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TABLE OF CONTENTS
Page
Table of Contents
i
1. Problem Statement
2
2. Objective 23. Introduction
2-34. ir Conditioning Process
4-!". #efrigeration C$cle
%-12&. Combustion C'amber for Internal Combustion (ngine
13-1%!. Conclusion
1)%. #eferences
1)). **endi+
1)
1
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PROBLEM STATEMENT
A simple ideal vapor-compression heat-pump system is shown in Figure 1 to provide a
heating effect for a building. The heat from evaporator of the refrigeration system is
dissipated to the outdoor air flowing at 20,000 cfm from 0! to condition it to "0! and the
relative humidity is "0#, respectively. An unmi$ed cross flow heat e$changer is used to
increase the temperature to 1%!. The heat e$changer is made of tubes through which flows
hot fluid from the compressor of the refrigeration system. Then, the air is humidified by the
in&ection of hot steam in the humidifying section.
OBJECTIVE
The ob&ective of this assignment is for students to conduct preliminary design calculations of
an air conditioning system combining all the relevant topics in thermal engineering.
INTRODUCTION
Air conditioning (often referred to as A/C or AC) is the process of altering the
properties of air (primarily temperature and humidity) to more comfortable conditions,
typically with the aim of distributing the conditioned air to an occupied space such as
a building or a vehicle to improve thermal comfort and indoor air quality. In common
use, an air conditioner is a device that lowers the air temperature. he cooling is
typically achieved through a refrigeration cycle, but sometimes evaporation or free
cooling is used. Air conditioning systems can also be made based on desiccants. In
the most general sense, air conditioning can refer to any form of technology that
modifies the condition of air (heating, cooling, (de!)humidification, cleaning,
ventilation, or air movement). "owever, in construction, such a complete system of
heating, ventilation, and air conditioning is referred to as heating, ventilation, and air
conditioning ("#AC !as opposed to AC ).
Principles
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$or an air conditioning system to operate with economy, the refrigerant must
be used repeatedly. $or this reason, all air conditioners use the same cycle of
compression, condensation, e%pansion, and evaporation in a closed circuit. he
same refrigerant is used to move the heat from one area, to cool this area, and to
e%pel this heat in another area.
&. he refrigerant comes into the compressor as a low!pressure gas, it is
compressed and then moves out of the compressor as a high!pressure
gas.
'. he gas then flows to the condenser. "ere the gas condenses to a
liquid, and gives off its heat to the outside air.. he liquid then moves to the e%pansion valve under high pressure.
his valve restricts the flow of the fluid, and lowers its pressure as it
leaves the e%pansion valve.. he low!pressure liquid then moves to the evaporator, where heat from
the inside air is absorbed and changes it from a liquid to a gas.*. As a hot low!pressure gas, the refrigerant moves to the compressor
where the entire cycle is repeated.
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A. Air Conditioning Process
Question 2: 'sing (ry mass balance ) m
1 *m
2+, ater mass balance ) m
1 ω
1 *m
2 ω
2+, nergy balance , Q m ∆ h
State 1
Pv1 ∅ 1Pg1/ P,sat0 T ℃ ,refer table -4
,.)",.&11
."%112Pa
Pa1 P1-Pv1
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)).41)Pa
1 RaT 1
Pa1 (0.287 k )(273)
99.419 k .!%%m
3
kgdry air
ω1
0.662 Pv1
P1− Pv1 0.662(0.58112k )100k −0.58112k 0.00364
kg h20
kgdry air / ,
ω1=ω2
State 2
ω1
=ω2
.3&4
kJ h20
kgdry
'1 ,C*T1 , ω 1 'g1 ,'g refer table -4 for 'g sat
,1.", ,.3&4,2".) 9.0925 kJ
kg
'2 ,C*T2 , ω 2 'g2 , 'g refer table -4 for 'g sat
,1.",1" ,.3&4,2"2%.3 24.278kJ
kg
5etermine t'e ṁ 1 /
61 V 1
V 1 566.337
0.788 /, V 1 from converting 2cfm to m37min
!1%.) kg
min / convert tokg
s !1%.)×1min
60 s 11.978 kg
s
T'erefore/
Q 61 ∆ ' ,11.)!%,24.2!% 8 ).)2" 18.1892 9 ,:eat
transfer
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Question 3: Analye rate of water vapour at t * 20 ℃
until 2/ ℃
)with increament 1 ℃
+
elative humidity * "0#-/# )assume 0#+
Water vapour at t20 ℃
6v 6a , ω 3- ω 2 / 62 11.)!%kg
s / ω 2 .3&4kg h20
kgdry air
5etermine ω 3/
ω 3
0.662∅ Pg
P−∅ Pg
0.662(0.6)2.3392 k
100k −(0.6)(2.3392 k ) %.%"42 ×10
−3
kg h20
kgdry air
m v ,11.)!%,%.%"42 ×10−3
8 .3&4 0.06246 kg
s
Water vapour at t21 ℃
6v 6a , ω 3- ω 2 / 62 11.)!%kgs / ω 2 .3&4
kg h20kgdry air
5etermine ω 3/
ω 3 0.662∅ Pg
P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 2.""32Pa
0.662(0.6)(2.50532k )
100k −(0.6 )(2.50532k ) .)4)2"
kg h20
kgdry air
m v ,11.)!%,.)4)2" 8 .3&4 0.07010 kg
s
Water vapour at t 22 ℃
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6v 6a , ω 3- ω 2 / 62 11.)!%kg
s / ω 2 .3&4kg h20
kgdry air
5etermine ω 3/
ω 3 0.662∅ Pg
P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 2.&!144Pa
0.662(0.6)(2.67144 k )100k −(0.6 )(2.67144k ) .113
kg h20
kgdry air
m v ,11.)!%,.113 8 .3&4 0.07774 kg
s
Water vapour at t 23 ℃
6v 6a , ω 3- ω 2 / 62 11.)!%kg
s / ω 2 .3&4kg h20
kgdry air
5etermine ω 3/
ω 3 0.662∅ Pg
P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 2.%3!&Pa
0.662(0.6)(2.3876 k )
100k −(0.6 )(2.3876 k ) .1!!kg h20
kgdry air
m v ,11.)!%,.1!! 8 .3&4 0.08540 kg
s
Water vapour at t 24 ℃
6v 6a , ω 3- ω 2 / 62 11.)!%kg
s / ω 2 .3&4kg h20
kgdry air
5etermine ω 3/
!
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ω 3 0.662∅ Pg
P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 3.3!Pa
0.622(0.6)(3.0037 k )
100k −(0.6)(3.0037 k ) .1142
kg h20
kgdry air
m v ,11.)!%,.1142 8 .3&4 0.09319 kg
s
Water vapour at t 25 ℃
6v 6a , ω 3- ω 2 / 62 11.)!% kgs / ω 2 .3&4 kg h
20
kgdry air
5etermine ω 3/
ω 3 0.662∅ Pg
P−∅ Pg !"# $ro% &'terpo(at&o' ta)(e A*4+/ Pg 3.1&)%Pa
0.622(0.6)(3.1698 k )
100k −(0.6)(3.1698 k ) 0.01206
kg h20
kgdry air
m v ,11.)!%,.12& 8 .3&4 0.10085 kg
s
%
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B. Refrigeration cycle
Select 2 working refrigerants for the system and explain the reasons of selection based onsafety and thermal properties.
First and foremost, -13"a do not contain chlorine atom so that it afford to undermine
the role of atmospheric oone4 besides, -13"a has a good safety performance )non-
flammable, non-e$plosive, non-to$ic, non-irritating no rot resistance+, in addition, -13"a is
easier to retrofit refrigeration system so that the heat transfer performance is closer. 5ast but
not least, -13"a heat transfer performance better than the -12 which can help the amount
of refrigerant greatly reduced.
As a refrigerant, ammonia offers three distinct advantages over other commonly usedindustrial refrigerants. First, ammonia is environmentally compatible. 6t does not deplete the
oone layer and does not contribute to global warming. 7econd, ammonia has superior
thermodynamic 8ualities, as result ammonia refrigeration systems use less electricity. Third,
ammonia9s recogniable odor is its greatest safety asset. 'nli:e most other industrial
refrigerants that have no odor, ammonia refrigeration has a proven safety record in part
because lea:s are not li:ely to escape detection.
The 2 important parameters that need to be considered in the selection of refrigerant
are the temperatures of 2 media )refrigerated space and the environment+ with which the
refrigerant e$change heat. Furthermore, the characteristics of the refrigerant include be non-
to$ic, non-corrosive, non-flammable and chemically stable.
6n the case of heat pump, the minimum temperature for the refrigerant maybe
considerably higher since heat is usually e$tracted from media that are well above the
temperature encountered in refrigeration systems.
Choose operating conditions for the refrigeration cycle such as the evaporator and
condenser pressure if the surrounding temperature is 0°C.
)
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The most common energy source for heat pumps is atmospheric air )air-to-air
system+, although water and soil also used. The ma&or problems with air source system isfrosting, which occurs in humid climates when the temperature falls below 2-/;!. the !<= of
heat pump usually range between 1./ and ", depending on the particular system used and the
temperature of the source. <ne system can be used as a heat pump in winter and an air
conditional in summer. This is accomplished by adding a reversing valve to the cycle, so it
can be in heating mode and cooling mode.
efrigerant Analytical Analysis>
The T-s diagram of the refrigeration cycle is drawn below.
This an ideal vapor compression refrigerant cycle, and thus the compressor are isentropic and
the refrigerants leave the condenser as a saturated li8uid and enter the compressor at saturated
vapor.
A refrigerator used refrigerant-13"a as the wor:ing fluid and the assuming design pressures
for the condenser are %%.3:=a while the evaporator is at 1.2?=a. =ressure of evaporator is
assumed below from pressure of condenser to allow heat transfer, Q L from surrounding
into the refrigerant andQ H from refrigerant into surrounding.
Assumptions>
1 7teady operating conditions e$ist
2 @inetic and potential energy changes are negligible.
1
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From the refrigerant -13"a tables, the enthalpies of the refrigerant at all four states are
determined as below>
State 1: Saturated vapor refrigerant-13a
Assumptions> =ressure * %%.3:=a
Temperature * 3/ !
eat transfer efficiently )100#+
By referring to the efrigerant-13"a tables>
To findhg and
sg @T 1 we have done the interpolation from table A-11 at
T 1=35° C
(35−34)(36−34)
= (hg – 268.57 )
(269.49 – 268.57)
h1=hg=269.03
kJ
kg
(35−34)(36−34)
= (s – 0.91743)
(0.91675 – 0.91743)
s=sg @ T 1
=0.91709 kJ
kg . K
State 2: Super!eated vapor refrigerant-13a
The chosen pressure is 1.2 ?=a due to the pressure on the condenser. e pic:ed this pressure
to allow heat transfer from refrigerant to surrounding. At state 2, it is isentropic process
S1=S
2 .
P2=1.20 Pa(ass!m"#i$n)
s2=s
1=sg@P
1
=0.91709kJ
kg. K
To find h, we have done the interpolation from table A-13 at P=1.2 Pa
(0.91709−0.9130)(0.9267 – 0.9130)
= (h2 – 273.87)(278.27 – 273.87)
h2 * 2/.21 :CD:g
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State 2S: Super!eated vapor refrigerant-13a !2s
The compressor efficiency is assumed at /#.
n%=h2s−h1
h2−h1
0.75=h2 s−269.03
kJ
kg
275.21 kJ /kg−296.03kJ
kg
h2s * 2%0."2 :CD:g
State : Saturated li"uid refrigerant-13a
)efer Table A-12+
P3=1.2 Pa
h3=h
4=h& @"3=117.77
kJ
kg
State !: #!rottling valve
'−(=h4−h
3
h4=h
3
h4=117.77
kJ
kg
Calculate the re"uired refrigerant mass flow rate to obtain the desired cooling effect.
´Q L=
´m(h1−h
4)
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Q L is obtained from 8uestion 1.Q L 6ndicate that the heat transfer rate from refrigerant
to surrounding.
´18.1892= ´m(269.03−117.77)
´m=0.12
kg
s
Calculate the maximum C#$ and actual C#$ of the cycle if the compressor efficiency is
assumed at %&'.
!<= defines the performance of the refrigeration cycle. To calculate !<=, we use this
formula
C)P= Q L
* ¿
?a$imum !<=
(¿= m(h2−h1)
( ¿=0.12 (275.18−269.03 )
(¿=780*
C)Pma+=18.1892
0.738=24.65
Actual !<=
( ¿= m(h2 s−h1)
(¿=0.12 (280.42−269.03 )
(¿=44.36k(
C)Pa%# =18.1892
1.366=13.31
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Suggest an innovative system that can improve the current C#$ i.e multistages or cascade
refrigeration cycle. $rove your suggestion using analytical analysis.
6t is obvious that the lower-temperature unit of the cascade system absorbs less power thanthe single stage system. This originates from the fact that the pressure ratio across the
compressor in the lower unit of the cascade system is less than that in the single-stage system
for the same refrigeration capacity. !<=s for the lower unit of the cascade system are higher
than those for the single-stage system.
(stimate the cost of running the system )single cycle and multistage or cascade* for a 12
hour operation )based only on the compressor work input* under steady conditions and
actual +alaysian daylight electrical tariff by ,- ),enaga -asional erhad*.
Tar&, - * Lo /o(ta#e 'utr&a( Tar&,
For Overa(( o't( Co'u%pt&o' Betee' 0*200 W%o't
;or all <' 3%. sen7<'
%$s# =!sag, ×38
%$s# =18.1892k ×38= R 691189.6 0
C. Co$%ustion C!a$%er for &nternal Co$%ustion 'ngine
14
Q H
To e+'aust
-C )2+. H 2)+/ 0 2
C 12
H 26
25℃
a ) +3.76 0
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Figure 2 > Eatural as !ombustor
Question 1: iven > G * 1./?att
Q= mair
m&!,1
C 12
H 26+1.2a#h ()2
+3.76 0 2) 2 - C )
2+ H
2)+/ 0
2
C :12= - (1 ) 3 - =12
H :26=. (2 ) 3. =13
) :1.2a (2 )= - (2 )+ (1)
1.2a (2 )=12 (2 )+13 (1 )
a=15.4167
0 :1.2a (3.76 )=/ (1 )
1.2(15.4167 ) (3.76)=/ 3 / =69.56
C 12
H 26+1.2 (15.4167) ()2
+3.76 0 2) 212C )
2+13 H
2)+69.56 0
2
Hrate of heat transfer
4¿= 4$!#
1"
h%=h R−h "
h R= 5 0 1
h&
h & =r,&,r Ta61, 7−26
h RC 12
H 26=¿
- 2)11
=97=mol
h R)2=0
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8ind h R¿ #a61,a# 273 K
1&
Standard
#eferences
C )2
2! %33"2!3 '2% %&)!273−270
280−270=
h−8335
8697−8335
)2
2! !%"%2!3 '2% %1"273−270
280−270=
h−7858
8150−7858
H 2
)
2! %)&12!3 '2% )2)&273−270
280−270=
h−8961
9296−8961
0 2
2! !%4)2!3 '2% %141273−270
280−270=
h−7849
8141−7849
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8ind h R¿ #a61,a# 273 K
4 R= 0 (h& 0+h−h
$)
¿ (1 ) (−291010+0−0)+18.5 (0+7945.6−8682 )+69.56 (0+7936.6−8669 )
¿−291010−13623.4−50945.744
¿−355579.144 *a##
1!
C )2
12& "!24412!3 '12% "%3%11273−1260
1280−1260=
h−57244
58381−57244
)2
12& 4")412!3 '12% 413121273−1260
1280−1260=
h−40594
41312−40594
H 2
)
12& 4!2212!3 :12% 4!)121273−1260
1280−1260=
h−47022
47912−47022
0 2
12& 3%"!12!3 '12% 3)4%%1273−1260
1280−1260=
h−38807
39488−38807
h=39249.65 97gC 12
H 26≫ 0 =1
)2≫ 0 =18.5
0 2≫ 0 =69.56
C )2≫ 0 =12
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4 "= 0 (h& 0+h−h1)
¿12 (−393520+57983.05−9364 )+13 (−2858.30+47600.5−9904 )+69.56 (0+39249.65−8669)
¿−4138811.4−3225735.5+2127190.014
¿−5237356.586 *a##
Q$!# = 4 R− 4 "
¿−3555790144+5237356.586
¿4881777.442 kJ
km$1
C 12
H 26=12 (12 )+26 (1 )
¿170 kg
km$1
'=4881777.442
170
¿28716.34 kJ /kg
Question 2: Assumptions> 1+ Tout of the saturated vapor is 100!
2+ 'se ! p average of water
3+ ṁv varied because of the temperature
Tout * 100!
1%
28716.34 × m&!,1=1.5 *a##
m =0.052k /s
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Tin * /
Average temperature * )Tout I Tin+D2
* )100 I /+D2
* /2./!
6nterpolation>
Temperature )!+ 7pecific heat, ! p )@&D:g. @+
/0 ".1%0
/2./ ".1%1
/ ".1J0
Q * ṁv ! p )Tout K Tin+
Temperature )!+ ?ass flow rate water vapor, ):gDs+ṁ
20 0.02"
21 0.0010
22 0.0"
23 0.0%/"0
2" 0.0J31J
2/ 0.100%/
At 20!,
Q * )0.02"+)".1%1 $ 10L3+)33 K 2%+ * 2"%0%.%0 :C
At 21!,
Q * )0.0010+)".1%1 $ 10L3+)33 K 2%+ * 2%"3.3 :C
At 22!,
Q * )0.0"+)".1%1 $ 10L3+)33 K 2%+ * 30%.J" :C
At 23!,
Q * )0.0%/"0+)".1%1 $ 10L3+)33 K 2%+ * 33J20."/ :C
1)
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At 2"!,
Q * )0.0J31J+)".1%1 $ 10L3+)33 K 2%+ * 301".0 :C
At 2/!,
Q * )0. 100%/+)".1%1 $ 10L3+)33 K 2%+ * "00/.12 :C
?ass flow rate water vapor, ):gDs+ṁ eat Transfer ate, Q
):C+
0.02" 24%%.%0.0010 2!%43.4
0.0" 3%!!.)
0.0%/"0 33)2."
0.0J31J 3!14.&
0.100%/ 4"!.1
2
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CONCLUSION
The ob&ective of this assignment is for students to conduct preliminary design calculations of
an air conditioning system combining all the relevant topics in thermal engineering has been
achieved.
REFERENCES
1. Jan F. Kreider. Handbook of heating, ventilation, and air conditioning.
2. Winnick, J (1996). Chemical engineering thermodnamic!. John Wile and "on!
3. #ao$, K% Wang, &ia ('). *#e!iccant cooling air conditioning+ arevie* . -eneable and "$!tainable nerg -evie!.
". /c#oall, -obert ('6). F$ndamental! of H0C "!tem!. l!evier. 2. 3.
/. 4a, C.K. ('6). Comb$!tion 5h!ic!. Cambridge, K+ Cambridge niver!it
5re!!.
. Arora, amesh !handra. */echanical va2o$r com2re!!ion
refrigeration*. -efrigeration and ir Conditioning. 7e #elhi, 8ndia+ 5H8 4earning.
2. 3.
APPENDIX
+or -oad istribution
o
.
ame o I 0ercentage
&. Ahmad Fai Fi:ri Abdul 5atif )5eader+ '1&2'&34' 15
'. ?ohd Fahmi Aiman bin Abdul ashid '1&4'44 '*5
. ?ohammad 7yahan bin EorMAman '1&'314' '*5. Abdul adi Bin . AB. ahim '1&'*'3'6 '15
21
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otal &115
22