Thermodynamics I Section 1 Test #3 Saturday 28August 2010 Time 1 hr 30 min.'" N / /1I:I'II't1J;Hm. • • • • • • • • ame Somnu:n Pnprem .Examiner: Assoc.ProfSommai Priprem, PhD.

1.An inventor claims to have developed a heat engine that receives 1,500 kJ/s ofheat from a source at 600 K and produces 900 kJ/s of net work while rejecting thewaste heat to a sink at 300 K. Is this a reasonable claim? Why? (5 Marks)

System: the claimed heat engine

Analysis: Use the Carnot principle

at the same operating conditions

High-temp. reservoir, TH= 600K

17th < 17carnot

•'\.NI\I\..tt- X 1" 0 0/6

•QH

qoo kW )( ~oo1o1,500 kW

The Carnot Principle: llirr < 1JirN

~<;i&£;Y" 0.. Cax'V\.ot ~cJ- &.AI\(J'IY\.Q 0~ cd- -feu .J.cvrne..Cl>

-rH ITL~c:! QH

300K1- -_--60{) K

C•.so

=

~-fI.-, > ~rl1V. ~ch. VI'b(atLS ~~rM+ ('J'<l''l!tCir'e-

T~, ~'j c\cJ~ tS vn res~ee

Thermodynamics 1Section I Test #3 Saturday 28 August 2010 Time 1 hr 30 min."" N / /m'llYltYtl1J. • • •• . • • • ame Sommt:n Pnprtm .Examiner: Assoc.ProfSommai Priprem, PhD.

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2) An air conditioning system is used to maintain a house at a constant temperatureof 26°C. The house is gaining heat from outdoors at a rate of 20,000 kJ/h, and theheat generated in the house from the people, lights, and appliances amounts to8,000 kJ/h. For a COP of 2.5, determine the required power input to this airconditioning system. (5 Marks)

, 0 e

Q = Q . + QtlDAd._f7.A1.- ~1Y\ed ~.J''''-'-......v

System: Air Conditioner: a heat pumpAnalysis: "0

COPR = Qd Wnet

Solution:

W,n=?kW

~~.ooo k2 -t- x.k 3bOQj)7~

t!J) /~

Wl\I\Lt =- Q\..--COp=

3.11 kW

Thermodynamics I Section I Test #3 Saturday 28 August 2010 Time 1 hr 30 min,"" N ' ,1il'll'nfltllJ. . . •.. . .. ame Sommllt Pnprtm .

Examiner: Assoc.ProfSommai Priprem, PhD.

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3. A piston-cylinder device contains refrigerant-R-134a initially as saturated vaporat 400 kPa, 0.5m3

• Heat is transferred from the refrigerant to the surroundings at35°C until all of the refrigerant becomes saturated liquid. If the process occurred atconstant pressure determine (a) the entropy change of the refrigerant, (b) theentropy change of the heat source, and (c) the total entropy change for this process.(d) does this process violates the principle of increase of entropy,why. (10 marks)

System: R-134a and the heat sourceAnalysis:

(a)(b)(e)(d)

Solution:t>~1Y ~ ~-1~4:-C\..-'7 Tctble A-1Sb

s+ctte "'\ .90.,+ vaf. o:t ;( ~?t:\

'&'\ :; \9-.5 = D.OObG n?/~.;

\'n -= V1 = o. Si W\ _ \;f l} o4--b ~S'\9-,\ D .OObQ ..;?l(cj

£.~-te-2. gC\,t, ~'~'d. £At- ~ = ~'\ == ~.o H PP\

n..l. = ~ = 1b<l.1'2. ~; /01.:>..,. At- -= O.5C€'.} k;J/~ K .

a) Gry\ltm~e-h~ ~ f(-\~'tC\

ti ~ - ty\/\ (fool. -h,,) = ~2.4-b kc;(o.5bR~-O.&~5~) !s28li~ - ...J ~ tc..

_ 2? .q4-ct3 ~J Ik ~\Aley-

LlSsystem = mts, - s1)LlSsource = Qc/To = -1Q21To

LlStotal = LlSsystem + LlSsource

LJStotal 2: 0

1t 1

R-J34a 2

· . . . . . . . . .

:::-sat v~~o~.:.:~JY''"2.·ijJ>d..·:· 1Q2· :Vj.:O 0.:5 m:3.:.:.

· .

bJ (;0'\m,fj ~ ~ ~ JutmtsncU~.rCl>V\lYP\ 'V\IlU~S R -H'tC\. \st- LAw .. Ctc!>sad. 9t{~\-e-''Y)-}p= CcMSI.

lQ.;l ~ 'VY\ ( ~.,1 +LLl~-LL,\ ) ") ~ 0""\ (ha. - h,\")= (-=t?..lt-b t:.3)C tb&.1!L--~~q.1~)!5J. = ,.-R,Oq.7.1 \:::J

~~S~~S :::: -(Pt;).) -= - C-%,04-~.llc.J) _ 6lb.1~.(::J A(vLsTo (?S~~'73 ') t: - K

c) ~ <PTotcUL =- (-.2.~.q~) +(~6,1~) = ~.1& .\:J Ik: ~S~

cl) 11u's ~~ cL0es 'not ~lJ lct& tk p,",·t}'\ciple.bf~'Y\cv-ease ~-.0Vt1n>p"1 Ce.UM.v\e A~To~ >- 0 ~ r:

Therr;t0dynamics [Section l Test #3 Saturday 28 August 20[0 Time I hr 30 min"" N ~''1''> ;!~'II'nffOU.. . . . . . . . ame ..:::>{)I1A..JtZ.U::nrnprtl1A.. ••••••••••••

Examiner: Assoc.Prof.Sommai Priprem, PhD.

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4. Air enters an adiabatic turbine at 800°C and 1.5 MPa at a rate of 1.2 kg/s andexhausts at 200 kPa. If the power output of the turbine is 370 kW. Assume air is anideal gas and uses constant specific heat at 900K, determine the isentropicefficiency of the turbine. (10 marks)System: Air, SSSF process and adiabaticAssumptions: Air is an ideal gas

constant specific heatneglect ~KE and ~PE

Analysis:'7isentropic = wt! w,

eW 3::JO-kU)QC\Uc&. WD'fk.) t}AS'~ = eo _ •

yY) 1.::2.k~Vs_ 30g, 2> \<:J/",~ <l

Solution:

T1'1 \ Actual

\ process

::::::iC \ ,

fi'1l\.d l's~p'c.. WDdc ) ws0vtst'd.erf- k0Y\k-ropic ~rt>dLS) CD -=)~

\':2.S T,- =(~\~/~ (~-I)/~

( Pg.) (?, ')

~~ : T1(~r)1k

p; j2S;

s

:::: ~OO +~~)~-::LO-:r~K. .~

2..0l> ~fOl _ 0.1'3'33\500~.fa..

T(;\ple A - c;;Lb C1.t~OOK

d.~ :::.1. t:2.1 ~J/~~Kt\2. -= 1.3 4-'t

•....~ - 0.'2500k -

D. .7.5bC,/c -r~A- _ ltD'1-2> K)( O.1?,??:» - b4-D. "+ K ..cd

c», Tovb\~ tST-law./ ~SS~ (-feY" tSe'Ylk¥o?i( ~cess)o

~ + h" - <tA\ + h~S~S - ~'\ - hJS ~ CfCTIl-T~S)

- ~,1<)"1 ~/~4c)( {O;'~ -b411,<7-)K ~ ~4-.1- k'J!.k5.o=:::J

I~'e ~'G\\eN\~) ~ =-( i~~Y)

~)(I()O·I. == 3~<6.3J(\OO/() ~63JD2%W~ 4{?4."::f htswer