Solution to Hw07

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I(NOWI\J: D",t-o. . , o .re provided for ~(.;l." carps involv:r'I~ "'eV~y'j',hle Fo~" ,

C'1 cl Pi" " p , " ( " - , , f- inj b- e frA,/epl1 he,t- a." c d Ce> I d r ese r:VOl',... S ' .

Fig. PS.20

(0) TH =: IlOO k) Tc =' 300". Find Y L '

~ - - - ( 0 . )

(I?) TH = SOO°C ,ITc=20oC) Wcyc1e = lOCO F :. J. F" . ...d C O e . a....d CYH .

\N . - + ~ E,.~. q o . . .. ..d 7 ) : : : W"IcJ ~/Q 1 M I w\",,·:cL.. . '1 ' p J . ' - e ~ ~ G < " ' . : J p o .. .. ~ . . . c .{ ~ _)

wcyde\ - Tc.

~ c 9 H ';;"' \ .J ' .Jc(c.le (000 ~:J

s: -(QH T~ I-IS. 1- 2 . c 1 ' 3 j < : :

I1t ~73' K

~)

\..u , . } t . . . _ 6~. ').9 J

~D.4()'::;- \-

Cd)



! p~cH~L.e·M 5.'7..(,r- - -

Ilr{..."a' ...u,.,,:

O...+~" " ' I ' " e . . p.r:l:JvIJ~d~,..

+ c . . v o "'V'eI",s.:bJe crc.l~s;r\. ser('es., I

c .o ~ v .a ·F C . + e ._ J. . bJ

·o .. . . "e.._~-:t res .. , . ..."'r cd: :I::,~R .... ~,...f¢T '- -.

D ~ ~_t:"~' ~-C.l ' _ 0 . ) _ . , . ) ~ P B , J . . . .~ ~ " ' ~ e +h~ .-" "" "" " ~ ff.·~'e' f!'l"y ;;{

. e l f e~pt'UJt' " tde1 (b) tk.c tk"r~_-t ...k·c.. . .~.H t.y ~f.'- 'Q.... f"140lt>I

pflt(' AP .p·~~~rr;yc.l e._ . .ir _ .,

I ~ C!.~ H\I'l til 1 " 1 .C ~ 'IIVIe fJ 1 > 1 \ 7Ao:

J , _ J

, . - - -, .C " < I Ii ,! : . " " cy. C":r (~ .s - '_ . -

1

T 1- 500g

" l ' \ . , = .. \ . . . '0000J() " Y } 2 . = - 1 - ; . .

Alsu IV\ I";.,.S v' \ ' 1 . . . _ c . . . H I ' c ·+ "" ' 3 , . . . . r u t T S -

( T ) l - (l - · ~ . r O O ).l-_ ,=.!S

. 'ClOO

"S o~".!I\"l t.( TJ", ,- 't"' .....V 1 : J . . . . . l:. ei ~ n · Q " 1 _ ~ e > , , · t l . I + . t

7"1, . - t - , 500 T- 'T-SQJOOO ~ 0

© 'Sol"'Ir\~.rNa E'~c;,&t h · d " . 1J

T:- ~SL· '3' o r e +--~

''¥\ . -. t-- 1 . 1 -

,"51.3'----

"VI - t-1 2 . _

(~l 11 ,,.. " "" ~I"~l c o QV-e,.A I t a't r : . : C C '

l(A . ....d T,_

op...,...b'14jb.efw.c-PVl r"'pS " ... " 0 , I " " "'I- i:..I

I- Sb.1J - C. $ ""

. IO ::!_O

(Sb%)

;~!1Tka 1 iI~..&:!I),.-h·(. ":1}.I~~a-,~_!:_O~A ¥'i'cr-h'~, ~o6t. - n . . . . , : , C-A ... ~'~'~ed.-

, be: ~ t"-,4 "'bs:ol ...e:tt~"" .. '- h V:t "J ," "'"F) ~""..., 0 " " ~ . "".P~h.·Itf.

'2, .~ ~~l;)I~.eg_ c~Jr.l~ +bfrM_M ~ffiufY\r7 ~.,;r'-'A..-iey - r l " . . . , _ U+h.:£;. .r-

·o·f -t4 ~1 o d . : " . ~ 4 u ~ , f)C . /',-r d :~,"1 "J i~"OZ,

I C - - - I ,



PRo8LEH 5.4D

/ < N O W N :S . J . e . . . . . d .1- . 1 ' + . + c = r - : - ~"Vlj c : L . . + c o . . . 0>.(<:: prov.- d . . e . : l . f c v - 0... PO!..AJe r f'l /d.~i:

~·'IiC.~r·"') E'.-)H~y h :J kLJ ·h"",sfe.- c . c . . . 1\.4.'~.

'D ~ \.c"'_'I<\~ t - h - o . . i",e,V"C:~" ~ 1-c. . . ~-pe,... ... ~ ' ' 7 ; t - t . . . . . rt'vf',..- 1rQ'f'a. J , . ' ~t. ~;" c. k. "'-.;:t- ~~ 1-- ~ e.,;ch0 6 f . . . . u o ~

River, r T l = 1.65 x 105 kgls

[ . r lo u toow.-c..r (->"" ... t. kI "'-'de Ird a . . . o ,

fOLve .....c.yc.{Q c:>pfU'o. t . 'nJ Q , . 1 - r~r

S-+ I)..k . .

2 , T4C.c>;" iov&' t» " '" r - - d c . . c . + r froV1h..

~V\e:r, ~ b " ~W-a. ...~ f a < - =i-kA p c ; > . . . . . -e e- cycle. E ,."e "'J ';J ...

J . . . . • c.-"'-.or, ed h : J ha..J:> ty-.,. ....r f . .r to

~ ",-,'ve..y- _ TkA..-t. o,..;."g + '-4. . """\y

: h . e . . : t - f v t , . . . i o - I . s f - en.

3. 7 1 . . a . . ("I ve r (.N...t < , . . . . . . ; w . . . . . .. d .. o .J ,, _ < I. -

: " '- C-40~"~' b-l!.. -1M . : - t c . . . Ce l o ' I s . f - r u . . t-o s I' f." c.. . r " c. ),-e-. t: C-

O- t~c r . ' ~ o f - t 1 - ' - - t ~ ~V) 4.., ~ ~ f-<-.._

ruhCe..i h~\re...

p . : , , . . . . . _ ~ " " t.-:z, I v- o Iv~ e. e. . . .do J I'Vl1

~ke f - U . . . , c...V\ e..., Y " " 'l ~ \0 4 lc. VI c . e

cq

Wetde

C O i "

~(.(c{e

Ui''td'C. + c J D ~(2)

- h - . c .

I...i~ Co:: 4 - - 1 . I ' - J / " , · j < f r , , _ T,.."cA~lq.

(a.) ¥ ' \ - = - Y\M''''~::[1- T~ttJ ~ [,- ~:~] = D.S-Or· J:...r.,d\",~ v ~ . . . . : . 6,.("f)

(_7~~f"Hv)1':~~ It~ -iJ = l.oS " K (~J .OSoC)C11~"1j):

(_l,!.\ X . 'o~~~l.s( +. 'l.K.J/~,,·K)

• .L ]We.)felt [ ' t \ . . -I

(1)

(b) 1 : : : . ~ Ylf-(-'tlC: 0.331

(0 (Tl-,.l-='2.lIK (_:::'2.'II~C)



PRoBLEM 5·4-3

5.43 A refrigeration cycle operating between two reservoirs receives energy Qc from a cold ..

reservoir at Tc =275 K and rejects energy Q H to a hot reservoir at TH = 315 K.F or each of the

following cases determine whether the cycle operates reversibly, operates irreversibly, or is

impossible:

(a) Qc =1000 kJ, Wcycle =80 kJ.

(b) Qc = 1200 kJ, Q H : = : ; 2000 kJ.(c) Q H - = 15751d, Wcycle =200 kJ.

.. (d) {J = 6.

KNOWN: A refrigeration cycle operates between two reservoirs with specified temperatures.

FIND: Whether each of four cycles operates reversibly, operates irreversibly, or is impossible.

SCHEMATIC AND GIVEN DATA:

~------I ,...----E=----.,

II

l Refrigeration WCYCle

II CycleI1

II '-----1-----

ENGINEERING MOnEL:

1. The control volume defined by the dashed line on the accompanying diagram undergoes a

refrigeration cycle.

ANALYSIS:The maximum coefficient of performance for a refrigeration cycle operating between two

reservoirs is

P m - Tc 275 K - 6.875ax- 1 ', - T = 315 K - 275 KH c



PR.O&LE 1'-1 5. 43 '" (0'lti~ued )

Coefficient of performance for any refrigeration cycle is

(a) Given Qc =1000 kl, WCYC l e = 80 kJ, the coefficient of performance determined using these

-.energy data is

jJ ~ 1000 kJ= 12.580k]

Since 8=12.5> 8 m B ! = 6.875 , the cycle is impossible. oE",~---

(b) Given Qc =1200 kJ, Q H =2000 kl, cycle work can. be determined from

Wcy c le :;; QH ~ Qc =2000 kJ - 1200 kJ=800 kJ

The coefficient ofperfonnance determined using these energy data is

p=1200kJ = 1.5

SOOkJ

Since f J ; ; : : : 1 . 5 <Bmax= 6.875 , the cycle operates irreve.rsibly. . -E<;"_---

(c) Given Q I f =1575 kJ, W c y c l e =200 kl.heat transferassociated with the cold reservoir can be

determined from .

W c y c l e =QH - Qc ---j. Qc = QH - Wcyc l e =1 5 7 5 kJ- 200 k1= 1 3 7 5 kJ

The coefficient of performance determined using these energy data is

/3_1375 kJ= 6.875200k]

Since p = Dmar . =6.875 , tbe cycle operates reversibly. . . . . , < , - . _

(d) Given p =6, the cycle is irreversible.

Since B =6 <D.ma! = 6.875, the cycle operates irreversibly.. """".-----.



I P t< .oaL. EM S.G:./t

! 1 - : - ' . . . . . -!

. As shown in, Fig P5.5.9,an air conditioner operating at

steady state maintains a dWelling at 70ap all a day when the

outside temperature is 90°F. If the rate of heat transfer into

the dwelling. through the walls and roof is 30,000 Btulh,

might a net power input to the air cOlrditionercompressor

of 3 hp be sufficie,nt? Ifyes, determine the coefficient@f per-

formance. If no, determine the minimum theoretical powerin pu t, in hp,

E'I\.)'~ ..m 0OEL. :[. ~ d...., """',( ':ne_ d. ~ f l . : VU."S' 1 :- h.o..

. ,;u ,. .. . ~ .~ . . Ii.o'f.'~'~ (~frl~e(""'_h'lNj fystt .......

2,. T~ f't de" lAo, ~ ..... t- ~f-C4ay ST-",-te.

'S.• Tk.A.I',,_ s . . ~ !?- ..u :i '" ....i-r.d.q_. Q.., ... P '7'"t . .

("Qh~., f'+4 'CG ,d. a.....o J h . c 1 t- rcj1e ....'-O .'·... 1....

ft:r I' C : c : . . r. 've 1 7 .

3 0 ,0 0 0 B [ ,l \I 'h Outside, 90· F

k::.._ Tjoj.=:5Sb0R _ .

-~jCondenser

F ig . P S .5 9 I

At- - l f iLYS( s : T o ~~,,-J ,..."k ~ d<.l . .>cl l .~. . . . : t 'ro < t I F ; _ 1'i- .::\,,...o"'d.,·h"." ..:.... Sys k ""--

~s-t 1"4 ............Q . ." 'Q , .. . . : l - ; - . . . ~ h . . . . : i . . . . _ i . ;t .. -- .. .. . . . ~ , .. ._ . o o .. . :: t" ' < > . . . . .N-b,_ ~~o.ooo \3h...h, -Tk...u, -dl!t,~ ;s. 0, OGlO e o h -j . 4 .

Atso.J

.Q,. , .t - 2(",5' ~".)W e.

>-

r-t I

""_T ,.".

i

II ~__L__

! -

I r In _ J



p ~() ISLE 1 '1 5. r e s

5.65 The refrigerator shown in Fig. P5.65 operates at steady state with a coefficient of

performance -of 5.0 within a kitchen at 23°C. The refrigerator rejects 4.8 kWby heat

transfer to its surroundings from metal coils located on its exterior. Determine

(a) the power input, in kW.

(b) the lowest.theoretical temperature inside the refrigerator, in K.

Refrigeratorp =

Fig. P5.65

KNOWN: Refrigeration system operates at known conditions with a known heat transfer

rate to surroundings.

FIND: Determine the power input, in kW and the lowest theoretical temperature inside

the refrigerator, inK.


R frigeration Wcycle

cycle

p~5.0

I

_ I . .. ._ - _ - _ _ - _ -_ - _ - t- -- _ -_ - _ - -' _ • J



ENGINEERING MODEL:

(1) The system shown on the accompanying diagram undergoes a refrigeration cycle in

steady state.

(2) The kitchen and refrigerator interior act as the hot and cold reservoirs, respectively.

ANALYSIS:

(aj.Determine the power input, in kW, where flo represents the energy into the

circulating refrigerant from the food and other contents within the refrigerator. With

E q. 5 .5 on a time rate basis:

f3 = ~c = . Q c.W cycle Q H - Qc

Rearrange to solve for Q c .

(1)

Substitute into Eq, (1) and rearrange to solve for power required for the cycle, inkW.

® #- l W = Q c = 4kW = O.8.kW• cycle f 3 5 -

(b) To determine the minimum theoretical temperature inside the refrigerator, begin by

noting that the actual coefficient ofperformance is less than or equal to the maximum

theoretical coefficient of performance:

f 3 >(,1=50max - p .

Then with Eg.5.10

Te ~ 5.0T H -Te

t; ~ ( % ) T H = (%)296K

t; ~246.7K

The minimum theoretical temperature is 246.7 K.

1. Alternatively, the net power for the cycle can also be found from an energy rate

balance: Q,H :. Q c . . • w~'t.cte. = t> \ i . J c1de ~ + . 9 K it v ' - 4 "\V = - O. g 1 0 1 -W.



PROBLEMS'ft-

5.77 A quantity of water within a piston-cylinder assembly executes a Carnot power cycle.

During isothermal expansion, th e water is heated from saturated liquid at 50 bar until it is a

saturated vapor. The vapor then expands adiabatically to a pressure of 5 bar while doing 364.31

kJ/kg of work.

(a ) Sketch th e cycle onp-v coordinates.·.(b) Evaluate the heat transfer per unit mass and work perunit mass fo r each process, in kJlkg .

.. (c) Evaluate the thermal ef f ic iency.

KNOWN: Water executes a closed system Carnot power cycle between two known pressures

and with specified work produced d urin g o ne process .

•" 1ND : S k etc h th e cycle o np -v c oo rd ina tes, d eterm ine beat and w o rk during e ac h p ro c e ss , andevaluate thermal ef f ic iency.


Carnot

Power

Cycle

50 1

3 5 151.9

4 5 151.9

ENGINEERING MODEL:1. The closed system defined by the dashed line on the accompanying diagram undergoes. a

Camet power cycle.2.. Kinetic and potential energy effects Can. be neglected.

ANALYSIS:

(a) The p-v diagram is shown. below. Process 1-2 is isothermal expansion from State 1,.a

saturated liquid, to State 2, a s a tu ra t ed vapor, both at 50 bar. In the saturated mixture region

temperature and pressure are dependent and thus constant during process 1-2. Temperature is



the saturation temperature at 5 0 bar.. Specific volume increases during process 1-2 expansion.

Process 2-3 is adiabatic expansion inwhich specific volume continues to increase while pressure

drops to 5 bar. Process 3-4 is isothermal compression during which specific volume decreases.

Since the water is still in the saturated mixture region, pressure and temperature are both

constant. The temperature is the saturation temperature at 5 bar. Process 4-1 is adiabatic

compression, during wh i c h specific volume continues to decrease while pressure increases back

to 5 0 bar.

p (bar)

50

5

p-v Diagram

v (JulJkg)

To determine the energy transfers during each process, the .energy balance for a closed system

and the expansion/compression work equation apply.

Process 1~2asothermal Expansion)State 1 is saturated liquid atpl = 50 bar, From Table A-3:

Vj =vn =0 . 0012859 m3/kg,

Ul = _ Un=1147.8 kJ/kg

State 2 is saturated vapor atp2 = 50 bar. From Table A-3:

V2 =v g 2 = 0.03944 m3/kg,

U 2 =U g2 =2597 J kJ/kg

Applying Eq. 2.17 for the constant-pressure expansion

Substi tuting values and applying appropriate conversion factors give



w : { 3 3 ) 10

5~ I kJ I

12 = (50 bar 0.03944 m. - 0.0012859 ill . ill 3 = 190.77 kJIkgm kg kg bar 10 N· m

..The positive sign associated with work indicates work is from the system.

Applying the closed system energy balance (neglecting kinetic and potential energy effects):

Solving for heat transfer per unit mass gives

(Qu/m) = (W12lm) + U2 - UL

Substituting values gives

(Qn/m) = 190.77 kJ/kg + 2597.1 kJ/kg -1147.8 kljkg =1640.07 kJ/kg

The positive sign associated with energy transfer by heat indicates heat is into the. system.

Process 2-3 (Adiabatic Expansion)

(Q6im) :::;; kJ/kg (adiabatic process)

(Wallm) = 364.31 kJfkgJgiven)

To fix State 3, a property in addition to pressure is required. Applyingthe closed system energy

balance (neglecting kinetic and potential energy effects):

Solving for internal energy at State 3 gives:

U3 = = (Q23lm) - ( W 1 . 3 lm ) + U2

U3 =0 kl/kg - 364.31 kJ/kg + 2597.1 kJ/kg =2232 ..8 kl kg

State 3 is a two-phase. mixture atp3 =5 bar. From Table A-3:

Ut3 = 639.68 kJ/kg

ug3 = 2561.2 kJ/kg

U3 = 2232.8 kJ/kg (computed from energy balance for process 2-3)

Vt3 = 0.0010926 m3/kg

.. 3Vg3 =0.3749 ill/kg

ho =640.23 kJlkg

hfg3 =2108.5 kJ/kg

Quality is determined from the relationship



U-Ufx,=-----==-

», -U[

Solving for quali ty at State 3 gives

kJ "k J2232.8-"', -639.68-

X3 = kg' kg - 0.8291

2561.2kJ _ 639.68 kJkg kg

Specific volume at State 3 can be determined from the quality relation

V3:::= 0.0010926 m3/k g + (0.8291)(0.3749 m 3/kg- 0.0010926 m 3

/kg) = 0.3110 m 3/kg

Specific enthalpy at State 3 is convenient later in the analysis, It can be determined from the

quality relation

h 3 =640.23 kl/kg + (0.8291)(2108.5 kl/kg) >2388.39 kJikg

Process 3~4(Isothermal Compression)

The heat transfer during Process 3-4 is the heat transfer of the cycle associated with the cold

reservoir while heat transfer during Process 1-2 is the heat transfer associated with the hot

reservoir. Since the cycle is a Camet cycle, and thus reversible, the following relationship

introduced in Sec. 5.8.1 is applicable:

Expressing heat transfer on a per w r i t mass basis and solving for (Q c1m ) yield

Qc =( Q u ) ( T c ) = ( 1 640.07k J , ' ) / 424.9 K ) =1297..70 kJ/kgm lm ' TH., kg~537.0K

Here Qc is a magni tude. Heat transfer during process 3-4 is from the system to the cold reservoir

and thus is out of the system. Consequently, {~ml = -1297.70 .kJIkg.

To determine the work per unit IDCj .SS during process 3 . .,4 . app ly the closed system energy balance

(neglecting kinetic and potential energy effects):



P~o13.LEMS. T':/- C0",- h ' ' - ' . t.t ed (5)

Applying Eq. 2..17 for the constant-pressure compression

W 34 =P _( v ~ v )3 4 3

m(1)

Substituting for work per unit mass in the energyequation gives

Rearranging terms gives

0= CQ3Jm) + (U3 + P3 V3) - (U4 + P4 V4)

eSubstituting enthalpy, h =u +pv, yields

Solving for the exit enthalpy, h4, gives

Substituting values yields

h4 = (-1 29 7.7 0. k J/k g) + 2 3& 8 .39 k Jlk g = 10.90. .69kJ lkg

State 4 is a two-phase mixture at P4 = 5 bar. . From Table A-3:

hf4 = 640 .. 23 k J /kg

hfg4 =210.8.5 kJ/kg

h4 = 1090 . . .69 kJ/kg (computed from energy balance for process 3-4). ·3

V r 4 =0. .0 .0 .10.926m /k gc 3 .

Vg4 =0. .3749 ill /kg

Ut~=639.68 kJ/kg

ug4 = = 2 5 61 .2 k .I /k g

State 4 now can.be fixed by h4 and P4 . Quality at State 4 is determined from the relationship

S o lv ing fo r quality a t S ta te 4 gives



1090.69 kJ_ 640.23 kJ

kg . kg= 0.2136

2108.5kJ

kg

Specific volume at Sta te 4 can be determined from the quality relation

V4= 0.0010926 mJ/kg + (0..2136)(0.3749 m3/kg - 0.0010926 m?/kg) =0.Q8094 m3/kg

S ubstitu tin g v alue s in to E q , (1) and app lying app ro pria te co nversio n fac to rs to so lve fo r w o rk per

unit mass give

W .

34

= (5 bar {.O.08094ill3. _

0.3110ill

3

J 10' ;Z I3k1

I = -115.03 k,J/kgm '\ kg kg. bar 10·N·m .

The negative sign associated with work indicates work is to the system.

Process 4-1. (Adiabatic Co'mpress.iim)

(Q,n: im) ;;::·0kJ/kg(adiaba.tic process)

To determine the work per writ mass during process 4-1, apply the closed system energy balance

(neglecting kineticand potential energy effects):

Substituting (Q4J im) = 0 ' and solving for work give

Applying the quality relation to determine U4g ives

U4 = = 639.68 kJ/kg + (0.2136)(2561.2 kJ /kg- 639..68 kl/kg) = 10 ' 50 ' . 12 kJlkg

Substituting values and solving for work per unit mass give

(W41/m)= 1050.12 kJ/kg - 1147..8 kJ/kg =-97.68 kJlkg

The negative sign associated with work indicates work is to the system.

Summarizing energy transfers



Process Qlm IkJ/kg] Wlm [kJ/kg]

1-2 1640..07 190.77

2-3 0 364.31

3-4 -1297.70 -115.034-1 0 -97.68

Sum 342.37 342.37

As expected for a cycle, the net heat transfer is equal to the net work.

(c) By definition, power cycle thermal efficiency is

(

Wcycle J . kJW -. - 342.37-

'1 7 = .cycle = m = kg = 0.2088

Q H ( Q .Ii) 1640.07 kJm kg

Alternatively, since the cycle is a Carnotcycle, the thermal efficiency can be determined from

the maximum thermal efficiency equation

. . = 1- Te = 1- 424.9 K- 0.2088'l7max l'a· 537.0K

ethe introduction of enthalpy here is only for computational convenience using table data.



j' •

_ ~ ~ , O ' p . . - . ; ~ ~ D"o\ "a.l"'K1~ 0.1; ~ te",,"oIla;~+~ e ",Lore'S'

~c:;( : .k .CI f -t~.-fa'v.r." c::.o~ p~,(1e .. t£

~2'. p ! W ~ Y '~ " '~ _ ~ I ! A + ~ " ' ; S f . . . . - ro.:fN ~r¢ p ; ; ; S , ' ' ' ' ' " r : :

~ iN. d .4 tc:c: f \ .""" !S f t'''''-4.'4('r 0 ! . o 4 I ' S ' .

,3 S'\'v~)' h e A - + : . n - - A t l ' ~ ".",-~~,.",'l""C.. ~r:I p6rt ... , , · 1 - f ,

,- ~ io Io« r:s 'j, -f fec :r t ~" " bel '~L<'4),.eo:l

IPRoBLEM~ 5.9'- - 5.88[ - - - - ~ ~ - - - - - - - - - - ~ - - - - ~ .

r '5.~6 Fi~ure P5.86 giv:s t~e schematic of a vapor power plant IIII which water steatl tly cIrculates through the fow- COmponents

shown. The waterflows through the boiler and con-denser at

cu?stan_ t . press~Fe and through the turbine and pump

adwtiatlcaUy._Kmetic ano potential energy effects can b~ignored. Process data lollow:

Precess 4--1: con.staln-pressure at 1 MPa from saturated liquidto saturated vapor

Process 2-:-3. constant-pressure at 3DkPa from X2' ='88% to

);;3 = l~%

~:!)Using Eq. 5.13 (:,~pressed on a time-rate basis determine

If the cycle is i r : . _ t e . r n a ~ Y reversib le , irrev!!!5:ijJle ,9r~i!!1p_o~'stble.

(b) D~lerl).1ine·thetherlt;al efficiency using Eq, 5.4 expressedon a Ilme-rale baSIS and steam table data.

(0)Compare ~!rere~UI1-,"0fpa:t(b) with the Carnoi efficiency

calculated usmg Eq .. ;).9 WI t h the boiler and condensertemperatures and comment. .

- ,

.QiL'$_ Q&.,

1i.3 T + J

L-__

w ,- 1

f

- ~t) ,

J



IPROBLEM t $8'- $',,9a ( (011""°, , , ~~J)

1-- ¥ . ; ~ , ) - ; - M . : _ . ~ .W c : r c.1 e - l.~. I (Q lV \

- ,1 .1. I

i, I I 1~r_,~r "L:1-

I) -I

.Oto~'~

( g " , , 1 o < \ )

(2.)

.. l ~ ..._

(O~8-0d8)(,,).S"-8•; 3 _ )~

~() lS.·.~

hI

T! I i

. _ , C , , )_ \ ~ "&1'

T 4 o ; 1

T

c o 1- 1,J'S.ZI_ O.LC.~-

+t3.CS'"

5.87 Repeat.Problem 5.86 for the following case:

Precess +,1: constant-pressureat 8 JVIPa from saturated l iquidt o s at ur at ed vapor

PtOCCss 2<-3: constant-pre,ssure at 8 kPa from - 1 : 2 = 675% to

X3 = 34.2%

'-T:"c~t:~~l1s 0 * ('Q) a e . .. d ee) L a _ ~ ! ~ - e ~ _ ~ _ I ; ~ " " : rov.l1d~~.~ T'1 .e C J·cI~ ....,. . . . . "'- . -. . . .- "' . ' -

......__-_~ __-

5.8S Repeat Problem 5.86 for the fo l lowing Case:

PtOl:e.sS 4,..,1: eo ns ta n t- p rs ss ur e a t QJj M P'l. IT em saturated liq uid

to saturated vapor

Pr()ces~ l--J: constant-pressure at 20 kPa from X2 = 90% to

x~= lQ%

IL__ I~

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Solution to Hw07