thermodynamics review problems

7/28/2019 Thermodynamics Review Problems

1/48

7TH ERMODYNAMICS

THERMO 1

Equa 't i ons o f s t a t e fo r a s ingle component can be an y o f the fol lowing, excep t :

(a ) th e i dea l gas law, Pv - RT.(b) the i dea l ga s la w modif ied by i nse r t i on o f a

co mp ress ib i l i ty fac to r, Pv = ZRT.(c ) any re la t ionsh ip i n t e r r e l a t i n g 3 or more

s t a t e fu n c t io n s .(d ) a mathematical express ion d ef in in g a path

between s t a t e s .(e ) r e l a t i onsh ips mathematica l ly i n t e r r e l a t i n g

thermodynamic p r o p e r t i e s o f th e mate r i a l .

All except (d ) are co rrec t . The id ea l ga s la w i s the s implest equat ion o fs t a t e ; i t i s o f ten appl ied to r e a l gases by us ing a co mp ress ib i l i ty f ac to r Z.Any r e l a t i o n s h i p s tha t i n t e r r e l a t e thermodynamic s t a t e function data ar eequations of s t a t e . Item (d ) expresses the path o f a process between s t a t e s

r a t h e r than a r e l a t i onsh ip between va r i ab l e s a t a single p o in t o r s t a t e .Answer i s ( d ) .

T HE RM O 2

The s t a t e o f a thermodynamic(a )(b )(c )(d )(e )

system i s always defined by i t sabsolu te tempera tureprocessprope r t i e stemperatu re an d p ressu rea v a i l a b i l i t y

Ava i l ab i l i t y i s th e amount o f work t h a t e x i s t s above completeequil ibr ium with ambient cond i t i ons . S t a t e i s always defined by i t sproper t i es . The number of prope r t i e s r eq u i r ed i s determined by theGibbs phase ru l e P + F = C + 2, where P = no . o f phases , F = degreeso f freedom or number of p rope r t i e s r eq u i r ed , an d C = no. ofd i s t i ngu i shab le components presen t .

Answer i s ( c ) .

131


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tH

138 / Thermodynamics

THERMO 3

s -

THERMO 4

On th e M ol l i e r diagram fo r stearn,which o f th e numbered l i n e s r ep re sen t sa l ine o f constan t p r es su re?

(a ) Line 1(b ) Line 2(c ) Line 3(d ) Line 4(e ) Line 5

(a ) c ons ta n t mois ture 100 - x% o rqua l i t y x%. (Line 1)

(b ) cons tant tempera ture , o f . (Line 2)

(c ) cons tant s upe rhe a t , o f . (Line 3)

(d ) constan t pressure, p s i a . (Line 4)(e ) cons tant entropy, BTU/lbOR.

(Line 5)

Horizonta l l i ne s a re constan t entha lpyor t o t a l heat co n ten t .

Answer i s (d ) . .

On the Moll ie r diagram fo r steam (above) which l e t t e r e d p o in t r ep re sen t s thec r i t i c a l poin t a t 3206 p s i a an d 705 e F?

The fo l lowing data ca n be read fo r these p o in ts on a large M ol l i e r diagram:Temp p H S

o f ps i a Remarks BTU/lb BTU/lboFa 705 3206 c r i t i c a l p o in t 910 1.065b 162 5 22% mois ture 910 1.490c 141 3 10% mois ture 1010 1.76fJd 440 375 dry sa tu ra t ed 0 1205 1.490e 780 200 superheated 40 0 1413 1.760

Answer i s (a ) ~

THERMO 5

Ma.thematical ly, a thermodynamic proper ty i s which of the following?(a ) a poin t funct ion(b ) a path funct ion(c ) an i ne xa c t d i f f e r e n t i a l(d ) discont inuous(e ) an e xa c t d i f f e r e n t i a l

Thermodynamic p r o p e r t i e s , such as P, v, T, U, H, and S, a re p o in t f u n c t io n s ,t h e i r de r iva t ive s a re e xa c t d i f f e r e n t i a l s . Heat , Q, and work, W, are pathfunct ions ex p ress ib le as areas t h a t are dependent upon path between p o in ts( s t a t e s ) , t h e i r de r iva t ive s are inexact d i f f e r e n t i a l s .

Answer i s ( a ) .


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Answer i s (c )

THERMO 10

For a ty p ica l r e f r i g e r a n t , draw a pre s s u re -e n tha lpy diagram which shows th eboundar ies of so l id , l iqu id , an d vapor s t a t e s . Iden t i fy on th e diagram th efollowing: (a ) superheated vapor zone

(b ) c r i t i c a l p ressu re(c ) compressed l i qu id zone(d ) sa tu ra t ed l i qu id l in e

I

THERMO 11

C'f"it\C:a.l pOint

COMPR.C.-5$eO i . . 1 ~ U . \ D

z o n e .

A c y l in d er f i t t e d with a w eig h t les s , f r i c t i o n l e s s p is to n conta ins m pounds o fa i r a t temperature T I, volume V I' an d ambient pre s s u re Pa. Heat i s added u n t i l

th e a i r in th e cy l in d er ha s a temperatu re T2 , a volume V2 , and ambient pre s s u re

Pa . The spec i f i c heat o f a i r a t cons tant pressure i s Cp , and th e s p e c i f i c heat

o f a i r a t cons tant volume i s Cv The heat t r ansfe r red during th e process i s :

(a ) mCp (T 2 - TI)

(b ) mCv (T2 - TI) - Pa (V 2 - VI )(c ) mC p (T 2 - TI) + Pa (V 2 - VI )(d ) mCv(T2 - T I) + Pa (V2 - VI )(e ) mCv (T2 - T I )~


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Thermodynamics / 141

Consider one o r more processes from PaV l Tl to Pa V2T2in a closed system and

apply th e f i r s t law o f thermodynami c s: A U " Q - W.

For an i d e a l ga s ( a i r ) : [).U = mCv [).T

Work done by a closed system a t co n s tan t p r es su re :

Combine th ese :

Therefore

Dimens ional i ty check:Compatibl e uni t s would be

V 2

W = JPdV = Pa (V 2 - VI)VI

Answer i s (d )

BTU = ( tb -wele . ) (0 BTU ) ( o n - + y ~( ft--3) ( BTUl'bZ: fJ>-J..-:.::r-~ + 7 7 8 " ' - -

THERMO 12 .

A nonflow (c losed) system conta ins 1 lb of an i d e a l gas (C p .. 0.24, Cv = 0 .1 7 ) .The ga s temperature i s increased by 10F while S BTU of work are done by th eg as . What i s heat t r a n s f e r in BTU?

(a ) -3 . 3

(b ) -2 .6(c ) +6.7(d ) +7.4(e ) none of these

The thermodynamic s ign convention i s + for heat in an d + fo r work out of asystem. Apply th e f i r s t la w for a closed system an d an i dea l ga s workingf l u i d :

AU = mC AT = 0 - \.\fv .0.17(10) - Q - (+S)

THERMO 13

1 .7 =Q - S/

Answer i s (c )

Q = 6.7

Sh af t work of -IS BTU/lb and heat t r a n s f e r of -1 0 BTU/ lb change entha lpy o f asystem by

(a ) -2S BTU/lb(b ) -IS BTU/lb(c ) -1 0 BTU/lb(d ) -S BTU/lb(e ) +S BTU/lb

The f i r s t la w appl ied to a flow system i s :

H = Q - \.\fs = -1 0 - ( - IS) '"' + S .

Answer i s (e )

THERMO 14

The f i r s t law of thermodynamics s t a t e s t h a t :(a ) heat energy cannot be completely t ransformed i n to work.

(b ) i n t e r n a l energy i s du e to molecular motion .(c ) h ea t ca n only be t r ansfe r red from a body of higher te rmpera ture to on e

of lower tempera ture .(d ) energy ca n be ne i the r created nor d es t ro y ed .

(e ) entropy of th e universe i s increased by i r r e v e r s i b l e processes .


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Use the thermodynamic s ign convention t h a t heat in an d work out are p o s i t i v e .

F i r s t law energy balance fo r th e f low system:

h2 + KE2 - hI - KEI - Q - Ws

Since th e working f lu id i s unspecif ied an d i n t e r n a l energy change i s des i red ,

us e th e d e f i n i t i o n : h - u + Pv .

Or

Calcu la t e numerica l va lues for a l l terms except u2 - u l .

P 20(144)(15) = 55.S BTU/lb2v 2 = 778 P 100(144)(4 .0)

a 74.0 BTU/lblV l .. 778

V2 (1000)2KE2 = 2g J z (64.4)(778) - 20.0 BTU/lb

V2 (500)2KEI - 2g J - (64.4)(778) 5 .0 BTU/lb

Ws195,000 f t - l b f

IbmBTU _ + 250.6 BTU/lb

778 f t - l b f

Therefore,

u 2 - ul~ -1 0 -(+250.6) + 74.0 + 5 .0 - 55.5 - 20.0 = -257.1 BTU/lb t t

decrease

T HE RM O 1 7

An engine developing 30 brake horsepower must be cooled by a rad ia to r throughwhich water i s pumped from the engine . Using the fol lowing fac to rs :

Heat t rans fe r red from engine to cooling water, expressed as horsepower,equals 40 HP.

Water tempera tures in rad ia to r :Top water = 200F.Bottom water = 190F.

Weight o f 1 gallon of water = 8.33 lb s .Calcu la t e th e required ra te o f flow of water in gal lons p er minute.

Basis of ca lcu la t ion : 1 minute.. 2545 BTU 1 h r

Heat loss to cooling water - 40 HP(HP-hour ) (60 min) - 1700 BTU/min

Heat loss m heat gain a t s teady s t a t e .

Heat gain by cooling water = 1700 BTU/min = mCpAT, where ~ i s water massflow r a t e in lb /min.

1700~ - ~ ( ~ 1 ~ . ~ 0~) ~( ~ 1 ~ 0 ~ )- 170 lb . water/min.

170 lb . galVolume flow r a t e = x - - 20.4 ga l /min. ttmin. 8.33 lb .


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144 I Thermodynamics

THERMO 1 8

Sa tu ra t e d steam i s suppl ied to a he a t exchanger. 500 gal lons per minute o fwater a re being heated from 6 i f t o 140F in the exchanger. Condensate i s beingdischarged to a r ece iv e r which i s vented to th e atmosphere.

A. I f th e en te r in g steam i s suppl ied to the he a t exchanger a t a p ressu re of120 p s i a , an d i s condens ing a t t h i s pressure:

(1 ) How many pounds pe r hour o f steam are used?

(2 ) What pe rc e n t o f condensate i s l o s t by f l a s h ing?

B. I f th e pre s s u re o f the steam suppl ied i s reduced by a pre s s u re reducingv


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Ther07odyna07ics / 145

A. (2 ) Percent condensate los t by 120 p s i a f l a sh ing (Basis : 1 l b . )

Mass balance: X - Weight f rac t ion f lashed o f f .

1 lb . s a t . l i qu id a t 120 ps ia R (X)lb. s a t . vapor a t 14.7 ps ia+ (1 - X) lb . s a t . l i qu id a t 14.7 ps i a

Heat balance:

l ( hf

a t 120 ps i a ) = X(hg

a t 14.7 ps i a ) + (1 - X)(hf

a t 14.7 psia)

1(312.5) = X(1150.4) + (1 - X)(lSO. l)312.5 = 1150.4X + lSO. l - lSO.lX

X = 0.1365 Therefore 13.657. o f condensate l o s t by f lash ing 4tB. Thro t t l e steam to 30 ps i a (a constan t en thalpy process)

During t h r o t t l i n g : h in - houtSatu ra ted steam a t 120 ps ia (h g 1190.6) becomes superheated steam a t

30 ps ia (h 1190.6).g

From sa tu ra t ed steam t ab l e a t 30 pSia , h f~ 21S.S BTU/lb.

Heat l ibe ra ted from steam: 1190.6 - 21S.8 = 971.S BTU/lb.

(3 ) Calcu la te steam flow ra te (Bas is : 1 hour)Use heat ba lance as i n (1 ) above: heat to wa te r z heat from steam

~ l C p ( T 2- T l ) = 2 ( 9 7 1 . 8 )

= 250,000(1)(140 - 60) 20,600 Ib . / hour . .m2 971.S . .

(4 ) Percent condensate l o s t by 30 ps i a f l a sh ing (Basis : 1 lb . )

Mass balance: Y = Weight f rac t ion f lashed o f f1 l b . s a t . l iquid a t 30 ps i a - (Y)lb . s a t . vapor a t 14.7 ps ia

+ (1 - Y) lb . s a t . l i qu id a t 14.7 ps ia

Heat balance:

l (hf

a t 30 psia)=

Y(hg a t 14.7 ps i a ) + (1 - Y) (hf

a t 14.7 p s ia )

1 (21S.8) = Y(1150.4) + (1 - Y)(180.1)Y = 0.0399 Therefore 3.99% of condensate l o s t by f lash ing 4t

C. Proposal Comparison

120 p s i a 30 ps iacondensa t ion condensa t ionan d f l a sh ing an d f lashing

Steam r a t e , Ib . /hour 22,800. 20,600.

Steam an d Condensate3,120. 820.l o s t , Ib /hour

Condensate r e tu rn ed , 19,680. 19,780.

Ib/hour

S u p e r f i c i a l l y, i t i s apparent t h a t th e 30 ps ia case loses much l e ss he a t tothe environment by f l a sh ing . Limitat ion on system o p e r a b i l i t y i s imposedby tempera ture on th e steam s ide of the heat exchanger. Temperature of120 ps i a sa tu ra t ed steam i s 341F, an d superheated steam a t 30 ps i a(h = 1190.6) i s determined from superheated steam t a b l e s to be 303F. Thelower tempera ture requires more condenser area .


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THERMO 2 0

The mass flow r a t e o f Freon 12 through a heat exchanger i s 10 pounds/minute.Enthalpy of Freon en t ry i s 102 BTU/lb. and o f Freon e x i t i s 26 BTU/lb .Water coolan t i s al lowed to r i s e 10F. The water f low r a t e in pounds/minute i s :

(a ) 24 .(b ) 76 .( c ) 83.(d ) l l 2 .( e ) 249.

Basis of ca l cu l a t i on : 1 minuteHeat gain by water - Heat l o s t by Freon = 0

mlCpAT - m 2 (h 2 - hI ) = 0

ml (1 )(10) - 10(26 - 102)~ 0 760ml = 1 0

Answer i s (b )

THERMO 21

76 . lb s . /min .

Five gallons pe r minute of hot water a t 180F i s produced in a flow system byi n j e c t i o n an d condensat ion o f low pressure steam a t 20 p s i (gage) an d 80i .q u a l i t y in to co ld water a t 60F. Calculate th e stearn flow ra te in pounds per

minute .

Calcu la t i on b a s i s : 1 minute

r - - - - - - - - ,I

I

1 - - - + - 5 gpm hot water a t 20 p s ig~ - - - ' I Q) an d 180 0 F .

Low pressure

20 pSig, x =- ~ ; : ~ \ . . ~ - '= 0

s

(1 ) F i r s t la w energy balance , ignoring an y PE and KE changes, for th e flowsystem.

rn3h3 - rnlh l - m2h2 = Q - Ws = 0 s ince no he a t i s l o s t o r work done.

(2 ) Mass flow ra te balance: ml = m3 - m2

rn3 = 5 m r ~ l8 ~ ; 1 ~ b .= 41.67 lb . /min

(3 )

A more prec i se ca l cu l a t i on would u t i l i z e sa tu ra t ed steam table data a t

180F where v f= 0.01605 f t 3 /1b.

3 = f t x lb . 41 .7 lb . /minm3 min 7.48 gal 0.01605 f t 3

(This al lows fo r th e t r i v i a l thermal expansion o f ...a t e r . )

Determine en th a lp ies from sa tu ra t ed steam t a b l e s :

hI entha lpy o f l i q . water a t 60F = hf = 28.07 BTU/lb.

h3 entha lpy o f l i q . water a t 180F = h f = 147.91 BTU/lb.

h2 entha lpy o f 34.7 ps i a ( 2 0 + 1 4 . 7 ) steam of 80i. qua l i t y :


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Satura ted Steam Table Data:

ps ia h f h fg h g

30 218.83 945.2 1164.040 236.02 933.7 1169.7

Double in te rpo la t ion i s required : fo r 807. qua l i ty, an d fo r 34.7 ps iapressure .

a t 807. q u a l i t y, (30 ps i a ) h - 218.83 + 0.80(945.2) = 975.0 BTU/lb.(40 ps ia ) h z 236.02 + 0.80(933.7) = 983.0 BTU/lb.

Linearly i n t e r p o l a t e between 30 and 40 for 34.7 ps i a a t 807. qua l i ty :

h2 = 975.0 + 107 (983.0 - 975.0) = 978.8 BTU/lb.

(4 ) Solve energy balance fo r ~ 2 ' us ing IDl C ID3 - ~ 2

m3h3 c mlh l + ID2h2

(41.7)(147.91) ., (41.7 - m2 ) (28 .07) + m2(978.8)

6167.8 z 1170.5 - 28.07m 2+ 978.8m 2

4997m2 E 950.7 = 5.26 Ib/min. 4t

THERMO 22

Exhaust steam from a turbine exhaus ts i n to a surface condenser a t a mass flowra te of 8000 Ib /h r, 2 ps ia an d 927. q u a l i t y. Cooling water en te r s th e condensera t 74F and leaves a t th e steam i n l e t tempera ture . What i s th e cooling water

massflow ra te in l b /h r?

Saturated steam t ab l e data a t 2 ps i a a re :

T OF hf

, BTU/lb h fg , BTU/lb h g , BTU/lb

126.08 93.99 1022.2 1116.2

hI entha lpy o f steam a t 927. qua l i ty = h f + 0.92h fg

h2

h3

94.0 + 0.92(1022.2) = 1034.4 BTU/lb.'" entha lpy of l iquid water a t 126.1F = 94.0 BTU/lb.

entha lpy o f l iquid water a t 74F ., 42.0 BTU/lb above reference o f 32F

Alternate: find hf in steam t ab l e sa t 74F ignoring neg l ig ib l e e f f e c t o f

p ressu re on entha lpy o f incompressib le l i q u i d s .

STEAM

2 ps ia ,~ l = 8000

74F COOLING WATER

m - mass flow r a te


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In the absence of da ta , assume steam condensate leaves a t l 2 6 . lo

F

Heat balance, bas i s : 1 hour

heat from steam = heat to cool ing water

8000(1034.4 - 94 .0) '" m2 (94.0 - 42.0) ~ 2 = 144,700. l b / h r "

THERM O 2 3

53.8 grams o f f ine ly divided aluminum i s heated to 98.3C an d dropped into76.2 grams o f water a t 18.6C contained in a ca lo r ime te r. The f i n a l temperature

o f the mixture i s 27.4C. The mass o f th e ca lo r ime te r i s 123 grams, and i t s

s p e c i f i c heat may be taken as 0.092 cal g - ldeg - l The combined thermal capacity

of the thermometer and metal s t i r r e r i s 6. 5 cal de g - l

Assuming no heat i s l o s t from th e system, ca lcu la te th e mean spec i f i c heat o faluminum fo r th e above temperature range .

MAICAl (T l - T 2 )= [mcalCcal + IDthermoCthermo + IDH20CH20] (T 2 - TO)

53.8CAl

(98.3 - 27 .4) '" [123(0.092) + 6.5 + 76 .2 (1 ) ] (27.4 - 18 .6)

3815C Al (11.31+ 6. 5 + 76.2)(8.8)

C _ (94 .01)(8 .8) c 0.217 cal/(gw C) . .Al 3815

T H E R M O 2 4

3000 cfm of 65F a i r i s required to mainta in a house a t 76F.

251M t o f a i r enter ing th e a i r condit ioner i s frOM out s ide a t 9()0!" and th eremainder i s recycled frow i ns ide a t 7 ~ o F .

REQUIRED: Determine th e ra t ing in tons o f the a i r cond i t i one r.(Assume sensible heat changes o n ly, i?,nore any dehumidif icat ion . )

65'" Air Density:~ _ 14.7 x 144 = 0.0756 Ibm/ft 3RT 53.3 x 525 R

Tota l mass flow r a te ~ = ~ Q= 0.0756(3000) = 227 Ibm/min= 13,620 Ibm/hr

= mass flow r a t e 90 Ai r 0.25(13,620) = 3405 Ibm/hr

'" mass f low r a t e 7 6 Ai r 0.75(13,620)=

10,215 Ibm/hr

C spec i f i c heat of Ai r = 0.24 BTU/lbmoFp

Qc

~ 9 0 C p A T+ ~ 7 6 C P o . T= 3405(0.24)(90- 65) + 10,215(0.24)(760

- 65)

= 20,430 + 26,970 = 47,400 BTU/hr1 to n of r e f r ige ra t ion = 200 BTU/min = 12,000 BTU/hr

47,400Tons of Air Conditioning = 12,000 3.95


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THERMO 25

In terms o f QH (heat from high temperatu re source) an d QL (hea t to low

temperatu re s ink) , th e net work o f a Carnot cycle i s :(a ) W .. QH - QL

(b ) W = QH - QLQL

(c ) W ..QL

QH - QL

(e ) none o f these

The work produced by a carn o t cycle hea t engine i s QH - QL tt

Answer i s (a )

THERMO 26

Selec t th e r e l a t i onsh ip de f in ing absolute temperature in terms of QH an d QL'

heats t r ansfe r red in a Carnot cy c le , and TH and T L, th e absolu te tempera tures

of th e r e se rvo i r s involved.

(a ) QH - QL = 1 (b )QH - QL TH (c )

QH TH- o a -

TH - TL QH TL QL TL

(d ) QH= TL QH - QL TH - TL

QL TH(e) QH TH

Statement (c) d ef in es ab so lu te thermodynamic temperature. Statement

(e ) i s numerica l ly equal to e ff i c i ency ~ of a Carnot cycle h ea t engine.

(c ) i s co rrec t , an d (e ) i s der ived therefrom. tt

THERMO 2 7

The maximum thermal eff i c iency tha t can be obtained in an i dea l r eve r s ib l e

heat engine operat ing between 15400 F an d 340F i s c lose s t to

'f\thermal

8001 - 2000 = 1

(a ) 100'10(b ) 607.( c ) 787.

(d ) 407.(e ) 22'70

TL .. 340F + 460 = 0 0 ~ R

TH .. 15400

F + 460 0 = 2000 0 R

0.40 = 0.60 = 607. ttAnswer i s (b )


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THERMO 2 8

A 3 HP r e f r i g e r a t o r o r heat pump o p era tes between OaF an d lOoc F .The maximum t h e o r e t i c a l heat t h a t ca n be t rans fe r red from the cold r e se rvo i r i sn e a r e s t to :

(a ) 7,600 BTU/hr(b ) 13,000 BTU/hr(c ) 23,000 BTU/hr(d ) 35,000 BTU/hr(e) 43,000 BTU/hr

Coeff i c i en t o f Performance o f a re f r ige ra tor or heat pump:

C.O.P.

T HE RM O 2 9

1 HP c 2545 BTU/hr

W = 7635 BTU/hr

QL = 35,100 BTU/hr t t

Answer i s (d )

A Carnot cycle heat engine o p era t in g between 15400

F and 440F has ane f f i c i e n c y of approx imately

(a ) 5570(b ) 4570(c ) 3570(d ) 2970(e ) 8270

Fo r a Carnot cycle heat engine , eff i c iency ~

Answer i s (a )

THERMO 3 0

20000

R - 900"R2000 oR

Second law l i m i t a t i o n on th e maximum h o r s e ~ o w e ro u tp u t from an y power uni tburning 1,000,000 BTU/hr of fuel with high and low tempera ture extremes of

l540F and 40F i s :

Bas is o f ca lcu la t ion : 1 hour

TH - TLCarnot eff i c iency yt a

TH - 0

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

98 .295.1140.3830.none of these

W= Q

H


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w 0.75QH = 0 . 7 5 x 10 6 '" 750 ,000 B.TU/hr

750 ,000 BTU/hr2545 BTU/(HP h r )

TH ERMO 31

2 9 5 . HP

Answer i s (b )

A h ea t pump i n s t a l l a t i o n i s used t o warm a s t r e a m o f c i r c u l a t i n g a i r i n a

r e s i d e n c e t o l lOoF. An o u t s i d e a m b i e n t o f 35F i s co n s id e red th e a v a i l a b l e

h ea t so u rce . A h e a t i n g r eq u i r em en t o f 90 ,000 BTU/hr i s n e c e s s a r y t o m a i n t a i n

a c o m f o r t a b l e t em p era tu re i n th e l i v i n g sp ace .

(a ) Dete rmine th e a b s o l u t e minimum e l e c t r i c a l r e q u i r e m e n t i n k i l o w a t t s an d i nHP to o p e r a t e a h e a t pump d e l i v e r i n g 9 0 , 0 0 0 BTU/hr u n d e r t h e s e c o n d i t i o n s .

(b ) Determine t h e e l e c t r i c a l r e q u i r e m e n t i n Kw t o accompl i sh th e same h e a t byr e s i s t a n c e h e a t i n g .

A C arn o t r ev e r sed c y c l e would p ro v id e t h e most e f f i c i e n t h e a t pump p o s s i b l e ,a l t h o u gh t h i s i s o n l y approached i n p r a c t i c e w i t h Rankine vapor compress ionc y c l e s .

s

( a )

Area '.1, work added

Area = QL ' h e a t from low t e m p e r a t u r e r e s e r v o i r

Both a r e a s = QH' h ea t r e j e c t e d a t h i g h t em p era tu re

F o r a h ea t pump: QH = QL + '.1

C o e f f i c i e n t o f Per fo rmance :

a s shown i n th e T-Sdiagram

QLC.O.P. = W

A b s o l u t e t em p era tu res a r e r e q u i r e d .

C . O . P. W 57 0 - 49 5 = 6 . 6 0

QH = 9 0 , 0 0 0 = QL + W

'.1 11 ,84Q BTU/hr

(b )

7.60 W

1 1~ 84 0 BTU/hr3413 BTU/Kwh

3 . 4 7 Kw0 . 7 4 6 Kw/HP

3 . 4 7 Kw

4 . 6 5 HP

Re s i s t a n c e h e a t i n g i s p u re e n e rg y d i s s i p a t i o n

90,0003 , 4 1 3

26.4 Kw


17/48


THERMO 3 2

In an y non-quas i s t a t i c thermodynamic p ro cess , th e o v e r a l l entropy ofan i s o l a t e d system w i l l :

(a ) Increase an d then decrease(b ) Decrease an d then in c r ease(c ) Stay th e same(d ) Increase only(e) Decrease only

q u a s i s t a t i c : i n f i n i t e l y slow, l o s s l e s s , hypothe t i ca l , by d i f f e r e n t i a lincrements

The o v e r a l l entropy w i l l in c r ease fo r an i so la ted system, or fo r th esystem plus surroundings

Answer i s (d )

T HE RM O 3 3

Entropy i s th e measure of(a ) th e change in enthalpy of a system(b ) th e i n t e r n a l energy o f a ga s(c) the he a t capaci ty of a subs tance(d) randomness or d i so rde r(e ) th e t o t a l heat co n ten t of a system

Answer i s ( d ) .

THERMO 3 4

A quan t i t y o f working f lu id undergoes th r ee sequen t i a l i r r e v e r s i b l e processes :(1 ) Receives heat from high temperature rese rvo i r (a t h i g h ~ T )

Reservoir entropy decreases 0 .240 BTU/oR an dFluid entropy increases 0 .245 BTU/oR.

(2 ) Does work in an ad iaba t i c t u rb ine .Fluid entropy increases 0 .013 BTU/oR.

(3 ) Discharges heat to low temperatu re rese rvo i r (a t high AT)

Flu id entropy decreases 0 .258 BTU/oR.Reservoir entropy in c r eases 0 .263 BTU/oR.

Entropy change fo r the universe i s BTU/oR i s :(a ) indeterminate(b ) -0 .023(c ) +0.010(d ) +0.023(e ) none of these

Ne t increase for the universe equals + . 023 , th e di ffe rence between hightempera ture rese rvo i r lo s s and low temperature g a in . This i s a consequenceo f th e i r r e v e r s i b i l i t i e s because th e heat t r a n s f e r i s not a t 6T c 0, an dt he re i s l o s t work i n th e tu rb ine

Answer i s (d )

THERMO 35

I f heat t rans fe rs o f th e previous problem were r e v e r s i b l e , and i f th e tu rb ineoperated i s e n t r o p i c a l l y, what i s th e entropy gained o r l o s t by th e lowtempera ture r e se rvo i r?

(a ) indeterminate(b ) 0(c ) +0.240 BTU/oR(d) +0.250 BTU/oR(e ) none o f these


18/48


Reservo i rs are cons idered i n f i n i t e an d no temperature r i s e oc c urs .Since AS for r eve r s ib l e heat t r a n s f e r = 0, an d th e tu rb ine i s i sen t rop i c( 6 S = 0 ) , en t ropy l o s s by the h igh temperature r e se rvo i r matches th een t ropy gain by th e low temperatu re r ese rv o i r +0.240 BTU/oR. tt

Answer i s (c )

THERMO 3 6

A Carnot cycle heat engine ope ra te s between l540 0 F an d 40F an d r e j e c t s250. BTU/lb to th e low temperatu re r e se rvo i r o r heat s in k . Maximum entropychange in BTU/lboR of th e working f l u id fo r an y process in th e cycle i s :

( a ) 0(b ) 0.25(c ) 0 .50(d ) 1 .0(e ) 4 .0

The T- S diagram fOt:' a Carnot heat engine shows i t s r e l a t i o n between he a ttemperature an d entropy.

T

Q = T ~ Sre v

~ work done by heat engine

- ' ; 1 - - - - a re ao R 1 . . - - - 1 . . . . . . . . , ; : - -

250 BTU heat r e jec ted to lo w temperatu rer e se rvo i r = OL

Both a re a s = heat in take from high temperature r e se rvo i r C QH.Since 250 = 500-AS : .AS "" 0 .50 .

Answer i s (c )

THERMO 3 7

Fo r spontaneously occurr ing na tu ra l processes in an i so l a t ed system, whichexpress ion b es t expresses dS?

dS = T

( a ) dO re vdS = - -T-- - only.

( a )

(b )

(c )

(d )

( e )

dS 0

d S ) . 0

dS < 0d S = C d t - R.

p T P

The r e v e r s i b l e requirement i s necessary to genera te the

exact height v s . r ec tan g u la r area equivalence on th e Carnot cycle T-Sdiagram.

(b ) Only a r eve r s ib l e ad i aba t i c process i s i s en t ro p ic by d e f i n i t i o n .(c) All na tu ra l l y oc c ur r ing spontaneous processes a re i r r e v e r s i b l e an d r e s u l t

in an en t ropy in c r ease . tt(d ) An energy input from surroundings i s required to reduce entropy.

(e ) This i s an express ion fo r en t ropy change in an id ea l gas.

Answer i s (c )


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"


T HE RM O 3 8

Which o f the fo l lowing s ta tements about entropy i s f a l se?

(a ) Entropy o f a mixture i s grea t e r than t h a t o f i t s components u n d e ~the .same co n d i t io n s .

(b )

(c )

(d )(e )

An i r r e v e r s i b l e process increases entropy o f the u n iv er se .Entropy has the u n i t s o f heat capac i ty.

Ne t entropy change i n an y closed cycle i s ze ro .oEntropy o f a c r y s t a l a t 0 F i s zero .

A ll a re t rue excep t ( e ) . The en t ropy o f a p e r f e c t c r y s t a l a t absolute zero(OoK o r OOR) i s z e ro . This i s th e 3r d la w o f thermodynamics. There i spresumably no randomness e x i s t e n t a t t h i s temperatu re in a c r y s t a l withoutf l a ws , imp u r i t i e s o r d i s l o c a t i o n s .

Answer i s ( e ) .

T HE RM O 3 9

Work or energy can be a funct ion of a l l of the fol lowing except :(a ) fo rce an d di s t ance(b ) to rque an d angular ro ta t ion

(c ) power and time(d ) force an d time(e ) temperatu re an d entropy

( a )

( b )

dW FdS fo r t rans la t ion

dW T d ~for ro t a t i on

( c )dWd t power : . dW = (Power)dt

( d ) Fdt = d(mV), th e impulse momentum e qua t ion . Impulse, Fdt , i s no t anenergy term.

(e) dQrev = TdS. Heat i s a form of energy

Al l excep t (d ) are c o r r e c t 4t

T HE RM O 4 0

Energy changes are represented by a l l excep t which on e of the fo l lowing?

(a) - ~ V d P

(a) i s sha f t work.(b ) dU

( c ) dH

TdS PdV

TdS + VdP.

(d ) dQrev-- - = d S .T

(b ) TdS - PdV

(c )

(d )

( e )

dQrev - W.

TdS + VdPdQrev

T

mCpdT

dU i s an i n t e r n a l energy change.

dH is an entha lpy change .

dS i s change in th e thermodynamic s t a t e funct ion, entropy.I t i s n ot an energy change.

(e ) dH = mCpdT for an i d e a l gas.Therefore (d ) r ep resen ts an en t ropy or s t a t e change , a l l o th er s r ep resen tenergy changes. tt

Answer is (d)


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THERMO 41

A high ve loc i ty flow o f ga s a t 800 f t / s ec possesses k i n e t i c energy nea re s t towhich of th e following?

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

1.03 BTU/lb4.10 BTU/lb9.95 BTU/lb

12.8 BTU/lb41.0 BTU/lb

Basis : 1 Ibm of f lu id flowing

V2

KE - - - - in f t - l b f , where V i s in f t / s e c , and gc 32.172g cUse J - 778 ft-lbf/BTU to convert to BTU.

800 2KE - - 12.8 B T U / l b . ~2(32.17)(778)

THERMO 4 2

(U + PV) i s a quant i ty ca l l ed

Answer i s (d )

(a ) flow energy(b ) shaf t work(c) entropy(d ) enthalpy(e) i n t e r n a l energy

Flow energy i s PV. Shaf t work, Ws ' i s - SVdP. Entropy i s S.

In te rna l energy i s U. Enthalpy H i s defined as U + PV, th e sum o f in te rna lenergy p Iu s flow energy

Answer i s (d )

THERMO 4 3

In flow process , neglec t ing KE and PE changes, -J dP represents which i tembelow?

(a) heat t rans fe r(b ) s h a f t work(c) c losed system work(d ) flow energy(e ) enthalpy change

Shaft work i s work or mechanical energy cross ing the fixed size boundary(control volume) of a flow (open) system. Shaft work Ws i s def ined , in th e

absence o f PE and KE changes, by dH ~ TdS + VdP, where TdS = dQrev and-VdP i s dW s . In i n t eg ra t ed form ~ Hmo STdS + )VdP = Qrev - WS ' where

Ws i s represented by - JVdP. Closed system work W i s defined by

dU u TdS - PdV, or ~ U- ~ T d S- JPdV C Qrev - W. Thus closed system

work i s + f dV Flow energy i s th e PV term, an d enthalpy change i s ~ H .

Answer i s ( b ) .


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THERMO 4 7

A Carnot heat engine cycle i s represented on th e T-S and P-V diagrams below:

- 2,

t / tT 4 3 P

5s - v - - -

Which of th e severa l a reas bounded by numbers or l e t t e r s represents th e amountof heat r e j ec t ed by th e f lu id dur ing on e cycle?

(a) area 1-2-6-5(b ) area B-C-H-G(c) area 3-4-5-6(d ) area D-A-E-F( e) area C-D-F-H

The table below gives th e s ign i f i cance o f each area of the diagrams:

T- S Diagram P-V DiagramPROCESS Area Represent ing Heat Area Represent ing Work

iso thermal expansion 1 -2-6-5 '" heat in from A-B-G-E 2 : work done by1 -2 and A-B high temp. r e se rvo i r f l u i d

i sen t rop ic expansion 2-3-6 = 0 heat t rans fe r B-C-H-G .. work done by2- 3 and B-C f lu id

i sothermal compression 3-4-5-6 D heat out to C-D - F-H ., work done on3-4 and C-D lo w temp. rese rvo i r f lu id

i sen t rop i c compression 4-1-5 = 0 he a t t rans fe r D-A-E-F .. work done on4- 1 and D-A f lu id

net r e s u l t o f process 1-2-3 -4 = n e t heat A-B-C-D .. n e t work doneconverted to work by process

Answer i s ( c ) .

THERMO 4 8

Which o f the following thermodynamic cycles i s the most e f f i c i e n t ?

(a) Carnot(b ) Brayton(c ) Otto(d ) Diesel(e) Rankine


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Answer i s ( a ) .

THERMO 4 9

A Carnot engine o p era t in g between 70F an d 20000

F i s modif ied so l e ly by ra i s ingthe high temperature by 150F and ra i s ing th e low tempera ture by 100F. Whichof the fol lowing s ta tements i s fa lse?

(a ) thermodynamic e f f i c i e n c y i s increased .(b ) more work i s done during th e isothermal

expansion .(c ) more work i s done during the i sen t rop i c

compression.(d ) more work i s done dur ing the r eve r s ib l e

ad i aba t i c expansion.(e ) more work i s done dur ing the isothermal

compression.

Carno t cycle e ff i c i ency i s o r i g i n a l l y

Afte r th e change

n = 2610 - 630 = 0.76-l 2610 Eff iciency i s t he re fo re reduced .

On th e T-S an d P-V diagrams below th e o r i g i n a l cycle i s shown as ABCD,

th e modif ied cycle as A 'B 'C 'D' .

, IA'

A r----,13I tt A B t

IC'T D

/ . P- - ' - ~ ~

- D . ,>< C

2

s - - v---


24/48


Compare work done during isothermal expansion (A to B, v s . A' to B')Orig inal: area A-B-8-4Modified: area A ' -B ' -7 -3 i s l a rge r

Compare work done during i sen t rop i c compression (D to A, v s. D' to A')Orig inal: area D-A-4-6Modified: area D ' -A ' -3 - 5 i s l a rge r

Compare work during r eve r s ib l e ( i sen t rop i c ) expansion (B to C, v s. B' to C')O r ig in a l : area B-C-lO-8Modif ied: area B '-C ' - 9-7 i s l a rge r

Compare work during isothermal compression (C to D, v s . C' to D ')Orig inal: area C-D-6-10Modified: area C'-D' -S - 9 i s l a rg e r

Statements (b) , ( c ) , (d ) , an d (e ) are c o r r e c t .

The f a l se answer i s ( a ) .THERMO 5 0

1

Expansionvalve

Wscompressor

4 . . . . - - . ; } - - - " 3evapora tor

tT

II

I

A

II

I

I EI

heat in

5 - - -

"

C

i s e n t r o p i ccompression

D

....

In th e i d e a l heat pump system out l i ned , the expansion valve 4 -1 performs th eprocess t ha t i s loca ted on the T-S diagram between p o in ts

(a ) A an d B(b ) Ban d C(c ) C and D(d ) D an d E(e ) E an d A

The vapor compression reversed Rankine cycle i s conducted counterclockwise onboth th e schematic and T-S diagrams.Numbers on th e schematic an d l e t t e r s on th e T-S diagram are re la ted :

1 = A, 2 = B, 3 = D, an d 4 = E.Process C-B-A occurs i n th e condenser between 2 an d 1 . The expansion processA - E occurs between 1 - 4 .

Answer i s ( e ) .

THERMO 51

Which a i r -s tandard power cycle do th e P-V and T-S diagrams on the next pagerepresent?

(a ) Otto cycle(b ) Reheat c ycle(c) Carnot cycle(d) Rankine cycle(e ) Brayton cycle


25/48


tp' - - ...... ~ 4

v --

tT z

cons t P

The Brayton cycle i s appl ied to th e s imple open-cycle ga s tu rb in e whereinintake a i r i s compressed (1-2) , combustion suppl ies thermal energy (2 -3 ) , an dcombustion products expand an d dr ive th e turbine (3 -4 ) , and exhaust a t 4 . tt

Answer i s (e )

THERMO 52

Equil ibr ium(a )

(b )( c )

(d )

condit ions e x i s t in a l l except which of th e fol lowing:in reve r s ib l e p ro cesses .in processes where dr iv ing forces are i n f i n i t e s i m a l s .along i dea l f r i c t i o n l e s s , nondi ss ipa t ive p a th s where forwardr ev er se processes occur a t equal r a t e s .in a s teady s t a t e flow process .

(e ) where nothing can occur without an e f f e c t on th e system'ssurroundings .

and

All except (d ) demonstrate equil ibr ium. Steady s t a t e o r time i nva r i en t flowp ro cesses may be conducted f a r from equil ibr ium cond i t i ons . tt

The fa l se answer i s (d )

THERMO 5 3

Data in th e t ab l e below descr ibe two s t a t e s of a working f l u i d t h a t e x i s t a ttwo l oca t ions i n a p iece of hardware.

p v T h sps ia f t 3 / lb o f BTU/lb BTU/ ( lboF)

S t a t e 1 25 . O . O l l 20 . 19.2 0.0424

St a t e 2 125. 0.823 180. 203.7 0.3649

Which of th e following s ta tements about the path from s t a t e 1 to 2 i sf a l se?

(a ) th e path r e s u l t s in an expansion.(b ) th e path determines th e amount of work done.(c ) th e path i s indeterminate from these da ta .(d ) the path r equ i re s t h a t energy be added in

the p ro cess .

(e ) the path i s revers ib le ad iaba t i c .


26/48


27/48


28/48


THERMO 58

I sen t rop i c compression o f 1 f t 3 of a i r , ~ .. 1.40,Cv

a t 20 ps i a to a p ressu re

of 100 ps i a gives a f i n a l volume o f :

(a ) 0 .16 f t 3

(b ) 0 .20 f t 3

( c ) 0.32 f t 3

(d ) 0.40 f t 3

(e ) 0.56 f t 3

An i sen t rop i c process fo r an i dea l ga s follows th e path:

1 .420 (1 ) 1 .4100(V 2 )1 .4V2 = 0.20 : .V2 =().317 f t

3

Answer i s ( c )

THERMO 5 9

Steam enters a tu rb ine a t 30 0 p s i a , 700F and exhaus ts to a low p ressu reprocess steam l i ne a t 30 p s i a . The tu rb ine de l ive r s 1000 Kw of power a t anengine e ff i c i ency o f 72%. Calculate th e mass flow ra te o f steam in l b / h r.

Engine e ff i c i ency i s th e r a t i o of output to th e avai lable work correspondingto an i sen t rop i c expansion .

Q = 0jf W = 1000 Kw a t 72% engine eff i c iency

700F r -&- - -/-!4300 ps i a --..L..J ~ 30 ps i aS u p e r h e a t e ~ I ~ - - q u a l ~ t yunknown

ill L_____ - - - - ' 01 Kwh = 3413 BTU

Calculat ion bas i s : 1 hour

1000 Kwh a t 72% eff i c iency i s equivalen t to 1389 Kwh, o r

Ws .. 4 .74 x 10 6 BTU from an i sen t rop i c expansion .

Incoming steam prope r t i e s from superheated t ab l e s :

hI = 1368.3 BTU/lb

Exhaust steam for i sen t rop i c expansion ha s s2 = 1.6751 BTU/(lboF) a t 30 p s i a .Consult sa tu r a ted steam t a b l e s a t 30 p s i a f inding:

T, o f h f h fg hg s f Sf g Sg

250.34 218.83 945.2 1164.0 0.3680 1 . 3312 1.6992

Linearly i n t e rpo la t e entropy to 1. 6751 and determine enthalTlY on the has i so f equal q u a l i t y.


29/48


s2 - s f = 1.6751 - 0 . 3680 1. 21191

fo r 100% q u a l i t y 1 . 3312 X = 1.2R91 0.9r.84Sfg 1. 3312h fg fo r

100% q u a l i t y 945.2

h2 a t . 9684 q u a l i t y - 218.83 + 0 . 9684(94 5.2) 1134.2 BTU/1h .

Let ~ l be steam flow r a t e in l b / h r.

~ l (h I - h 2 ) ~ Ws

IDl(1368.3 - 1134. 2 ) K 4 .7 4 x 10 6 BTU

.=

64.74. ~ = 20 , 20 0 l b / h r. 4t

23 4 . 1

THERMO 60

An i dea l ga s a t a pre s s u re o f 500 ps i a an d a tempera ture o f 75F i s

conta ined ina

cy l in d er witha volume of 700

cubicf e e t . A

c e r t a i namount of th e ga s i s released so t ha t the p ressu re in th e cy l in d er dropsto 250 p s i a . The expans ion of the ga s i s i s e n t r o pic . The h ea tcap ac i ty r a t io i s 1.40 an d the ga s co n s tan t i s 53.3 f t lbf/lbmoR.

What i s the weight of the ga s remaining in the cy l in d er?

Given:

B a s i s :

Ck = ~ = 1.40CvPI 50 0 p s i a

VI 700 cu f t

Tl 75F + 46 0 535R

R = 53.3 f t lb fIbm OR

P 2 250 p s i a

V2 700 cu f t

T2 ?w2 ?

Id ea l ga s la w PlVl

= wRT l an d th e b as ic equat ions for r eve r s ib l ead i aba t i c ( i sen t rop i c ) expansion

and ; : =(::)\'The ga s r e m a ~ n ~ n gin the t ank cools as i t expands

T, = T,(:: ) ~ '1 . 4 -1 0.2857

535 ( ~ ~ ~ )1 .4 = 535 ( ~ ) =

Check dimensions to use the proper conversionf a c t o r.

(250) (144) (700)(53 .3 )(439)


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THERMO 61

Determine the t h e o r e t i c a l horsepower required for th e isothermal compression

o f 800 f t 3 /min of a i r from 14.7 to 120 p s i a .

At th i s volume flow r a t e , th e process i s a flow process . Work of i sothermalcompression of an i dea l ga s i s numerica l ly the same i n a s teady flow process2 S in a closed system.

PV c constan t = PIV I - P 2 V 2 = RT

closed system:

flow system:

= mRT In V 2 = mRT In P l

Basis o f ca l cu l a t i on : 1 minute

W ~ PIVI In

:J- c 14.7(144)(800) In ( ~ ~ o 7 ) =-3 ,560,000 f t . l b / mi n , work i ns 2

Ws

= 3,560,000 f t . l b / mi n33,000 f t . l b /HP -min

108. HP

THERMO 6 2

Enthalpy o f an i dea l ga s i s a function only of:

(a) i n t e r n a l energy(b ) entropy(c) the product of pressure and spec i f i c volume(d ) pressure(e) temperature

H - U + PV, PV - RT. Enthalpy an d i n t e r n a l energy of i dea l gases are a

function only o f tempera ture . dh - CpdT a nd d u - CvdT.

Answer i s ( e ) .

THERMO 6 3

Which o f th e following s tatements i s fa l se concerning the devia t ions of r e a lgases from ideal ga s behavior?

(a) Molecular a t t r a c t i o n in te rac t ions are compensated for in the i dea lgas law.

(b ) Devia t ions from i d e a l ga s behavior a re large near the s a t u r a t i o ncurve.

(c ) Devia t ions from i dea l ga s behavior become s i g n i f i c a n t a t pressuresabove th e c r i t i c a l p o i n t .

(d ) Molecular volume becomes s i g n i f i c a n t as spec i f i c volume i sdecreased.

(e ) Compressib i l i ty fac to r Z i s used to modify the i dea l ga sequat ion of s t a t e to f i t r e a l ga s behavior.


31/48


All s tatements except (a ) are t r u e . The i dea l ga s la w does not cons idervolume of the molecules o r an y i n t e r a c t i o n othe r than e l a s t i c co l l i s ion . . .

The f a l se s t a t e me n t i s (a)

T HE RM O 6 4

There are 3 Ib s . o f a i r in a r i g i d conta iner a t 25 ps ia an d lOOOF.Given the ga s constan t for a i r i s 53.35(a ) Determine th e volume of the co n ta in e r(b ) I f th e temperatu re i s ra ised to l80oF, what i s the r e s u l t i n g

absolu te pressure?

PV = mRTlb . f t 3 '" Ibmf t 2

f t - lb ORIbm OR

P = 25 ps i a x 144 = 3600 I b / f t 2R 53.35 f t- lb / lbmoRTl lOOoF + 46 0 560 0 R

(a )PV = mRT3600(V) = 3(53.35)(560) v

(b ) PIVl P2V2 For constan tTl T2

PI P2- c

Tl T2

3(53.35) (560)3600

volume

24.9 f t 3 . .

P2PIT 2 25(640) 28.6 p s i a "

Tl 560

T HE RM O 65

A mixture a t 14.7 ps i a and 68F t h a t i s 30% weigh t CO2 (m wt= 44 ) and

70% weight N2 (m wt = 28) ha s a p a r t i a l pressure o f CO2 in ps i a t h a t i sn e a r e s t to :

(a ) 2.14(b ) 3.15(c ) 6.83(d ) 7.86( e ) 11.55

B a s i s of ca lcu la t ion : 1 l b . mixed gases

(1 ) Calcu la te weight of each component, number o f moles o f each presen t .

(2 ) Compute mole f rac t ion o f each, an d p ro p o r t io n t o t a l pressure according tomole f r a c t i o n .

Number o f P a r t i a l PressureComponent Weight , lb . lb . moles Mole Frac t ion ps i a

CO2

0.30 0.30 .00683 .00683 .. 0 .214 3.15~ = .03183

N2 0.70 0.70 .. 0.02500.0250 = 0.786 11.55

28 o 03183Total 1.00 0.03183 1.000 14.70


32/48


Since mole f rac t ion of a ga s = volume f rac t ion ,Composition o f the mixture i s 21.4% vo l . CO 2 and 78.6% vo l . N2

From th e t ab le , th e co rrec t p a r t i a l pressure of CO 2 i s 3.15 p s i a . 4tAnswer i s (b )

THERMO 6 6

Given tha t molar Cp ofCO

2 i s 8.92 BTU/(lb.moleoR), and molar Cp ofN2 i s

6.95 BTU/(lb.moleoR), th e ca l cu l a t ed Cp~ pound o f mixture contain ing

25% vol . CO 2 and75% vol . N2 i s :

(a ) 0.23 BTU/(lb.oR)(b ) 2.23 BTU/(lb.oR)( c ) 5.21 BTU/(lb.DR)(d ) 7.44 BTU/(lb.oR)

(e ) none o f these

Note t ha t t h e o r e t i c a l molar heat capac i t i es o f i dea l gases a t zero p ressu re

are for monotomic: Cv 3, Cp= 5. BTU/ (l b .mo leoR)

dia tomic: Cv 5, Cp 7.

t r ia tomic: Cv 7, Cp 9.

Ba s i s o f ca lcu la t ion 1 lb . mole of mixtureWeight of Molar Cp

Component Mol. Wt. % v o l . Mole f r ac t i on each component con t r ibu t ion

CO2 44 25 . 0.25 11.0 0.25(8 .92) .. 2.23

N2 28 75 . 0.75 21.0 0.75(6 .95) .. 5.21

Total 100. 1.00 32.0 7.44Average Mol.

Weight

One mole of mixture weighs 32.0 lb . and ha s a Molar Cp of 7.44 BTU/(lb.moleoR)

The mixture ha s a spec i f i c Cp of 7.44 = 0.23 2 BTU/(lb.oR) 4t32 .0Answer i s (a )

THERMO 67

Dry a i r has an average molecular weight o f 28.9, cons i s t i ng o f 21 mole-% 02 '

78 mole-% N2 and 1 mole-% (Argon, t r a ce s o f C02). I t s calcula ted wt.% 02 i s

nea re s t to :(a) 21.0(b ) 22.4(c) 23.2(d ) 24.6(e) 28.0


33/48

Thermodynamics / 169B a s i s of ca lcu la t ion : 1 lb . mole dry a i r

Component Mol. Wt. Mole Frac t ion Weight . lb . % Weight

2 32.0 0.21 6.72 23.2

N2 28.0 0.78 21.80 75.4

Ar 40.0 0.01 0.40 1.4

To ta l 1.00 28.92 100.0

Answer i s ( c ) .

T HE RM O 68

A ll of the fol lowing s ta tements about we t bulb tempera ture a re t rue EXCEPT:

(a ) we t bulb temperature equals a d i a b a t i c s a t u r a t i o n temperatu re(b ) we t bulb temperature l i e s numer ical ly between dewpoint an d

dry bu lb tempera tures for unsatura ted systems(c ) we t bulb tempera ture equals both dry bulb an d dewpoint

tempera ture a t 100% r e l a t i v e humidity(d ) we t bulb tempera ture i s the only tempera ture ne c e s s a ry to

determine grains of wa te r p er l b . o f dry a i r(e ) we t bulb tempera ture i s the lowes t tempera ture a t t a i n a b l e by

evaporat ive cool ing a t ambient pressure

Wet bulb tempera ture , as commonly measured, an d ad iaba t i c sa tu ra t iontempera ture are equal fo r the a i r-wate r system. Fo r unsaturated systems,we t bulb temperature i s approximately equal to dewpoint p lu s 1/3 o f th edi ffe rence between dry bulb and dew poin t temperatures . At sa tu ra t ion ,we t b u lb , dr y bulb and dewpoint a re equal . Wet bulb tempera ture alone i si n s u f f i c i e n t to determine s t a t e on a psychrometr ic c h a r t , hence insuff ic ien tfo r determining spec i f i c humidity r a t i o (g ra ins of water / lb . dry a i r ) o r an yothe r p rope r ty. Both we t and dry bulb tempera tures a re normally used toloca te a poin t on a psychrometric char t . Wet bulb ( ad i aba t i c sa tu ra t i on)tempera ture i s normally the lowest tempera ture obtained by evaporat ivecooling a t ambient pressure .

Answeri s

THERMO 69

Line E of the usual psychrometric cha r t above r ep re sen t s a l ine o f co n s tan t :(a) r e la t ive humidity(b ) dew point o r s p e c i f i c humidity(c ) entha lpy or wet bulb temperature(d ) s p e c i f i c volume(e ) dry bulb tempera ture


34/48


Sloped l i ne s A are l ines o f c o n s tan t we t bulb temperature an d a re almostl i ne s o f cons tant entha lpy.

Steeper l i ne s B are l ines o f cons tant sp e c i f i c volume, cu f t we t a i r / l b or y a i r .

Ver t ica i l i ne s C a re l i n e s o f cons tant dry bulb temperatu re .Curved l i ne s D a re l i n e s o f cons tant r e l a t i v e humidity.

Horizonta l l ines E a re l i n e s o f c ons ta n t dew poin t , s p e c i f i c humidityan d wa te r vapor p ressu re .

: Note: Psychrometr ic cha r t s a re only va l id a t on e p ressu re , u su a l ly 14 .7 p s i a ,though they are usable with l i t t l e e r r o r for barometr ic p ressu resbetween 29-31 " Hg. Answer i s ( b ) .

THERMO 7 0

A smal l p l a s t i c ba g i s f i l l e d with moist a i r a t 50% r e l a t i v e humid i ty,sea led an d p la c ed in an envi ronmental chamber whose temperatu re an d pres s uremay be independent ly va r i ed . The r e l a t i v e humidity i n the ba g w i l l belowered under which of th e fol lowing condi t ions?

(a ) p ressu re i s increased

(b ) p ressu re i s decreased(c ) tempera ture i s decreased(d ) tempera ture i s decreased an d pre s s u re i s increased(e ) none of th e above c ond i t ions

Unless water condensat ion occurs a t 100% R.H. , the sea led ba .g co n ta in s af ixed mole f r ac t i on of water vapor i n a i r .

~ m o l ef rac t ion of water vapor i n the mixtureRela t ive Humidity = < . - - - - - - - - - - - - - - - - - - - - - - - - - - -

, mole f rac t ion of water vapor in a sa tu ra t ed mixture ',L a t th e same temperatu re an d p ressu re . I

Vapor pressure of water i s tempera ture dependent only, an d in c r eases withtemperatu re . The denominator term of the above R.H. def in i t ion i s increasedwith elevated temperatu re an d decreased with e lev a ted p r es su re . Thus:

pressure increase r a i s e s R.H. , p ressu re decrease lowers R.H. , temperatu redecrease ra i ses R.H. , while s imul taneous temperatu re decrease an d p ressu reincrease both ra i se R.H. Note t h a t a temperatu re in c r ease lowers R.H. ,but t ha t choice was no t g iven in the problem.

Answer i s (b )

THERMO 71

Which of the fo l lowing s ta tements about s tag n a t io n p ro p er t ie s i s l e a s t co r r ec t?

(a ) Fo r a compressib le f l u i d , KE change i s converted to s tag n a t io ntemperatu re r i s e , an d pressure r i s e i s i s e n t r o p i c a l l y c a l c u l a te d .

(b ) Stagnat ion temperatu re i s r e l a t ed to aerodynamic he a t ing of leadingedges o f a i r c r a f t wings.

(c ) For an incompress ible f l u id impacting a p i t o t tube , KE change may beconverted to pressure r i s e .

(d ) Fo r gases a t low v e l o c i t y, the s impli fying assumption of incompressi b i l i t y y ie ld s s tag n a t io n preSSU4es with on ly a s mall percentagee r r o r.

(e ) In a f l u id a t Mach number g rea te r than 1, an i sen t rop i c ca l cu l a t i onof s tagnation p r o p e r t i e s i s va l id .

Whenever a flow i s brought to r e s t i s en t rop ica l ly, a s tag n a t io n p r es su re a r i s e s .At Mach ~ 1 a shock wave forms in f ro n t of th e body and d i s s i p a t e s energy by aprocess which i s no t i s en t rop ic . Stagnat ion pressures produced a t the body arelower than an t i c ipa t ed .

Answer (e ) i s false.


35/48


THERMO 72

Which of th e fol lowing s ta tements about Mach number (M) i s f a l se?

(a ) Mach number i s the r a t i o of ve loc i ty to sonic v e l o c i t y.

(b ) Mach 1 i s th e maximum a t t a i n a b l e v e l o c i t y in a nozzle t h r o a t .

(c ) Supersonic v e l o c i t i e s (M > l ) are achievable in th e d i f f u s e r s ~ c t i o nof a roc ke t nozzle i f expansion r a t i o i s grea t enough.

(d ) Mach angle 0 ( , or angle the shock f ron t makes with th e ve loc i tyvec to r o f a moving source , an d Mach number are re la ted by MsinO( = 1 .

(e ) Mach no . ha s th e dimensions o f v e l o c i t y.

Mach number, being a r a t i o of v e l o c i t i e s , i s d ime ns ion le s s . All o th ers tatements are c o r r e c t .

Answer (e ) i s f a l s e .

THERMO 7 3

Sonic v e l o c i t y (Mach 1) a t 50,000 f ee t a l t i t u d e i n the s tandard upperatmosphere (-67.6F and 1.68 ps i a ) i s nea re s t to :

Sonic v e l o c i t y fo r i dea lca l cu l a t ed from

C =VkgcRT where

(a ) 880. f t / s e c(b ) 971. f t / s e c(c ) 995. f t / s e c(d ) 1064. f t / s e c(e ) 1117. f t / s e c

gases i s temperatu re dependent only and may be

C i s sonic ve loc i ty i n f t / s e c .

k = ~Cv

gc = 32.17 Ibm f t / ( l b f sec2 )

R ga s constan t f t Ibf/(lbmoR)

T a temperatu re in oR.

C =,(1 .40(32 .17)(53 .34)(392 .4) = 971 f t / s e c . .Answer i s (b )

T HE RM O 7 4

A ir f lows through a I f t diameter c i r c u l a r duct a t 20 p s i a and 80F.A manometer connects a p i t o t tube , loca ted on th e cen te r l ine o f th e duct , witha s t a t i c pressure tap upstream and reads a d i f f e r e n t i a l p ressu re o f 3 incheso f water. Determine mass flow r a t e of th e a i r in Ib / sec .

Air ...BOoF (540 o R)

20 ps i a

- - - - 1 ' D - -

i


36/48


(1 ) Solut ion based on assumed incompressibi l i ty

Since d i f f e r e n t i a l p ressu re i s t r i v i a l , th i s i s acceptab le .(3 " water = 0.25 ' water = 0.25(62 .4) = 15.6 p s f = 15.6/144 0.108 ps i )

In th e incompressible so lu t ion , KE i s converted to an add i t i ona l p ressu reenergy ind icated by th e p i t o t tube , an d ve loc i ty i s determined.

Basis : 1 Ibm o f f lu id flowing.

V2-- = APh = where h i s add i t i ona l head i n f ee t = f t Ih / IbTgc ~ f m 3i s dens i ty o f th e f l u id flowing in lb / f t,a P i s d i f f e r e n t i a l pressure in I b f / f t

2 !1'

V i s v e l o c i t y in f t / s e c

gc i s 32.17 Ibm f t / ( l b f sec)

f t lb fFo r

a i r , R'"

53.34 Ibm oR anddensi ty i s th e rec iprocal o f s p e c i f i c volume,

1v ' " f20(144)

53.34 (540)0.10 l b / f t 3 densi ty of a i r in the duct

V2

= 05.1.06 = 156. = h (),. P2(32.17) == TThis pressure energy (flow energy) i s equivalen t to a 156. foot (head) columnof a i r a t th e T an d P e x i s t i n g i n th e d u c t .

v = ~ 2 ( 3 2 . l 7 ) ( 1 5 6 )= 100 f t / s e c .

Duct c ross - sec t iona l area = ". d 2 = 0.785 f t 2 = A4

Volume flow ra te = 0.785 f t 2 (100 f t / s e c ) = 78.5 c f s . = AV = (1

Since.

78.5f t 3 0.10 ~ = 7.85 lb / sec . = q f m z - - x, sec f t 3 1'1

(2 ) Solut ion based on compress ible f lu id

KE i s converted to heat , r e su l t ing in a s tag n a t io n tempera ture , T t an d t h i ss agha s a corresponding i s en t rop ic pressure r i s e , AP to stagnat ion pressure , Ps tag

Basis : 1 Ibm o f f l u id flowing

Ts tag

Ps tag

For an i sen t rop i c compression:

where k = ~Cv

Tl + AT - T 2

P l + 6p O: P2

Fo r a i r : cp ~ 0.240 BTU/lboRCv = 0.171 BTU/lboR

k = 1.40(k i s of t en re fe r red t o as r )

These r a t ios may be handled using an ycons i s t en t absolu te u n i t s .


37/48

0.2857(1.00540)

or


To avoid inaccuracy, i t i s necessary to us e as many d i g i t s as poss ib l e .

V2

'" 2g c JC pAT where J = 778 f t Ib/BTU

V = l /2 (3 2 .1 7 ) (7 7 8 ) (0 .2 4 0 ) (0 .8 3 2 ) '" 100. f t / s e c .

Note t h a t t h i s r e s u l t agrees with the incompress ible e a s i e r case wherev e l o c i t i e s are low an d Ts tag an d Ps t ag are no t much d i f f e r e n t from the

duc t tempera ture an d pressure . The compressib le so lu t ion i s the onlyv a l i d on e a t ve ry high v e l o c i t i e s .

Summary: v '" 100 f t / s e c m '" 7 .8 5 I b / s e c .

T HE RM O 7 5

During complete s to ich io met r ic combustion of 1 lb . mole o f methane (CH4 ) witha i r , th e number of l b . moles of n i t ro g en and othe r i n e r t s t h a t pass through th ecombust ion zone i s n eares t to :

(a ) 0 .79(b ) 3. 9( c ) 5. 6

( d ) 7. 5( e ) 16.1

CH4 + 2 0 2 ~CO2 + 2H 20

1 lb . mole o f methane r eq u i r es 2 lb . moles of 02 fo r complete combust ion.

Air i s 21% v o l . 02 and 79% vol . (N 2+ i n e r t s ) .

2 lb . moles o f 02 i s suppl ied by 0 . ; 1 = 9.52 l b . moles of a i r , whi ch contains7.52 lb . moles o f (N 2

+ i n e r t s )

Answer i s (d )

T HE RM O 7 6

During combust ion o f hydrocarbon fu e l s with l e s s than s to ich io met r ic a i r , th eproducts depend on th e amount o f a i r suppl ied an d could be a l l o f thefol lowing except :

(a) Unburned f u e l , C, CO 2 an d H2 0

(b ) Unburned fue l , C, CO and H 2 0

(c ) C, CO, CO2 and H 2 0

(d ) C, CO an d H20(e) CO, CO2 and H20

Combustion o f hydrocarb on fu e l s occurs s tepwise i n the fol lowing sequence:f i r s t : a l l hydrogen to H20 , leaving carbon

next:

l a s t :

a l l carbon to CO

a l l CO to CO2


38/48


This i s summarized in th e t ab l e below:

Products from Combustion Zone

UnburnedAmount of Air Fuel Carbon CO CO2 H2O

excess a i r X X

s to ich io met r ic X X

s l i g h t l y fuel r i ch X X

less1 than X X Xs to ich iometr ic a i r

+ X X X X

very fuel r i ch X X X

Asa

p r a c t i c a l matter i t i s d i f f i c u l t to obtaina

CO-free s t ack ga s du eto slow k ine t i cs o f CO combustion to C02, and due to equil ibr ium inthe water ga s s h i f t reac t ion .

Choice (a ) i n th e problem does no t occur . .THERMO 7 7

( a )

(b )

( a )

(b )

Write th e combust ion equation for a heavy o i l whose composit ion averagesC17 H36

with complete combust ion in theo re t i ca l a i r .

Determine the pounds o f t h e o r e t i c a l a i r p er pound o f C17H

36 f u e l .

During stoichiometr ic combust ion a l l hydrogen i s burned to water an dcarbon to CO2 ,

+ +

From the ca lcu la t ion below, 97.8 (N 2+ i n e r t s ) pass through unreacted .

Basis of ca l cu l a t i on : 1 lb . mole o f C17H

36

26 lb . moles o f 02 a re r eq u i r ed . Air i s 21% vo l . (21 mole%) 02 an d

79% vol . (79 mole%) N2 + i n e r t s .

26 lb . moles 0221

X lb . moles (N 2+ i n e r t s )

79Y lb . moles a i r

10 0

x = 97.8 lb . moles (N2 + ine r t s ) tha t pass through the combust ion zone

and Y = 123.8 lb . moles a i r r eq u i r ed .

Since mol. wt . o f C17 H36= 17(12) + 1(36) - 240

an d th e mol. wt . of a i r i s 28.95

Ib a i rlb fuel

123.8(28.95) = 14.9 a i r / f u e l wt. r a t i o . .1(240)


39/48


THERMO 7 8

Gasol ine ha s a heat o f combustion o f 21,000 BTU/lb and a spec i f i c weight o f6.17 l b / g a l . Food has an average heat of combustion of roughly 6000 BTU/lb,an d th e average human's ra te o f energy comsumption i s approx imately 12,000BTU/day. The automobile ' s average r a te of fuel consumption in c i ty drivingi s 3 gallons per hour.

Given t h a t th e equat ion for th e combust ion of both gasoline an d food i sFuel + Oxygen = Carbon Dioxide + Water,

an d given t h a t the chemical formulas fo r gasoline an d food are CSH1S and

C6 H12 0 6 , r e spec t ive ly, compute th e r a t i o o f th e amount o f a i r used pe r hour

by th e automobile to th e amount used by an average human.

Basis of ca l cu l a t i on : 1 hour

( 1) Automob i l eStoich iometr ic equat ion:

1 lb . mole of fue l , mol. wt . = S(12) + l S ( l ) = 114 lb . fue l ,r equ i re s 12.5 lbmoles of oxygen

Fuel consumption = 3 ga l /h r x 6.17 Ib /ga lOxygen requ i rement i s obta ined by r a t i o :

l S . 5 l I b / h r.

l S . 5 l- - - =11 4

(2) Human

lhmoles oxyp::en12 .5

Jxygen = 2.030 lb moles/hr.

Sto ich iometr ic equat ion:

1 lb . mole o f food, mol. wt . = 6(12) + 12(1) + 6(16) = ISO lb . food,requires 6 lb . moles o f oxygen.

12,000 BTU/dayFood consumption = 6,000 BTU/lb. ~ = 0.0833 lb /h r.x 24 hrOxygen requirement i s obta ined by r a t i o :

(3 ) Comparison

THERMO 79

0.0833180

lbmoles oxvgen6

Oxygen E 0.00278 lb moles /hr.

Auto Air ~ Auto OxygenHuman Air Human Oxygen

= 2.030 .O.0027 B 731.

Propane (C 3H8 ) i s burned in a hea t e r with 20% excess a i r a t ambient

co n d i t io n s (68F, 14.7 p s i a , 307. r e l a t i v e humidity) . Assume complete

combust ion to water an d carbon dioxide.

Per pound mole of propane, ca lcu la te :

(a ) volume o f s t ack gases cooled to 800F.

(b ) composit ion of s t ack gases (voL i. )

(c ) dew poin t o f s t ack gases .


40/48


Ca l cu l a t i on bas i s : 1 lb . mole o f C3Hs

(1 ) Stoichiometr ic equat ion:

5 lb . moles 02 a re r e quired for complete combust ion. At 20% excess a i r ,

6 lb . moles en te r combust ion zone accompanied by some moistu re an d much

ni t rogen plus i n e r t s .

(2 ) Calcula t ion of mois ture an d N2 plus ine r t s car r ied by in take a i r

From th e sa tu r a ted steam t a b l e , water vapor pre s s u re a t 6SoF i s 0.339 ps i a ;

t h i s corresponds to 100% r e l a t i v e humid i ty. At 30% R.H. th e water vapor

pressure i s (0 .30)(0 .339) = 0.102 p s i a .0.102Mole f r ac t i on of water vapor i s ~ = 0.0069 5 , or 0.695% and mole

f r ac t io n of dry a i r i s 0 .993 .

Al terna t e l y, from a psychrometr ic cha r t , th e water contained in intake a i r

i s 31 . g r a i n s / ( l b . dr y a i r ) . At 7000 g ra i n s / l b . , t h i s i s0.0044 lb . H2

0 / ( l b . dry a i r ) .

moles water vaporMole f r ac t i on of water vapor moles dr y a i r + moles water vapor

0.0044IS

_ 1 _ + 0 . 00442S.9 - l - S -

0.070

Mole f r ac t io n of dry a i r 1 - 0.070 0.993

6 lb . moles 02 a re ca r r i ed by 1 ~ ~ ( 6 )2S.6 lb . moles dr y a i r or by

~ = 2S.S lb . moles moist a i r conta ining 0 2 lb moles H .993 . 2 .Ni t rogen + i n e r t s c a r r i ed = 2S.6 - 6 .0 = 22.6 lb . moles .

(3 ) Summary lb moles en te r ing an d leaving combustion zone

N2 plusC3 HS O2 i n e r t s H2O CO2 Tota l

Enter ing 1 .0 6.0 22 .6 0 .2 0 29.S

Leaving 0 1.0 22 .6 0 .2+4 . 0 3 .0 30.S= 4 .2

Note from th e s to ich io met r ic equat ion t ha t 6 moles of r eac tan ts form7 moles of products - j u s t i f y i n g change in moles in an d out above.

(4 ) Compu t a t i on of s tack ga s volume1545 f t lb f

Us e id ea l ga s law PV = nRT, where R = OR ' to ca l cu l a t e volumeIbm moleof 30.S moles s t a c k gases a t SOO F (1260 o R) and 14.7 p s i a .

l4.7(144)V = 30.S(1545(1260) : . V 2S,300 f t 3

Alterna te ca l cu l a t i on based on 1 Ib.mole 359 f t 3 a t 32c F, 14.7 ps i a :

2S,400 f t 3


41/48


(5 ) Stack ga s composit ion (vo l . % mole %)~

N2 p lusComponent Excess 2

i n e r t s CO2 H2O Tota l

l b . moles 1 .0 22.6 3 .0 4 .2 30.8

v o l . % 3.25 73.35 9.75 13.65 100.00 (6 ) Dew p o in t of s tack ga sDetermine temperatu re where vapor pressure of wa te r equals p a r t i a lp ressu re of water in s t a c k gas .

From above t a b l e , vo l . % H20= 13.65 mole f r a c t i o n = 0.1365

Wat e r p a r t i a l pre s s u re in s tack ga s = (14.7 p s ia ) (0 .1 3 6 5 ) = 2 .01 p s i a .Consul t sa tu ra t ed steam t ab le for vapor pressure of water a t va r iouste mpe ra tu re s . Find:

120F 1.692 ps i a130F 2.221 p s i a

Although vapor pre s s u re does no t vary l i n e a r l y with tempera ture , l i n e a r

i n t e r p o l a t i o n i s adequate over a sh o r t temperatu re range an d i s as a c c ura teas the problem input d a ta j u s t i f y . Conclude t h a t dew p o in t = 126F

Summary of r e s u l t s

(a ) Stack ga s volume = 28,300 cf / lb .mo le C3H8 (b ) Stack ga s composit ion, v o l . % (see t ab l e a b o v e ) "(c ) Stack ga s dew p o in t = 1 2 6 F .

THERMO 8 0

Theore t i ca l l y, how many pounds of a i r a re re qu i re d fo r completecombust ion of 10 pounds o f ethane ga s (C2H6), an d how many pounds of

water vapor are produced in th e products of combust ion?Assume th e a i r c on ta ins 237. oxygen by weight .

Write and balance th e combustion equat ion

The molecu lar weigh ts must be ca lcu la ted

C = 1230

H = 132

16

44

Thus 2 moles of CZH6 (2 x 30 Ibs) plus 7 moles o f 02 (7 x 32 lbs )

y ie ld 4 moles of CO2 (4 x 44 Ibs) plus 6 moles of H20 (6 x 18 l b s ) .

Therefore 10 Ib s of C2H6 require

10(7 x 32 ) = 37.33 Ib s of oxygen60

Since a i r i s 231. oxygen by weight,

pounds of a i r required 37.33G . T I 162 pounds

10 Ib s of C2 H 6 wi l l produce

1060(6 x 18) = 18 lb s o f water vapor 4t


42/48


THERMO 81

I f the heat of combust ion of kerosene (ClO

H22

) i s 1625 K c al /mo1e

and kerosene co s t s 25 cents per g a l lo n , what i s th e cos t o f 25 K ca l

ob ta ina b le by burn ing kero s ene? Assume complete combustion of

kerosene, th e d en s i ty of kerosene i s 0.80 grams per c . c . , an d t h a t

on e f lu id ounce i s equal to 30 c . c .

Atomic 1,.Teights - 16mole = gram - molecu lar weight

25number of moles burned = 1625

C - 12 H - 1

165 mole kerosene

CIO H22 ha s a molecu lar weight of 10(12)+ 22(1) = 142

1kerosene burned = 65 x 142 2.185 gram s

kerosene co s t2.185 gms

0 .8 gmscc

x1 1 1 x 25 cents

30 cc x 32 f l oz x 4 q t -gay-f l oz qua r t gal

0.0178 c e n t s .

THERMO 8 2

Given the following d a t a fo r propane:

Formula C3H8

Weight ( l i qu id )

Sp ec i f ic Volume

REQUIRED: Compute the fol lowing:

4 . 24 l b /ga l

8.55 f t 3 ga s /1 b

(a ) Ib s O2 pe r lb of ga s fo r complete combust ion

(b ) lbs CO2 produced per 1b of ga s burned

(c ) lbs H20 produced per lb of ga s burned

F i r s t , write th e balanced chemica l e qua t ion .

Knowing th e atomic weights , compute the molecu lar weigh ts .

44 32 44 18 (C = 12 , = 16 , H = 1)So , 44 pounds o f C3H8 combine with 5 x 32 = 160 pounds of oxygen to form

3 x 44 132 pounds of CO2 plus 4 x 18 = 72 pounds of H2O.

(a )160

3.64 pounds o f 02 per pound of C3H8 4 =132

(b ) 44 3.0 pounds o f CO2 pe r pound of C3H8 (c ) 72 1.64 pounds of H20 per pound o f C3H8 44


43/48


At s teady s t a t e the same Q flows a c ros s each mate r i a l an d tempera tures descendin d i r e c t r e l a t i o n to thermal r e s i s t ance ( rec ip roca l o f conduc t iv i t y ) .

R e s i s t a nc e o f br i ck a ~ = 0.667 f tk 0 .40 BTU/(ft 2 F / f t )

Resis tance o f Ce l lo tex = = 0.OB3 f t 2 .77k 0.03

Res is tance o f p l a s t e r = = 0.042 f tk 0.30

0.14

~ T o v e r a l lQ - - -- - ~- - ~ - - - -- - -t o t a l r e s i s t a n c e

T4 - T3

To ta l Resis tance

6 T fo r p l a s t e rxk fo r p l a s t e r

4.58 ftZoF/(BTU)

T3 - TZ2.77

TZ - Tl1.67 - 10.B5 BTU/hr.

1 .5F Since T4 = 70 . o f , T3 = 6B.5F

T3 - T Z = 30.3-F Since T =3 6B.SoF, T Z = 38.Z-FT Z - Tl '" lB.2"F Since T2 3B.ZoF, Tl - 20F ( i n agreement

wi th given d a ta )

Answer i s (e )

THERM O 8 5

The ra te o f heat t r a n s f e r through a given sec t io n o f a uniform wall fo r agiven temperature di ffe rence i s :

Qk A t1 T)

whereL

(a ) Di r e c t l y p ro p o r t io n a l to th e thermal

(b )

(c )

Cd)

(e )

k

AflT

L

co n d u c t i v i t y and to th e th ickness o f th e w al l .

In v e r se ly p ro p o r t io n a l to the thermalconductiv i ty and d i r e c t l y propor t i ona l toth e th ickness o f th e wall .

Direct ly p ro p o r t io n a l to the thermalconductiv i ty and in v er se ly propor t i ona l toth e th ickness of th e w al l .

I n v er se ly p ro p o r t io n a l to th e thermalconductiv i ty and to th e th ickness of th e wall .

Independent of th e th ickness of th e w al l .

conductiv i tysurface areatemperatu re difference

wal l t h i c kne s s

Answer i s ( c ) .


44/48

Thermodynamics I 179

THERMO 83

Th e temperatu re di ffe rence between th e two s id es of a so l id r ec tan g u la r s lab of

a r ea A and th ickness L as shown below i s ~ T . The he a t t r ansfe r red through thes lab by conduct ion in time t i s propor t ional to :

(a ) ALAT t

(b ) AL ..AIt -

(c ) AL _ t_~ + < s - < S -

AT

(d )A ~ Tt

L

(e) A( A T) 2 tL

Heat t r ansfe r red through s lab by conduction in time t i s :

Q = kAAT where k

(BTU f t ) ( f t 2 ) (AT)oF t hrs = BTU = k ( A AT t )

Q = k hr f t2 OF A L f t L

k i s the c o e f f i c i e n t of thermal conductiv i ty o f the mate r i a l , hence he a tt rans fe r in a given mate r i a l i s propor t ional to th e othe r v a r i a b l e s .

Answer i s (d ) Gt

THERMO 8 4

- - - - - 1 ' - - 1 "

Given the composi te wall shown with s teady s t a t e o u te r temperatu re Tl = 20Fand inne r tempera ture T4 = 70F, which of the following i s nea re s t to T3 inoF?

(a ) 27 .(b ) 3 8 .(c ) 46 .(d ) 5 8 .(e ) 69 .


45/48


THERMO 86

The h ea t l o s s pe r hour through 1 sq f t of furnace wa l l 18 " th icki s 520 Btu. The in s id e wal l temperatu re i s 1900F, an d i t s averagethermal co n d u c t iv i ty i s 0 . 61 Btu/hr f t o f .

The o u ts id e su r f ace temperatu re of th e wall i s n eares t to

(a ) 100F

A t 2 )k r ;( t i-

Solving fo r t2

THERM O 8 7

(b ) 300F(c ) 600F(d ) 1000F(e ) 1900F

where t l 1900F t =2 o u ts id e temp.

k 0 . 61 Btu/hr f t of

g 520 Btu/hrA

L 1 .5 f t

- 520 ~ + 1900 = - 1279 + 19000.61Answe r i s (c )

6 2 1 F .

Heat i s t rans fe r red by conduct ion from l e f t to r i g h t through the compo s i t ewal l shown in th e drawing below:

OUTER

Q

~ X l= 5 " - - - - t - - - x 2 = 3" - - - I X 3 = I""I : . : . : . : . : . : . : . : . : . : - : . } : . ~ o n cre e

;k l '" 0 .8 0Rock Woolk2 = 0.04

INNER

3 ,., 0.07BTU/hr ftOF

Q

Assume the t h ree ma te r i a l s are in good thermal co n tac t and no s i g n i f i c a n t f i lmc o e f f i c i e n t s e x i s t a t any o f th e i n t e r f a c e s . Determine which o f the fo l lowing

numbers most nearly represents th e overa l l c o e f f i c i e n t U in BTU/hr f t 2 F .

(a ) 0.04

( b ) 0.13(c ) 0.35(d ) 0.91( e ) 1.92=

Ove ra l l c o e f f i c i e n t U, thermal co n d u c t iv i ty, k an d film c o e f f i c i e n t , h, a re;Z,

rec ip roca l s of t h e i r thermal res i s t ances . Thermal res i s t ances in se r ie s a re


46/48

182 / Thermodynilmics

handled analogously to se r ie s e l e c t r i c a l res i s t ances , hence

U in BTU/(hr f t 2 oF)1

Overall c o e f f i c i e n t U i s used in a simpl i f ied conduction equat ion Q = A ~ T .In th i s problem:

1 1U = . , .------,--, ,----, ,--:, . . . ,--5/12 + 3/12 + 1/12 0.52 + 6.25 + 1.19

0.126 BTUI (hr f t 2o F ) .

O.BO 0.04 0.07

Answer i s (b )

THERMO 88

Given an inner wall a tsurroundings a t 40F.

a 15 mph wind i s about

BOOF and an oute rFilm c o e f f i c i e n t ,

7 BTU/(hrf t 2 ~ F ) .

wall exposed to ambient wind andh , for convect ive hea t t r a n s f e r a t

:::: ::::::::

T 1!!!!!!!!!!!!!!!!!!!!!!!!!!!1T2

BOoF [!i!!!!!!!!!!!!!!!!!!!!!!!!,1fo r wind and

Q surroundings

k ~ 0.30 BTU/(ft 2 hrOF/f t)

Ignoring any rad ia t ion l osse s ca l cu l a t e an ove ra l l coeff i c ienc t fo r th econduction an d convect ion lo s ses .

(a ) 0.14(b ) O.BO( c ) 1.25(d ) 7. 1(e ) B. 2

Since conduction an d convect ion a re based o n 6 T, absolu te tempera tures are notr eq u i r ed . Fo r s teady s t a t e , heat conducted through wal l = heat l o s t byconvection:

Q

Q ca n s imi la r ly be expressed by an ove ra l l coeff i c ien t

Q = UA(TI - T 3 )u i s calcula ted in a manner analogous to tha t used for thermalconduc t iv i t i e s i n s e r i e s .

U -=1

1 xl+ - +

hI k l

U = -::--_1,-;--;c:-=-.!. + 4 /127 0.30

Alternate Solu tion

-::--::-:-:::----,1=-:--:-:- = O. BO BTUI (h r f t 2 o f ) 0.143 + 1.11

(1 )

( 2 )

( 3 )

I f the quest ion ha d asked fo r T2 , equat ion (1 ) would have given i t . Simi la r ly

th e so lu t ion fo r Q from (1 ) would yie ld U from (2):


47/48


Q - 0430 (80 - T2 ) = 7(T 2 - 40 )IT

Q - 72 - 0.9T 2 = 7T2 - 280 an dQ = 32 BTU/(hr f t 2 )

Q = UA(T l - T3 ) - 32 8 U(1)(80 - 40 ) U = ; = 0.80 BTU/(hr f t 2 o F ) t tNote t h a t t h i s a l te rna te approach becomes in o p era t iv e when more than one filmc o e f f i c i e n t an d on e conductiv i ty are involved, but lack ing informat ion on th ei n t e r f a c e temperatures .

Answer i s (b )

T HE RM O 8 9

A metal object a t 120 0F i s s e t on an in su la t ing pad to cool . The temperaturei n i t i a l l y f a l l s from 120 0F to 1000F in 12 minutes . Surroundings a re a t 6S oF.Find time required for tha t o b jec t to cont inue to cool from 98 0F to 80 0F.Assume neg l ig ib le conduction and rad ia t ion lo s ses i n both cases.

ra te o f heat loss from block = ra te of convective heat t r a n s f e rde

-fCV .- .. hA Sd t

, where f - dens i ty, C = heat capac i ty, V = volume, te= temperatu re di ffe rence with surroundings ,

t ime,

h .. convective he a t t r a n s f e r c o e f f i c i e n t , and A - area .Rearrange an d i n t e g r a t e us ing th e f i r s t cool ing l i m i t s :

]e,=3s

0F i h A ' ] t2=12min. )

In e - ) t In(s3sS) = f.. ~ V( 12 - 0 )

G,=SSoF f C V tl=Ornin. ~ r ~

Solve fo r th e unspecif ied constan ts : ( - ~ ~ V )- -.0377 min- lIn se r t the new problem data and re-evaluate th e i n t e g r a l . 8

3= 98-65, 49

4=80-65,

t 4 - t 3- unknown.

In e ~ ) - .0377 ( t4 - t 3) t 4 - t 3 = 20.9 min This demons tra tes th e exponential nature o f tempera ture change with t ime, whereth e heat t r a n s f e r r a t e by conduction o r by convect ion v a r i e s so le ly withtempera ture d i ffe rence , i . e . in th e absence of heat genera ted in , o r los t fromth e objec t i n o the r ways.

THERMO 9 0

A s h e l l an d tube br ine cooler cools 150 gallons of b r ine p er minute from16F to 12F, using ammonia a t SOF. The e f f e c t i v e out s ide area of th etubes i s 31 0 f t 2 . The brine ha s a s p e c i f i c grav i ty of 1 .2 an d a spec i f i c

heat o f 0 .7 0 .(a ) Compute th e r a t i n g o f the cooler

in tons o f r e f r i g e r a t i o n .

(b ) Compute the c o e f f i c i e n t of heatt r a n s f e r , U, in BTU/(hr f t 2 F) .


48/48


(a) Heat of Cool in g = Qc = ~ epB(T l - T2 )

where ~ = br ine mass flow r a te

(b )

CpB = br ine spec i f i c heat = 0 .70

Tl br ine i n l e t temperature = 16F

T2 b r i ne o u t l e t temperature = 12F

150 gal xmin

cu f t7 .4 8 ga l 20 f t

3 /min

1 to n o f r e f r i g e r a t i o n = 200 Btu/min = 12,000 Btu/hr 62.4 Ib x 1.2 x 20 cu f t~ = cu f t min

x 60 min = 90,000 lbm/hrhr

Q = 90,000 x 0 .70(16c 12) = 252,000 Btu/min

Tons o f Ref r ige ra t i on = 252,00012,000 21 .

u where q he a t t r ansfe r in Btu/hr

e ffec t i ve area of the tubes in sq f t

~ t m lo g mean temp difference

~ tm

16 _ 5 = 11F

12 - 5 = 7F

Weare us ing the logari thmic mean temperatu re di ffe rence whichap p l ies to co u n te r cu r r en t he a t exchange; i d e a l l y a cor rec t ion

should be applied as t h i s i s a s h e l l and tube exchanger.

U = q = 252,000 BTU/hrAr ~ t m 31 0 f t 2 (8 . 9F)

91 .4 BTU/(hr f t 2 Op)

TH ER MO 91

Which o f th e following i s not a usual express ion o f th e power/unit areaStefan-Boltzmann co n s tan t fo r black body radia t ion?

(a ) 1.36 x 10 - 12 c a l / ( s e c cm2oK4)

(b ) 5 .68 x 10 - 5 e rgs / ( sec cm2oK4 )

(c) 5.68 x 10 -8

watts / (m2o

K4

)(d ) 0.171 x 10 - 8 BTU/(ft 2 h r ~ 4 )

(e) 5.68 x 10 - 8 coulombs/(sec m 2 ~ 4 )

All a re numerica l ly c o r r e c t conversion of th e constan t in terms of power pe ru n i t area

thermodynamics review problems

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