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Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com
ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120
www.ijera.com 116 | P a g e
CFD Analysis of Manipulator Cabin by Selecting Proper Air
Conditioning System
Umesh S. Ghorpade1*
, Manik A. Patil2**
,Amit S. Shelake3**
,
Himanshu S. Ghodake4**
1
Assistant Professor, Sanjeevan Engineering & Technology Institute, Panhala, India
2
Assistant Professor, Sanjeevan Engineering & Technology Institute, Panhala, India
3
Assistant Professor, Sanjeevan Engineering & Technology Institute, Panhala, India
4
Student, Sanjeevan Engineering & Technology Institute, Panhala, India
ABSTRACT
Manipulator is a machine which is used to transfer heavy objects, to reduce human efforts in many industrial
applications. Some of them are fully automated while some are manually operated. It is difficult to work in
cabin of manipulator for the operator under hot conditions. So it is essential to provide comfort to operator using
air-conditioning system. For such applications standard air conditioners are not compatible, so we have to
develop assembled system. In this paper we will be dealing with design, selection and fabrication of components
like compressor, condenser, expansion device and evaporator. In order to do so, we have calculated heat load.
Keywords: Heat load, Compressor, heat transfer coefficient, CFD.
I. INTRODUCTION
Air conditioning is concerned with the
absorption of heat from where it is objectionable
plus its transfer to and rejection at a place where it
is unobjectionable and keeps the place comfortable
for occupants. Air conditioning comprises
maintaining temperature, humidity level, noise
control, cleanliness. There are so many methods of
air conditioning; according to situation a particular
system is adopted. Usually after estimating the load
standard manufactured unit is installed. Regardless
of availability of various systems we need to
develop a system fitting to the specific application
due to co-occurrence of multiple factor affecting
the system. As the environment of casting industry
is hot, dusty, noisy, affecting the efficiency of
workers. Now days casting industries are
automated industries. They have fully mechanized
conveyors for transportation of hot casting to the
fettling section. To place casting from vibrator to
conveyor manually operated manipulators are used.
The working condition for the operator is largely
influenced by the temperature and dust of castings.
The outer temperature is around 500
C during
working hours. The manipulator is not equipped
with the air conditioning facility; so it’s not
comfortable to the operator. The cabin of
manipulator is so compact that only a person can fit
in it and hence can’t be equipped with standard AC
system. Even we apply standard system; cabin
vibrations with high amplitude causes system to
fail due to unsuitable mounting. So we have
developed an air conditioning system which
maintains temperature, humidity level, noise
control and cleanliness.
II. CALCULATION
Calculations are required to determine the total
load inside the manipulator cabin. The dimensions
of the manipulator cabin are shown in the Fig. 1.
Figure 1.Manipulator Cabin
Heat Load calculation.
 Indoor condition: 25°C, with RH 25% - 70%
and selected 60%,
Therefore 25° DBT and 60% RH selected.
 Outdoor condition: 50°C, with RH 30% - 40%.
We will consider 42°C as the maximum limit
on psychrometric chart.
 Volume of cabin:
Area 1= A1 = 1.7×1.3 = 2.21m2
; Area 2= A2 =
0.22×1.7×0.8 = 2.27m2
Therefore, Total Area: A = A1+A2 = 2.27+0.187
=2.457m2
(1)
Volume, V =2.457×0.950 = 2.3341 m3
(2)
 Thermal conductivities:
RESEARCH ARTICLE OPEN ACCESS
Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com
ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120
www.ijera.com 117 | P a g e
For fiber glass, k = 0.04 W/m-K; for steel, k =
43 W/m-K
 Heat transfer coefficient: Between air and
glass, h = 25 W/m2
-K
 Sensible heat and Latent heat; At 25° C i.e.
78° F seated very light work,
Sensible heat = 215 BTU/hr = 62.995 = 63 W
and
Latent heat = 185 BTU/hr = 54.25 = 54 W
 Overall heat transfer coefficient for glass U:
Hence, U =4.88 W/m2
-K
(3)
 Heat estimation:
Q = UADT = 4.88*(50-25)* A
Through front glass window, Qf = 4.88×25×
(1.615×1/sin83) = 198.51 W (4)
Through side doors, Qs = 2*Qs1
And Qs1 = Qg + Qsteel
Qg = UADT = 4.88× [(0.65×0.850) +
(0.5×150×0.001×0.850)] ×25 = 75.152 W (5)
 Overall heat transfer coefficient for steel and
glass interface U; For glass,
( 6)
For steel: From MS steel Engineering Tool box
we get, Ust =7.9 i.e. 8 Appr. (7)
 Heat estimation
Area As = (2.397-0.616) = 1.781 m2
(8)
Q for steel, Qst = 8×1.781×25 = 356.2 W. (9)
Therefore, Qs1 = 75.152+356.2 = 431.352 W
And Qs = 2*Qs1 = 2×431.352 = 862.74 W. (10)
 Through the rear side there is negligible heat
gain due to various outer equipments mounting
and noise reducing coating layer.
 Occupancy Load:
The heat emitted from the bodies of the people
also constitutes a major portion of a summer
cooling load. The heat quantities given up by the
occupants are dependent on activity of the persons,
sex, age and indoor dry bulb temperature. From
Standard table,
Latent heat = 54 W and Sensible heat = 63 W
Therefore, Total occupancy load Qo = 117 W.
(11)
 Other loads;
For other heat source 10 % to 15 % of total heat
load is taken.
 Fresh air load estimation:
Air change per hour = 2, considering office private
and no smoking.
Taking outside air 0.75 m3/min-person
Therefore, Fresh air load = 3×2 = 6 m3
/hr.
 Infiltration load :
For infiltration load considering 10 % of the total
load
 Total heat :
Q total = Qf + Qs +Qo = 198.51+ 862.70+ 117
=1178.214 W (12)
 Load due to light and equipment load :
Considering light load and equipment load as 20 %
of total heat load i.e.,
Light load and equipment load = 20 % Qtotal =
0.2×1178.214 = 235.64 W. (13)
 On Psychrometric Chart :
Following points were plotted,
A] 42° C - DBT, 40 % RH, B] 25° C - DBT, 60
% RH,
Specific volume of air at 1,
Vs1 = 0.92+8×10-4×3 = 0.9224 m3/kg (14)
Infiltrated air at point 1 = 20% (Vol)/hr = 0.2×3/60
= 0.02 m3
/min.
Mass, ma = 0.01= 0.0108 Kg/min (15)
Fresh air =2×V =2×3 =6 m3
/hr
i.e. equal to 0.1 m3
/min
Enthalpy at 1 i.e. h1 = 97.5 KJ/Kg of dry air,
Enthalpy at 2 i.e. h2 = 5 6 KJ/Kg of dry air,
Enthalpy at A i.e. hA = 74 KJ/Kg of dry air.
 Heat due to infiltration:
i. Sensible heat gain due to infiltration air = ma
(ha-h2) = 0.0108 (74-56)
= 0.1944 KJ/min i.e. equal to 3.24 W. (16)
ii. Latent heat due to infiltration air = ma (h1-hA)
= 0.0108 (97.5-74)
= 0.2538 KJ/min i.e. equal to 4.23 W. (17)
 Total latent heat gain in room, RLH = 4.23+54
= 58.23 W. (18)
 Total sensible heat gain in room, RSH =
3.24+63+235.64+117.214 = 1480.004. (19)
 Room sensible heat factor,
RSHF = (RSH/RSH) + RLH =
(1480.094/1480.094) +58.23 = 0.9621. (20)
 Alignment Circle: Alignment circle points are,
(20% DBT, 50% RH)
Suppose for 70% return air and 30% fresh air then
by trial and error method drawing line from 30 to
15 on saturation curve we get,
a) Bypass factor =0.157, b) td3 =28.2° C c) td4
=17.5° C d) td6 =15.5° C
Enthalpy at point 4 =47.3 KJ/Kg of air; Enthalpy at
point 3 =63 KJ/Kg of air
 Mass entering room: Mass entering room, ma
= RSH+RLH = 1480.09 +58.23
Therefore, ma = 0.1768 Kg/s = 10.61 kg/min
(21)
Capacity of plant = ma (h3-h4)
Therefore, Capacity of plant = 10.61 (63-47.3) =
166.577 KJ/min = 0.793 TR (22)
 Total load; Assuming factor of safety i.e. F.S
=1.25
Hence, Total Load = Capacity of plant × F.S =
0.793×1.25 = 0.9875≈ 1TR. (23)
Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com
ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120
www.ijera.com 118 | P a g e
III. CFD analysis
The Analysis of Manipulator cabin is carried out
using Ansys14.0 software. This process consists of
Three Primary Steps:
3.1 Pre-Processing:- This is the first step of CFD
simulation process which helps in describing the
geometry in the best possible manner. One needs to
identify the fluid domain of interest.
Figure 2. Geometry of Cabin
Figure 3. Meshing of Cabin
3.2 Solver: - Once the problem physics has been
identified, fluid material properties, flow physics
model, and boundary conditions are set to solve
using a computer.
 Indoor condition: 25°C.
 Outdoor condition: 50°C.
 Capacity of air conditioner:1 TR
3.3 Post-Processing:- The next step after getting
the results is to analyze the results with different
methods like contour plots, vector plot, streamlines,
data curve etc. for appropriate graphical
representations and report.
Figure 4. Velocity Streamlines
Figure 5. Temperature Velocity Rendering
Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com
ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120
www.ijera.com 119 | P a g e
Figure 6. Temperature Volume Rendering
Figure 7. Pressure Volume Rendering
IV. CONCLUSION
By the study of heat load estimation
formulae’s, some considerations and from the basic
laws of thermodynamics the total heat load is
calculated as 1 TR.
From the temperature volume rendering diagram
and temperature velocity rendering diagram we
conclude that the temperature distribution inside
the cabin is uniform and is under comfort zone.
REFERENCES
[1] Liu Sheng jun et.al, “Thermodynamic
analysis of actual air cycle refrigeration
system”, System Engineering, Procedia1
(2011) 112-116.
[2] K. Nagalakshmi et.al, “The design and
performance analysis of refrigeration system
using R-12 and R134a refrigerants”, ISSN:
2248-9622. VOL. Issue 2(Version 1),
February 2014, pp.638-643.
[3] Xing Xue et.al, “Modeling and simulation of
an air cooling condenser under transient
conditions”, Procedia Engineering 31
(2012) 817-822.
[4] T. S. Jadhav, “Theoretical energy saving
analysis of air conditioning system using
heat pipe for Indian climatic zones”,
Engineering Science and Technology,
Journal (15)669-673.
[5] Xu Song et.al, “Cooling and
dehumidification capacity chart of surface
air cooler in air conditioning”, Procedia
Engineering 121 (2015) 2014-2020.
Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com
ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120
www.ijera.com 120 | P a g e
[6] A. V. Domkundwar, “Refrigeration and Air
Conditioning”, DhanpatRai Publishing
House Pvt. Ltd. Reprint 2011 ISBN 978-81-
7700-000-9.
[7] Abdullah A. A. a. Al-Rashid “Effect of
evaporator temperature on vapour
compression refrigeration system”,
Alexandria Engineering Journal (2011) 50,
283-290
[8] Isacco Stiaccini et.al “A hybrid time-
frequency approach for numerical modeling
of reciprocating compressors” Procedia
Engineering 31 (2012) 817-822
[9] Liu Jing1, PEI Qing-quing2 “Numerical
Simulation And Experimental Study of
Indoors Thermal Environment in Summer
Air-Conditioned Room” Procedia
Engineering 52 (2013) 230-235
[10] Qi Cheng et.al. “Robust optimal design of
chiller plants based on cooling load
distribution” Energy Procedia 75 (2015)
1354-1359
[11] R.K.Rajput “Air conditioning and
refrigeration” Eurasia Publishing House
(P).Ltd Reprint, 2015, ISBN 978-81-21-
2781-9.

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CFD Analysis of Manipulator Cabin by Selecting Proper Air Conditioning System

  • 1. Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120 www.ijera.com 116 | P a g e CFD Analysis of Manipulator Cabin by Selecting Proper Air Conditioning System Umesh S. Ghorpade1* , Manik A. Patil2** ,Amit S. Shelake3** , Himanshu S. Ghodake4** 1 Assistant Professor, Sanjeevan Engineering & Technology Institute, Panhala, India 2 Assistant Professor, Sanjeevan Engineering & Technology Institute, Panhala, India 3 Assistant Professor, Sanjeevan Engineering & Technology Institute, Panhala, India 4 Student, Sanjeevan Engineering & Technology Institute, Panhala, India ABSTRACT Manipulator is a machine which is used to transfer heavy objects, to reduce human efforts in many industrial applications. Some of them are fully automated while some are manually operated. It is difficult to work in cabin of manipulator for the operator under hot conditions. So it is essential to provide comfort to operator using air-conditioning system. For such applications standard air conditioners are not compatible, so we have to develop assembled system. In this paper we will be dealing with design, selection and fabrication of components like compressor, condenser, expansion device and evaporator. In order to do so, we have calculated heat load. Keywords: Heat load, Compressor, heat transfer coefficient, CFD. I. INTRODUCTION Air conditioning is concerned with the absorption of heat from where it is objectionable plus its transfer to and rejection at a place where it is unobjectionable and keeps the place comfortable for occupants. Air conditioning comprises maintaining temperature, humidity level, noise control, cleanliness. There are so many methods of air conditioning; according to situation a particular system is adopted. Usually after estimating the load standard manufactured unit is installed. Regardless of availability of various systems we need to develop a system fitting to the specific application due to co-occurrence of multiple factor affecting the system. As the environment of casting industry is hot, dusty, noisy, affecting the efficiency of workers. Now days casting industries are automated industries. They have fully mechanized conveyors for transportation of hot casting to the fettling section. To place casting from vibrator to conveyor manually operated manipulators are used. The working condition for the operator is largely influenced by the temperature and dust of castings. The outer temperature is around 500 C during working hours. The manipulator is not equipped with the air conditioning facility; so it’s not comfortable to the operator. The cabin of manipulator is so compact that only a person can fit in it and hence can’t be equipped with standard AC system. Even we apply standard system; cabin vibrations with high amplitude causes system to fail due to unsuitable mounting. So we have developed an air conditioning system which maintains temperature, humidity level, noise control and cleanliness. II. CALCULATION Calculations are required to determine the total load inside the manipulator cabin. The dimensions of the manipulator cabin are shown in the Fig. 1. Figure 1.Manipulator Cabin Heat Load calculation.  Indoor condition: 25°C, with RH 25% - 70% and selected 60%, Therefore 25° DBT and 60% RH selected.  Outdoor condition: 50°C, with RH 30% - 40%. We will consider 42°C as the maximum limit on psychrometric chart.  Volume of cabin: Area 1= A1 = 1.7×1.3 = 2.21m2 ; Area 2= A2 = 0.22×1.7×0.8 = 2.27m2 Therefore, Total Area: A = A1+A2 = 2.27+0.187 =2.457m2 (1) Volume, V =2.457×0.950 = 2.3341 m3 (2)  Thermal conductivities: RESEARCH ARTICLE OPEN ACCESS
  • 2. Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120 www.ijera.com 117 | P a g e For fiber glass, k = 0.04 W/m-K; for steel, k = 43 W/m-K  Heat transfer coefficient: Between air and glass, h = 25 W/m2 -K  Sensible heat and Latent heat; At 25° C i.e. 78° F seated very light work, Sensible heat = 215 BTU/hr = 62.995 = 63 W and Latent heat = 185 BTU/hr = 54.25 = 54 W  Overall heat transfer coefficient for glass U: Hence, U =4.88 W/m2 -K (3)  Heat estimation: Q = UADT = 4.88*(50-25)* A Through front glass window, Qf = 4.88×25× (1.615×1/sin83) = 198.51 W (4) Through side doors, Qs = 2*Qs1 And Qs1 = Qg + Qsteel Qg = UADT = 4.88× [(0.65×0.850) + (0.5×150×0.001×0.850)] ×25 = 75.152 W (5)  Overall heat transfer coefficient for steel and glass interface U; For glass, ( 6) For steel: From MS steel Engineering Tool box we get, Ust =7.9 i.e. 8 Appr. (7)  Heat estimation Area As = (2.397-0.616) = 1.781 m2 (8) Q for steel, Qst = 8×1.781×25 = 356.2 W. (9) Therefore, Qs1 = 75.152+356.2 = 431.352 W And Qs = 2*Qs1 = 2×431.352 = 862.74 W. (10)  Through the rear side there is negligible heat gain due to various outer equipments mounting and noise reducing coating layer.  Occupancy Load: The heat emitted from the bodies of the people also constitutes a major portion of a summer cooling load. The heat quantities given up by the occupants are dependent on activity of the persons, sex, age and indoor dry bulb temperature. From Standard table, Latent heat = 54 W and Sensible heat = 63 W Therefore, Total occupancy load Qo = 117 W. (11)  Other loads; For other heat source 10 % to 15 % of total heat load is taken.  Fresh air load estimation: Air change per hour = 2, considering office private and no smoking. Taking outside air 0.75 m3/min-person Therefore, Fresh air load = 3×2 = 6 m3 /hr.  Infiltration load : For infiltration load considering 10 % of the total load  Total heat : Q total = Qf + Qs +Qo = 198.51+ 862.70+ 117 =1178.214 W (12)  Load due to light and equipment load : Considering light load and equipment load as 20 % of total heat load i.e., Light load and equipment load = 20 % Qtotal = 0.2×1178.214 = 235.64 W. (13)  On Psychrometric Chart : Following points were plotted, A] 42° C - DBT, 40 % RH, B] 25° C - DBT, 60 % RH, Specific volume of air at 1, Vs1 = 0.92+8×10-4×3 = 0.9224 m3/kg (14) Infiltrated air at point 1 = 20% (Vol)/hr = 0.2×3/60 = 0.02 m3 /min. Mass, ma = 0.01= 0.0108 Kg/min (15) Fresh air =2×V =2×3 =6 m3 /hr i.e. equal to 0.1 m3 /min Enthalpy at 1 i.e. h1 = 97.5 KJ/Kg of dry air, Enthalpy at 2 i.e. h2 = 5 6 KJ/Kg of dry air, Enthalpy at A i.e. hA = 74 KJ/Kg of dry air.  Heat due to infiltration: i. Sensible heat gain due to infiltration air = ma (ha-h2) = 0.0108 (74-56) = 0.1944 KJ/min i.e. equal to 3.24 W. (16) ii. Latent heat due to infiltration air = ma (h1-hA) = 0.0108 (97.5-74) = 0.2538 KJ/min i.e. equal to 4.23 W. (17)  Total latent heat gain in room, RLH = 4.23+54 = 58.23 W. (18)  Total sensible heat gain in room, RSH = 3.24+63+235.64+117.214 = 1480.004. (19)  Room sensible heat factor, RSHF = (RSH/RSH) + RLH = (1480.094/1480.094) +58.23 = 0.9621. (20)  Alignment Circle: Alignment circle points are, (20% DBT, 50% RH) Suppose for 70% return air and 30% fresh air then by trial and error method drawing line from 30 to 15 on saturation curve we get, a) Bypass factor =0.157, b) td3 =28.2° C c) td4 =17.5° C d) td6 =15.5° C Enthalpy at point 4 =47.3 KJ/Kg of air; Enthalpy at point 3 =63 KJ/Kg of air  Mass entering room: Mass entering room, ma = RSH+RLH = 1480.09 +58.23 Therefore, ma = 0.1768 Kg/s = 10.61 kg/min (21) Capacity of plant = ma (h3-h4) Therefore, Capacity of plant = 10.61 (63-47.3) = 166.577 KJ/min = 0.793 TR (22)  Total load; Assuming factor of safety i.e. F.S =1.25 Hence, Total Load = Capacity of plant × F.S = 0.793×1.25 = 0.9875≈ 1TR. (23)
  • 3. Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120 www.ijera.com 118 | P a g e III. CFD analysis The Analysis of Manipulator cabin is carried out using Ansys14.0 software. This process consists of Three Primary Steps: 3.1 Pre-Processing:- This is the first step of CFD simulation process which helps in describing the geometry in the best possible manner. One needs to identify the fluid domain of interest. Figure 2. Geometry of Cabin Figure 3. Meshing of Cabin 3.2 Solver: - Once the problem physics has been identified, fluid material properties, flow physics model, and boundary conditions are set to solve using a computer.  Indoor condition: 25°C.  Outdoor condition: 50°C.  Capacity of air conditioner:1 TR 3.3 Post-Processing:- The next step after getting the results is to analyze the results with different methods like contour plots, vector plot, streamlines, data curve etc. for appropriate graphical representations and report. Figure 4. Velocity Streamlines Figure 5. Temperature Velocity Rendering
  • 4. Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120 www.ijera.com 119 | P a g e Figure 6. Temperature Volume Rendering Figure 7. Pressure Volume Rendering IV. CONCLUSION By the study of heat load estimation formulae’s, some considerations and from the basic laws of thermodynamics the total heat load is calculated as 1 TR. From the temperature volume rendering diagram and temperature velocity rendering diagram we conclude that the temperature distribution inside the cabin is uniform and is under comfort zone. REFERENCES [1] Liu Sheng jun et.al, “Thermodynamic analysis of actual air cycle refrigeration system”, System Engineering, Procedia1 (2011) 112-116. [2] K. Nagalakshmi et.al, “The design and performance analysis of refrigeration system using R-12 and R134a refrigerants”, ISSN: 2248-9622. VOL. Issue 2(Version 1), February 2014, pp.638-643. [3] Xing Xue et.al, “Modeling and simulation of an air cooling condenser under transient conditions”, Procedia Engineering 31 (2012) 817-822. [4] T. S. Jadhav, “Theoretical energy saving analysis of air conditioning system using heat pipe for Indian climatic zones”, Engineering Science and Technology, Journal (15)669-673. [5] Xu Song et.al, “Cooling and dehumidification capacity chart of surface air cooler in air conditioning”, Procedia Engineering 121 (2015) 2014-2020.
  • 5. Umesh S. Ghorpade.et.al. Int. Journal of Engineering Research and Application www.ijera.com ISSN: 2248-9622, Vol. 7, Issue 1, (Part -1) January 2017, pp.116-120 www.ijera.com 120 | P a g e [6] A. V. Domkundwar, “Refrigeration and Air Conditioning”, DhanpatRai Publishing House Pvt. Ltd. Reprint 2011 ISBN 978-81- 7700-000-9. [7] Abdullah A. A. a. Al-Rashid “Effect of evaporator temperature on vapour compression refrigeration system”, Alexandria Engineering Journal (2011) 50, 283-290 [8] Isacco Stiaccini et.al “A hybrid time- frequency approach for numerical modeling of reciprocating compressors” Procedia Engineering 31 (2012) 817-822 [9] Liu Jing1, PEI Qing-quing2 “Numerical Simulation And Experimental Study of Indoors Thermal Environment in Summer Air-Conditioned Room” Procedia Engineering 52 (2013) 230-235 [10] Qi Cheng et.al. “Robust optimal design of chiller plants based on cooling load distribution” Energy Procedia 75 (2015) 1354-1359 [11] R.K.Rajput “Air conditioning and refrigeration” Eurasia Publishing House (P).Ltd Reprint, 2015, ISBN 978-81-21- 2781-9.