1
INTRODUCTION
Thermal engineering is a field of engineering which is deal with the heat transfer; heating and cooling process. It’s become the important field of engineering that give a lot of benefit to mank mankin ind. d. The The basic basic thre threee laws laws of therm thermod odyn ynam amics ics is a based based foun founda dati tion on in therm thermal al engineering. Every calculation and solution must obey to these laws. Heat conduction, heat convection, heat echanger, and refrigeration cycle are topics which are discuss in thermal thermal engineering. engineering. Heat conduction conduction is heat transfer between the more energetic particles to the less energetic particle in contact. There are three ways to solve the problem in heat conduction. There are thermal circuit which is used to solve one dimensional problem, differential e!uations and numerical e!uations. "et, heat convection is heat transfer between fluids in motion with a surface. Heat convection is divided into two types which are f orce convection and natural convection. #orce convec convectio tion n can be divide divided d into into two situat situation ions; s; flat surface surface and pipe. pipe. "ussel "usselt’ t’ss "umber "umber,, $eynold’s "umber, %ranatl’s "umber, &rashof’s "umber, and $eyleigh "umber are widely used in this chapter. Heat echanger is a device that facilitates the echange of heat between two fluids that are at different temperatures while keeping them from miing with each other. In other word, fluid is used to heating or cooling the other fluid. 'og mean temperature difference (')T*+ and effectiveness "T- method are used in heat echanger. $efrigeration cycle is the gas refrigeration cycle in which the refrigerant remains in the gaseous phase throughout. The main purpose is to drop the temperature from high to low temperature. This situation will not obey the eroth 'aw of thermodynamics which is heat will transfer from high temperature to low temperature. Thus, $everse /arnot /ycle by totall y reversible from /arnot /ycle which consist isothermal and isentropic process invented to archive this purpose. 0ir conditioning system is another chapter in thermal engineering. 0ir conditioning process is a series of processes of treating air to simultaneously control the air temperature, humidity, cleanliness, and distribution to meet the comfort re!uirement of the occupants in a space. &enerally, the comfort temperature range is 112/ to 132/ while the comfort humidity range is 456$H to 756$H.
Therm Thermal al Engi Engine neeri ering ng ()E/ ()E/88 889+ 9+ sub: sub:ect ect is re!u re!uire ired d us as a stude student nt to do the the assignment or case study about air conditioning. e are divided into several groups of < persons to complete the task. This task needs us to cooperate and do it as a team. e had discussed together and find out the solution for this problem to conduct preliminary design calculations of an air conditioning system. =ased on syllabus that we had studied all the topics mention above in the class, we had referring other thermal books, handbooks, and internet sources to finish this assessment.
2.0 2.0
APPL APPLIC ICA ATION ION IN ENG ENGIN INEE EERI RING NG PRI PRINC NCIP IPLE LE AND CON CONCE CEPT PT
$efrigeration and air conditioning is used to cool products or a building environment. The refrigeration or air conditioning system ($+ transfers heat from a cooler lowenergy reservoir to a warmer highenergy reservoir.
There are several heat transfer loops in a refrigeration system as shown in #igure 1. Thermal energy moves from left to right as it is etracted from the space and epelled into the outdoors through five loops of heat transfer>
i
Indoor air loop . In the left loop, indoor air is driven by the supply air fan through a cooling coil, where it transfers its heat to chilled water. The cool air then cools the building space.
ii
Chilled water loop. *riven by the chilled water pump, water returns from the cooling coil to the chiller’s evaporator to be recooled.
iii
Refrigerant loop . -sing a phasechange refrigerant, the chiller’s compressor pumps heat from the chilled water to the condenser water.
iv
Condenser water loop . ater absorbs heat from the chiller’s condenser, and thecondenser water pump sends it to the cooling tower.
v
Cooling tower loop. The cooling tower’s fan drives air across an open flow of the hot condenser water, transferring the heat to the outdoors.
Air-Conditioning Systems
*epending on applications, there are several options ? combinations of air conditioning, which are available for use> a b c d
0ir conditioning (for space or machines+ @plit air conditioners #an coil units in a larger system 0ir handling units in a larger system
Reriger!tion Systems "or #ro$esses%
The following refrigeration systems eists for industrial processes (e.g. chilling plants+ and domestic purposes (modular units, i.e. refrigerators+>
i
@mall capacity modular units of the direct epansion type similar to domestic
ii
refrigerators. /entraliAed chilled water plants with chilled water as a secondary coolant for a
iii
temperature range over typically 8 o/. They can also be used for ice bank formation. =rine plants, which use brines as a lower temperature, secondary coolant for typically sub Aero temperature applications, which come as modular unit capacities as well as
iv
large centraliAed plant capacities. The plant capacities up to 85 T$ (tons of refrigeration+ are usually considered as small capacity, 85 B 185 T$ as medium capacity and over 185 T$ as large capacity units.
0 large company may have a bank of units, often with common chilled water pumps, condenser water pumps, cooling towers, as an offsite utility. The same company may also have two or three levels of refrigeration and air conditioning such as a combination of> i ii iii
/omfort air conditioning (15 B 18 o/+ /hilled water system (C5 B 955 /+ =rine system (subAero applications+
Dapour /ompression $efrigeration @ystem /ompression refrigeration cycles take advantage of the fact that highly compressed fluids at a certain temperature tend to get colder when they are allowed to epand. If the pressure change is high enough, then the compressed gas will be hotter than our source of cooling (outside air, for instance+ and the epanded gas will be cooler than our desired cold temperature. In this case, fluid is used to cool a low temperature environment and re:ect the heat to a high temperature environment.
Dapor compression refrigeration cycles have two advantages. #irst, a large amount of thermal energy is re!uired to change a li!uid to a vapor, and therefore a lot of heat can be removed from the airconditioned space. @econd, the isothermal nature of the vaporiAation allows etraction of heat without raising the temperature of the working fluid to the temperature of whatever is being cooled. This means that the heat transfer rate remains high, because the closer the working fluid temperature approaches that of the surroundings, the lower the rate of heat transfer.
Dapour 0bsorption $efrigeration @ystem
The vapour absorption refrigeration system consists of> i
0bsorber> 0bsorption of refrigerant vapour by a suitable absorbent or adsorbent,
ii
forming a strong or rich solution of the refrigerant in the absorbent? adsorbent %ump> %umping of the rich solution and raising its pressure to the pressure of the
iii
condenser &enerator> *istillation of the vapour from the rich solution leaving the poor solution for $ecycling
The absorption chiller is a machine, which produces chilled water by using heat such as steam, hot water, gas, oil etc. /hilled water is produced based on the principle that li!uid (i.e. refrigerant, which evaporates at a low temperature+ absorbs heat from its surroundings when it evaporates. %ure water is used as refrigerant and lithium bromide solution is used asabsorbent.
Heat for the vapour absorption refrigeration system can be provided by waste heat etracted from the process, diesel generator sets etc. In that case absorption systems re!uire electricity for running pumps only. *epending on the temperature re!uired and the power cost, it may even be economical to generate heat ? steam to operate the absorption system.
&.0
INTEGRATION 'AT(E'ATICAL SOLUTIONS
A. Air-conditionin proc!"" Evaporator T3 = 20C
T2 = 10 2 = 100%
Condensate 3
2
1 Condensate Water
T1 = 35C 1 = 70% P = 100 kPa 38,000 cfm
Provide a sketch of the air – conditioning processes with the ambient pressure of 100kpa.
Determine the reuired heat e!traction rate at the cooling rate and heating rate when the ambient air enters at "#$C and %0$C of relative humidit& and leave the s&stem at '0$C.
0ssumptions> a+ This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process. b+ *ry air and the water vapor are ideal gases. c+ The kinetic and potential energy changes are negligible. @T0TE 9> #rom the %sychrometric chart and 04 table, h9 955.5 kF ? kg dry air G9 5.518< kg H 1 ? kg dry air
v 1=
RT P 1 − Ø 1 Pg 1
( 0.287∗10 ) ( 35 +273 ) v 1= [ ( 100∗10 )−( 0.7 ) ( 5.6291∗10 ) ] 3
3
3
v9 5.151 m < ? kg dry air 9 cfm 5.51C<9 m
@T0TE 1> %sat 9.11C9 k%a h1 1.7 kF?kg dry air G1 5.553C kg H 1?kg dry air.
@T0TE <> @ince G< G1 5.553C kg H 1?kg dry air
/0'/-'0TI"> mw ma (G9G1+ ma D9 ? v9
m a=
18 09202
m! ) 1*.+, gs
mw =
19.56
0.0253
− 0.0078
m/ ) 0.&2& gs
Jdot.out ma (h9h1+ B mwhw hw hf K Tcondensate hf 41.511 kF?kg at T 95L/
( 0.3423 )∗( 42.022 ) Jdot.out [( 19.56 )∗( 100−29.6 )]/ ¿ M
dot.ot ) 1&,2.,2 3
Jdot.in ma (h<h1+ h< (9.558+(15+ N (5.553C+(18<3.4+ h< <.C1 kF?kg Jdot.in 9.87 (<.C11.7+ dot.in ) 201.&1 3
The re!uired heat etraction rate at the cooling coil is 1&,2.,2 3 and the heating section is 201.&1 3.
(nal&)e the cooling rate and heating rate when the ambient temperature changes from '*$C to +0$C if the e!it temperature will maintain at '0$C
@ince the value of T1 is greater than the T 9, the cooling rate will become smaller than the value from 0(1+. It can be proved when the value of h 1 O h 9 on the psychrometric chart, the value of Jdot.out is increase. -sing the
Q= Ma ( h 1 −h 2 ) e!uation, we can calculate the new J dot.out.
@ince the Jdot.out (ve+ B (Nve+ sign, the value of J dot.out will be greater for the cooling rate. @o, the cooling cannot be operated. #or the heating rate, the value is good for heating purpose. @ince the value of heat transfer is bigger, the heating rate can be operated.
#. $!fri!ration cc&!
,elect ' refrigerants for the s&stem and e!plain the reasons of selection based on safet& and thermal properties.
#irst and foremost, r9<4a do not contain chlorine atom so that it afford to undermine the role of atmospheric oAone; besides, r9<4ahas a good safety performance (nonflammable, non eplosive, nontoic, nonirritating no rot resistance+, in addition, r9<4ais easier to retrofit refrigeration system so that the heat transfer performance is closer. 'ast but no least, r9<4aheat transfer performance better than the $91 which can help the amount of refrigerant greatly reduced. 0s a refrigerant, ammonia offers three distinct advantages over other commonly used industrial refrigerants. #irst, ammonia is environmentally compatible. It does not deplete the oAone layer and does not contribute to global warming. @econd, ammonia has superior thermodynamic !ualities, as result ammonia refrigeration systems use less electricity. Third, ammoniaPs recogniAable odor is itPs greatest safety asset. -nlike most other industrial refrigerants that have no odor, ammonia refrigeration has a proven safety record in part because leaks are not likely to escape detection.
Choose operating conditions for the refrigeration c&cle such as the evaporator and condenser pressure if the surrounding temperature is "#$C.
0nalysis> The Ts diagram of the refrigeration cycle is drawn below.
This an ideal vapor compression refrigerant cycle, and thus the compressor is isentropic and the refrigerant leaves the condenser as a saturated li!uid and enter the compressor at saturated vapor. 0 refrigerator used refrigerant9<4a as the working fluid and the assuming design pressures for the condenser are 5.CC3)%a while the evaporator is at 9.1)%a. %ressure of evaporator is Q L assumed below from pressure of condenser to allow heat transfer, from surrounding
Q H
into the refrigerant and
from refrigerant into surrounding.
0ssumptions> 9 1
@teady operating conditions eist Qinetic and potential energy changes are negligible.
#rom the refrigerant 9<4a tables, the enthalpies of the refrigerant at all four states are determined as below> State 1>
@aturated vapor refrigerant9<4a
0ssumptions> %ressure
5.CC3 )pa
Temperature <8 L/ Heat transfer efficiently (9556+ =y referring to the $efrigerant9<4a tables> To find
hg
and
s g @T
(<8<4+ ? (<7<4+ (
h =hg =269.03
1
hg
we have done the interpolation from table 099 at B 17C.83+ ? (17.4 B 17C.83+
kJ kg
(<8<4+ ? (<7<4+ (
s
B 5.934<+ ? (5.9738B 5.934<+
T 1 =35 ° C
s = s g @ T =0.91709 1
State 2>
kJ . K kg
@uperheated vapor refrigerant9<4a
The chosen pressure is 9.1 )%a due to the pressure on the condenser. e picked this pressure to allow heat transfer from refrigerant to surrounding. 0t state 1, it is isentropic process
S 1= S2
.
P2=1.20 MPa ( assumption ) s 2= s1= s g@P =0.91709 1
kJ . K kg
To find h,we have done the interpolation from table 09< at
P=1.2 MPa
(5.9355.9<5+ ? (5.173 B 5.9<5+ (h 1 B 13<.C3+ ? (13C.13 B 13<.C3+ h1 138.9C<7 kF?kg
@tate 1@> @uperheated vapor refrigerant9<4a h 1s The compressor efficiency is assumed at C56.
nc =
h 2 s −h 1 h 2− h 1
h 2 s −269.03 0.8
=
kJ kg
/ kg −296.03
275.18 kJ
h1s 13<.8 kF?kg
State 3>
@aturated li!uid refrigerant9<4a
(refer Table 091+
P3=1.2 MPa
kJ kg
h3= h4 =hf @p 3 =117.77
State 4>
kJ kg
Throttling valve
− w = h 4− h 3 h4 =h3 h4 =117.77
kJ kg
Calculate the reuired refrigerant mass flow rate to obtain the desired cooling effect.
´
Q L =m ( h 1´−h4 ) Q L
is obtained from !uestion 9.
Q L
indicate that the heat transfer rate from refrigerant
to surrounding. 1362.62
´ =m ( 229.03´−117.77 )
´ kg ´ =9.01 m s
Calculate the ma!imum C-P and actual C-P of the c&cle if the compressor efficienc& is assumed at *0.
/% defines the performance of the refrigeration cycle. To calculate /%, we use this formula
C!P=
Q L " ¿
)aimum /%
w ¿= ´ m ( h2−h1 ) w ¿= 9.01 ( 275.18 −269.03 )
w ¿=55.44 k w
C!Pma# =
1362.62 55.44
=24.58
0ctual /%
w ¿= ´ m ( h2 s − h 1 ) w ¿= 9.01 ( 273.95 −269.03 ) w ¿= 44.36 kw C!Pact =
1362.62 44.36
=30.72
,uggest an innovative s&stem that can improve the current C-P i.e multistages or cascade refrigeration c&cle. Prove &our suggestion using anal&tical anal&sis.
It is obvious that the lowertemperature unit of the cascade system absorbs less power than the 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 singlestage system for the same refrigeration capacity. /%s for the lower unit of the cascade system are higher than those for the singlestage system.
/stimate the cost of running the s&stem single c&cle and multistage or cascade for a 1' hour operation based onl& on the compressor work input under stead& conditions and actual 2ala&sian da&light electrical tariff.
Tari' ( - )o* +o&ta! nd"tria& Tari' or /v!ra&& ont& Con"mption #!t*!!n 0-200 kmont For all kWh 34.50 sen/kWh
cost =usag$ % 34.5
cost =1362.62 % 34.5 = RM 46995.9
C. Com4stor or t5e (e!t e6$5!nger Determine the mass flow rate of the hot gases
´ hotgas =m´ ai& + m ´ fu$' m
0.75CH4 + 0.1 N2 + 0.07O2+0.05CO2 +0.03H2 25°C To !at!r t4!" "CO2 +#H2O+ $N2 2000°C ath (O2 +3.76N2 320!
/ombustion e!uation, (theoretical+ 5.38/H 4 N 5.9 " 1 N 5.531N5.58/1 N5.5
N ath (1 N<.37"1+ R/1 NSH1N "1
=alance the e!uation> The unknown coefficients ,y ,A and a th are determined from mass balances />
5.38 N 5.58 R R 5.C
" 1 >
5.9 N <.37 a th 8.8<<1
H>
5.38 (4+ N 5.5<(1+ 1S
1>
5.53 N 5.58 N a th R NS?1
S 9.8<
ath 9.448
/ombustion e!uation for the system is> 5.38/H 4 N 5.9 " 1 N 5.531N5.58/1 N5.5
Qout =
'8 gmo7 97.54< 1C.59< <9. 44.59 1.597 9C.598
5 o8 9mo7 34C85 5 5 <<815 5 149C15
5&20: 8 9mo7 <57 <18
52*;: 8 9mo7 C77 C7C1 <74 C47C 54
∑ ( ( h´ ° + h−h ° )−∑ ( (h´ ° + h−h ° ) &
f
p
f
Jout 5.38(34C85 N 5 5+ N 5.9(5 N <57 C77+ N 5.53(5 N <18 C7C1+ N 5.58(<<815 N 959C7 <74+ N 5.5<(5 N 955 C47C+M B 5.38( <<815 N 993
´ out Q ´ ( fu$' = Qout
´ fu$' = (
201.31 48385.1
= 4.161 % 10−
3
kmo' fu$' s
´ " fu$' =0.75 ( 16 ) + 0.1 ( 28.01 )+ 0.07 ( 32 )+ 0.05 ( 44.01 ) + 0.03 ( 2 % 1 ) M ´ " fu$'=19.302 M
kg kmo'
522<&: 8 9mo7 347<.1 3C13.34 993C43.47 35CC.94 7338.51
´ " ai& =1.445 (! 2 + 3.76 ( 2 )=1.445 # 4.76 # 29=199.4678 kg / kmo' M
´ fu$' M ´ " fu$' ´ fu$'= ( m ´ fu$'=( 4.161 % 10 m
−3
)* =
)* =
) ( 19.302 ) =0.0803
´ " ai& ´ ai& M m = ´ " fu$' ´ fu$' M m 199.4678 19.302
=10.33
´ ai& =m´ fu$' % )* =( 0.0803 ) ( 10.33 ) m ´ ai& =0.8298 m
´ hotgas =m´ ai& + m ´ fu$' m ´ hotgas =0.8298 +0.0803 =0.9101 m
kg s
3he operating cost for 1' hours if the 4atural 5as cost is R21*.''6mmbtu
cost =usag$ ( hou&s )+ th$&ms + p&ic$ p$& th$&m cost =12 % 43.385 % 18.22= RM 9485.67
(. nmi6!d Cro"" &o* !at !6can!r for t! !atin !'!ct Determine the heat convection coefficient in the heater tube and at the outer flow of the tube.
25 %% 15 %%
*2 ,2
0.5 % Heater *'&e 0.5 % -r Condtonn Cond't Heater *'&es )ha'st Co%&'stor *3 ,3 )a%le o Heater *'&es ao't
c + ∈¿= 10 ℃ T ¿ T c+out =20 ℃ h + ∈¿=25 ℃ T ¿ T h+ out =2000 ℃
´ co',=19.56 kg / s m ´ hot =0.9101 kg / s m
0ssuming n C
C pc =1.005 kJ / kg C p =2.226 kJ / kg c + ∈¿ T c+out −T ¿ ´ c =m´ C pc ¿ Q
´ c =( 19.56 ) ( 1.005 ) (20 −10 )= 196.578 kw Q
%roperties of air at 95 L/ (9 atm+>
- =1.246 kg / m
2
k =0.02439 ( / m. k −5
=1.778 % 10 kg / m. s −5
2
/ =1.426 % 10 m / s P& ¿ 0.7336
0 1=
´ m 19.56 = -) ( 1.246 ) ( 0.5 % 0.5 )
0 1=62.79 m / s
ℜ= 01 = /
( 62.79 )( 0.025 ) −5
1.426 % 10
5
ℜ= 110080.64 <5 % 10 ('aminar flow+
1
1
h1 =0.664 ℜ2 P& 3 (u = k
(u c2'in,$& = 0.3 +
[ [ ( )] ] ] 0.62 ℜ
1+
1
1
2
P& 3
0.4
P&
2 3
1 4
[ ( 1+
ℜ 282000
)] 5 8
4 5
(u c2'in,$& = 0.3 +
[
1
1
3 2
3
0.62 ( 110.08 % 10 1+
[[ (
) ( 0.7336 )
)] ] 2
0.4 0.7336
3
1 4
]
[ ( 1+
3
110.08 % 10 282000
)] 5 8
4 5
(u =249.0 249.0 =
h0 ( 0.025) 0.02439
h0= 242.92 " / m. k
%roperties of /H4 at 18L/
−90 + 100 0.0797 −0.0967 = k −0.0967 25 + 100 k =−0.1158 " / m. k
249.0
=
h0 ( 0.025 −0.015 )
−0.1158
h1=−2883.42 " / m. k
Calculate the overall heat transfer coefficient7 8 neglect the conduction effect in the heater tube
3 =
[
3 =
[−
1 1 + h 1 h0
−1
]
1 2883.42
+
1 242.92
3 =265.27 " / m. k
]
−1
(nal&)e the reuired number of heater tubes for heat e!changing process using 923D or :438 method.
´ h c ph C h =m C h =( 0.9101 ) ( 2.226 ) =2.026 K" / k
´ c c pc C ch=m C c =( 19.56 ) ( 1.005 ) =19.66 K" / k C c > C h h$nc$C h=min
(T3 =
´= Q
) s 3 C min
=
8 ( 41L)( 3 )
C min
5 T Rtota'
Rtota'=
2000−25 3
196.578 % 10
ln
()
=0.01
&2
&1 1 Rtota'= + h0 ) 0 2 4kL
+
1 h1 ) 1
ln 0.01=
1
( 242.92 ) ( 8 4 % 0.025 ) L
+
(
0.025 0.015
)
2 4 (−0.1158 ) L
+
1
(−2883.42)( 8 4 % ( 0.025− 0.015 )) L
L=7.1893 m
Hence the obtained
(T3 =
8 4 ( 0.025 % 7.1893 )( 265.27 ) 3
2.026 % 10
(T3 =0.53
Effectiveness, ɛ /ross flow for both fluid unmied
C =
C =
Cmin Cma# 2.026 19.66
=0.115
{
0.22
(T3 6 =1−exp C
{
0.22
[ exp (−C (T3 )−1 ] 0.78
[
}
0.53 0.78 6 =1−exp exp (−0.115 ( 0.53 ) ) −1 0.115
]
}
6 =0.4
EN=IRON'ENTAL ASPECTS AND SOCIETAL I'PACT
)ostly refrigerant has a dangerous particle that can affect environmental aspect. $efrigerant systems are producing H/#/ gases that are dangerous to oAone layer and may have other negative factors. =asically the H/#/ gases can reduce the thickness of oAone layer. Thus, the depletion problem may introduce to other problem such as the radiation light from the sun that may harmful.
"owdays, for safety purposes and the awareness of environmental issues, government and private sector have taken a serious matter. The uses of H/#/ gases are recommended replace by natural refrigerant. These natural refrigerants may help to reduce the affect of oAone layer thickness. The uses of / 1 or $344 have become an alternative way to replace the H/#/ gases. These gases may able to avoid the negative affect during production. )oreover, blended H#/ also has been introduced. These gases are not natural gases, but it also has !uite same purpose for reducing environmental issues. These are some eample of blended H#/> •
C>C +02 is usually used in low temperature commercial and small industrial cooling
installations (e.g. supermarket froAen food systems, small cold stores and small blast freeAers+. In the -Q /#/ 851 became scarce !uite !uickly after the 98 phase out of /#/ production, so it is believed that there are relatively few /#/ 851 systems still in use. •
(C>C 22 is a very commonly used refrigerant. It is widely used in commercial,
industrial and airconditioning systems. It is currently used in many applications that cannot be manufactured using H/#/s after 9st Fanuary 1559. It is also the most likely refrigerant to be used in the airconditioning and heat pump applications •
C>C 12 is used for a wide variety of refrigeration and airconditioning applications.
0ll domestic refrigerators and freeAers built before 94 used /#/ 91. )any are still in use. @imilarly /#/ 91 is used for many other small hermetic systems such as retail display cases, icemakers and etc. /#/ 91 is used in many medium and large siAed systems in commercial and industrial refrigeration.
Onroad and la&orator e)er%ents th a 200 Ford )lorer and a 200 *oota Corolla ere ond'ted to assess the 'el ons'%ton enalt assoated th ar ondtoner (-/C 'se at dle and hha r'se ondtons. 8ehle data ere a9'red onroad and on a hasss dna%o%eter. :ata ere athered or ;aro's -/C settns and th the -/C o< and the ndos oen. -t stead seeds &eteen 64.4 and 113 kh (40 and 70 %h &oth ;ehles ons'%ed %ore 'el th the -/C on
at %a)%'% ooln load (o%ressor at 100= d't le than hen dr;n th the ndos don. *he )lorer %antaned ths trend &eond 113 kh (70 %h hle the Corolla 'el ons'%ton th the ndos don %athed that o r'nnn the -/C at 121 kh (75 %h and e)eeded t at 12 kh (>0 %h. *he nre%ental 'el ons'%ton rate enalt d'e to ar ondtoner 'se as nearl onstant th a slht trend o nreasn ons'%ton th nreasn ;ehle (and o%ressor seed. - loer 'el enalt d'e to -/C oeraton s o&ser;ed at dle or &oth ;ehles lkel d'e to the lo o%ressor seed at ths oeratn ont altho'h the erentae nrease d'e to -/C 'se s hhest at dle.
+.0
CONCLUSION
In conclusion, we have achieved the ob:ective of this assignment that is to conduct preliminary design of an air conditioning. Through the calculations, we also has indicates the possible value for the correct air conditioning system. The calculations of problem solving also have teach and improved our fundamental of calculus and thermal principles. ther than that, we have identified the basic principle and the relationship between theoretical and practical value of thermal engineering especially in air conditioning. =ased on the result and calculations, it is proved that high //% value can reduce the work in needed for the system. In air conditioning application the electrical source can be reduced by manipulated the /% value. The selection of air conditioning product also must take in care because there is !uite different between the energy consumption needed. 0lways prefer to select the product that least energy consumption. #or the old refrigerant and air conditioning system, we need to regularly maintenance for better performance by cleaning and service the coil of cooling and heating. Thus, Thermal Engineering has been proven to teach us on how to apply the knowledge on our daily life. It also can be reference and guidance for us to reduce the uses and sources of natural environment.
,
UTILI?ATION O> RESOURCES
9 1 < 4 5 6 7
)E/889, Thermal Engineering, 159<, #irst edition, )c &raw Hill Education alter T. &rondAik, 0irconditioning system *esign )anual, 1553, Elsevier http>??www.gsa.gov?portal?content?95913 http>??www.epa.gov?ia!?schooldesign?hvac.html htt?//.enneerntool&o).o%/anssteeles
[email protected]%l htt?//.;es%a.o%/t'toral/hr@rnles.ht% htt?//en;ron%ent.natonaleorah.o%/en;ron%ent/reen
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FACULTY OF MECHANICAL ENGINEERING MEC551 – THERMAL ENGINEERING
A9:E:T
Group Members:MOHAMAD NURHAFIZ BIN ANUAR
2012741441
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AMIRUL HAKIM BIN MOHD ALL!H
2012"101#1
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AFI$ A%MAN BIN MOHD &ILU
2012'1###1
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