College of Engineering and Computer Science Mechanical Engineering Department

Mechanical Engineering 370 Thermodynamics

Unit Twelve Homework Solutions, December 7, 2010

T rottl!n" #alv e Ev aporator 4

Compressor

T e state po!nts $or t e cycle are s o%n on t e d!a"ram& For t e !deal cycle t ere !s no pressure drop or %or' !n a eat trans$er dev!ce so (1 ) (4 ) 0&12 *(a and (2 ) (3 ) 0&+ *(a& T e compressor !s assumed to be !sentrop!c so s2 ) s1& ,tate 1 !s a saturated vapor and state 3 !s a saturated l!-u!d& .e apply t e $!rst la% $or open systems assum!n" steady $lo% and ne"l!"!ble c an"es !n '!net!c and potent!al ener"!es/ eac dev!ce as one !nlet and one outlet& T us our $!rst la% =W +m ( hout hin ) $or eac dev!ce !s Q

.e can $!nd t e ent alpy values $rom tables 0111 to 0113&

"

2(evap ) 120 '(a3 ) 234&9+ '56'"

s1 ) s"2(evap ) 120 '(a3 ) 0&94++9 '56'"7

) 2(2 ) (cond ) +00 '(a8 s2 ) s1) 0&94++9 '56'"73 ) 2+3&94 '56'"7 by !nterpolat!on& ) $2(cond ) +00 '(a3 ) ::&:2 '56'"
4

For t e t rottl!n" valve8

so8

) $2(cond ) +00 '(a3 ) ::&:2 '56'"

,!nce t ere !s no use$ul %or' !n t e evaporator8 %e can $!nd t e eat removed $rom t e re$r!"erated space 2t e evaporator3 us!n" t e $!rst la% as $ollo%s&

0.05 kg 236.97 kJ 88.82 kJ ( h1 h4 ) = Q evap = m s kg kg

kW s 1 kJ

,!nce t ere !s no eat trans$er to t e compressor8 %e can $!nd t e compressor po%er !nput us!n" t e $!rst la% as $ollo%s&

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0.05 kg 236.97 kJ 273.54 kJ ( h1 h2 ) = W comp = m s kg kg

kW s 1 kJ

T e eat re=ected to t e env!ronment !s t e evaporator eat trans$er& .e can $!nd t !s !" temperature eat trans$er $rom t e "eneral cycle relat!ons !p: Q H = Q L + W & For t e re$r!"erat!on cycle8 t !s becomes&
Q H = Qcond = Q L + W = Qcond + Wcomp = 7.41 kW +1.83 kW
Q H = 9.24 kW

F!nally8 %e $!nd t e coe$$!c!ent o$ per$ormance t e values computed above $or %or' and eat&
cop = Q evap W comp = 7.41 kW 1.83 kW

cop / 4$#)

0f the throttling al e in 'ro"$ 1 is replaced "! an isentropic tur"ine- determine the percentage increase in the C1' and in the rate of heat remo al from the refrigerated space$ >e$er to t e d!a"ram $or problem 1& I$ t e t rottl!n" valve !s replaced by an !sentrop!c turb!ne8 only t e value o$ 4 %!ll c an"e& T e values o$ 18 28 and 38 %!ll rema!n t e same as t ey %ere !n problem 1& For an !sentrop!c process $rom po!nt 3 to po!nt 48 s 4 ) s3 ) s$2(cond ) 0&+ *(a3 ) 0&33230 '56'"7& 0t t e evaporator pressure o$ 120 '*(a8 t !s entropy value !s !n t e m!<ed re"!on& ?ere %e compute t e ent alpy8 4 $rom t e -ual!ty !n t e usual $as !on&

x4 =

s 4 s f ( P4 = 120kPa) s fg ( P4 = 120kPa)

0.33230 kJ 0.09725 kJ kg K kg K = = 0.2749 0.85503 kJ kg K

22.49 kJ 214.48 kJ 81.45 kJ + (0.2749) = kg kg kg .!t t !s value o$ 48 2and t e value o$ 1 $rom problem 101113 t e evaporator eat trans$er becomes h4 = h f ( P4 =120kPa ) + x 4 h fg ( P4 =120kPa ) =

0.05 kg 236.97 kJ 81.45 kJ ( h4 h1 ) = Q evap = m s kg kg

kW s 1 kJ = 7.78 kW

7.78 7.41 ) 4$2&3 $or t e eat removed $orm t e re$r!"erated space 7.41

above t e value !n problem 1& T e !sentrop!c turb!ne %or' can be used to reduce t e %or' !nput to t e compressor& T e ava!lable %or' $rom t e turb!ne !s $ound as $ollo%s&
0.05 kg ( h3 h4 ) = W turb = m s 88.82 kJ 81.454 kJ kg kg kW s 1 kJ
W 0.368 kW turb =

T e net cycle %or' !s =ust t e d!$$erence bet%een t e compressor !nput and t e turb!ne output: 5acaranda 2En"!neer!n"3 3919 E1ma!l: lcaretto;csun&edu *a!l Code :34: ( one: :1:&4++&444: Fa<: :1:&4++&+042

 W 1.828 kW 0.368 kW net = Wcomp Wturb =

W net

= 1.46 kW

For t !s cycle8 t e coe$$!c!ent o$ per$ormance !s t e rat!o o$ Q to W net 8 evap

cop =

Q evap 7.78 kW = 1.46 kW Wnet

cop / )$3%

T e percent !ncrease !n t e cop !s 29&32 @ 4&09364&09 ) 31$)3& 3 A 3## k4*min refrigeration s!stem operates on a apor-compression refrigeration c!cle with refrigerant 134a as the working fluid and an isentropic efficienc! of 5) percent for the compressor$ (he refrigerant enters the compressor as a saturated apor at 14# k'a and is compressed to 5## k'a$ Determine +a, the 6ualit! of the refrigerant at the end of the throttling process- +", the coefficient of performance and +c, the power input to the compressor$
3 Condenser 2

T rottl!n" #alv e Ev aporator 4

Compressor

T e state po!nts $or t e cycle are s o%n on t e d!a"ram& For t e !deal cycle t ere !s no pressure drop or %or' !n a eat trans$er dev!ce so (1 ) (4 ) 140 '(a and (2 ) (3 ) :00 '(a& T e !deal compressor !s assumed to be !sentrop!c so s28s ) s1& ,tate 1 !s a saturated vapor and state 3 !s a saturated l!-u!d& .e apply t e $!rst la% $or open systems assum!n" steady $lo% and ne"l!"!ble c an"es !n '!net!c and potent!al ener"!es/ eac dev!ce as one !nlet and one outlet& T us our $!rst la% =W +m ( hout hin ) $or eac dev!ce !s Q

.e can $!nd t e ent alpy values $rom tables 0111 to 0113&

"

2(evap ) 140 '(a3 ) 239&14 '56'"

s1 ) s"2(evap ) 140 '(a3 ) 0&94494 '56'"7

28s

) 2(2 ) (cond ) :00 '(a8 s28s ) s1) 0&9322 '56'"73 ) 2+9&39 '56'"7 by !nterpolat!on&
2

,!nce t e compressor !s a %or' !nput dev!ce8 %e can $!nd t e actual value o$ 2 by t e e-uat!on ) 1 @ 2 1 @ 28s36s ) 239&14 '56'" @ 2239&14 '56'" @ 2+9&39 '56'"7362:9A3 ) 2+:&40 '56'"&
3

) $2(cond ) :00 '(a3 ) 93&42 '56'"

For t e t rottl!n" valve8

so8

) $2(cond ) :00 '(a3 ) 99&4+ '56'"

.e can $!nd t e -ual!ty at t e t rottl!n" valve e<!t as $ollo%s&

95.47 kJ 27.08 kJ h4 h f ( P4 = 140 kPa) kg kg x4 = = 212 . 08 kJ h fg ( P4 = 140 kPa) kg

74 / 3%$%3&

.e $!nd t e CB( $rom t e usual e-uat!on !n terms o$ t e cycle ent alpy values&

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COP =

q evap wcomp

239.16 kJ 95.47 kJ h h4 kg kg = 1 = 281 . 78 kJ 239 . 16 kJ h2 h1 kg kg

C1' / 3$3&&

T e re-u!red po%er !nput to t e compressor !s $ound $rom t e re$r!"erant eat load and t e cop&

W comp
4

300 kJ min kW s min 60 s 1 kJ = = cop 3.37 Q evap

W comp

= 1.48 kW

An ideal 1tto c!cle has a compression ratio of 5$ At the "eginning of the compression process- air is at 2) k'a and %&oC- and &)# k4*kg of heat is transferred to air during the constant- olume heat addition process$ (aking into account the ariation of specific heats with temperature- determine +a, the pressure and temperature at the end of the heataddition process- +", the net work output- +c, the thermal efficienc!- and +d, the mean effecti e pressure for the c!cle$ To andle t e var!able eat capac!ty8 use t e !deal "as tables $or a!r 2Table 011+8 start!n" on pa"e :493& 0t t e !n!t!al temperature o$ 2+oC ) 300&19 7 %e $!nd u1 C u2300 73 ) 214&0+ '56'" and vr2300 73 ) 421&2& T e !sot ermal compress!on process $rom po!nt one to po!nt t%o !s computed as $ollo%s us!n" t e vr values $rom Table 011+&

v2 vr ( 2 ) = v1 v r ( 1 )

vr ( 2 ) = vr ( 1 )

v2 1 = 621.2 = 77.65 v1 8

= 673.1 K and u2 =

491.2 kJ kg

T e values o$ T2 and u2 are $ound by !nterpolat!on& For t e constant volume eat add!t!on8 %e apply t e $!rst la% to "et u3 ) u2 D -? ) 491&2 '56'" D +90 '56'" ) 1241&2 '56'"& 0t t !s value o$ !nternal ener"y8 %e $!nd T3 ) 1939 and vr2T33 ) 4&9::& T e pressure at po!nt 3 can be $ound $rom t e !deal "as la%&

P3 =

! 3 ! 3 ! 3 v1 ! 3 P (C! ) = = = (C!) = 1 v3 v2 v1 v2 ! 1 1 P1

(95 kPa )(8)(1539 K ) = 3,898 kPa (300.15 K )

T e !sentrop!c e<pans!on $rom po!nt 3 to po!nt 4 !s computed !n t e same %ay as t e !sentrop!c compress!on $rom po!nt 1 to po!nt 2&

v4 vr ( 4 ) = v3 v r ( 3 )

vr ( 4 ) = vr ( 3 )

v4 = (6.588)(8) = 52.70 v3

= 774.5 K and u 4 =

571.69 kJ kg

T e lo% temperature eat re=ect!on !s $ound $rom t e $!rst la% $or closed1system8 constant1volume processes: E-LE ) Eu1 @ u4E ) E214&0+ '56'" @ 9+1&49 '56'"E ) 39+&42 '56'"& .e can t en $!nd t e ne% %or' as t e d!$$erence bet%een t e t%o cycle eat trans$ers: E%E ) E-?E 1 E-LE ) +90 '56'" @ 39+&42 '56'" or 8w8 / 32%$35 k4*kg& T e cycle e$$!c!ency !s $ound as $rom t e usual e-uat!on& ) E%E6E-?E ) 2392&3: '56'"3 6 2+90 '56'"3 or / )%$33& 5acaranda 2En"!neer!n"3 3919 E1ma!l: lcaretto;csun&edu *a!l Code :34: ( one: :1:&4++&444: Fa<: :1:&4++&+042

T e mean e$$ect!ve pressure !s de$!ned by t e e-uat!on mep ) E%E62v ma< @ vm!n3& For t e Btto cycle8 vma< ) v1 and vm!n ) v2 ) v16C>& T us8 %e can $!nd our mean e$$ect!ve pressure as $ollo%s&

mep =

w 1 v1 1 C!

w ! 1 P 1 1 1 C!

392.38 kJ kg 0.287 kJ mep / 42) k'a& (300.15 K ) 1 kg K 1 95 kPa 8

Air enters the compressor of a gas-tur"ine engine at 3## k and 1## k'a- where it is compressed to &## k'A and )5# 9$ :eat is transferred to air in the amount of 2)# k4*kg "efore it enters the tur"ine$ ;or a tur"ine efficienc! of 5<3 determine +a, the fraction of the tur"ine work ouput used to dri e the compressor and +", the thermal efficienc!$ Assume aria"le specific heats for air$ For eac component !n t e "as turb!ne cycle %e assume t at t e component !s steady $lo% %!t ne"l!"!ble '!net!c and potent!al ener"y c an"es& Eac component as one !nlet and one outlet so t at t e "eneral $!rst la% $or eac component !n t !s system !s - ) % u D out @ !n& .e assume t at t ere !s no %or' !nput !n t e eat add!t!on step and t at t e compressor and turb!ne are ad!abat!c& .e also assume t at t ere !s no pressure loss dur!n" t e eat add!t!on step so t at t e compressor e<!t pressure o$ +00 '(a !s t e same as t e turb!ne entrance pressure& .e %!ll use t e a!r tables 2Table 011+8 start!n" on pa"e :493&to compute t e a!r propert!es %!t var!able eat capac!ty& For t e compressor8 t e $!rst la%8 %!t - ) 08 becomes % ) 300&19 '56'" @ 9:4&04 '56'" ) 12:9&:9 '56'"&
!n

out

& ) 2300 73 @ 29:0 73 )

For t e !deal8 !sentrop!c turb!ne8 %e $!nd t e end state $rom t e !deal "as table relat!on $or !sentrop!c processes: (r2Tout8s3 ) (r2T!n32(out6(!n3 % ere (r2T!n3 ) 4+4&11 and (out6(!n ) 2100 '(a36 2+00 '(a3 so t at (r2Tout8s3 ) 4+4&116+ ) 4+&+3& Interpolat!on !n t e !deal "as tables "!ves out8s ) 909&:3 '56'"& T e actual turb!ne %or' t en !s $ound as % ) s2 !n @ out8s3& T e !nlet ent alpy to t e turb!ne !s $ound $rom t e $!rst la% $or t e eat add!t!on step: - ? ) turb8!n @ comp8out8 so t at turb8!n ) -? D comp8out8) 990 '56'" D 9:4&04 '56'" ) 1934&04 '56'"& .e can t en $!nd t e turb!ne %or'& .turb ) s2
!n

out8s

3 )2:4A32934&04 '56' 1 909&:3 '56'"3 ) 942&0 '56'"

T e compressor %or' !nput as a $ract!on o$ t e turb!ne %or' output !s no% $ound& E%compE6E%turbE ) E12:9&:9 '56'"E 6 2942&0 '56'"3 T e e$$!c!ency !s $ound $rom t e usual e-uat!on& ) E%netE6E-?E ) 2942&0 '56'" 1 2:9&:9 '56'"3 6 2990 '56'"3 or / %&$#3& 8wcomp8*8wtur"8 / )%$&3&

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