Fig.).
nvolving no
a constant
stationary
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First Law of ThermodynamicsS K Mondal’s Chapter 2
32
u = specific internal energy, kJ/kg
dv = change in specific volume, m3/kg.
Specific heat at constant volumeThe specific heat of a substance at constant volume C
v is defined as the rate of change of
specific internal energy with respect to temperature when the volume is held constant, i.e.,
∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠v
v
uC
T
For a constant volume process
( )2
1
.
T
vv
T
u C dT Δ = ∫
The first law may be written for a closed stationary system composed of a unit mass of a pure
substance.
Q = Δu + Wor d Q = du + d W
For a process in the absence of work other than pdv work
d W = pdvTherefore d Q = du + pdv
Therefore, when the volume is held constant
( ) ( )
( )2
1
v v
v
u
.
T
v
T
Q
Q C dT
= Δ
= ∫
Since u, T and v are properties, Cv is a property of the system. The product m⋅
Cv is called the
heat capacity at constant volume (J/K).
Specific heat at constant pressureThe specific heat at constant pressure pC is defined as the rate of change of specific enthalpy
with respect to temperature when the pressure is held constant.
PC P
h
T
∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠
For a constant pressure process
( )2
1
.
T
P P
T
h C dT Δ = ∫
The first law for a closed stationary system of unit mass
dQ = du + pdv
Again, h = u + pv
Therefore dh = du + pdv + vdp
= d Q + vdp
Therefore dQ = dh – vdp
Therefore ( dQ )P = dh
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First Law of ThermodynamicsS K Mondal’s Chapter 2
33
or ( ) ( h) p pQ = Δ
∴ Form abow equations
( )2
1
P .
T
P
T
Q C dT = ∫
pC is a property of the system, just like Cv. The heat capacity at constant pressure is equal to m
pC (J/K).
Application of First Law to Steady Flow Process S.F.E.E
S.F.E.E. per unit mass basis
(i)
+ + + = + + +2 2
1 2
1 1 2 2
C Cd Q d Wh g z h g z
2 d m 2 d m [h, W, Q should be in J/Kg and C in m/s and g in m/s2]
(ii)+ + + = + + +
2 2
1 1 2 21 2
gZ Q gZ Wh h
2000 1000 dm 2000 1000 dm
C d C d
[h, W, Q should be in KJ/Kg and C in m/s and g in m/s2]
S.F.E.E. per unit time basis
⎛ ⎞+ + +⎜ ⎟τ⎝ ⎠
⎛ ⎞= + + +⎜ ⎟
τ⎝ ⎠
211 1 1
2
22 2 2
w z2
w z2
x
C dQ h g d
dW C h g
d
Where, w = mass flow rate (kg/s)
Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of the Control
Volume
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S
M
W
E
S
Th
en
N
A
dr
be
K Mon
ss balance
ere v = spec
ergy balan
me examp
e following
gineering sy
zzle and D
ozzle is a d
p, whereas
ow a nozzle
Firdal’s
1 1
1
A C
v
ific volume (
ce
⎛ +⎜
⎝
⎛ = ⎜
⎝
1 1
3 3
w h
w h
e of steady
xamples ill
tems.
ffuser:
evice which
a diffuser in
which is ins
1
2
C h +
st La
+1
2 2
2
w
A C
v
m3/kg)
+
+
2
11
2
33
2
2
C Z g
C Z g
flow proc
strate the a
increases t
creases the
lated. The s2
1
Q Z g
dm+ +
d
of T
34
=
=
2
3 3
3
w w
A C
v
⎛ +⎟ ⎜
⎝
⎞ ⎛ +⎟ ⎜
⎝ ⎠
2 2
4 4
w h
w h
sses:-
pplications o
e velocity
pressure of
teady flow e2
2
22
C h= + +
ig.
herm
+
+
3 4
4
4
w
A C
v
+
+ +
2
22
2
44
2
2
C Z g
C Z
f the steady
r K.E. of a
fluid at th
ergy equati
2
x W
g dm
+ d
odyn
4
⎞+⎟
τ ⎠
⎞+⎟ τ ⎠
x
dQ
d
dW
d
flow energy
fluid at the
expense of
on of the con
micsCh
equation in
expense of
its K.E. Fig
trol surface
pter 2
some of the
its pressure
ure show in
gives
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First Law of ThermodynamicsS K Mondal’s Chapter 2
35
Here 0; 0,x dW dQ
dm dm= = and the change in potential energy is zero. The equation reduces to
2 2
1 21 2
2 2
C C h h+ = + (a)
The continuity equation gives1 1 2 2
1 2
A A w = =
C C
v v (b)
When the inlet velocity or the ‘velocity of approach’ V1 is small compared to the exit velocity V2,
Equation (a) becomes2
2
1 2
2 1 2
2
2( ) /
C h h
or C h h m s
= +
= −
where (h1 – h2) is in J/kg.
Equations (a) and (b) hold good for a diffuser as well.
Throttling Device:
When a fluid flows through a constricted passage, like a partially opened value, an orifice, or a
porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled. Figure
shown in below, the process of throttling by a prettily opened value on a fluid flowing in an insulated
pipe. In the steady-flow energy equation-
0, 0x W dQ
dm dm= =
d
And the changes in P. E. are very small and ignored. Thus, the S.F.E.E. reduces to2 2
1 21 2
2 2
C C h h+ = +
(Fig.- Flow Through a Valve)
Often the pipe velocities in throttling are so low that the K. E. terms are also negligible. So
1 2h h=
or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling.
Turbine and Compressor:
Turbines and engines give positive power output, whereas compressors and pumps require power
input.
For a turbine (Fig. below) which is well insulated, the flow velocities are often small, and the K.E.
terms can be neglected. The S.F.E.E. then becomes
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First Law of ThermodynamicsS K Mondal’s Chapter 2
36
(Fig.-. Flow through a Turbine)
1 2
1 2
x
x
dW h h
dm
W or h h
m
= +
= −
The enthalpy of the fluid increase by the amount of work input.
Heat Exchanger:
A heat exchanger is a device in which heat is transferred from one fluid to another, Figure shown in
below a steam condenser where steam condense outside the tubes and cooling water flows through
the tubes. The S.F.E.E for the C.S. gives
c 1 s 2 c 3 s 4
s 2 4 c 3 1
w w w w
, w ( ) w ( )
h h h h
or h h h h
+ = +
− = −
Here the K.E. and P.E. terms are considered small, there is no external work done, and energy
exchange in the form of heat is confined only between the two fluids, i.e. there is no external heat
interaction or heat loss.
Fig. -
Figure (shows in below) a steam desuperheater where the temperature of the superheated steam is
reduced by spraying water. If w1, w2, and w3 are the mass flow rates of the injected water, of the
steam entering, and of the steam leaving, respectively, and h1, h2, and h3 are the corresponding
enthalpies, and if K.E. and P.E. terms are neglected as before, the S.F.E.E. becomes
1 1 2 2 3 3w h w h w h+ =
and the mass balance gives
w1 + w2 = w3
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S
Th
pr
un
(1)
(2)
(b)
(c)
K Mon
e above law
ctical situa
ity.
Work dev
(a) Water
In this ca
1 1 1v +z g p
(b) Steam
In this ca
(hW =
Work abs
(a) Centrifu
The system
In this sy
1 1 1v +z g p
Centrifug2
11
h2
C + +
Blowers –
Firdal’s
is also call
ions as wor
loping sys
turbines
e Q = 0 and2
12
z g2
C + =
or gas turbi
e generally
)2
11 2 – h
C ⎛ + ⎜
⎝
rbing syst
al water pu
is shown in
tem Q = 0 a
2W z g+ = +
l compress2
2 –2
C W Q =
In this case
st La
ed as stea
developing
ems
ΔU = 0 and
2 2p v W+
nes
ΔZ can be as2
2 Q2
C ⎞−+ Δ⎟
⎠
ms
p
the Figure b
Fig.
nd ΔU = 0; t2
22 2
p v 2
C +
or – In this
2h +
we have Δ z
of T
37
y flow en
system and
equation bec
sumed to be
elow
e energy eq
system Δz =
= 0,1 1v p =
herm
rgy equati
work absor
omes
zero and the
uation now
0 and the eq
2 2p v and Q
odyn
on. This ca
tion system
equation be
ecomes,
uation beco
= 0; now th
micsCh
n be applie
. Let the m
comes
es,
energy sim
pter 2
to various
ss flow rate
lifies to
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S
(d)
(e)
(3)
(a)
(b)
(c)
(vi
M
ca
un
wisu
K Mon
1u +W
Fans – In
and hence t
2
2
2
C W =
Reciproca
equation ap
or
Non-work
Steam boi
equation for
Steam conare very sm
(h1 – h2) and
heat lost by
Steam noz
In this syst
possible hea
The en
i) Unstead
ny flow pro
be analyze
der non-stea
hin the conface, as giv
Firdal’s
2
22
u as2
C = +
ans the tem
e energy eq
ing compr
lied to a rec
(1
2
h – h
h
Q
W Q
=
= +
developin
ler – In thi
a boiler bec
denser – Ill. Under st
this heat is
steam will b
le:
m we can a
t loss also ze
rgy equatio
1
2
h
or C
+
Flow Ana
esses, such
by the con
dy state con
rol volume in below:
st La
2 1 C C
perature ris
ation for fa
ssor – In a
iprocating c
)2
1
–
– h
W
and absor
system we
mes Q = (h2
this systeady conditi
also equal t
equal to he
ssume ΔZ aro.
for this cas2
1 2
2
2
1 1
2 2
2(
C C h
C h
= +
+
ysis:-
s filling up
rol volume t
itions (Figu
accumulat
of T
38
e is very sm
s becomes,
reciprocatin
mpressor is
ing syste
neglect ΔZ, – h1)
the work dns the chan
o the chang
at gained by
d W to be
e becomes.
2)h
and evacuati
echnique. Co
re-shown in
d is equal to
herm
all and heat
compresso
s
KE and W
ne is zero age in enthal
in enthalp
the cooling
ero and hea
ng gas cylin
nsider a dev
below). The
the net rate
odyn
loss is negl
ΔKE and Δ
(i.e.) ΔZ = Δ
nd we can ay is equal t
of cooling
ater.
t transfer w
ers, are not
ice through
rate at whic
of mass flo
micsCh
cted (i.e.) Δ
E are negli
E = W = 0;
lso assumeheat lost b
ater circula
hich is noth
steady, Suc
hich a flui
the mass o
across the
pter 2
h = 0, q = 0
ibly energy
the energy
Z and ΔKEsteam. Q =
ted (i.e.) the
ing but any
processes
is flowing
fluid
ontrol
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S
W
Th
ac
Ra
W
Fo
an
Or
K Mon
ere vm is t
Over a
vmΔ =
e rate of acc
oss the cont
te of energyv
dE=
d
⎛ = ⎜
⎝ v
E
τ
ere m is th
11
dE
d
h +2
∴
⎛ ⎜⎝
v
C
τ
llowing Figu
vEΔ =
Equati
d the equati
vdE
dτ =
dEv =
Firdal’s
1v
dmw
dτ = −
e mass of fl
y finite peri
1 2m – mΔ Δ
umulation o
rol surface. I
increase = R2
11 1
2
h + +2
mC + +
2
⎛ ⎜⎝
C w
mass of flui
11
d mU +
d
dm+Z g
dτ
⎛ ⎜⎝
⎞⎟
⎠
τ
re shows all
1h⎛
− + ⎜⎝
∫W
n (A) is gen
vdE = 0dτ
n reduces
dQ dW
d dτ τ −
Q – dW or
st La
12
dm d
dτ = −
id within th
od of time
f energy wit
f Ev is the e
ate of energ
1
v
dQg +
dτ
gZ
⎞−⎟
⎠
⎞⎟
⎠
w
d in the con2
v
2
+ mgZ
dQh +
dτ
⎞⎟
⎠
⎛ − ⎜
⎝
these energ2
11+Z g dm
2
⎞⎟ ⎠
C
eral energy
or a closed s
dQ = dE +
of T
39
ig.
2m
τ
e control vol
in the cont
ergy of fluid
inflow – Ra2
22 2 2h + +Z
2
⎛ ⎜⎝
C
rol volume a
2
2 22
dm+Z g
2 d
⎞⎟
⎠ τ
flux quanti2
21 2h +
2
⎛ − ⎜
⎝ ∫
C
quation. Fo
ystem w1 =
W
herm
ume at any i
ol volume is
within the
te of energydW
gd
⎞−⎟
⎠ τ
t any instan
dW
d−
τ
ties. For any
2 2+Z g dm ⎞⎟ ⎠
Fig.
steady flow
, w2 = 0, the
odyn
nstant.
equal to th
ontrol volu
outflow.
equati
t
(equ
time interv
,
n from equa
micsCh
net rate of
e at any ins
...on A
)........tion B
l, equation
tion (A),
pter 2
energy flow
tant,
B) becomes
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SFl
Ex
Va
tec
th
an
su
p
Sy
w
W
be
vol
U
K Monw Processes
ample of a
riable flow p
hnique, as il
Consid beginning
d gas flows i
ply to the p
P P P,T , v , h ,
stem Tech
ich would e
Energy
1E m=
ere ( 2m –
2E
E
=
Δ =
The P.E.
n omitted.
Now, the
ume. Then t
ing the firs
Q = ΔE
= 2m
Firdal’s
variable fl
rocesses ma
lustrated be
r a processhe bottle co
nto the bottl
ipeline is ve
P Pu and v .
ique: Assu
entually en
of the gas b
(1 1 2u m –+
)1 is the m
2 2
2 1
u
– E m=
erms are ne
e is a chan
he work don
W =
=
for the proc
+ W
2 1 1u –m u –
st La
w proble
be analyze
low.
in which a gtains gas of
e till the ma
y large so t
e an envelo
er the bottle
efore filling.
)2
1m2
P P
C u
⎛ +⎜
⎝ ss of gas in
(2 1 1u – m u
glected. The
ge in the vo
e
( 2p – V p V
( p 0 –
p ⎡⎣
( 2 – m – m
ess
)2 1m – m⎡⎢⎣
of T
40
:
d either by t
s bottle is fi mass m1 at
s of gas in t
at the state
pe (which is
, as shown i
⎞⎟
⎠
he pipeline
)2
1 1m – u
2
P C ⎛
⎜⎝
gas in the b
lume of gas
)
)2 1 – m v
)1 Pp v p
(2
PP
+ u2
⎤−⎥
⎦
herm
e system te
lled from a ptate p1, T1,
e bottle is
of gas in the
extensible)
Figure abo
nd tube wh
P u
⎞+ ⎟
⎠
ttle is not i
because of
⎤⎦
)2 1 – m P P p v
odyn
chnique or t
ipeline (Fig1, h1 and u1.
2 at state p2 pipeline is c
f gas in the
e.
ich would en
motion, an
the collapse
micsCh
e control vo
re shown inThe valve is
, T2, v2, h2 a
onstant at
pipeline and
ter the bottl
so the K.E.
of the enve
pter 2
lume
below). Inopened
d u2. The
the tube
.
terms have
lope to zero
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S
w
CFi
wr
Si
No
DLe
ap
As
A
or
or
w
qu
K Mon
= 2m
ich gives th
ntrol Voluure above,
itten on a ti
ce hP and C
w
v 2
2 2
E U –
m u –
=
=
ischar t us conside
plying first l
suming K.E.
ain
ich shows
asi-static.
For ch
Firdal’s
2 1 1u – m u –
energy bal
e Techniqpplying the
e rate basi
vdE dQ
dτ dτ= +
are consta
vE Q ⎛
Δ = + ⎜⎝
1 2 2
1 1 P
m u –
u – h
=
⎛
⎜⎝
ing a a tank dis
aw to the co
and P.E. of
d(mmdu+ udm
( )
=
= =
+
= −
= −
+ == 0
dm du
m pv
V vm c
vdm mdv
dm dv
m v
du dv
pv v
d u pv
dQ
hat the pro
rging the ta
st La
)2 1m – m2
C ⎛ ⎜⎝
nce for the
ue: Assumeenergy equa
-2
PP
dh +
2 dτ
C ⎛ ⎞⎜ ⎟⎝ ⎠t, the equat
(2
PP 2+ m
2
⎞⎟ ⎠
C
(
1 1
2
P
2
m u
m2
C ⎞
−⎟ ⎠
d Chaharging a fl
trol volume,
= V dU d
the fluid to
u) = hdm= udm+ pv
=
.
0
0
nst
cess is adia
nk
of T
41
P+ h
⎞⎟ ⎠
rocess.
a control voltion in this
ion is integr
)1m
)1
ging auid into a s
⎛ + +⎜
⎝ h
e small and
m
batic and
herm
ume boundecase, the foll
ted to give f
Tankpply line (F
⎞+ ⎟
⎠
2
o
gz2
C
dQ = 0
Chargi
odyn
d by a controwing energ
or the Total
igure). Since
out
ut
dm
ng and Dis
micsCh
l surface asbalance m
process
x dW = 0 a
charging a
pter 2
shown iny be
d dmin = 0,
Tank
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First Law of ThermodynamicsS K Mondal’s Chapter 2
42
( ) = Δ = −
= −
∫ 2 2 1 1in2 2 1 1
V
p p
hdm U m u m u
m h m u m u
where the subscript p refers to the constant state of the fluid in the pipeline. If the tank is initially
empty, m1 = 0.
= 2 2 p pm h m u
Since
=
=2
2
p
p
m m
h u
If the fluid is an ideal gas, the temperature of the gas in the tank after it is charged is given by
=
= γ2
2
p p vc T c T
or T T
PROBLEMS & SOLUTIONSExample 1
The work and heat transfer per degree of temperature change for a closed system is given by
1 1/ ; /
30 10
dW dQ kJ C kJ C
dT dT ° ° = =
Calculate the change in internal energy as its temperature increases from 125ºC to 245ºC.
Solution:
( ) ( )
( )
2
1
2
1
2 1
30
1 1245 125
30 30 30
10
1245 125 12
10 10
T
T
T
T
dT dW
dT W T T
dT dQ
dT Q kJ
=
= = − = −
=
= = − =
∫
∫
Applying First Law of Thermodynamics
Q = W + ΔUΔU = Q – W = 12 – 4 = 8kJ.
Example 2
Air expands from 3 bar to 1 bar in a nozzle. The initial velocity is 90 m/s. the initial temperature is
150ºC. Calculate the velocity of air at the exit of the nozzle.
Solution:
The system in question is an open one. First Law of Thermodynamics for an open system gives2 2
1 21 1 1 2 2 2
2 2
C C w h Z g Q w h Z g W
⎡ ⎤ ⎡ ⎤+ + + = + + +⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
Since the flow is assumed to be steady.
1 2w w=
Flow in a nozzle is adiabatic flow.
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First Law of ThermodynamicsS K Mondal’s Chapter 2
44
( )
( )
( )
i i e e 2 2 1 1
e
1 2
2
0
0
0 2
2 0
2 0
Q m h = m h m u – m u W
Here Q 0, W 0, m 0 no mass leaving from control vol.
0 evacuated
0.718 0.287 273
0.718 25 0.287 298
103.48 /
103.48 /
0.71
i
i
i i i i
m m m
h u
h u pv u T T
u
u kJ kg u
or u u kJ kg
u u
+ + +
= = =
= ∴ =
∴ =
= + = + + +
= + × + ×
= + =
− =
= +2
2 0
2
8
103.48144.2
0.718 0.718
T
u uT C
−= = = °
Example 4
A system whose mass is 4.5 kg undergoes a process and the temperature changes from 50° C to
100°C. Assume that the specific heat of the system is a function of temperature only. Calculate theheat transfer during the process for the following relation ship.
801.25 /
160n
c kJ kg C t
= + °+
[t is in oC]
Solution:
[ ] ( )
[ ] ( ) ( )( )
100 100
1 2
50 50
100 100
50 50
100100
50
50
804.5 1.25
160
4.5 1.250.0125 2.0
14.5 1.25 ln 0.0125 2.0
0.01251
4.5 1.25 50 1.25 2.0 0.625 20.0125
4
⎛ ⎞= = +⎜ ⎟+⎝ ⎠
⎧ ⎫⎪ ⎪= +⎨ ⎬
+⎪ ⎪⎩ ⎭
⎧ ⎫⎪ ⎪⎡ ⎤= + +⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭
⎧ ⎫⎡ ⎤= × + + − +⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭
=
∫ ∫
∫ ∫
nQ mc dt dt
t
dtdt
t
t t
ln ln
1 3.25.5 62.5 358
0.0125 2.625
⎧ ⎫+ =⎨ ⎬⎩ ⎭
ln kJ
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S
A
CT
st
ar
no
G
G
G
K Mon
SKED
pplicati
mmon De inlet an
am for a
e as indi
tations are
TE-1. If
th
(a)
TE-2. As
wa
en
(a)
TE-3. Th
th
Ac
(a)
Firdal’s
OBJE
revio
on of
ata for Qd the outl
adiabatic
ated in t
as usually
ass flow
turbine (i
12.157
ume the a
ter at the
rgy effects
0.293
followin
rmodynam
ording to
Figures 1an
st La
TIVE
s 20-
irst La
estionst conditio
steam tu
he figure.
followed.
ate of stea
MW) is:
ove turbi
inlet to th
, the specif
(
four fig
ic cycle, on
he first la
2 (b) Fig
of T
45
UEST
ears
to St
1 and Qns of
bine
The
m through
(b) 12.941
e to be pa
pump is
c work (in
b) 0.35 1
ures have
the p-v an
of thermo
ures 1and 3
herm
IONS (
GAT
ady Fl
:
the turbin
(c
t of a simp
1000 kg/m3.
kJ/kg) sup
(c)
been dr
T-s plane
dynamics,
(c) Figure
odyn
ATE
Que
w Pro
e is 20 kg/
) 168.001
le Rankine
Ignoring
lied to the
2.930
wn to r
.
qual areas
1and 4
micsCh
, IES,
tions
ess S.
[G
the powe
[G
(d)
cycle. The
inetic an
pump is:
[ (d)
present a
[
are enclos
(d) Figures
pter 2
IAS)
.E.E
TE-2009]
output of
TE-2009]
168.785
density of
potential
ATE-2009]3.510
fictitious
ATE-2005]
ed by
2 and 3
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First Law of ThermodynamicsS K Mondal’s Chapter 2
46
Internal Energy – A Property of SystemGATE-4. A gas contained in a cylinder is compressed, the work required for
compression being 5000 kJ. During the process, heat interaction of 2000 kJ
causes the surroundings to the heated. The change in internal energy of the
gas during the process is: [GATE-2004](a) – 7000 kJ (b) – 3000 kJ (c) + 3000 kJ (d) + 7000 kJ
GATE-4a. The contents of a well-insulated tank are heated by a resistor of in
which 10 A current is flowing. Consider the tank along with its contents
as a thermodynamic system. The work done by the system and the heat
transfer to the system are positive. The rates of heat (Q), work (W) and
change in internal energy during the process in kW are [GATE-2011]
(a) Q = 0, W = –2.3, = +2.3 (b) Q = +2.3, W = 0, = +2.3
(c) Q = –2.3, W = 0, = –2.3 (d) Q = 0, W = +2.3, = –2.3
Discharging and Charging a TankGATE-5. A rigid, insulated tank is initially
evacuated. The tank is connected with a
supply line through which air (assumed to
be ideal gas with constant specific heats)
passes at I MPa, 350°C. A valve connected
with the supply line is opened and the tank
is charged with air until the final pressure
inside the tank reaches I MPa. The final
temperature inside the tank
(A) Is greater than 350°C
(B) Is less than 350°C
(C) Is equal to 350°C
(D) May be greater than, less than, or equal to
350°C, depending on the volume of the tank
Previous 20-Years IES Questions
First Law of Thermodynamics
IES-1. Which one of the following sets of thermodynamic laws/relations is directlyinvolved in determining the final properties during an adiabatic mixing
process? [IES-2000]
(a) The first and second laws of thermodynamics
(b) The second law of thermodynamics and steady flow relations
(c) Perfect gas relationship and steady flow relations
(d) The first law of thermodynamics and perfect gas relationship
23Ω
( U)
Δ U Δ UΔ U Δ U
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First Law of ThermodynamicsS K Mondal’s Chapter 2
47
IES-2. Two blocks which are at different states are brought into contact with each
other and allowed to reach a final state of thermal equilibrium. The final
temperature attained is specified by the [IES-1998]
(a) Zeroth law of thermodynamics (b) First law of thermodynamics
(c) Second law of thermodynamics (d) Third law of thermodynamics
IES-3. For a closed system, the difference between the heat added to the system and
the work done by the system is equal to the change in [IES-1992]
(a) Enthalpy (b) Entropy
(c) Temperature (d) Internal energy
IES-4. An ideal cycle is shown in the figure. Its
thermal efficiency is given by
3 3
1 1
2 2
1 1
1 11
(a)1 (b) 11 1
γ
⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
− −⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
v v
v v
p p
p p
( )
( )
( )
( )3 1 3 11 1
2 1 1 2 1 1
1(c)1 (b) 1γ
γ
− −− −
− −
v v v v p p
p p v p p v
[IES-1998]
IES-5. Which one of the following is correct? [IES-2007]
The cyclic integral of )( W Q δ δ − for a process is:(a) Positive (b) Negative (c) Zero (d) Unpredictable
IES-6. A c