Chapter 8
Air Refrigeration Systems 8.1 Introduction Introduction The advent of high-speed passenger aircraft, jet aircraft and missiles has introduced the need for compact, and simple refrigeration systems, capable of high capacity, with minimum reduction of pay load. When the power requirements, needed to transport the additional weight of the refrigerating system are taken into account, the air cycle systems usually prove to be the most efficient. The cooling demands per unit volume of space, are heavy. An ordinary ordinary passenger passenger aircraft aircraft requires a cooling system capable of T! capacity and a super constellation requires a cooling system of more than T! capacity. A jet fighter fig hter traveling tra veling at "#$ "# $ km%h needs a cooling coo ling system sy stem capable ca pable of &$ to '$ T! T! capacity. To dissipate the heat load from &$ kW of electronic equipment in a missile missile or other other high speed( flight system, appro)imately * T! of cooling capa city are required. The miniaturi+ation miniaturi+ation of electronic equipment concentrates a heavy cooling load in a small area. t creates difficulty in transferring heat to air at high altitudes. oreover, low pressure of air further complicates the refrigeration design requirements.
8.2 Methods of Air Refrigeration Systems The various methods of air refrigeration systems used for aircrafts these days are as follows 1. /imple air cooling system, 2. /imple air evaporative cooling system, 3. 0oot strap air cooling system, 4. 0oot strap air evaporative cooling system, 5. !educed ambient air cooling system, and . !egenerative air cooling system.
121
8.3 Simp!e Air Coo!ing System A simple air cooling system system for aircrafts is shown in 1ig. .&. The main main components components of this this system are the main co mpressor driven by a gas turbine, a heat e)changer, a cooling turbine and a cooling air fan. The air required for refrigeration refrigeration system is bled off from the main compressor. This high pressure and high temperature air is cooled initially in the heat e)changer where ram air is used for cooling. t is further cooled in the cooling turbine by the process of e)pansion. The work of this turbine is used to drive the cooling fan which draws cooling air through the heat e)changer. This system is good for ground surface cooling and for low flight speeds.
"ig.
8.1 Simp!e air coo!ing system.
The T-s diagram for a simple air cooling system is shown in 1ig. .&. The various processes are discussed below 1. Ramming process . 2et the pressure and temperature of ambient air is p1 and T 1 respectively. respectively. The ambient air is rammed isentropically from pressure p ressure p1 and temperature T 1 to the pressure P pressure P 2 and temperature T 2.2. This ideal ramming action is shown by the vertical line &-' in 1ig..'. n actual practice, because of internal friction due to irreversibilities, the temperature of the rammed rammed air is is more more than than T 2.2. Thus the actual ramming process is shown by the curve &-'( which is adiabatic but not isentropic due to friction. The pressure and temperature of the rammed air is now p 2 '
122
and T 2' respectively. respectively. 3uring the ideal or actual ramming process, the total energy or enthalpy remains constant i.e., h2 h2 ' and T 2 T 2 ' .
"ig. 8.2 #$s diagram for simp!e air cyc!e coo!ing system .
f 4 is the aircraft velocity or the velocity of air relative to the aircraft in metres per second, then kinetic energy of outside air relative to aircraft, K . E .
v2 2000
kJ/kg
5.&6 1rom the energy equation, we know that h2
h1
v
2
2000 2
c p . T 2
c p . T 1
or and 5.'6 We now that, c p
V
2000 2
T 2
T 2 T 1 T 2 ' T 1
cv
T 1
V
2000 c p 2
1
V
2000 c p . T 1 2
1
V
2000 c p . T 1
R
123
7 ( T 2 T 2 ' )
c p 1
cv
R
c p
or
R
R
c p
1
c p 1
-1
/ubstituting /ubstituting the value of c p in equation 5.'6, we have T 2
T 2 '
T 1
T 1
V 2 (
1
1 )
2000 R T 1 2
V (
1
Where conditions.
2a
Also
Where
5.*6
1 )
5.86
2
a 9 2ocal sonic or acoustic velocity at the ambient air 9
R T 1 , where R where R is is in :%kg ;.
T 2
T 2'
T 1
T 1
1
1 2
2
M
M 9 9 uch number of the flight. flight.
The temperature T 2 = T 2 ' is called the sta the stagna gnation tion temper tem peratu ature re of the ambient air entering the main compressor c ompressor.. The stagnation pressure after isentropic compression 5 p 2 ) is given by p 2
T 2
p1
T 1
1
3ue to the irreversible compression in the ram, the air reaches point 2' instead of point 2 at the same stagnation temperature but at a reduced stagnation pressure p 2 ' The pressure p 2 ' may be obtained from the e)pression of ram efficiency (η R ) .
which is given as #cua$ rise i pressure R
!se"ropic rise i pressure
p 2 ' p1 p 2
5.#6
p1
2. Compression process . The isentropic compression of air in the main compressor is represented represented by the line 2' 3 . n actual practice, because of internal friction, due to irreversiblities, the actual compression is represented by the curve 2' 3 . The work done during this compressi compression on process is given by & % a c p ( T ' T 2 ' ) 5.<6 '
where
m a 9 ass of air bled from the main compressor for refrigerationpurposes.
124
3. Coo!ing process . The compressed air is cooled by the ram air in the heat e)changer. This process is shown by the curve 3' 4 in 1ig. .'. n actual practice, there is a pressure drop in the heat e)changer which is not shown in the figure. The temperature of air decreases from T 3 ' to T 4 . The heat rejected in the heat e)changer during the cooling process is given by R % a c p ( T ' T * ) 5.=6 4. E+pas E+p asio io proces pro cesss . The cooled air is now e)panded isentropically in the cooling turbine as shown by the curve 4-5. n actual practice, because of internal friction due to irreversibil irreversibilities ities,, the actual e)pansion in the cooling turbine is shown by the curve 4-5' . The work done by the cooling turbine during this e)pansion process is given by & R % a c p ( T * T , ' ) 5.6
The work of this turbine is used to drive the cooling air fan which draws cooling air from the heat e)changer. 5. Rerigera"io 5. Rerigera"io process . The air from the cooling turbine 5 i.e. after e)pansion6 is sent to the cabin and cock pit where it gets heated by the heat of equipment and occupancy. This process is shown by the curve 5'-6 in 1ig. .. The refrigerating effect produced or heat absorbed is given by R E % a c p ( T T , ' ) 5."6 T 9 nside temperature of cabin, where
We know that >.?.@. of the air cycle rerigera" ig eec" prouce ork oe
% a c p (T T , ' ) % a c p (T ' T 2 ' )
T T , ' T ' T 2 '
5.&$6 f Q tonnes of refrigeration is the cooling load in the cabin, then the air required for the refrigeration purpose, %a
210 c p ( T
T , ' )
g!min
@ower required for the refrigeration refrigeration system,
125
5.&&6
p
% a c p ( T ' T 2' ) 0
5.&'6
"
and >.?.@. of the refrigerating system 210
210
% a c p ( T ' T 2' )
0
5.&*6
%ote& The value of c p for air is taken as 1 #!g $. '(amp!e 8.1
A simple air cooled system is used for an aeroplane having a load of1% tonnes. The atmospheric pressure and temperature are %.& bar and 1% respectively. The pressure increases to 1.%13 bar due to ramming. The temperature of the air is reduced by 5% in the heat e)changer. The pressure in the cabin is 1.%1 bar and the temperature of air leaving the cabin is 25 . 3etermine & @ower required to take the load of cooling in the cabin and '. ..P . of the system. Assume that all the e)pansions and compressions are isentropic. The pressure of the compressed air is 3.5 bar.
So!ution Biven Q * 1% TR + p1 * %.& ,ar + T * 1% * 1% 2/3 * 203 $ + p2 * 1.%13 ,ar + p5*P 6 * 1.%1 ,ar + T 6 *25*252/3*2&0 $+ p3 *3.5 ,ar &. Poer reuire to tae the oa o cooing in the ca,in 1irst of all, let us find the mass of air (m a ) required for the refrigeration purpose. /ince the compressions and e)pansions are isentropic, therefore the various processes on the T-s diagram are as shown in 1ig. .*. 2et T 2 9 Temperature of air at the end of ramming or entering the main compressor, T 3 9 Temperature of air leaving the main compressor after isentropic compression, T 4 * Temperature of air leaving the heat e)changer, and T 5 9 Temperature of air leaving the cooling turbine
12
"ig. 8.3
. p We know that 2 T 1 p1
1
T 2
1.%13
%.&
1.4 1 1.4
( 1.125 )%.206 1.%34
T 2 T 1 1.%34 203 1.%34 2&2.6 $
/imilarly,
p 3 T 2 p 2 T 3
1
3.5 1.%13
1.4 1 1.4
( 3.45 )%.206 1.425
T 3 T 2 1.425 203 1.425 41/ $ 144
/ince the temperature of air is reduced by #$C > in the heat e)changer, therefore temperature of air leaving the heat e)changer, T 4 144 5% &4 36/ $
We know that
p 5 T 4 p4 T 5
1
1.4 1
1.%1
3.5
1.4
( %.200 )%.206 %./
T 5 T 4 %./ 36/ %./ 25/ $
We know that mass of air required for the refrigeration purpose,
12)
ma
21% Q c p ( T 6 T 5 )
21% 1% 1( 2&0 25/ )
51.2 g ! min
75Taking c p for air * 1 #!g $ 6 @ower required to take the load of cooling in the cabin, m a c p ( T 3 T 2 ) 51.2 1( 41/ 2&2.6 ) 1%6 " ns. p 6%
6%
'. ..P. o the sstem We know that >.?.@, of the system
21% Q p 6%
21% 1% 1%6 6%
%.33 ns.
'(amp!e 8.2
An aircraft refrigeration plant has to handle a cabin load of 3% tonnes. The atmospheric temperature is 1/ . The atmospheric air is compressed to a pressure of %.&5 ,ar and temperature of *$C > due to ram action. This air is then further compressed in a compressor to 4./5 ,ar , cooled in a heat e)changer to 6/ , e)panded in a tu rb in e to bar pressure and supplied to the cabin. The air leaves the cabin at a temperature of 2/ . The isentropic efficiencies of both compressor and turbine are %.&. >alculate the mass of air circulated per minute and the ..P . 1or air, c p * 1.%%4 #!g $ and c p ! c 7 1.4. So!ution
Biven Q*3% TR + T 1*1/ *1/23/*2&% $ + p2*%.&5 ,ar + T 2*3% *3%2/3*3%3
$ + p3*p3'*4./5 ,ar + T 4*6/ *6/2/3*34% +p5*p5'*1 ,ar +T 6 *2/ *2/2/3*3%% $ + η ηT %.& + c p*1.%%4 #!g $+ c p !c7* *1.4 The T-s diagram for the simple air refrigeration cycle with the given conditions is shown in 1ig. .8 2et T 3 9 Temperature of the air after isentropic compression in the compressor, T 3(9 Actual temperature of the air leaving the compressor, T 59 Temperature of the air leaving the turbine a fter isentropic e)pansion, T 5' 9 Actual temperature of the air leaving the turbine. We know that for isentropic compression process '-*,
128
1
p 3 T 2 p2 T 3
1.4 1
4./5
%.&5
1.4
( 5 )%.206 1.504
T 3 T 2 1.504 3%3 1.504 40% $
and isentropic efficiency of the compressor,
8sentropic increase in temperature
%.&
ctua- increase in t emperature 40% 3%3 T 3 ' 3%3
T 3 ' 3%3 1// ! %.& 1&6 ./ or
T 3 T 2 T 3 ' T 2
1// T 3 ' 3%3
T 3 ' 3%3 16 ./ 4&&./ $
"ig. 8.4
Dow for the isentropic e)pansion process 8-#, p 4 T 5 p 5 T 4
1
1.4 1
4./5 1
1.4
( 4./5 )%.206 1.561
T 5 T 4 ! 1.561 34% ! 1.561 21/ .0 $
and isentropic efficiency of the turbine, T T 5 ' ctua- increase in temperatur e T 4 8sentropic increase in temperaure T 4 T 5
12*
34% T 5 '
%.&
34% T 5 '
34% 21/ .0 122.2 T 5 ' 34% %.& 122.2 23% $
9ass o air circuate per minute
ma
21% Q c p ( T 6 T 5 ' )
. P . .
21% 1% 1.%%4( 3%% 23% )
21% Q ma c p (T 3' T 2 )
0&.64 g ! min ns. 21% 3%
0&.64 1.%%4( 4&&./ 3%3 )
%.356 ns.
'(amp!e 8.3
An aircraft moving with speed of &$$$ km%h uses simple gas refrigeration cycle for air-conditioning. The ambient pressure and temperature are $.*# bar and -&$C > respectively. The pressure ratio of compressor is 8.#. The heat e)changer effectiveness is $."#. The isentropic efficiencies of compressor and e)pander are $. each. The cabin pressure and temperature are &.$< bar and '#C >. 3etermine temperatures and pressures at all points of the cycle. Also find the volume flow rate through compressor inlet and e)pander outlet for &$$ T!. Take c p 9 &.$$# k:%kg ; ! 9 $.'= k:%kg ; and c p ! c7 1.4 9 for air. So!ution
Biven : * 1%%% m ! h * 2//.0 m!s + p1 *%.35 ,ar + T 1 * 1% * 1 % 2/3 * 263 $ + p3 !p2 * 4.5 + ; * %.&5 + T * %.0 + p5 *p 5'* 1.%6 ,ar + T = 25 * 25 2/3 * 2&0 $ + Q * 1%% TR + c p * 1.%%5 #!g $ + R * %.20/ #!g $ * 20/ #!g $ + c p !cv = * 1.4
13+
"ig. 8.5 #emperatures and pressures at a!! points of the cyc!e
The T-s diagram for the simple gas refrigeration cycle with the given conditions is shown in 1ig. .#. 2et T 2 and p29 /tagnation temperature and pressure of the ambient air entering the compressor, T 3 and P 39Temperature and pressure of the air leaving the compressor after isentropic compression, T 3' * Actual temperature of the air leaving the compressor, T 49 Temperature of the air leaving the heat e)changer or entering the e)pander, p4 * @ressure of the air leaving the heat e)changer or entering the e)pander T 5 9 Temperature of the air leaving the e)pander after isentropic e)pansion, T 5' 9 Actual temperature of the air leaving the e)pander. : 2 ( 2// .0 ) 2 263 We know that T 2 T 1 2%%% c p 2%%% 1.%%5
p 2
and
p1
263
T 2 T 1
30 .4
3%1 .4 $ ns.
1
1.4 1
3%1.4
263
1.4
( 1.146 )3.5 1.611
p 2 p 1 1.611 %.35 1.611 %.564 ,ar ns.
131
/ince
p 3 ! p 2 4.5 5Biven6, therefore p3 p2 4.5 %.564 4.5 2.54 ,ar ns.
We know that for isentropic compression process '-*, 1
1.4 1 p T 3 3 4.5 1.4 ( 4.5 )%.206 1.53/ T 2 p 2 T 3 T 2 1.53/ 3%1.4 1.53/ 463.3 $
We also know that isentropic efficiency of the compressor, 8sentropic temperature rise T 3 T 2 ctua- tem perature rise T 3 ' T 2 %.0
463.3 3%1.4
T 3 ' 3%1.4
161.& T 3 ' 3%1.4
T 3 ' 3%1.4 161.& ! %.0 2%2.4
T 3 ' 3%1.4 2%2.4 5%3.0 $ ns.
Effectiveness of the heat e)changer ( < ) , T ' T 4 5%3 .0 T 4 5%3 .0 T 4 %.&5 3 T 3 ' T 2 5%3 .0 3%1 .4 2%2.4
T 4 5%3.0 %.&5 2%2.4 311 .5 $ ns. p 4 p3 2.54 ,ar ns.
and
Dow for isentropic efficiency of the e)pander,
;
% .0
ctua- tem perature rise 8sentropic temperature rise
311 .5 T 5 '
T 4 T 5 ' T 4 T 5
311 .5 T 5 '
311 .5 243 60.5 T 5 ' 311 .5 %.0 60.5 256 ./ $ ns.
4olume flow rate 2et
7 2 9 4olume flow rate through the compressor inlet, and 7 5 ' 9 4olume flow rate through the e)pander outlet.
We know that mass flow rate of air,
132
ma
21% Q c p ( T 6 T 5 ' )
21% 1%% 1.%%5( 2&0 256 ./ )
5%6 g ! min
p 2 7 2 ma R T 2
and
72
ma R T 2 p 2
5%6 20/ 3%1.4 %.564 1%
5
//6 m 3 ! min ns.
75 R is taken in :%kg ; and p2 is taken in '
D%m 6
/imilarly
p5 ' 75 ' ma R T 5 '
7 5 '
ma R T 5 ' p5 '
5%6 20/ 256 ./ 1.%6 1%
5
351./ m 3 ! min ns.
'(amp!e 8.4
The cock pit of a jet plane flying at a speed of 12%% m!h is to be cooled by a simple air cooling system. The cock pit is to be maintained at 25 and the pressure in the cock p it is 1 ,ar . The ambient air pressure and temperature are %.05 ,ar and 3% . The o ther data available is as follows& >ock-pit cooling load 9 1% TR ain compressor pressure ratio 9 4 !am efficiency 9 &% = Temperature of air leaving the heat e)changer and entering the cooling turbine 9 <$C > @ressure drop in the heat e)changer 9 %.5 ,ar @ressure loss between the cooler turbine and cockp it 9 %.2 ,ar. Assuming the isentropic efficiencies of main compressor and cooler turbine as 0%=> find the quantity of air passed through the cooling turbine and ..P. of the system. Take *1.4 and c p * 1 #!g $. So!ution
Biven ? : * 12%% m!h * 333.3 m!s +T 6 * 25 * 25 2/3 * 2&0 $ + P 6 * 1 ,ar + p1* %.05 ,ar + T * 3% * 3% 2/3 * 3%3 $ ? Q*1% TR + P 3 !P 2 * 4 + R &%= *%.& + T 4* 6 % * 6 % 2 / 3 * 3 3 3 $ + p4 * ( p 3 ' %.5 ) ,ar + p5 p 5 ' p6 %.2*1%.2*1. 2 ,ar + T * 0%= * %.0 The T-s diagram for the simple air cooling system with the given conditions is shown in 1ig. .<. 2et T 2' 9 /tagnation temperature of the ambient air entering the main compressor 9 T 2 , p2 9 @ressure of air after isentropic ramming, and
133
p2' 9 /tagnation pressure of air entering the main compressor. We know that
T 2 T 2 ' T 1
: 2 ( 333 .3 ) 2 3%3 2%%% c p 2%%% 1
3%3 55.5 350 .5 $ T p 2 2 p1 T 1
and
1
1.4 1
350.5 3%3
1.4
( 1.103 )3.5 1.0
p 2 p1 1.0 %.05 1.0 1.53 ,ar
We know that ram efficiency, R
ctua- pre ssure rise 8sentropic pressure rise
% .&
p 2 ' %.05 1.53 %.05
p 2 ' p1 p 2 p1
p 2 ' %.05 %.60
p 2 ' %.& % .60 %.05 1.46 ,ar
Dow for the isentropic process 2'-3 , 1
1.4 1
p T 3 3 ( 4 ) 1.4 ( 4 )%.206 1.406 T 2 ' p2 ' T 3 T 2 ' 1.406 350.5 1.406 532./ $
134
"ig. 8.
and isentropic efficiency of the compressor, 8sentropic temperature rise T 3 T 2 ' ctua- tem perature rise T 3 ' T 2 ' 532./ 350.5 1/4.2 %.0 T 3 ' 350.5 T 3 ' 350.5 T 3 ' 5/6 $ /ince the pressure ratio of the main compressor ( p3 ! p 2 ' ) is 8, therefore pressure of air leaving the main compressor, p3 *p3'*4p2'*4 1.46*5.04 ,ar @ressure
drop in the heat e)changer * %.5 ,ar
@ressure of air after passing through the heat e)changer or at entrance to the cooling p4 p3 ' %.5 5.04 %.5 5.34 ,ar turbine, Also there is a pressure loss of $.' bar between the cooling turbine and the cock pit. Therefore pressure of air leaving the cooling turbine, p 5 p 5 ' p6 %.2 1 %.2 1.2 ,ar
Dow for the isentropic process 8-#, p 4 T 5 p 5 T 4
1
1.4 1
5.34
1.2
1.4
( 4.45 )%.206 1.53
T 5 T 4 ! 1.53 333 ! 1.53 21/ .6 $
We know that isentropic efficiency of the cooling turbine, T T 5 ' ctua- tem perature rise 4 T 8sentropic temperature rise T 4 T 5 %.0
333 T 5 ' 333 21/ .6
135
333 T 5 ' 15.4
T 5 ' 24%./ $
Quantit o air passe through the cooing tur,ine We know that quantity of air passed through the cooling turbine, ma
21% Q c p ( T 6 T 5 ' )
21% 1% 1( 2&0 24%./ )
36 .6 g ! min ns.
..P. o t he sstem We know that >.?.@. of the system
21% Q ma c p ( T 3 ' T 2 ' )
21% 1% 36 .5 1( 5/6 350.5 ) %.264
ns.
'(amp!e 8.5
n an aeroplane a simple air refrigeration is used. The main compressor delivers the air at 5 ,ar and 2%% . The bled air taken from compressor is passed through a heal e)changer, cooled with the help of ram air so that the temperature of air leaving the heat e)changer is 45 and the pressure is 4.5 ,ar . The cooling turbine drives the e)haust fan which is need to force the ram air through the heat e)changer. The air leaving the heat e)changer pass through the cooling turbine and then supplied to cabin at & bar. The pressure loss between the cooling turbine and cabin is %.2 ,ar . f the rate of flow of air through the cooling turbine is 2% g!min, determine the following
1. The temperature of the air leaving the e)pander 2. The power delivered to the ram air which is passed through the heat e)changer,and 3. The refrigeration load in tonnes when the temperature of the air % leaving the cabin is limited to 25 .
Assume that the isentropic efficiency of the cooling turbine is /5= and no loss of heat from air between the cooling turbine and cabin. Take 9 1.4 and c p 1 # ! g $ .
.
So!ution
13
Biven p 3* 5 ,ar + T 3 *2%% *2%%2 /3*4/3$+ T 4*45 * 452/3 * 310 $ + P 4 * 4.5 ,ar + ma * 2% g!min + T 6 * 25 *252/3*2&0 $ + *1.4 + c p 1 #!g $
T
/5'=. *%./5 +
The schematic diagram for the simple air refrigeration system is shown in 1ig..&. The various processes on the T-s diagram are shown in 1ig. .=. The point * represents the air delivered from the compressor to heat e)changer and the point 8 shows the condition of air leaving the heat e)changer. The vertical line 4-5 represents the isentropic e)pansion of air in the cooling turbine and the curve 4-5' shows the actual e)pansion of air in the cooling turbine due to internal friction .The line 5'-6 represents the refrigeration process. 1. Temperature o air ea7ing the e@paner 2et
T 5 9 Temperature of air at the end of isentropic e)pansion in the cooling turbine or e)pander, and T 5' 9 Actual temperature of air leaving the cooling turbine or e)pander.
p We know that 5 T 4 p4 T 5
1
1.2 4.5
1.4 1 1.4
( %.26/ )%.206 %.605
T 5 T 4 %.605 310 %.605 21/ .0 $
sentropic efficiency of the cooling turbine, T T 5 ' ctua- tem perature rise 4 T 8sentropic temperature rise T 4 T 5 %./5
310 T 5 '
310 21/ .0 T 5 ' 242.05 $
310 T 5 ' 1%%.2 ns.
13)
"ig. 8.)
2. Poer ei7ere to the ram air hich is passe through the heat e@changer We know that work delivered to the ram air which is passed through the heat e)changer, ma c p ( T 4 T 5 ' ) 2% 1( 310 242.05 ) 15%3 # ! min @ower
delivered 9 15%3!6%*25.%5 " #s.
*. Rerigeration oa We know that the refrigeration load taken from the cabin m a c p ( T 6 T 5 ' ) 2% 1( 2&0 242 .05 ) 11%3 # ! min
11%3 ! 21% 5.25 TR ns.
8.4 Simp!e Air ',aporati,e Coo!ing system A simple air evaporative cooling system is shown in 1ig. .. t is similar to the simple cooling system e)cept that the addition of an evaporator between the beat e)changer and cooling turbine. The evaporator provides an additional cooling effect through evaporation of a refrigerant such as water. At high altitudes, the evaporative cooling may be obtained by using alcohol or ammonia. The water, alcohol and ammonia have different refrigerating effects at different altitudes. At '$ $$$ metres height, water boils at 8$$ >, alcohol at "$ > and ammonia at - =$C >.
138
"ig. 8.8. Simp!e air e,aporati,e coo!ing system.
The T-s diagram for simple air cycle evaporative cooling system is shown in 1ig. .". The various processes are same as discussed in the previous article, e)cept that the cooling process in the evaporator is shown by 4 4 in 1ig. .". f a Q tonnes of refrigeration is the >ooling load in the cabin then the air required for the refrigeration purpose, %a
210 c p ( T
T ,' )
kg / %i
13*
5.&86
"ig. 8.*. T-s diagram for simp!e e,aporati,e coo!ing system.
@ower required for the refrigerating system, m a c p ( T 3 ' T 2 ' ) p " 6%
And > ?.@. of the refrigerating system 210
210
% a c p ( T ' T 2' )
0
5.
The initial mass of evaporant ( me ) required to be carried for the given flight time is given by where
me
Qe.t h g
Qe Feat to be removed in evaporation in k:%min, and h g 2atent heat of vaporisattion of evaporant in k:%kg.
%otes& 1 . n. T-s diagram as shown in 1ig. .", the thick lines show the ideal condition of the process, while e dotted lines show actual conditions of the process.. 2. f cooling of 8# minutes duration or less is required, it may be advantageous to use evaporative cooling alone. '(amp!e 8.
/imple evaporative air refrigeration system is used for an aeroplane to take '$ tonnes of refrigeration load. The ambient air conditions are '$C> and $." bar. The ambient air is rammed isentropically to a pressure of &bar. The air leaving the main compressor at pressure *.# bar is .first cooled in the heat e)changer having effectiveness of $.< and then in the evaporator where its temperature is reduced by #C>. The air from the evaporator is passed through the cooling turbine and then it is supplied to the cabin which is to be maintained at a temperature of '#C> and at a pressure of &.$# bar. f the internal efficiency of the compressor is $G and that of cooling turbine is =#G determine
14+
l . ass of air bled off the main compressor '. @ower required for the refrigerating system and *. >.?.@. of the refrigeration system. So!ution
Biven? Q*2% TR + T 1 * 2% * 2% 2 /3 * 2&3 $ + p 1 * %.& ,ar + p2 * 1 ,ar + p 3 p 3 ' 3.5 ,ar + < %.6 + T 6 * 25 * 25 2/3 * 2&0 $ + p6 * 1.%5 ,ar + * 0%= * %.0 + T /5=*%./5 The T-s diagram for the simple evaporative air refrigeration system with the given conditions is shown in 1ig. .&$ , 2et T 2 9 Temperature of air entering the main compressor, T 3* Temperature of air after isentropic compression in the main compressor T 3' 9 Actual temperature of air leaving the main compressor and T 4 9 Temperature of air entering the evaporator. We kno w tha t for an isen trop ic ramming process &-', p T 2 2 T 1 p1
1
1.4 1
1 %.&
1.4
( 1.11 )%.206 1.%3
75Taking
T 2 T 1 1.%3 2&3 1.%3 3%1 .0 $
Dow for the isentropic compression process '-*, 1
1.4 1
p 3.5 1.4 3 ( 3.5 )%.206 1.43 T 2 1 p 2 T 3 T 2 1.43 3%1.0 1.43 431.6 $ T 3
141
9&.86
"ig.8.1+
We know that efficiency of the compressor, T T 2 8sentropic increase in temperatur e 3 ctua- increase in temperatur e T 3 ' T 2 431.6 3%1.0 12&.0 %.0 T 3 ' 3%1.0 T 3 ' 3%1.0
T 3' 3%1.0 12&.0 ! %.0 464 $
Effectiveness of the heat e)changer ( < )> T ' T 4 646 T 4 464 T 4 %.6 3 T 3 ' T 2 ' 464 3%1 .0 162 .2 ...( T 2 ' T 2 )
T 4 646 %.6 162.2 366 ./ $ 3&./
/ince the temperature of air in the evaporator is reduced by #C >- therefore the temperature of air leaving the
Dow for the isentropic e)pansion process 8(-#,
142
p 3 T 5 p6
T 4 '
1
3.5 1.%5
1.4 1 1.4
( 3.33 )%.206 1.41
T 5 T 4 ' ! 1.41 361 ./ ! 1.41 256 .5 $
Efficiency of the cooling turbine, T ' T ' ctua- increase in temperatur e 4 5 T 8sentropic increase in temperatur e T 4 ' T 5 %./5
361./ T 5 '
361 ./ T 5 '
361./ 256 .5 1%5.2 T 5 ' 361./ %./5 1%5.2 202.0 $
&. 9ass o air ,e o the main compressor We know that mass of air bled off the main. compressor, 21% Q 21% 2% 2/6 g ! min ns. ma c p ( T 6 T 5 ' ) 1( 2&0 202 .0 ) 2. Poer reuire or the rerigerating sstem> We know that power required for the refrigerating system, m a c p ( T 3 ' T 2 ' ) 2/6 1( 464 3%1 .0 ) p /46 " ns. 6% 6% *. ..P. o the rerigerating sstem We know that >.?.@. of the refrigerating system 21% Q 21% 2% %.%&4 ns. P 6% /46 6%
8.5 oot$strap Air Coo!ing System A boot-strap air cooling system is shown in 1ig. .&&. This cooling system has two heat e)changers instead of one and a cooling turbine drives a secondary compressor instead of cooling fan. The air bled from the main compressor is first cooled by the ram air in the first heat e)changer. (This cooled air, after compression in the secondary compressor, is led to the second heat e)changer where it is again cooled by the ram air before passing to the cooling turbine. This type of cooling system is mostly used in transport type aircraft. The T-s diagram for a boot-strap air cycle cooling system is shown in 1ig. .&. The various processes are as follows
143
1. The process &-'( represents the isentropic ramming of ambient air from pressure p1, and temperature T to pressure P 2 and temperature T 0. The process &-'( represents the actual ramming process because of internal friction due to irreversibilities. 2.The process '(-* represents the isentropic compression of air in the main compressor and the process '(-*( represents the actual compression of air because of internal friction due to irreversibilities. 3. The process *(-8 represents the cooling by ram air in the first heat e)changer. The pressure drop in the heat e)changer is n eglected. 4. The process 8.# represents compression of cooled air, from first heat e)changer, in the secondary compressor. The process 8-#( represents the actual compression process because of internal friction due to irreversibilities. 5.The process #(-< represents the cooling by ram air in the second heat e)changer. The pressure drop in heat e)changer in neglected. .The process <-= represents isentropic e)pansion of the cooled air in the cooling turbine upto the cabin pressure. The process <-=( represents actual e)pansion of the cooled air in the cooling turbine. ). The process =(- represents the heating of air upto the cabin temperature T 0.
f Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will .be %a
210 c p ( T 3
T ' )
kg / %i
144
5.&<6
"ig. 8.11. oot$strap air coo!ing system.
"ig. 8.12 #$s diagram for /oot strap air coo!ing system.
@ower required for the refrigerating system,
145
p
% a c p ( T ' T 2 ' ) 0
5.&=6
k&
and >.?.@. of the refrigerating system 210
210
% a c p ( T ' T 2 ' )
0
5.&6
'(amp!e 8.)
A boot-strap cooling system of &$ T! capacity is used in an aeroplane. The ambient air temperature and pressure are '$C> and $.# bar respectively. The pressure of air increases from $.# bar to & bar due to ramming action of air. HThe pressure of air discharged from the main compressor is * bar. The discharge pressure of air from the au)iliary compressor is 8 bar. The isentropic efficiency of each of the compressor is $G, while that of turbine is #G. #$G of the enthalpy of air discharged from the main compressor is removed in the first heat e)changer and *$G of the enthalpy of air discharged from the au)iliary compressor is removed in the second heat e)changer using rammed air. Assuming ramming action to be isentropic, the required cabin pressure of $." bar and temperature of the air leaving the cabin not more than '$C >, find 1. the power required to operate the system, and 2. the >.?.@. of the system. 3raw the schematic an d temperature -entropy diagram of the system. Take 9&.8 and c p 9& k:%kg ;. So!ution
Biven Q 9 &$ T! T 1 9 '$C > 9 '$ I '=* 9 '"* ; p1 9 $.# bar p2 9 & bar P 3 *P 3 '* P 4 *3 ba r p 5 * p 5 ' * p 6 *8 bar 1 2 $G 9$. T 9 #G 9 $.# p/ p/ ' p0 $." bar T 0 9'$$>9 '$I '=* 9' "* ; y9 &.8 c p 9 & k:%kg ; The schematic diagram for a boot-strap cooling system is shown in 1ig. .". The temperature- entropy 5T-s6 diagram with the given conditions is shown in 1ig. .&*. We know that for isentropic ramming process. &-' p 2 T 1 p1
T 2
1
1.4 1
1 %.05
1.4
( 1.1/6 )%.206 1.%4/
T 2 T 1 1.%4/ 2&3 1.%4/ 3%6 .0 $ 33.0
14
"ig. 8.13
Dow for isentropic process '-*, 1
1.4 1
p 3 1.4 3 ( 3 )%.206 1.3/ T 2 1 p 2 T 3 T 2 1.3/ 3%6 .0 1.3/ 42%.3 $ 14/ .3 T 3
We know that isentropic efficiency of the compressor, T T 2 8sentropic increase in temperatur e 3 1 ctua- increase in temperatur e T 3 ' T 2 42%.3 3%6 .0 113 .5 %0 T 3 ' 3%6 .0 T 3 ' 3%6 .0
T 3 ' 440./ $ 1/5./
/ince #$G of the enthalpy of air disc8arged.,from the main compressor is removed in the first heat e)changer 5 i.e. during the process *(-86, therefore temperature of air leaving the-first heat e)changer, T 4 %.5 1/5./ 0/ .05 36%.05 $
Dow for the isentropic process 8-#,
14)
1
1.4 1
p 4 1.4 5 ( 1.33 )%.206 1.%05 T 4 3 p4 T 5 T 4 1.%05 36%.05 1.%05 3&1.5 $ 110.5 T 5
We know that isentropic efficiency of the au)iliary compressor,
c2
%.0
T 5 T 4 T 5 ' T 4 3&1.5 36%.05 T 5 ' 36%.05
3%.65 T 5 ' 36%.05
T 5 ' 3&&.16 $ 126 .16
/ince *$G of the enthalpy of air discharged from the au)iliary compressor is removed in the second heat e)changer 5 i.e. during the process #(-<6, therefore temperature of air leaving the second heat e)changer, T 6 %./ 126 .16 00.3 361.3 $
1or the isentropic process <-=, 1
p T / / T 6 p6
%.& 4
1.4 1 1.4
( %.225 )%.206 %.653
T / T 6 %.653 361.3 %.653 236 $ 3/
We know that turbine efficiency T
%.05
ctua- increase in temperatur e 8sentropic increase in temperatur e
361.3 T / '
T 6 T / ' T 6 T /
361.3 T / '
361.3 236 125.3 T / ' 254.0 $ 10.2
&. Poer reuire to operate the sstem We know that amount of air required for cooling the cabin, 21% Q 21% 1% ma 55 g ! min c p ( T 0 T / ' ) 1( 2&3 254.0 )
148
and power required to operate the system, m a c p ( T 3 ' T 2 ) 55 1( 440 ./ 3%6 .0 ) 13% " ns. P 6% 6% '. .. P. o the sstem We know that >.?.@. of the system
21% Q ma c p ( T 3 ' T 2 )
21% 1% 55 1( 440./ 3%6 .0 )
8. oot$strap Air ',aporati,e Coo!ing System
"ig. 8.14& oot$strap air e,aporati,e coo!ing system
14*
"ig. 8.15. T-s diagram for /oot$strap air e,aporati,e coo!ing system.
A boot-strap air cycle evaporative cooling system is shown in 1ig. .&8. t is similar to the boot-strap air cycle cooling system e)cept that the addition of an evaporator between the second heat e)changer and the cooling turbine.The T-s diagram for a boot-strap air evaporative cooling system is shown in 1ig .&#. The various processes of this cycle are same as a simple boot-strap system e)cept the process #((-< which represents cooling in the evaporator using any suitable evaporant. f Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be ma
21% Q c p ( T 0 T / ' )
g ! min
@ower required for the refrigeration system is given by ma c p ( T 3 ' T 3 ' ) P " 6%
and >.?.@. of the refrigerating system 21% Q
21% Q
9 m c ( T ' T ' ) P 6% a p 3 2
15+
%ote& /ince the temperature of air leaving the cooling turbine in boot-strap evaporative system is lower than the simple boot-strap system, therefore mass of air (ma ) per tonne of refrigeration will be less in boot- strap evaporative system. '(amp!e 8.8
The following data refer to a boot strap air cycle evaporative refrigeration system used for an aeroplane to take '$ tonnes of refrigeration load Ambient air temperature 15 o %.0 ,ar Ambient air pressure 1.2 ach number of the flight &%= !am efficiency 4 ,ar @ressure of air bled off the main compressor 5 ,ar @ressure of air in the secondary compressor &%= sentropic efficiency of the main compressor 0%= sentropic efficiency of the secondary compressor 0%= sentropic efficiency of the cooling turbine Temperature of air leaving the first heat e)changer 1/% o Temperature of air leaving the second heat e)change 9155 o Temperature of air leaving the evaporator 1%% o >abin temperature 25 o 1 ,ar >abin pressure
1ind 1. ass of air required to take the cabin load, 2. @ower required for the refrigeration system, and 3. >.?.@. of the system. So!ution
Biven Q*2% TR + T 1*15* 1523/*200 $ + p1*%.0 ,ar + 9*1.2 + R &%= * %.&+P 3*p3'*p4*4 ,ar + p5*p5'*p5''*p6 *5 ,ar + 1 &%=*%.& + 2 0%= + T 0%=*%.0 + T 4*1/% *1/%2/3*443 $ + T 5''*155 *1552/3*420 $ + T 6 *1%% *1%%2/3*3/3 $ + T 0*25 *252/3*2&0 $ + p0*p/ *p/ ''*1 ,ar The T-s diagram for the boot-strap air cycle evaporative refrigeration system, with the given conditions, is shown in 1ig. .&<.
151
2et
We know that
T 2 (9 /tagnation temperature of ambient air entering the mai n compressor, p 2 9 @ressure of air at the end of isentropic ramming, and p 2 ' 9 /tagnation pressure of ambient air entering the main compressor T 2 ' T 1
1
1
2
2 9 1
1.4 1 2
2 ( 1.2 ) 1.200
T 2 ' T 1 1.200 200 1.200 3/1 $
1or isentropic process &-', 1.4
T 2 p1 T 1
p 2
1
3/1 1.4 1 ( 1.200 )3.5 2.425 200
p 2 p1 2.425 %.0 2.420 1.&4 ,ar
"ig. 8.1
We know that ram efficiency, R
ctua- pressure rise 8sentropic pressure rise
152
p 2 ' p1 p 2 p1
% .&
p 2 ' %.0
p 2 ' %.0
1.&4 %.0 1.14 p 2 ' %.& 1.14 %.0 1.026 ,ar
Dow for the isentropic process '(-*, 1
1.4 1
p 4 1.4 3 ( 2.1& )%.206 1.25 T 2 ' p 2 ' 1 . 026 T 3 T 2 ' 1.25 3/1 1.25 463.0 $ T 3
We know that isentropic efficiency of the main compressor 8sentropic increase in temp. T 3 T 2 ' 1
ctua- increase in temp. 463.0 3/1 &2.0 %.& T 3 ' 3/1 T 3 ' 3/1 T 3 ' 4/4 $
T 3 ' T 2 '
Temperature of air leaving the first heat e)changer, T 4 443 $
1or the isentropic process 8-#, 1
1.4 1
p 5 1.4 5 ( 1.25 )%.206 1.%66 T 4 p 4 4 T 5 T 4 1.%66 443 1.%66 4/2 $ T 5
sentropic efficiency of the secondary compressor, T 5 T 4 2
T 5 ' T 4 4/2 443 2& %.0 T 5 ' 443 T 5 ' 443 T 5 ' 4/& $
Temperature of air leaving the second heat e)changer, T 5 ' ' 420 $
Temperature of air leaving the evaporator, T 6 3/3 $
Dow for the isentropic process <-=,
153
1
p 6 T / p/
T 6
1.4 1
5
1
1.4
( 5 )%.206 1.504
T / T 6 ! 1.506 3/3 ! 1.504 235.5 $
We know that isentropic efficiency of the cooling turbine, T T ' ctua- increase in temp. 6 / T 8sentropic increase in temp. T 6 T / % .0
3/3 T / '
3/3 235.5 T / ' 263 $
3/3 T / ' 13/ .5
1. 9ass o air reuire to tae the ca,in oa ma
21% Q c p (T 0 T / )
21% 2% 1( 2&0 263 )
12% g! min ns.
'. Poer reuire or the rerigeration sstem We know that power required for the refrigeration system, m a c p (T 3' T 2') 12% 1( 4/4 3/1 ) 2%6 " ns. P 6% 6% 3. ..P. o the sstem We know that >.?.@. of the system
21% Q ma c p (T 3' T 2')
21% 2% 12% 1( 4/4 3/1 )
8.) Reduced Am/ient Air Coo!ing System
154
%.34 ns.
"ig. 8.1). Reduced am/ient air coo!ing system .
155
"ig. 8.18. T-s diagram for reduced am/ient air cyc!e coo!ing system.
The reduced ambient air cooling system is shown in 1ig. .&=. This cooling system includes two cooling turbines and one heat e)changer. The air reduced for the refrigeration initially in system is bled off from the main compressor. This high pressure and high temperature air is c ooled the heat e)changer. The air for cooling is taken from the cooling turbine which lower the high temperature of rammed air. the cooled air from the heat e)changer is passed through the second cooling turbine from where the air is supplied to the cabin. The work of the cooling turbine is used to drive the cooling fan 5through reduction gears6 which draws cooling air from the heat e)changer. The reduced ambient air cooling system is used for very high speed aircrafts. The T-s diagram for the reduced ambient air cycle cooling system is shown in 1ig. .&. The various processes are as follows 1. The process &-' represents isentropic ramming of air and the process &-'( represents actual ramming of air because of internal friction due to irreversibilities. 2. The process '(-* represents isentropic compression in the main compressor and the process '(-*( represents actual compression of air, because of internal friction due to irreversibilities.
15
3. The process *(-8 represents cooling of compressed air by ram air which after passing through the first cooling turbine is led to the heat e)changer. The pressure drop in the heat e)changer is neglected. 4. The process 8-# represents isentropic e)pansion of air in the second cooling turbine upto the cabin pressure. The actual e)pansion of air in the second cooling tur bine is represented by the curve 8-#(. 5. The process #(-< represents the heating of air upto the cabin temperature T 6 .
f Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose wilt be ma
21% Q c p ( T 6 T 5 ' )
g ! min
@ower required for the refrigeration system is given by m a c p ( T 3 ' T 2 ' ) P " 6% and >.?.@. of the system
21% Q ma c p ( T 3 ' T 2 ' )
21% Q P 6%
'(amp!e 8.*
The reduced ambient air refrigeration system used for an aircraft consists of two cooling turbines, one heat e)changer and one air cooling fan. The speed of aircraft is &#$$ km%h. The ambient air conditions are $. bar and &$C >. The ram efficiency may be taken as "$G. The rammed air used for cooling is e)panded in the first cooling turbine and leaves it at a pressure of $. bar. The air bled from the main compressor at < bar is cooled in the heat e)changer and leaves it at &$$C > The cabin is to be maintained at '$C > and & bar. The pressure loss between the second cooling turbine and cabin is $.& bar. f the isentropic efficiency for the main compressor and both of the cooling turbines are #G and $G respectively, find 1. mass flow rate of air supplied to cabin to take a cabin load of &$ tonnes of refrigeration 2. quantity of air passing through the heat e)changer if the temperature rise of ram air is limited to $ ; 3. power used to drive the cooling fan and 4. >.?.@. of the system.
15)
So!ution
Biven :*15%% m!h*41/ m!s + p1*%.6 ,ar + T 1*1% *1%2/3*203 $+ R &%=*%.& + p3*p4*6 ,ar + T 4*1%% *1%%2/3*3/3 $+ T 6 *2% *2%2/3*2&3 $ + p6 * 1 ,ar + 05=*%.05 + T 1 T 2 0%=*%.0 + Q*1% TR The T-s diagram for the reduced ambient air refrigeration system with the given conditions is shown in 1ig. .&". 2et T 2 ' /tagnation tempera ture of ambie nt air enteri ng the main compressor, p 2 @ressure of air at the end of isentropic ramming, and p 2 ' /tagnation pressure of air entering the main compressor.
We know that T 2 ' T 1
: 2 ( 41/ ) 2 203 3/% $ 2%%% c p 2%%% 1
"ig. 8.1*
1or the isentropic ramming process &-',
158
1.4
T 1 3/% 1.4 1 2 ( 1.31 ) 3.5 2.5/ p1 203 T 1
p 2
p 2 p1 2.5/ %.0 2.5/ 2.%6 ,ar
We know that ram efficiency, p ' p1 ctua- rise in pressure 2 R 8nsentropic rise in pressure p 2 p1 p 2 ' %.0 p 2 ' %.0 % .& 2.%6 %.0 1.26 p 2 ' 1.&3 ,ar The T-s diagram for the e)pansion of ram air in the first cooling turbine is shown in 1ig. .'$.the vertical line '(-&( represents the isentropic cooling process and the curve '(-'(( represents the actual cooling process. Dow for the isentropic cooling process '(-&( , p ' 2 T 1' p1'
T 2 '
1
1.&3 %.0
1.4 1 1.4
( 2.4 )%.206 1.204
...( p1' p 1 ) T 1' T 2 ' ! 1.204 3/% ! 1.204 200 $
sentropic efficiency of the first cooling turbine, T ' T ' ' ctua- increase in temp. 2 2 T 1 8sentropic increase in temp. T 2 ' T 1' %.0
3/% T 2 ' '
3/% 200 T 2 ' ' 3%4 $
3/% T 2 ' ' 02
1or the isentropic compression process '(-*, 1
1.4 1
p 6 1.4 3 ( 3.11 )%.206 1.30 T 2 ' p 2 ' 1 . &3 T 3 T 2 ' 1.30 3/% 1.30 511 $ T 3
15*
"ig. 8.2+ T-s diagram for the first coo!ing tur/ine.
We know that isentropic efficiency of the compressor, 8sentropic increase in temp. T 3 T 2 ' ctua- increase in t emp. T 3 ' T 2 ' 511 3/% 141 %.05 T 3' 3/% T 3' 3/% T 3 ' 536 $ /ince there is a pressure drop of $.& bar between the second cooling turbine and the cabin, p 5 p 5 ' p 6 %.1 1 %.1 1.1 ,ar Dow for the isentropic e)pansion of air in the second cooling turbine 5process 8-#6, 1
1.4 1
p 6 1.4 4 ( 5.45 )%.206 1.62 T 5 1.1 p5 T 5 T 4 ! 1.62 3/3 ! 1.62 23% $ T 4
We know that the isentropic efficiency of the second cooling turbine, T T 5 ' ctua- increase in temp. 4 T 2 8sentropic increase in temp. T 4 T 5
1+
%.0
3/3 T 5 '
3/3 T 5 '
3/3 23% T 5 ' 250.6 $
143
&. 9ass o rate o air suppie to ca,in We know that mass flow rate of air supplied to cabin, ma
21% Q c p (T 6 T 5')
2% 1% 1( 2&3 250.6 )
61 g! min ns.
2. Quantit o ram air passing through the heat e@changer
m R Juantity of ram air passing through the heat e)changer.
2et
The compressed air bled off at temperature T 3' 9 #*< ; is cooled in the heat e)changer to a temperature T 4 9 *=* ; by the ram air from the first cooling turbine at a temperature T 2A 9 *$8 ;. The temperature rise of ram air in the heat e)changer is limited to $ ;. >onsidering perfect heat transfer in the heat e)changer, m R c p 0% ma c p ( T 3 ' T 4 )
m R 1 0% 61 1( 536 3/3 ) &&43
m R &&43 ! 0% 124.3 g ! min ns.
*. Poer use to ri7e the cooing an /ince the work output of both the cooling turbines is used to drive the cooling fan, therefore work output from the first cooling turbine,
" T 1
m R c p ( T 2 ' T 2 A )
124.3 1( 3/% 3%4 ) 02%4 # ! min and work output from the second cooling turbine, " T 2 ma c p ( T 4 T 5 ' )
61 1( 3/3 250 .6 ) 6&/0 # ! min >ombined
work output from both the cooling turbines, " T " T 1 " T 2 02%4 6&/0 15102 # ! min
and power used to drive the cooling fan
15102 ! 6% 253 " #s.
11
8. ..P. o the sstem We know that >.?.@. of the system
21% Q ma c p ( T 3 ' T 2 ' )
21% 1% 61 1( 536 3/% )
%.21 ns.
'(amp!e 8.1+
The reduced ambient system of air refrigeration for cooling an aircraft cabin consists of two cooling turbines, one heat e)changer and one fan. The first cooling turbine is supplied with the ram air at &.& bar and &# C > and delivers after e)pansion to the heat e)changer at $." bar for cooling the air bled off from the main compressor at *.# bar. The cooling air from the heat e)changer is sucked by a fan and discharged to the atmosphere. The cooled air from the heat e)changer is e)panded upto bar in the second cooling turbine and discharged into air cabin to be cooled. The air from the cabin is e)hausted at ''C >. The refrigerating capacity required is &$ tonnes. f the compression inde) for the main compressor is &.# and the e)pansion inde) for both the cooling turbines is &.*#, determine 1. ass flow rate of the cabin air 2. >ooling capacity of the heat e)changer and flow rate of the ram air when compressed air is to be cooled to <$C > in the heat e)changer and temperatue rise in the heat e)changer for the ram air is not to e)ceed *$ ; 3. >ombined output of both cooling turbines driving the airfan with transmission efficiency of <$G and 4. >.?.@. of the refrigerating system considering only power input to the compressor . So!ution
Biven p2*1.1 ,ar + T 2*15 *152/3*200 $ + p2'*%.& ,ar + p3*p4*3.5 ,ar + p5*p6 *1 ,ar + T 6 *22 *222/3*2&5 $ + Q*1% TR + 1 *1.5 + 2 *1.35 + T 6%=*%.6 + T 4*6% *6%23/*333 $ The T-s diagram for the reduced ambient system of air refrigeration for cooling an aircraft cabin with the given conditions is shown in 1ig. .'&. The process of cooling the ram air in the first cooling turbine is shown by the curve '-'(. 2et T 2 ' temperature of ram air after e)pansion in the first cooling turbine, and T 3
temperature of air bled off from the main compressor
12
(.
"ig. 8.21
We know that for the e)pansion in the first cooling turbine 5 process '-'(6, 2 1
1.35 1
p 1.1 1.35 2 ( 1.22 )%.206 1.%53 T 2 ' p 2 ' %.& T 2 ' T 2 ! 1.%53 200 ! 1.%53 2/3.5 $ T 2
2
1or the process '-*,
p 3 T 2 p 2 T 3
1 1 1
3.5
1.1
1.5 1 1.5
( 3.10 )%.333 1.4/
T 3 T 2 1.4/ 200 1.4/ 423.4 $
and for the process 8-#,
p 4 T 5 p 5 T 4
2 1
2
3.5 1
1.35 1 1.35
( 3.5 )%.26 1.305
T 5 T 4 ! 1.305 333 ! 1.305 24% .4 $
&. 9ass o rate o ca,in air We know that mass flow rate of cabin air,
13
ma
21% Q 21% 1% 30 .5 g ! min c p ( T 6 T 5 ' ) 1( 2&5 24% .4 )
...( c p or air 1#!g $)
'. ooing capacit o the heat e@changer an o rate o ram air We know that cooling capacity of the heat e)changer
ma c p ( T 3 T 4 ) 30.5 1( 423.4 333 ) 340% # ! min ns.
n order to find the flow rate of ram air (9R)> equate the enthalpy lost by compressed air to the enthalpy gained by ram air. We know that enthalpy lost by compressed air ma c p ( T 3 T 4 ) 30.5 1( 423.4 333 ) 340% # ! min B
(i) /ince the temperature rise for the ram air is not to e)ceed *$ ;, therefore enthalpy gained by ram air m R 1 3% 3% m R # ! min B(ii6 Equating equations (i) and 5ii6, m R 116 g ! min ns. *. om,ine output o ,oth cooing tur,ines We know that output of first cooling turbine ma c p ( T 2 T 2 ' ) 116 1( 200 2/3.5 ) 1602 #! min
1602 ! 6% 20 " and output of second cooling turbine ma c p ( T 4 T 5 ) 30.5 1( 333 24%.4 ) 3565 # ! min
3565 ! 6% 5&.4 " /ince the transmission efficiency9<$G, therefore combined output of both the cooling turbines ( 20 5&.4 ) %.6 52.44 " ns.
8. .. P. o the rerigerating sstem
14
We know that >.?.@. of the refrigerating system
21% Q ma c p ( T 3 T 2 )
21% 1% 30.5 1( 423.4 200 )
%.4 ns.
8.8 Regenerati,e Air Coo!ing System The regenerative air cooling-system is show in 1ig. .. t is a modification of a simple air cooling system with the addition of a regenerative heat e)changer .The high pressure and high temperature air from the main compressor is first cooled by the ram air in the heat e)changer. This air is further cooled in the regenerative heat e)changer with a portion of the air bled after e)pansion in the cooling turbine. This type of cooling system is used for supersonic aircrafts and rockets. The T-s diagram for the regenerative air cooling system is shown in 1ig..'*. The various processes are as follows 1. The process &-' represents( isentropic ramming of air and process &-'( represents actual ramming of air because of internal friction due to irreversibilities. 2. The process '(-* represents isentropic compression of air in the main compressor and the process '(-*( represents actual compression of air because of internal friction due to irreversibilities. 3. The process *(-8 represents cooling of compressed air by ram air in the heat e)changer. 4. The process 8-# represents cooling of air in the regenerative heat e)changer. 5. The process #-< represents isentropic e)pansion of air in the cooling turbine up to the cabin pressure and the process #-<( represents actual e)pansion of air in the cooling turbine. .The process <(-= represents heating of air upto the cabin temperature T / . f Q tonnes of refrigeration is the cooling load in the cabin, then the quantity of air required for the refrigeration purpose will be %a
210 c p ( T
T . ' )
kg / %i
5.&"6
1or the energy balance of regenerative heat e)changer, we have % 2 c p ( T 3
T ' )
%2
% 1 c p ( T *
T , )
% 1 ( T *
T , )
( T 3
T . ' )
15
5.'$6
Where T 0 * Temperature of air leaving to atmosphere from the regenerative heat e)changer.
"ig. 8.22 Regenerati,e air coo!ing system.
@ower required for the refrigeration system,
% 1 c p ( T ' T 2 ' ) .0
k&
5.'&6
and >.?.@. of the refrigerating system 210
210
% 1 c p ( T ' T 2 ' )
0
1
5.'&6
"ig.8.23. T-s diagram for regenerati,e air coo!ing system
'(amp!e 8.11
A regenerative air cooling system is used for an airplane to take '$ tonnes of refrigeration load. The ambient air at pressure -$. bar and temperature &$C> is rammed isentropically till the pressure rises to &.' bar. The air bled off the main compressor at 8.# bar is,. cooled by the ram air in the heat e)changer whose effectiveness is <$G. The air from the heat e)changer is further cooled to <$C> in the regenerative heat e)changer with a portion of the air bled after e)pansion in the cooling turbine . The cabin is to be maintained at a temperature of '#C> and a pressure of bar. f the isentropic efficiencies of the compressor and turbine are "$G and $G respectively, find 1. ass of the air bled from cooling turbine to be used for regenerative cooling 2. @ower required for maintaining the cabin at the required condition and 3. >. ?. @. of the system.
Assume the temperature of air leaving to atmosphere from the regenerative heat e)changer as & $$C >.
So!ution
1)
Biven ? Q * 2% TR + p 1 *%.0 ,ar + T 1 *1% *1% 2/3 * 203 $ + p 2 * 1.2 ,ar + p 3 p 4 p 5 4.5 4,ar + < 6 % = * % . 6 + T 5 * 6% * 6%2/3 * 333 $+ T / *25 * 25 2 /3 * 2&0 $ + p/ * p 6 * p 6 ' *1 ,ar + 0%= * %.0 + T 0 *1%% *1%% 2/3 * 3/3 $
&%&6 * %.& + T *
The T-s diagram for the regenerative air cooling system with the given conditions is shown in 1ig. .'8.
"ig. 8.24.
2et T 2 9 Temperature of air at the end of ramming and entering to the, main compressor, T 3(9 Temperature of air leaving the main compressor. We know that for the isentropic ramming of air 5 process &-' 6, p 2 T 1 p1
T 2
1
1.4 1
1.2 %.0
1.4
( 1.5 )%.206 1.123
T 2 T 1 1.123 203 1.123 31/ .0 $
and for the isentropic compression process '-*,
18
1
1.4 1
p 4.5 1.4 3 ( 3./ )%.206 1.46 T 2 1.2 p 2 T 3 T 2 1.46 31/ .0 1.46 464 $ T 3
sentropic efficiency of the compressor,
8sentropic increase in temp. ctua- increase in t emp.
%.&
464 31/ .0 T 3' 31/ .0
T 3 T 2 T 3 ' T 2
146 .2 T 3 ' 31/ .0
T 3 ' 40% $
We know that effectiveness of the heat e)changer ( < ) , T ' T 4 40% T 4 40% T 4 %.6 3 T 3 ' T 2 40% 31/ .0 162.2
T 4 302./ $
Dow for the isentropic cooling in the cooling turbine 5 process #-<6, p 5 T 6 p6 T 5
1
1.4 1
4.5 1
1.4
( 4.5 )%.206 1.54
T 6 T 5 ! 1.54 333 ! 1.54 216 $
and isentropic efficiency of the cooling turbine,
T
%.0
ctua- increase in t emp. 8sentropic increase in temp. 333 T 6 '
333 216 T 6 ' 23&.4 $
T 5 T 6 ' T 5 T 6
333 T 6 ' 11/
&. 9ass o air ,e rom the cooing tur,ine to ,e use or regenerati7e cooing 2et
1*
ma* ass of air bled from the cooling turbine to be used for regenerative cooling, m 9 Total mass of air bled from the main compressor, and m' 9 ass of cold air bled from the cooling turbine for regenerative heat e)changer. We know that the mass of air supplied to the cabin, ma m1 m2
m2
21% Q c p ( T / T 6 ' )
m1 ( T 4 ( T 0
T 5 )
T 6 ' )
and
m1
1( 2&0 23&.4 )
m 1 ( 202 ./ 333 )
/1./ g ! min
1rom equation 5i6, we find that m1 m 2 /1./ or
21% 2%
( 3/3
23& .4 )
%.32/ m1
B(i) B(ii)
m1 %.3/2 m1 /1./
/1./
113.4 g ! min 1 %.3/2 m 2 %.3/2 m1 %.3/2 13.4 42.2 g ! min ns.
%ote& 1rom equation 5ii6, m2 ! m1 %.3/2 . therefore we can say that the air bled from the cooling turbine for regenerative cooling is *=.'G of the total air bled from the main compressor.
'. Poer reuire or maintaining the ca,in at the reuire conition We know that the power required for maintaining the cabin, at the required condition, m1 c p ( T 3 ' T 2 ) 113 .4 1( 40% 31/ .0 ) 3%/ " ns. P 6% 6% *. ..P. o the sstem We know that >.?.@. of the system
8.*
21% Q m1 c p ( T 3 ' T 2 )
21% 2% 113 .4 1( 40% 31/ .0 )
Comparison of 0arious Air Coo!ing Systems used for Aircraft
1)+
%.23 ns.
The performance curves for the various air cooling systems used for aircraft are shown in 1ig. .'#. These curves show the dry air rated turbine discharge temperature 53A!T6 against the ach number. 1rom 1ig. .'#, we see that the simple air cooling system gives ma)imum cooling effect on the ground surface and decreases as the speed of aircraft increases. The boot strap system on the other hand, requires the air plane to be in flight so that the ram air can be used for cooling in the heat e)changers. ?ne method of overcoming this drawback of boot strap system is to use part of work derived from turbine to drive a fan which pulls air over the secondary heat e)changer, thus combining the features of a simple and boot strap system .As the speed of aircraft increases, the temperature of ram cooling air rises and the ram air becomes less effective as a coolant in the heat e)changer. n such cases, a suitable evaporant is used with the ram air so that the cabin temperature does not rise.
"ig. 8.25. erformance cur,es for ,arious air coo!ing systems.
1or high speed aircrafts, the boot strap evaporative or regenerative systems are used because they give lower turbine discharge temperature than the simple cooling system. n some cases, aeroplanes carry an au)iliary gas-,turbine for cabin pressurisation and, air conditioning . 1rom 1ig. .'#, we see that the turbine. 3ischarge
1)1
temperature of the air is variable. Therefore, in order to maintain the content temperature of simply air to the cabin, it requires some control system.
8.1+ ro/!ems 8.1+.1. A simple air refrigeration system is used for an aircraft to take a load of '$ T!. The ambient pressure and temperature are $." bar and ''C> respectively.
1)2
The pressure of air is increased to & bar due to isentropic ramming action. The air is further compressed in a compressor to *.# bar and then cooled in a heat e)changer to ='C>. 1inally, the air is passed through the cooling turbine and then it is supplied to the cabin at a pressure of &.$* bar. The air leaves the cabin at a temperature of '#C >. Assum ing the isentropic efficiencies of the compressor and turbine as $ percent and =# percent respectively, find &. @ower .required to take the load in the cooling cabin and '. >.?.@. of the system. Take cp * &.$$# k:%kg ; and 9 &.8. Cns. 3&% " + %.10 D 8.1+.2. The cockpit of a jet plane is to be cooled by a simple air refrigeration system. The data available is as follows
>ock-pit cooling load /peed of the plane Ambient air pressure Ambient air temperature !am efficiency @ressure ratio in the main compressor @ressure drop in the heat e)changer sentropic efficiencies of main compressor and cooling Temperature of air entering the cooling turbine @ressure of air leaving he cooling turbine @ressure in the cock-pit f the cock-pit is to be maintained at '#C >, find
9 '$ T! 9 &$$$ km%h 9 $.*# bar 9-& #C> 9 "$G 9* 9 $.& bar 9 $G 9 *$C > 9 &.$< bar 9 bar
&. /tagnation temperature and pressure of air entering the main compressor '. ass flow rate of air to cock-pit *. @ower required to drive he refrigerating system and 8. >.?.@. of system. Cns. 21% g!min + 4// " + %.14/ D 8.1+.3. A boot-strap air refrigeration system of '$ T! capacity is used for an aeroplane flying at an altitude of '$$$ m. The ambient air pressure and temperature are $. bar and $C>. The ram air pressure and temperature are &.$# bar and &=C>. The pressure of air after isentropic compression in the main compressor is 8 bar. This air is now cooled to '=C> in another au)iliary heat e)changer and then e)panded isentropically up to the cabin pressure of &.$& bar. f the air leaves the cabin at '#C> and the efficiencies for the main compressor, au)iliary compressor
1)3
and the cooling turbine are $G =#G and $G respectively find &, @ower required to operate the system and '. >.?.@. of the system. Cns. 1%6.2 " + %.66D 8.1+.4. A boot strap air refrigeration system is used for an a eroplane to take &$ tonnes of refrigeration load. The ambient air conditions are &#C> and $." bar. This air is rammed isentropically to a pressure of &.& bar. The pressure of the air bled off the main compressor is *.# bar and this is further compressed in secondary compressor to a pressure of 8.# bar. The isentropic efficiency of both the compressors is "$G and that of coo ling turbine is #G. The effectiveness o f both the heat e)changers is $.<. f the cab in is to be maintained at '#C> and the pressure in the cabin is & bar, find & &. mass of air passing through the, cabin '. power used for the refrigeration system and *. >.?.@. of the system. Cns. 55.3 g!min + 125 " + %.20 D
8.1+.5. The following data refer to a reduced ambient refrigeration system
Ambient pressure @ressure of ram air Temperature of ram air @ressure at the end of main compressor Efficiency of main compressor Feat e)changer effectiveness @ressure at the e)it o the au)iliary turbine. Efficiency of au)iliary turbine Temperature of air leaving the cabin @ressure in the cabin 1low rate of air through cabin
9 $. bar 9 &.& bar 9 '$o> 9 *.* bar 9 $G 9 $G 9 $. bar 9 #G 9 '#G 9 &.$&* bar 9 <$ kg%min
1ind & &. >apacity of the cooling ling system required '. @ower needed to operate the system *.. >.?.*. >.?.@. of the system. ns. 1&.5 TR + 136 " + %.5%4 D
8.1+.. The following data refers to a reduced ambient air refrigeration system used for an air- craft
/peed of aircraft
9 &#$$ km%h
1)4
