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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
VTU-NPTEL-NMEICT
Project Progress Report
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The Project on Development of Remaining Three Quadrants to
NPTEL Phase-I under grant in aid NMEICT, MHRD, New Delhi
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Subject Matter Expert Details
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Dr.A.R.ANWAR KHAN
Prof & H.O.D
Dept of Mechanical Engineering
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Applied Thermodynamics
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Course Name:
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SME Name :
web 
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Type of the Course
Module
I
DEPARTMENT OF MECHANICAL
ENGINEERING,
GHOUSIA COLLEGE OF ENGINEERING,
RAMANAGARA -562159
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 1 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
CONTENTS
Sl.
No.
DISCRETION
1.
Quadrant -2
a. Animations.
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b. Videos.
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c. Illustrations.
2.
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b. Open Contents
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a. Wikis.
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Quadrant -3
3.
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Quadrant -4
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a. Problems.
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b. Assignments
c. Self Assigned Q & A.
d. Test your Skills.
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 2 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
MODULE-I
PROPERTY RELATIONSHIP
QUADRANT-2
Animations
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1 )http://www.youtube.com/watch?v=t0sjFKPdvIc
2) http://www.youtube.com/watch?v=0PkOEHMNOLk
3)
https://www.google.co.in/search?q=animation+related+to+air+conditioning&tbm=isch
&tbo=u&source=univ&sa=X&ei=FtseU6rmOMX9rAeVgIGQAg&ved=0CDMQsAQ&
biw=1440&bih=809
4) http://www.dnatube.com/video/7911/Animation-of-how--Air-Conditioning-works
5) http://www.dnatube.com/video/7939/An-animation-of-how--Air-Conditioning-works
6) http://www.dnatube.com/video/9079/Principles-Of-Air-Conditioning
7) http://www.yazaki-airconditioning.com/products/absorption_cooling.html
8) http://www.wisegeek.org/how-does-air-conditioning-work.htm#didyouknowout
9) http://www.youtube.com/watch?v=_lFUlA1PZ8U
10) http://educypedia.karadimov.info/education/mechanicsjavathermo.htm
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Videos:
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1 )http://www.youtube.com/watch?v=t0sjFKPdvIc
2) http://www.youtube.com/watch?v=0PkOEHMNOLk
3) http://www.dnatube.com/video/7911/Animation-of-how--Air-Conditioning-works
4) http://www.dnatube.com/video/7939/An-animation-of-how--Air-Conditioning-works
5) http://www.dnatube.com/video/9079/Principles-Of-Air-Conditioning
6) http://www.youtube.com/watch?v=_lFUlA1PZ8U
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 3 of 32
2014
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
ILLUSTRATIONS
Pure Substance
A pure substance is defined as a substance having a constant and uniform chemical
composition.
Homogeneous Mixture of Gases
Any mixture of gases in which the constituents do not rennet chemically with one
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another and they are in a fixed proportion by weight is referred to as homogeneous mixture
of gases and is regarded as a single substance. The properties of such a mixture can be
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determined experimentally just as for a single substance, and they can be tabulated or related
algebraically in the same way. Therefore, the composition of air is assumed invariable for
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most purposes and air is usually treated as a single substance.
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Mixture of Gases
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Mixture of gases is generally imagined to be separated into its constituents in such a
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way that each occupies a volume equal to that of the mixture and each is at the same
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temperature as the mixture.
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EMPIRICAL LAW FOR MIXTURES OF GASES
Consider a closed vessel of volume V at temperature T, which contains a mixture of
perfect gases at a known pressure. If some of the mixture were removed, then the pressure
would be less than the initial value. If the gas removed was the full amount of one of the
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 4 of 32
2014
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
constituents then the reduction in pressure would be equal to the contribution of that
constituent to the initial total pressure. This is not only applicable to pressure but also to
internal energy, enthalpy and entropy.
Dalton’s Law
In a mixture of gases, each constituent contributes to the total pressure by an amount
which is known as the partial pressure of the constituent. The relationship between the
partial pressures of the constituents is expressed by Dalton’s law, as follows: The pressure
of a mixture of gases is equal to the sum of the pressures of the individual constituents when
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each occupies a volume equal to that of the mixture at the temperature of the mixture.
(1.1)
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m = mA + mB + mC + ……. or m = ∑mi
T
mixtures at low pressures. By the conservation of mass:
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Dalton’s law is based on experiment and is found to be obeyed more accurately
by gas
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By Dalton’s law
P = PA + PB + PC + …..…. or P = ∑Pi
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(1.2)
where mi = mass of a constituent; and Pi = partial pressure of a constituent.
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GIBBS-DALTON LAW
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Dalton’s law was reformulated by Gibbs to include a second statement on the
properties of mixtures. The combined statement is known as the Gibbs-Dalton law, and is
stated as follows:
The internal energy, enthalpy and entropy of a mixture of gases are respectively equal to
the sums of the internal energies, enthalpies and entropies of the individual constituents
when each occupies a volume equal to that of the mixture at the temperature of the mixture.
This statement leads to the following equations.
mu = mAuA + mBuB + mCuC + ……. or mu = ∑miui
(1.3)
mh = mAhA + mBhB + mChC + …… or mh = ∑mihi
(1.4)
and ms = mAsA + mBsB + mCsC + ……. or ms = ∑misi
(1.5)
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 5 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
VOLUMETRIC ANALYSIS OF A GAS MIXTURE.
The analysis of a mixture of gases is often quoted by volume as this is the most
convenient for practical determinations. Considered a volume V of a gaseous mixture at a
temperature T, consisting of three constituents A, B and C. Let each of the constituents be
compressed to a pressure P equal to the total pressure of the mixture, and let the temperature
remain constant.
Partial pressure of A, P = m A RAT
or
= PAV
m
A
RA T
A
or
∑
P
i
since ∑Pi = P,
∑Vi = V
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C
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V
P
= PA V
P
(1.6)
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P
∑Vi =
A
Vi = Pi
V
P
In general, Vi = Pi V or
and
V
A
A
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P V = PV
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mA = PVA
T
Also, the total pressure, P = mA RAT or
V
je
c
RA T
V
Therefore,
t
A
(1.7)
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Therefore the volume of a mixture of gases is equal to the sum of the volumes of the
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mixture.
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individual constituents when each exists alone at the pressure and temperature of the
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
Mixtures of Perfect Gases
Each of the constituents in the mixture is assumed to obey the perfect gas equation of
state:
for the mixture,
or P =
PV = mRT
or P =mi RiT
for the constituent, P V = m R T
i
i
mRT
V
i
i
V
From dalton’s law, P = ∑ Pi = ∑ mi RiT
V
V
T
∑
m R
i
t
=
i
je
c
mRT
V
Mi
=∑
, therefore, mRo = ∑ mi Ro and Eq. (1.8) becomes:
M
mi
M
i
Mi
(1.9)
N
M
M
M
m
(1.8)
T
R = Ro
EI
C
since Ri = Ro and
Pr
o
that is, mR = ∑miRi
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The ratio of mass (m) of a gas to its Molar mass (M) is defined as number of moles
contain by the volume ot the gas, that is m/M = n and mi/Mi = ni. Substituting these
or
n = nA + nB + nC + ……
(1.10)
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n = ∑ni
-N
in Eq. (1.9) we have:
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Therefore, the number of moles of a mixture is equal to the sum of the moles of the
constituents.
THE MOLAR MASS AND SPECIFIC GAS CONSTANT
PiV = miRiT
or
PiV = niRoT
(1.11)
V ∑Pi = RoT ∑ni
(1.12)
since ∑Pi =
P,
and ∑ni = n
therefore, PV =
nRoT
(1.13)
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 7 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
This is a characteristic equation for the mixture, which shows that the mixture
also acts as a perfect gas. It can be assumed that a mixture of perfect gases obeys all
the perfect gas laws. The relationship between the volume fraction and pressure
fraction is obtained from ratio of Eq. (1.12) to Eq. (1.13):
Pi V
n R T
= i o
nRoT
PV
or
Pi
=
P
ni
(1.14)
n
Combination of Eq. (1.14) with Eq. (1.6) gives:
Pi ni Vi
=
=
P
n V
je
c
t
(1.15)
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In order to find the specific gas constant for the mixture in terms of the specific gas
constants of the constituents, consider the following equations both for the mixture
T
and for a
for a constituent
PiV = miRiT
or∑Pi V = ∑miRiT
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then V ∑Pi = T ∑miRi
M
PV = mRT
N
for mixture,
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C
constituent:
since ∑Pi = P, PV = T ∑miRi
or
mRT = T ∑miRi
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therefore, mR = ∑miRi
mi R
R=∑ m i
VT
(1.16)
where mi/m is the mass fraction of a constituent.
SPECIFIC HEAT CAPACITIES OF A GAS MIXTURE
From the Gibbs-Dalton law, mu = ∑miui
at constant volume, u = CVT
Therefore, mCVT = ∑miCViT
mCV = ∑miCVi orCV = ∑[(mi /m)Cvi]
(1.17)
Similarly, mh = ∑mihi
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 8 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
at constant pressure, h = CpT
Therefore, mCp = ∑miCpi
or
Cp = ∑[(mi /m)Cpi]
(1.18)
From Eqs. (1.17) and (1.18)
Cp – CV = ∑[(mi /m)Cpi] – ∑[(mi /m)Cvi]
Cp – CV = ∑[(mi /m)(Cpi – Cvi )]
Also, Cp – CV = Ri,
therefore,
therefore, for the
mixture:
je
c
Ri ,
Pr
o
R=∑
Recall Eq. (1.16),
mi
m
t
Cp – CV = ∑[(mi /m)Ri
T
Cp – CV = R
;
R
Cv =
; and C p =
γ −1
.
γ−1
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Cv
γR
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Cp
N
γ=
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C
This shows that the following equations can also be applied to a mixture of gases:
mRoT
U
=
-N
UNIVERSAL GAS LAW
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pV ~
N
where Ro is the universal constant 8314.4 J/kmol K
Ñ is the relative molecular mass which is 18 for water vapour treated as a gas and
28.96 for dry air treated as a single gas.
PARTIAL PRESSURES
The pressure exerted by a gas on the surface of containment is due to the
bombardment of the surface by the molecules. The relative distance between
molecules is very large so if two or more gases exist in the same space, their
behaviour is unaffected by the others and so each gas produces a pressure on the
surface according to the gas law above. Each gas occupies the total volume V and
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 9 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
has the same temperature T. If two gases A and B are considered, the pressure due
to each is :
ma RoT
pa =
~
pb =
Na V
mb RoT
~
NVb
p = pa + pb
The total pressure on the surface of containment is
This is Daltons Law of partial pressures.
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Now let’s see how these laws are applied to mixtures of vapour and air.
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AIR - VAPOUR MIXTURES
T
In the following work, water vapour is treated as a gas.
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Consider a mixture of dry air and vapour. If the temperature of the mixture is cooled
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until the vapour starts to condense, the temperature must be the saturation temperature
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(dew point) and the partial pressure of the vapour ps must be the value of ps in the
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fluids tables at the mixture temperature.
If the mixture is warmed up at constant pressure so that the temperature rises, the
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vapour must become superheated. It can be shown that the partial pressure of the
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vapour and the dry air remains the same as at the saturation temperature.
Let condition (1) be at the saturation condition and condition (2) be at the higher
temperature. p is constant so it follows that :
V
1
T
=
V
2
T
1
2
The initial partial pressure of the vapour is:
m RoT
ps1 = ~s
1
Ns V1
The final pressure of the vapour is :
m RoT
ps2 = ~s
2
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 10 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
NV
s
2
V1 = V2 then p
Since
T
T
1
s1
=p s2
2
By the same process it can be shown that pa1 = pa2
If p is constant then the partial pressures are constant and the partial pressure of the
vapour may easily be found by looking up the saturation pressure at the dew point if it
is known.
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When the air is contact with water, it will evaporate the water and the water will cool
down until it is at the saturation temperature or dew point. This idea is used in wet
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bulb thermometers for example, which measure the dew point. When stable conditions
are reached, the air becomes saturated and equal to the temperature of the water and so
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its temperature is the dew point (ts) in fluids tables.
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HUMIDITY
There are two ways to express humidity SPECIFIC AND RELATIVE.
SPECIFIC HUMIDITY ω
U
-N
ω = mass of water vapour/mass of dry air
VT
Starting with the gas law
~
m = pVN
RoT
~
~
p VRoTN s p N
p
p
18
s
ω=
= 0.622 s
~ = s~s
= sx
p
pa VRoTNa pa Na p
p
a
a
28.96
s
ω = 0.622 p − ps
Dr.A.R.ANWAR KHAN,Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 11 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
RELATIVE HUMIDITYφ
φ = mass of vapour/maximum possible mass of vapour
The maximum possible mass of water vapour which can be held by air is when the vapour is
saturated and the temperature of the mixture is the saturation temperature.
mass = Volume/specific volume = V/v
When saturated,
v = vg at the mixture temperature.
m V V v
φ = s = ÷ = a mg vs va vs
vs =
p
oj
ec
t
v = V/m
Ng pg
and vg =
RoT
RoT
Np
s
s
Pr
Alternatively
C
T
φ = s pg
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ps = partial pressure of the actual vapour
ps
p
and φ = s
p − ps
pg
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ω = 0.622
-N
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pg = partial pressure when saturated.
PT
ωp−p
s )
φ= (
VT
U
-N
0.622pg
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 12 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
QUADRANT-3
Wikis:
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1) http://en.wikipedia.org/wiki/Air_conditioning
2) http://en.wikipedia.org/wiki/HVAC
3) http://simple.wikipedia.org/wiki/Air_conditioner
4) http://z32.wikispaces.com/Air+Conditioning+System
5) http://whirlpool.net.au/wiki/aircon_faq
6) http://commons.wikimedia.org/wiki/File:Air_conditioning_unit-en.svg
7) http://wiki.hometech.com/tiki-index.php?page=HVAC+Control+Tutorial
8) http://climatetechwiki.org/technology/efficient-air-conditioning-systems
9) http://home.howstuffworks.com/ac1.htm
10) https://www.ashrae.org/resources--publications/free-resources/top-ten-things-aboutair-conditioning
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Applied Thermodynamics by R. K. Rajput
C
T
Open Contents:
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Applied Thermodynamics for Engineering Technologists by Eastop
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Applied Thermodynamics by B. K. Venkanna B. V. S
VT
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Basic and Applied Thermodynamics by Nag
Applied Thermodynamics by D. S. Kumar
A textbook of applied thermodynamics, steam and thermal ... by S. K. Kulshrestha
Applied thermodynamics by Anthony Edward John Hayes
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 13 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
QUADRANT-4
Problems
1) The atmospheric conditions are; 20°C and specific humidity of 0.0095 kg/kg of dry air.
Calculate the following: (i) Partial pressure of vapour (ii) Relative humidity (iii) Dew point
temperature.
Solution:
Dry bulb temperature, tdb = 20ºC
Specific humidity, W = 0.0095 kg/kg of dry air
The specific humidity is given by
oj
ec
t
(i) Partial pressure of vapour, p v:
W=
Pr
0.0095 =0.622 PV/ (1.0132- PV)
C
T
0.0095(1.0132 – p v) = 0.622 p v
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0.009625 – 0.0095 p v = 0.622 p v
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pv = 0.01524 bar.
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(ii) Relative humidity φ :
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Corresponding to 20ºC, from steam tables, pvs = 0.0234 bar
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∴ Relative humidity, φ = pv / pvs = 0.01524/0.0234= 0.65 or 65%. (Ans)
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(iii) Dew point temperature, tdp :
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0.01524 bar,
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The dew point temperature is the saturation temperature of water vapour at a pressure of
t dp [from steam tables by interpolation]
=
= 13.24°C.
2) 0.004 kg of water vapour per kg of atmospheric air is removed and temperature of air after
removing the water vapour becomes 20°C. Determine :(i) Relative humidity (ii) Dew point
temperature. Assume that condition of atmospheric air is 30°C and 55% R.H. and pressure is
1.0132 bar.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 14 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
Solution:
Corresponding to 30ºC, from steam tables, pvs = 0.0425 bar
∴ Relative humidity (R.H.), φ =
i.e., 0.55 = PV/0.0425 ∴ Pv = 0.02337 bar.
Also the specific humidity,
W=
= (0.622*0.02337)/ (1.0132-0.02337)
= 0.01468 kg/kg of dry air.
The specific humidity after removing 0.004 kg of water vapour becomes,
0.01468 – 0.004 = 0.01068 kg/kg of dry air and the temperature t db is given as 20ºC.
oj
ec
t
The partial pressure of water vapour, pv , at this condition can be calculated as follows :
W=
Pr
0.01068 = 0622Pv/(1.0132-Pv)
C
T
or, 0.01068 (1.0132 – p v ) = 0.622 p v
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or, 0.01082 – 0.01068 p v = 0.622 p v
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0.6327 p v = 0.01082
-N
∴ p v = 0.0171 bar
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Corresponding to 20ºC, from steam tables, pvs = 0.0234 bar.
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(i) Relative humidity, φ =Pv/Pvs = 00171/00234 = 0.73 or 73%. (Ans)
(ii) Dew point temperature, tdp :
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U
-N
Corresponding to 0.0171 bar, from steam tables, t dp = 15°C.
3) The sling psychrometer in a laboratory test recorded the following readings:
Dry bulb temperature = 35°C
Wet bulb temperature = 25°C.
Calculate the following :
(i) Specific humidity (ii) Relative humidity
(iii) Vapour density in air (iv) Dew point temperature
(v) Enthalpy of mixture per kg of dry air
Take atmospheric pressure = 1.0132 bar.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 15 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
Solution:
For finding the partial pressure of vapour, using the equation:
Corresponding to 25ºC (from steam tables),
(Pvs )wb = 0.0317 bar
Substituting the values in the above equation, we get
Pv=0.0317 = 0.0317 – 0.0065 = 0.0252 bar.
(i) Specific humidity,
= (0.622*0.0252)/ (1.0132-0.0252)
oj
ec
t
W=
Pr
= 0.01586 kg/kg of dry air. (Ans)
(ii) Relative humidity, φ = Pv/Pvs
C
T
[Pvs = 0.0563 bar corresponding to 35ºC, from steam tables]
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= 0.447 or 44.7%. (Ans.)
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PvVv = M v Rv T v
-N
From characteristic gas equation
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(iii) Vapour density :
PT
Pv = M v R v T v / Vv = ρ v Rv T v
-N
where ρ v =vapour density, (characteristic gas constant Rv= Universal gas constant
8314.3/18
VT
=
U
/Molecularweight of H2O
∴ 0.0252 × 10 5 = ρ v × (8314.3/18)× (273 + 35)
∴ ρ v = (0.0252 ×105 ×18) /(8314.3× 308) = 0.0177 kg/m3 (Ans)
(iv) Dew point temperature, tdp :
Corresponding to 0.0252 bar, from steam tables (by interpolation),
t dp =
= 21.2°C. (Ans)
(v) Enthalpy of mixture per kg of dry air, h :
h = Cp tdb + W hvapour
= 1.005 × 35 + 0.01586 [h g + 1.88 (t db – tdp )]
= 35.175 + 0.01586 [2565.3 + 1.88 (35 – 21.2)]
(where hg = 2565.3 kJ/kg corresponding to 35ºC t db ) = 76.27 kJ/kg of dry air.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 16 of 32
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MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
4) One kg of air at 35°C DBT and 60% R.H. is mixed with 2 kg of air at 20°C DBT and 13°C
dew point temperature. Calculate the specific humidity ofthe mixture.
Solution:
For the air at 35°C DBT and 60% R.H. :
Corresponding to 35ºC, from steam tables,
Pvs = 0.0563 bar
Relative humidity, φ = Pv/Pvs
∴ p v = φ p vs = 0.6 × 0.0563 = 0.0338 bar
= (0.622*0.0338)/ (1.0132-0.0338) = 0.0214 kg/kg of dry air
oj
ec
t
W=
Corresponding to 0.0338 bar, from steam tables, tdp =
Pr
26.1ºC
=
Enthalpy, h = c p t db + Wh vapour
C
T
= 1.005 t db + W [h g + 1.88 (t db – t dp )]
EI
= 1.005 × 35 + 0.0214 [2565.3 + 1.88 (35 – 26.1)]
M
= 90.43 kJ/kg of dry air.
-N
For the air at 20°C DBT and 13°C dew point temperature : Pv is the vapour pressure
EL
corresponding to the saturation pressure of steam at 13ºC.
= (0.622*0.015)/(1.0132-0.015)= 0.00935 kg/kg of dry air
-N
W=
PT
∴Pv = 0.0150 bar
U
Enthalpy, h = c p t db + Wh vapour
VT
= 1.005 × 20 + 0.00935 [h g + 1.88 (t db – t dp )]
= 20.1 + 0.00935 [2538.1 + 1.88 (20 – 13)]
= 43.95 kJ/kg of dry air
Now enthalpy per kg of moist air
= 58.54 kJ/kg of moist air
Mass of vapour/kg of moist air
= 0.01316 kg/kg of moist air
Specific humidity of mixture
= 0.01316/(1-0.01316)
= 0.01333 kg/kg of dry air.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 17 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
5) 90 m 3 of air per minute at 20°C and 75% R.H. is heated until its temperature becomes
30°C. Calculate : (i) R.H. of the heated air. (ii) Heat added to air per minute.
Solution:
(i) For air at 20°C and 75% R.H. :
Pvs = 0.0234 bar (from steam tables, at 20ºC)
Pv = φ ×Pvs = 0.75 × 0.0234 = 0.01755 bar
tdp =
W1 =
= 15.5ºC
= (.622*0.01755)/(1.0132-0.01755) = 0.0109 kg/kg of dry air
Enthalpy, h 1 = c p t db + Wh vapour
oj
ec
t
= 1.005 × 20 + 0.0109 [h g + 1.88 (t db – t dp)]
= 1.005 × 20 + 0.0109 [2538.1 + 1.88(20 – 15.5)] = 47.85 kJ/kg of dry air
Pr
(i) Relative humidity of heated air :
C
T
For air at 30°C DBT :
Since the saturation pressure of water vapour at 30ºC is higher than the saturation pressure of
EI
water vapour at 20ºC so it is sensible heating, where p v is same after heating.
-N
M
∴ Relative humidity, φ = Pv/Pvs =0.01755/0.0425 = 0.412 or 41.2%
Pvs = 0.0425 bar, corresponding to 30ºC)
EL
i.e., Relative humidity of heated air = 41.2%
PT
(ii) Heat added to air per minute :
-N
Enthalpy, h 2 = c p t db + Wh vapour
U
= 1.005 × 30 + 0.0109 × [h g + 1.88 (t db – t dp )]
VT
= 1.005 × 30 + 0.0109 [2556.3 + 1.88 (30 – 15.5)]
= 58.31 kJ/kg of dry air
Mass of dry air in 90 m 3 of air supplied
m a = PV /RT = (Pt-Pv)V/RT =((1.0132- 0.01755) × 105 ×90) / (287×(273+20)) = 106.5
kg/min.
Amount of heat added per minute
= 106.5 (h2 – h1) = 106.5 (58.31 – 47.85)  1114 Kj.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 18 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
6) 40 m 3 of air at 35°C DBT and 50% R.H. is cooled to 25°C DBT maintaining its specific
humidity constant. Determine : (i) Relative humidity (R.H.) of cooled air ; (ii) Heat removed
from air.
Solution:
For air at 35°C DBT and 50% R.H. :
Pvs = 0.0563 bar (At 35ºC, from steam tables)
∴P v = φ × Pvs = 0.5 × 0.0563 = 0.02815 bar
W1 =
= 0.0177 kg/kg of dry air
h1 = Cp t db 1 + W 1 [ hg 1 + 1.88 ( t db1 – t dp1 ]
oj
ec
t
t dp 123ºC (corresponding to 0.02815 bar)
∴ h1 = 1.005 × 35 + 0.0177 [2565.3 + 1.88 (35 – 23)] = 80.98 kJ/kg of dry air
Pr
For air at 25°C DBT :
C
T
(i) R.H. of cooled air :
EI
Since the specific humidity remains constant the vapour pressure in the air remains constant.
M
φ = Pv/Pvs = 0.02815/0.0317 = 0.888 or 88.8%
-N
i.e., Relative humidity of the cooled air = 88.8%. (Ans.)
EL
(ii) Heat removed from air :
h 2 =Cp t db 2 + W 2 [ h g 2 + 1.88 ( tdb 2 – t dp 2 )]
-N
= 70.27 kJ/kg of dry air.
PT
= 1.005 × 25 + 0.0177 [2547.2 + 1.88 (25 – 23)]
U
To find mass of dry air (m a ), using the relation :
W1=W2 tdp2=tdp1
VT
PaVa=maRaTa
∴ m a = pa va /R a Ta=(1.0132- 0.02815) ×105 ×40 /287×(273+35)= 44.57 kg
∴ Heat removed from 40 m 3 of air
= ma (h 1 – h 2) = 44.57 (80.98 – 70.27) = 477.3 kJ. (Ans)
8) 150 m 3 of air per minute is passed through the adiabatic humidifier. The condition of air
at inlet is 35°C DBT and 20 per cent relative humidity and the outlet condition is 20°C DBT
and 15°C WBT. Determine the following :
(i) Dew point temperature (ii) Relative humidity of the exit air
(iii) Amount of water vapour added to the air per minute.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 19 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
Solution:
For air at 35°C DBT and 20% relative humidity.
Pvs = 0.0563 bar (At 35ºC from steam tables)
Pv = φ ×P vs = 0.2 × 0.0563 = 0.01126 bar
W1 =
= 0.00699 kg/kg of dry air
(i) The dew point temperature of air which is the saturation temperature of steam
corresponding to the pressure 0.01126 bar is
8 + (9 – 8) × ((0.01126- 0.01072)/(0.01150-0.01072))= 8.7ºC
oj
ec
t
i.e., Dew point temperature = 8.7°C. (Ans.)
M
= 0.00852 kg/kg of dry air
-N
W2 =
EI
= 0.0137 bar
C
T
Pr
(ii) Relative humidity of the exit air : For air at 20ºC DBT and 15ºC WBT.
EL
Relative humidity = φ = Pv/Pvs = 0.585 or 58.5%. (Ans.)
(Pvs = 0.0234 bar, corresponding to 20ºC, from steam tables)
PT
The dew point temperature of air which is the saturation temperature of steam correspond-
-N
ing to 0.0137 bar is 11°C (from steam tables). (Ans.)
VT
kg
U
The amount of water vapour per kg of dry air = W 2 – W 1 = 0.00852 – 0.00699 = 0.00153
The mass of dry air in 150 m 3 of mixture
∴ m a = pa va /R a Ta=(1.0132- 0.01126) ×105 ×150 /287×(273+35)= 170 kg
(iii) The amount of water vapour added to air per minute = m a (W2 – W1) = 170 × 0.00153 =
0.26 kg/min.
9) An air-water vapour mixture enters an adiabatic saturation chamber at 28°C and leaves at
18°C, which is the adiabatic saturation temperature. The pressure remains constant at 1.0 bar.
Determine the relative humidity and humidity ratio of the inlet mixture.
Solution:.
The specific humidity at the exit W2S =
=
= 0.01308 kg/kg of dry air
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 20 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
The specific humidity at the inlet (equation 10.18)
W1 =
=
= 0.01704 kg/kg of dry air
W1 =
∴ 0.01704 =
Or 0.01704 (1.00 – Pv1) = 0.622 Pv1
Or 0.01704 – 0.01704 Pv1= 0.622 Pv1
Or 0.0639 Pv1= 0.01704
∴ Pv1= 0.02666 bar
∴ Relative humidity = Pv1/Pvs1 =
oj
ec
t
0.02666/0.03782 = 0.7 or 70%.
Pr
10) An air-water vapour mixture enters an air-conditioning unit at a pressure of 1.0 bar. 38°C
C
T
DBT, and a relative humidity of 75%. The mass of dry air entering is 1 kg/s. The air-vapour
mixture leaves the air-conditioning unit at 1.0 bar, 18°C, 85% relative humidity. The
EI
moisture condensed leaves at 18°C. Determine the heat transfer rate for the process.
M
Solution:
EL
t db2 = 18ºC, R.H., φ 2 = 85%
-N
tdb1 = 38ºC, R.H., φ 1 = 75%
PT
The flow diagram and the process are shown in Figs respectively.
-N
At 38°C
U
From steam tables : Pvs = 0.0663 bar, hg1 = 2570.7 kJ/kg
W1 =
VT
∴ Pv = φ × Pvs = 0.75 × 0.0663 = 0.0497 bar
= 0.0325 kg/kg of dry air
At 18°C
From steam tables : Pvs = 0.0206 bar, hg2 = 2534.4 kJ/kg hf2 = 75.6 kJ/kg
Pv = 0.85 × 0.0206 = 0.01751 bar
W2 =
= 0.01108 kg/kg of dry air
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 21 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
Heat transfer rate,
q = (W2hg2 – W1 hg1) + Cp(tdb2 – tdb1)+(W1–W2)hf2
= (0.01108 × 2534.4 – 0.0325 × 2570.7) + 1.005 (18
– 38) + (0.0325 – 0.01108) × 75.6
oj
ec
t
= – 55.46 – 20.1 + 1.62 = – 73.94 kJ/kg of dry air.
for the following hot and wet summer conditions :
C
T
Outdoor conditions ...... 32ºC DBT and 65% R.H.
Pr
11) It is required to design an air-conditioning system for an industrial process
EI
Required air inlet conditions ...... 25ºC DBT and 60% R.H.
M
Amount of free air circulated ...... 250 m3/min.
-N
Coil dew temperature ...... 13ºC.
EL
The required condition is achieved by first cooling and dehumidifying and then by heating.
PT
Calculate the following :
(i) The cooling capacity of the cooling coil and its by-pass factor.
-N
(ii) Heating capacity of the heating coil in kW and surface temperature of the heating coil if
U
the by-pass factor is 0.3.
VT
(iii) The mass of water vapour removed per hour.
Solve this problem with the use of psychrometric chart.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 22 of 32
2014
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
Solution. Refer Fig.

Pr
oj
ec
t
Locate the
points ‘1’, ‘5’
and ‘3’ as
shown on
psychrometric
chart.
 Join the line 15.
 Draw constant specific humidity line through ‘3’ which cuts the line 1-5 at point ‘2’.
The point ‘2’ is located in this way.
From psychrometric chart :
C
T
h1 = 82.5 kJ/kg, h2 = 47.5 kJ/kg
h3 = 55.7 kJ/kg, h5 = 36.6 kJ/kg
EI
W1 = 19.6 gm/kg, W3 = 11.8 gm/kg
EL
The mass of air supplied per minute,
-N
M
tdb2 = 17.6ºC, vs1 = 0.892 m3/kg.
PT
ma =250/ 0.892 = 280.26 kg/min.
-N
(i) The capacity of the cooling coil
VT
= 42.04 TR. (Ans)
U
=( ma (h1 − h2) × 60 )/14000 =(280.26(82.5-47.5) × 60)/14000
The by-pass factor of the cooling coil is given by :
BF =
= 0.237 (Ans)
(ii) The heating capacity of the heating coil
= ma (h3 – h2) = 280.26 (55.7 – 47.5) = 2298.13 kJ/min = 2298.13/60 kJ/s = 38.3 kW (Ans)
The by-pass factor of the heating coil is given by
BF =
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 23 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
∴ tdb6 = 28.2ºC.
Hence surface temperature of heating coil = 28.2ºC. (Ans)
(iii) The mass of water vapour removed per hour
=
=
= 131.16 kg/h.
oj
ec
t
12) It is required to design an air-conditioning plant for a small office room
for following winter conditions :
Required conditions ...... 20ºC DBT and 60% R.H.
Pr
Outdoor conditions ...... 14ºC DBT and 10ºC WBT
C
T
Amount of air circulation ...... 0.30 m3/min./person.
EI
Seating capacity of office ...... 60.
M
The required condition is achieved first by heating and then by adiabatic humidifying.
-N
Determine the following :
EL
(i) Heating capacity of the coil in kW and the surface temperature required if the by pass
PT
factor of coil is 0.4.
-N
(ii) The capacity of the humidifier.
VT
U
Solve the problem by using psychrometric chart.
Solution. Refer Fig.
_ Locate the points ‘1’ and ‘3’ on the psychrometric chart.
_ Draw a constant enthalpy line through ‘3’ and constant specific humidity line through‘1’.
_
Locate the point ‘2’ where the above two lines intersect
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 24 of 32
2014
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
oj
ec
t
From the psychrometric chart :
h1 = 29.3 kJ/kg, h2 = h3 = 42.3
Pr
kJ/kg
tdb2 = 24.5ºC, vs1 = 0.817 m3/kg
EI
C
T
The mass of air circulated per minute,
ma =(0.30*60)/0.817= 22.03 kg/min.
M
(i) Heating capacity of the heating coil
-N
= ma(h2 – h1) = 22.03 (42.3 – 29.3) = 286.4 kJ/min.
EL
= 4.77 kJ/s or 4.77 kW. (Ans)
PT
The by-pass factor (BF) of heating coil is given by:
-N
BF=
U
∴ 0.4 = ( tdb4 – 12) = tdb4 – 24.5
VT
i.e., tdb4 (coil surface temperature) = 32.8ºC (Ans)
(ii) The capacity of the humidifier
= ma (W3 −W1)/1000 × 60 kg/h = 22.03(8.6-6.8)/1000 *60 = 2.379 kg/h.
12) 200 m 3 of air per minute at 15ºC DBT and 75% R.H. is heated until its temperature is
25ºC.
Find : (i) R.H. of heated air.
(ii) Wet bulb temperature of heated air.
Solution:
(iii) Heat added to air per minute.
Locate point 1 on the psychrometric chart on intersection of 15ºC DBT and 75% R.H.lines.
Through point 1 draw a horizontal line to cut 25ºC DBT line and get point 2.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 25 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
Read the following values from the psychrometric chart :
h 1 = 35.4 kJ/kg
h 2 = 45.2 kJ/kg
vs1 = 0.8267 m 3 /kg.
(i) R.H. of heated air (read from chart) = 41%. (Ans)
(ii) WBT of heated air (read from chart) = 16.1ºC. (Ans)
(iii) Mass of air circulated per min., m a = 200/0.8267 = 241.9 kg.
PT
EL
-N
M
EI
C
T
Pr
oj
ec
t
∴ Heat added to air/min. = m a (h 2 – h 1) = 241.9 (45.2 – 35.4) = 2370.6 kJ.
-N
Frequently asked Questions.
VT
U
1. What is an ideal gas ?
2. What is the difference between an ideal and a perfect gas ?
3. What are semi-perfect or permanent gases ?
4. Define ‘Equation of state’.
5. Write a short note on Van der Waals’ equation.
6. Define the following terms :
(i) Saturated air (ii) Dry bulb temperature
(iii) Dew point temperature (iv) Relative humidity (v) Specific humidity.
7. State ‘Dalton’s law of partial pressure’.
8. Describe briefly any two of the following processes :
(i) Sensible heating (ii) Cooling and dehumidification
(iii) Heating and humidification (iv) Heating and dehumidification.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 26 of 32
2014
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
9. A vessel of 0.03 m3 capacity contains gas at 3.5 bar pressure and 35°C temperature.
Determine the mass of the gas in the vessel. If the pressure of this gas is increased to 10.5 bar
while the volume remains constant, what will be the temperature of the gas ?
For the gas take R = 290 J/kg K.
[Ans. 0.118 kg, 650°C]
10. A vessel of spherical shape is 1.5 m in diameter and contains air at 40°C. It is evacuated
till the vacuum inside the vessel is 735 mm of mercury. Determine : (i) The mass of air
pumped out ;
(ii) If the tank is then cooled to 10°C what is the pressure in the tank ? The barometer reads
760 mm of mercury. Assume that during evacuation, there is no change in temperature of air.
oj
ec
t
[Ans. (i) 1.91 kg, (ii) 3 kPa]
11. 100 m3 of air per minute at 35ºC DBT and 60% relative humidity is cooled to 20ºC DBT
C
T
(ii) Amount of water vapour removed per hour, and
Pr
by passing through a cooling coil. Find the following : (i) Capacity of cooling coil in kJ/h
(iii) Relative humidity of air coming out and its wet-bulb temperature.
EI
[Ans. (i) 1037088 kJ/h, (ii) 465.36 kg/h, (iii) 100%, 20ºC]
M
12. Atmospheric air at 38ºC and 40 per cent relative humidity is to be cooled and
-N
dehumidified to a state of saturated air at 10ºC. The mass rate of flow of atmospheric air
EL
entering the dehumidifier is 45.4 kg/h. Neglecting any pressure drop, determine : (i) The
PT
mass of water removed ; (ii) The quantity of heat removed.
-N
[Ans. (i) 0.397 kg/h, (ii) 2287 kJ/h
U
13. The atmospheric conditions are 30ºC and specific humidity of 0.0215 kg/kg of air.
VT
Determine : (i) Partial pressure of air (ii) Relative humidity (iii) Dew point temperature.
Atmospheric pressure = 756 mm Hg.
[Ans. (i) 14.89 mm of Hg, (ii) 46.8%, (iii) 17ºC]
14. A mixture of air and water vapour at 1 bar and 25ºC has a dew point temperature of 15ºC.
Determine the relative humidity and specific humidity.
[Ans. 53.8%, 0.01078 kg/kg of dry air]
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 27 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
Assignment:
1) The air supplied to a room of building in winter is to be at 17ºC and have a relative
humidity of 60%. If the barometric pressure is 1.01325 bar, calculate the specific humidity.
What would be the dew point under these conditions ?
[Ans. 0.00723 kg/kg of dry air, 9.18ºC]
2) A mixture of air and water vapour at 1.013 bar and 16ºC has a dew point of 5ºC.
Determine the relative and specific humidities.
[Ans. 48%, 0.0054 kg/kg of dry air]
3) 100 m3 of air per minute at 35ºC DBT and 60% relative humidity is cooled to 20ºC DBT
oj
ec
t
by passing through a cooling coil. Find the following : (i) Capacity of cooling coil in kJ/h
(ii) Amount of water vapour removed per hour, and
Pr
(iii) Relative humidity of air coming out and its wet-bulb temperature.
C
T
[Ans. (i) 1037088 kJ/h, (ii) 465.36 kg/h, (iii) 100%, 20ºC]
4) Atmospheric air at 38ºC and 40 per cent relative humidity is to be cooled and dehumidified
EI
to a state of saturated air at 10ºC. The mass rate of flow of atmospheric air entering the
-N
M
dehumidifier is 45.4 kg/h. Neglecting any pressure drop, determine :
EL
(i) The mass of water removed ; (ii) The quantity of heat removed.
[Ans. (i) 0.397 kg/h, (ii) 2287 kJ/h]
PT
5. 1 kg of air at 24ºC and a relative humidity of 70% is to be mixed adiabatically in a steady
-N
state, steady flow device with 1 kg of air at 16ºC and a relative humidity of 10%. Assuming
U
that the mixing is to be carried out at a constant pressure of 1.0 atm, determine the
VT
temperature and relative humidity of the stream leaving the device.
[Ans. 19.5ºC, 50%]
6) A balloon of spherical shape is 8 m in diameter and is filled with hydrogen at a pressure of
1 bar abs. and 15°C. At a later time, the pressure of gas is 95 per cent of its original pressure
at the same temperature. (i) What mass of original gas must have escaped if the dimensions
of the balloon are not changed ? (ii) Find the amount of heat to be removed to cause the same
drop in pressure at constant volume.
[Ans. (i) 5 per cent, (ii) 3.26 MJ]
7) A constant volume chamber of 0.3 m33 capacity contains 1 kg of air at 20°C. Heat is
transferred to the air until its temperature is 200°C. Find : (i) Heat transferred ;
(ii) Change in entropy and enthalpy.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 28 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
[Ans. (i) 128.09 kJ, (ii) 0.339 kJ/kg K, 180.8 kJ]
8) 1 kg of air at 20°C occupying a volume of 0.3 m333 undergoes a reversible constant
pressure process. Heat is transferred to the air until its temperature is 200°C. Determine :
(i) The work and heat transferred.
(ii) The change in internal energy, enthalpy and entropy.
[Ans. (i) 51.5 kJ, 180.8 kJ ; (ii) 128.09 kJ, 180.8 kJ, 0.479 kJ/kg K]
9) Air expands in a cylinder in a reversible adiabatic process from 13.73 bar to 1.96 bar. If
the final temperature is to be 27°C, what would be the initial temperature ?
Also calculate the change in specific enthalpy, heat and work transfers per kg of air.
Pr
oj
ec
t
[Ans. 524 K, 224.79 kJ/kg, zero, 160.88 kJ/kg]
Self Answered Question & Answer
C
T
1) 100 m3 of air per minute at 35ºC DBT and 60% relative humidity is cooled to 20ºC DBT
EI
by passing through a cooling coil. Find the following : (i) Capacity of cooling coil in kJ/h
M
(ii) Amount of water vapour removed per hour, and
-N
(iii) Relative humidity of air coming out and its wet-bulb temperature.
EL
[Ans. (i) 1037088 kJ/h, (ii) 465.36 kg/h, (iii) 100%, 20ºC]
PT
2) 1 kg of air at 24ºC and a relative humidity of 70% is to be mixed adiabatically in a steady
-N
state, steady flow device with 1 kg of air at 16ºC and a relative humidity of 10%. Assuming
U
that the mixing is to be carried out at a constant pressure of 1.0 atm, determine the
VT
temperature and relative humidity of the stream leaving the device.
[Ans. 19.5ºC, 50%]
3) The atmospheric conditions are 30ºC and specific humidity of 0.0215 kg/kg of air.
Determine : (i) Partial pressure of air (ii) Relative humidity (iii) Dew point temperature.
Atmospheric pressure = 756 mm Hg.
[Ans. (i) 14.89 mm of Hg, (ii) 46.8%, (iii) 17ºC]
4) Air expands in a cylinder in a reversible adiabatic process from 13.73 bar to 1.96 bar. If
the final temperature is to be 27°C, what would be the initial temperature ?
Also calculate the change in specific enthalpy, heat and work transfers per kg of air.
[Ans. 524 K, 224.79 kJ/kg, zero, 160.88 kJ/kg]
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 29 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
2014
5) A vessel of spherical shape is 1.5 m in diameter and contains air at 40°C. It is evacuated
till the vacuum inside the vessel is 735 mm of mercury. Determine : (i) The mass of air
pumped out ;
(ii) If the tank is then cooled to 10°C what is the pressure in the tank ? The barometer reads
760 mm of mercury. Assume that during evacuation, there is no change in temperature of air.
[Ans. (i) 1.91 kg, (ii) 3 kPa]
6) 1 kg of air at 27°C is heated reversibly at constant pressure until the volume is doubled and
then heated reversibly at constant volume until the pressure is doubled. For the total path find
:(i) The work ; (ii) Heat transfer ;(iii) Change of entropy.
[Ans. (i) 86.14 kJ, (ii) 728.36 kJ, (iii) 1.186 kJ/kg K]
oj
ec
t
7) A mass of air initially at 260°C and a pressure of 6.86 bar has a volume of 0.03 m3. The
air is expanded at constant pressure to 0.09 m3, a polytropic process with n = 1.5 is then
Pr
carried out, followed by a constant temperature process which completes the cycle. All
C
T
processes are reversible. Find (i) The heat received and rejected in the cycle, (ii) The
efficiency of the cycle.Show the cycle on p-v and T-s planes.
-N
M
EI
[Ans. (i) 143.58 kJ, – 20.3 kJ ; (ii) 38.4%]
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Test Your Skills
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Choose the Correct Answer :
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1. In an unsaturated air the state of a vapour is
(a) wet (b) superheated (c) saturated (d) unsaturated.
2. For saturated air
(a) Wet bulb depression is zero (b) Wet bulb depression is positive
(c) Wet bulb depression is negative (d) Wet bulb depression can be either positive or
negative.
3. Which one of the following statements is correct ?
(a) Dew point temperature can be measured with the help of thermometer
(b) Dew point temperature is the saturation temperature corresponding to the partial pressure
of the water vapour in moist air.
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 30 of 32
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
(c) Dew point temperature is the same as the thermodynamic wet bulb temperature.
(d) For saturated air, dew point temperature is less than the wet bulb temperature.
4. During sensible heating of moist air, enthalpy
(a) increases (b) decreases (c) remains constant (d) none of the above.
5. During sensible cooling, wet bulb temperature
(a) decreases (b) increases (c) remains constant (d) can decrease or increase.
6. Which one of the following statements is correct ?
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(a) Evaporative cooling and sensible cooling is the same
(b) Evaporative cooling is a cooling and humidification process
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(c) Evaporative cooling is a cooling and dehumidification process
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7. An air washer can work as a
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(d) Evaporative cooling is not effective for hot and dry climates.
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(a) filter only (b) humidifier only (c) dehumidifier only (d) all of the above.
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8. The relative humidity, during sensible heating,
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(a) can increase or decrease (b) increases (c) decreases (d) remains constant.
9. The vapour pressure, during sensible heating of moist air,
(a) increases (b) decreases (c) can increase or decrease (d) remains constant.
10. The relative humidity, during heating and humidification,
(a) increases (b) decreases (c) may increase or decrease (d) remains constant.
11. The relative humidity, during cooling and dehumidification of moist air
(a) increases (b) decreases (c) can increase or decrease (d) remains constant.
12. (a) A perfect gas does not obey the law pv = RT
(b) A perfect gas obeys the law pv = RT and has constant specific heat
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 31 of 32
2014
MODULE-I --- Property Relationships for Pure Substances and Mixtures
APPLIED THRMODYNAMICS
(c) A perfect gas obeys the law pv = RT but have variable specific heat capacities.
13. Boyle’s law states that, when temperature is constant, the volume of a given mass of a
perfect gas
(a) varies directly as the absolute pressure (b) varies inversely as the absolute pressure
(c) varies as square of the absolute pressure (d) does not vary with the absolute pressure.
14. Charle’s law states that if any gas is heated at constant pressure, its volume
(a) changes directly as it absolute temperature (b) changes inversely as its absolute
temperature
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(c) changes as square of the absolute temperature
(d) does not change with absolute temperature.
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(a) pressure and volume (b) pressure and temperature
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15 The equation of state of an ideal gas is a relationship between the variables :
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(c) pressure, volume and temperature (d) none of the above.
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16. Joule’s law states that the specific internal energy of a gas depends only on
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(a) the pressure of the gas (b) the volume of the gas
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(c) the temperature of the gas (d) none of the above.
ANSWERS
1. (b) 2. (a) 3. (b) 4. (a) 5. (a) 6. (b) 7. (d) 8. (b) 9. (d) 10. (a) 11. (c). 12. (b) 13. (b) 14. (a)
15. (c) 16. (c)
Dr.A.R.ANWAR KHAN,Prof & HOD ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
Page 32 of 32
2014