1

FE Thermodynamics Review

Outline of Thermodynamics topicsSuggestions for FE Exam preparationThermo context of exam & reference handbookExample test questions– morning session– afternoon session

Exam Pointers

Familiarize yourself with the reference handbookMake educated guesses if you can’t solve the problem (or don’t have the time)– cross out answers you know are incorrect– work problems backwards using the available

answers

If a problem seems difficult– guess and mark it– work on it later if you have the time

2

FE pointers - Thermo

Know how to use ideal gas law & steam tables to evaluate propertiesBe able to apply 1st Law to open/closed systemsKnow basic definitions/concepts (e.g., adiabatic, isentropic, Carnot efficiency)Know how to calculate ∆u, ∆h, and ∆s for ideal gases & liquids/solids

FE Exam Thermo Problems

Morning session– 11 questions out of 120

Afternoon session (general)– 6 questions out of 60

Afternoon session (ME discipline)– 6 questions out of 60

Afternoon session (ChE discipline)– 6 questions out of 60

3

SAMPLE MORNINGQUESTIONS

A liquid boils when its vapor pressure is equal to

a) one atmosphere of pressureb) the gage pressurec) the absolute pressured) the ambient pressure

Which thermodynamic property is the best measure of molecular activity of a substance?

a) enthalpyb) internal energyc) entropyd) external energy

4

What is the maximum thermal efficiency possible for a power cycle operating between 600oC and 110oC?

a) 47%b) 56%c) 63%d) 74%

All real gases deviate to some extent from the ideal gas law. For which of the following conditions are the deviations smallest?

a) high temperatures and low volumesb) high temperatures and low pressuresc) high pressures and low volumes

d) high pressures and low temperatures

For every gas there is a particular temperature above which the properties of the gas cannot be distinguished from the properties of a liquid no matter how great the pressure. This temperature is the

a) absolute temperatureb) saturation temperaturec) standard temperatured) critical temperature

Which of the following statements is the closest interpretation of the first law of thermodynamics for a closed system?

a) Mass within a closed control volume does not change.b) Net energy crossing the system boundary is the change in

energy inside the system.c) Change of total energy is equal to the rate of work performed.d) All real processes tend toward increased entropy.

5

The net entropy change of the universe during an adiabatic, irreversible process is:

a) positiveb) negativec) zerod)

A Carnot refrigerator has a coefficient of performance of 10. If the refrigerator’s interior is to be kept at -45oC, the temperature of the refrigerator’s high temperature reservoir is most nearly?

a) 250 Kb) 270 Kc) 300 Kd) 350 K

∫∂TQ

If both the volume and the pressure of an ideal gas contained in a cylinder are doubled, the temperature is:

a) constantb) doubledc) quadrupledd) halved

A container contains half liquid water by volume and half vapor by volume. Select the best estimate of the quality if the pressure is atmospheric:

a) x = 0b) x = 0.0006c) x = 0.1d) x = 0.5

6

Steam initially at 1 MPa and 200oC expands in a turbine to 40oC and 83% quality. What is the change in entropy?

a) -0.35 kJ/kg-Kb) 0.00 kJ/kg-Kc) 0.26 kJ/kg-Kd) 0.73 kJ/kg-K

What is the change in internal energy of air (assume to be an ideal gas) cooled from 550oC to 100oC?

a) 320 kJ/kgb) 390 kJ/kgc) 450 kJ/kgd) 550 kJ/kg

Air is compressed in a piston-cylinder arrangement to 1/10th of its initial volume. If the initial temperature is 35oC and the process is frictionless and adiabatic what is the final temperature?

a) 350Kb) 360Kc) 620Kd) 770K

Air (an ideal gas) undergoes an adiabatic process in a closed rigid container. The reversible work done is:

a) 0b) (P2 – P1)Vc) cp∆Td) T∆S

7

SAMPLE AFTERNOON QUESTIONS

The next 3 problems refer to a power plant operating on an ideal Rankine steam cycle. The plant operates between the pressure limits of 1 MPa and 5.628 kPa. The temperature of the steam entering the turbine is 500oC . The total gross power generated is 300 MW.

What is the quality of the steam after it has expanded in the turbine?

a) 78%b) 92%c) 97%d) 100%

8

What is the enthalpy of the steam as it enters the condenser?

a) 2380 kJ/kg

b) 2420 kJ/kgc) 2560 kJ/kgd) 2600 kJ/kg

What is the total steam flow rate through the turbine?

a) 10 kg/sb) 110 kg/sc) 200 kg/sd) 270 kg/s

if

EIT EIJH E£VIE W(

~pple Problems on ~ermodyn&mics

1.

li'ift l.bm of saturated stem .at hooor is contained in a closed ,piston-cylinderarrangement. The steam expmds reversibly and isothermally to a final.pressure of 60 psia. Dete~e: (i) the initial volume.. (ii) the heattranster1'ed" and (iii) the work done by the steam.(i) (ii) (11.1)(A.) (A) 0 Btu (A) 0 Btu(B) (B) -800 Btu (B) 8()Q Btu(C) (0) +800 Btu (C) + 32., Btu(D) (D) + 32., Btu (D) +67, Btu(E) (E) +12, Btu (E) +12, Btu

0.37.3

!t39.3'15 .f't35.805 .f't39.222fi30.093f't3

2.

~)team ente~ a turbine at 300 psia and 700or \lith a veloi:1ty of 200 !'t/sec.The mass. .flow rate of steam is 10,,000 l'bm/hr" and the heat transfer ratefrom the tUrbine is .30,,000 Btu./hr. The stesm .flows steadily through theturbine and leaves the tm-bine at 1; psia as a saturated vapor with -avelocity of 600 rt/sec. Detem.1ne: (i) the powr output of the turbine,and (11) the irreversibility for the process" if the SU!"rOtmd1ng temperatureis ;37~. .(i) (11)(A) 890 hp (A.) u;6,hOO Btoa/hr(B) ~8" 700 hp (B) .30,000 ~(C) 209 hp .(C) h23,400 Btu/hr(D) 82.3 hp (D) .30,800 Btu/hr(E) 6~ hp (E) 0 Btoa/hr .

J~reon-12 at 180 psia and lOOor enters an expansion valva in. a va.por-compre.ssionrefrigeration system. The flow through the valve is adiabatic, steady", andwith negligible changes in kinetic energy. Dete~e the temperature ofthe Freon (if in the suJ=erheated vapor regime) or the quality of the Freon(if in the two-phase region) leaving the expansion valve.

3.

-8.3or0.650

(D)(E)

B3oro.hOB0.350

(1)(B)(C)

It. ~l Oarnot cycle ref'rigerator operates in a roOD. in which the temperatureis 80~. It is required to transfer 10,000 Btwmin. fro~ the cold spacebeing held at -20or. Determine:. (1) th~ rate of change of' the entroP7of the room, and (ii) the power requirement of the refrigerator.(1) (11)(A) 125.0 Btu/min.-~ (A) 53.5 hp

..(B) -22. 7 Btu/min.-~ (B) 289.4 hp(0) -500 Btu/min.-~ (0) 235.8 hp(D) +22.7 Btu/min.-~ (D) 525.2 hp(E) 100.0 Btu/min. ~ (E) h2.4 hp

-2

s.. Air is c"anpressed in a reversible steady'-flow system from 1.5 psi a andl00~ to anrexit pres~ of 100 psia. The compression process ispolytropic, with a polytropic exponent of n -1.2.50 The air may- beas5'4ed to behave as an ideal ga.s with constant specific heats.Determine: (i) the change in entroPY'" (ii) the work of compression-oer \mit ma.ss, and (iii) the heat -ttansferred per unit ma8s.i) -(ii) (iii)(A) -0.1)0 Btu/lbm-~ (A) +62.0 Btu/lbm (A.) -8.9 Btu/l.bm,(B) -101.1 Btu/lbm~~ (B) -70.9 Btu/lbm (B) -26.6 Btu/lbm(C) +0.221 E~~~-~ (C) -88.6 Btu/lbm (C) -44.3 Btu/lbm(D) -0.039 Btu/l~-~ (D) +70.9 Btu/lbm (D) -+62.0 Btu/lbm(E) +0.039 Btu/lbm-~ (E) +hU.3 Btu/lbm (E) +44.3 Btu/lbm

A nozzle is designed-to "expand -air- reversibly and adiabatican.,. ~m 'l00 psiaand 80~ to an exit pressure of 20 psia. The iDlet kinetic ene~~ isnegligible, and the mass flow rate of air is 3 l~eec. Dete~e: (i) theexit velocity" and (ii) the exit cross eectional area.(i) (ii)(A) 1)10 it/sec. (A)(B) 342 it/sec. (B):C) 2.5.50 it/sec. (C),D) 1.547 rt/st.;~. (D;E) 43,100 it/sec. (E:

6.

22.9lu9.l0.391.491.76

)

0..5160..3450.4840.1990.250

13,100 kW19,500 kW

Uh7 kW7570 kW

17,600 kW

))

.).'jI'.:)J)

:11):A)

,B),C),D):E)

65.2$10.5%89.5%46.8%53.2$

39 psi14.7 psi

331 psi197 psi173 psi

(

.r"

(:(.(:( I

-3 -.

Methane (qH4) is burned with atmospheric air. The analysis of the products.of combustion" as determined by m Orsat apparatus (dry- analysis on amolal bas1s~ "is as folloW's z

9.

-84.8%.7.7$ ; Na a 2

The products of' combustion are at 1, psia. Determine: I(i) the air-fuelratio on a mass basis, (1.1) the percent theoretical air, and (ill) the dewpoint temperature or the products or combustion.(1) 1:..":'1 (11)(A.) 11.3l1bm, air/lbm fuel (J.(B) 9.,2 lbm, air/lbm, 1'u21 (B(c) 7.93 lbm air/lbm .fuel (C(D) 17.2 lbm, air/lbm, .fuel (D(E) 2,.9 lbm, air/lbm ruel (E

co .7.1% ; co .O.~ ; 0

iii)

A)BC)D,)E)

213~12S~

h2or63~84~

lS~lootS~83%

27zt

((((((

S:tesm is condensed in & surface condenser at 0.50 psia. Cooling vaterat a mass .flow rate or 26 x 105 J.bm/hr is supplied at an inlet tetlperatureof 68°r. The steaM enters the condenser at a quality or 0.900 and & mass£low rate of 8O,000 J.bm/hr. No sub-cooling of the condensate occurs.The overall coefficient of heat transfer is U -490 Btu/br-fta_~.Determine the square feet of condenser surface required. .

(A) J.8,900 fi2(B) 53,000 .t't2.(C) 16,200 .ft2(D) J.U,OOO .ft2, .(E) 32,400 ft2. .

10.

-4 -

.ANSWERS to Sample Problems on 'l'hermod3'namics.-

The system and the process on the T-s plane are shown below. PropertyvaJ.ues are taken .tram. the CO!nbustion Engineering Steam Tables.

1..

TQ

lLOOOY

8

'The initial specific volume is:

-v -1.8630 ft3/lbm at 400O:rJ, gJ,

'l'he .1.nj.tiaJ. vo1'ame or the steam is found from:

v

ms. (B)v -my3. 3.

Fc)r a reversible isothermal process, the energy transfer. as heat is given by:

.1.7134 Btu/lbm- ~ at 400°F and 60 psia82

.(.5.0)(860~)(1.7134 -1..5274) c (.5.0)(860°)(0.1860)ThE~nJ Q

ARS. (c)-+ 799.8 Btu

~

Jpp:lying the First Law of Thermodynaaics for a closed system, we find:

Q -W -U -U .m(u].2 ].2 a ]. a ].

The internal energy- values are .fO\md as .follows.

-u )

.1201.0 -(247.259)(144 1n~/ft2)(1.8630)/(778 ft-1bf/BtU)1

-1201.0 -85.3 .1115.7 Btu/lbm

-hu -pv 1 11.

u2.

-s -

(No.-l, cont'd).

w~ 2

W~ 2. ANS. CD)

2. ~rhe sy-stem is shown below. Propen,. values are taken from the CombustionEngineering $team Tables. .

1mh -1368.9 Btu/lbm (300 psia,

J. .700Cj-)

h2 -1150.9 Btu/lbm (15 psia,.sat.vap.)

s -1.6758 Btu/lb -~J. .m

s , -1. 7552 Btu/J.bm-~

= ---[~. -30,000 Btu/hr

:;t; I .II .

W

@

r

2

IJ

!@App1y:ing the First Law of Thermodyn~csfor steady flow, we obtain:

Q -W + m(h + ~/2g -h -~/2g ) .o1 ~ C 2 2 C

..~ ';;'2v. W/m .q. h -h .(y- -V )/2g~ 2 4 2 c.

V .(-30,000/10,000) + (1368.9 -1150.9) .(2002-6002)/(2)(32.2)(778)

The powr output is:

ms. CD)For stesq no " the irreversibility is given by:

...0I .R T (8 -S ) -Q .(10,,000)(537 )(1.7552 -1.6756) -(-30,,000)0 2 1 0.

I .426,,360 + 30,,000 .456,,360 Btu/hr ms. (A)

~

-799.8 -($.0)(2,.0) .799.8 -12$.0 -+674.8 Btu

<!>,-- -, @p

@180psia

~~

/

'- ---20psia ---J.. @

hApplying the First Law of '!'hemodynamics .f'or stead3-."f1ow" on a unit mass

basis"

q -w'" (h -h ) ...(~- V2)/2g ...(gig )Ez -z ) .0.12 12 C .c 1 2

T]le system. is adiabatic (q .0)" there is no mechanism .f'c:r doing work (w .0)"C11ange s in Jdnetic and potential ene rgies are negligible; therefore.. theFjLrst Law reduces to:

h -h.-

x2

~

ANS. (C)

-7

Th!~ Carnot. ref'rigeration system is shoiln below.-r ROOM The entroPY' change for the roan is

TH .80~ .,40~ found from the f'ollowing, since the.rooQ, is an isothermal 87stem:

h.

Qg--'l'g

6,Sroom~.

C JPJl O'l'REF.

The heat transfer rat..e to the roan. is:w

10,000

4400)~~ .10,,000 Btu/min.

QL-T (-) -(540°)(H T

L

-J.2,270 Btu/min.~r .OOLD SPACETL .-20or .hhO~ Then,

.6,Sroom12,270

5400-

.+ 22.7 Btu/min.-~~Sroom ANS. (D)

Note that the entropy of the room incre ases. because hea't is ~ to the room.

'I'he power requirement .for the Carnot re.frigerator is .round !'ran:

w -~ -~ -12,270 -10,000 .2270 Btu/min.

w -(2270 Btu/min.)(60 min./br.)/(2545 Btu!bp-hr.)

ANS. (A)w -53..5 hp

~

Jon alternate solution is as follows. The COP for a Carnot refrigerator is:

~-w

-5hO -b4oTH -TL

'lben,w .10,000/4.40 .2270 Btu/min. .53.5 hp

-B

T2

p (n-1)/n 100 ()-T (-!). -(5600,&)( -) J..2~-J. IJ..215 -(560°)(6.667)°.2.00

J. P 151

-(';-60°)(1.161) -R1A_),~T

8 -s2 1

s -82 J.

s -s2

~

ANS. CD)1

n(p v2 2

n R(T -T ).a ].

-- ~2 V dp -

, 1 -n

~

(1.25)(53.34)(818.4 -560)

1 -n

-ANS. (0)(778)(1 -1.25)

~

p (k-l)/k-T (-a.)J. .

pJ.

-(540°)(0.631) .340.9~

T -(S40~)( ~ )(~.~o-l)/~.~O100

., (5hOO)(O.20)O.2862

T2

or

-9

ANS. (D)

~~e specific volume at the exit is found frOM the equation of stete,

(S.3..3u)(.3hO.90)RT--=.-P2

v2. (20)(lh4)

.-m .J. V Iv

2 2~['he required e.xit cross sectional. area. is fo1md from:

2.mv

--2-v

2

.0.01224 ft2(3.0)(6.314)

(1547)

ANS. (E)

~

7. TJle system and the cycle on the temperature-entropy plane are sho n below..Ul property. values are taken from the Canbustion Engineering S~am. Tables.

T@

700~m .100,,000

lbm/hr

~~ t~--

@

@ ".~~@

POMP

-.~~ '4 -

.~~]:1 . 69.73 Btu/1~ ; V

1

P ) .69.73 + (0.016136)(400 -1)(144)/(778) .69.73 + 1.191

1.

-h + v ( p -~ ~ 2-

-70.92 Btu/l~

h2

]~2.

]:13 3

(No.6, cont'd)

-10 -(No.7, cont'd) The expansion through the turbine is isentr~icj therefore,the quality ~t the exit of the turbine (point 4) is found from:

The enthalPY' at the en t of the turbine .is:

-69.73++ (O.8l7l)(lO36~) -69.73. 646.6h -h +x hr" ".fg"-.916.11 Btu/lbm

"h

4

Jlpplying the First Law to the turbine, for steady now with negligiblec:hanges in kinetic and potential energies,

...Wt .m (h -h ) -(lOO,OOO)(l363.h -9l6JL) .(100,000) (4h7.0).

3 4

.4.47 X 10? Btu/hr .(4.47)(10?)/(34l3 Btu/kW-hr).Wt.Wt -13,100 kW ANS. (A)

~

T.t1e pump work is found frOm:

W .m(h -h ) .(100,000)(-1.19) --119,000 Btu/hrP 1. 2

T11e net work output of the cycle is: .

Wnet .(U47.0 -1..19)(105) .4.4S8 x 1.0'7 Btu/hr .13,060 kW

The heat tre.nsfer rate in the steam boiler is found by appl;;ing the FirstLaw to the boiler.

Th!! thermal efficiency for the cycle is .found trau:

..'7~et 4.4,8 x .10'? 0.34, -34.$%

~ANS. (B)

~12.925

X 10"

-11

8.

Tbe cy-cle 'on the p-v and T-s planes is sho\o'n below. Properties or air are:'~ .'O.2hO Pt.u/lbm-~ ; Cv .O.l71lL Btu/lbm-Op. ; R .53.34 rt-lbr/l~-Op.:k .J..400 .

4500Dr : @@ @

@p @T

@@)

@---@

lh.7ps:1a (;)

v s

The temperature at the end of the compression process is found from the

isentropic relationship,

v .J.

The pres~ at the end of the cornpression stroke is:

k

The spe cific 'VOlume at point 3 is:

(S3.3U) (4960°)RT-~-P3

.3.107 .ft3/J.bmV3 (591.h)(~) .

The temperature at point 4 is fO'1nd frm the isentropic relationship,

k-l

qH .h -h3 2

The speci1'ic vol1me at point 1 is:

The heat added during the constant-pressure caa.bustion process is:

AHS. (E)

net work(442.9)(778)Wnet

--v -v

1 2

mep -

.

.pi~on displacement

mep .196. 7 1b,rl1n~AIlS. CD)

~

-~

7.1

CO2

+ 0.4 co + 7.,7 02

+ 84.8 N + C H 02. 2.

2

a -7.1 + 0.4 -7.50

4a.

2c , or c .2a. (2)(7.50) -15.00

(1 + 3.76) bAFR.-~ -- -

a (7.50)

..ANS. (E)

co-~" 2 2 2

1.3 -

(No. 9, c~t'd) The theoretical air-fuel ratio is:.AFRtheo.

'~e percent theoretical air is:

-(2 + 7.52)/(1) .9.52 lbmole air/lbmOle fuel

AFR 14.31ANS. (A)%TA. . -1..50 -1.50 % theoretical air.

AFRtheo. 9.$2

The mole fraction of vater in the products is:

y(H 0) .(1,.00)/(100.0 + 1,.00) .0~3042;

The partial pressure of the water vapor in the products of c~bustion is:

p .P ,.(H 0) .(1, psia.)(O.l304) -1.9,6, psiav m 2

From the Combustion Engineering Steam Tab1es~ the sa.turation tempera-turecorresponding to this pressure is:

ANS. (B)

~10.

:From the Combustion" Engine:ring Steam TablesJ we find the .following propertie~:

T -79.586°F at 0.50 psia Ste am0.50 psis.

47 .62 Btu/l~h -:r

T.68°rT 21.

The change in entha.1w of the stem: is:Condensate

L::\.h .(h f + x h ) -h f .x hfg fg

~h .(O.900)(lOh8.6) .991.36 Btu/lbm

The heat transfer rate from the steam is:

The exit temperature of the cooling water is:

-660 + (7.93)(107)/(26)(108)(1.0) .660 + 3.0So. 7l.0S~T2

The log~ean temperature difference is:

3.050

0.3055

(79.586 -68) -(79.586 -71.05)-9.9SoFIRTD . -

lnCn.586/8.536)

ANS. (c)I -DIU tKTD -(7.93)(lO?)/(490)(9.9BO) -16,200 !t2'T'\,.ft"

k""'"T -.~-' -~-V OT':"{'" '\'- C"' ~ ...~ ;':(s::'. ~ne..., -~r--=-"_. :.J--'.:_.2. :> ,;>~ C ~:.~n

~~ Steam enters a stem condenser at 1. psia and a qucl.1ity' of 9~, with amass now rate of hO,OOO 1bmJhr. The condensate" leaves as a sa.turatedl1qu:id. -Liquid water. at an inlet temperature of 60~ is used forcondensing the steam. The minim\m .flow rate of coolj.ng water wotlld be:

(A) 1270 gpD(B) 76,196 gpn(C) 81,775 gpm .,

(D). 1825 gpIt(E) 1959 ~

!s~ In B. steam de superheater, stem at 100 psia end 1000~ enters at a massnow rate or 100 1bm/min. Liquid water at 80"i' enters and is mixed withthe stemn to produce steam at 60 psia and 100~ supe1"heat. The mass nowrate o.r liquid water is:

A. n .82 lbM/Min.B 26.:;11 lbm/Ddn.C 25.511bM/min.D 29.02 lbrll/MiD.E 16.92 1 bm/lI'&.in. .

(((((

)))))

fu~ Steam enters a throttling calorimeter at 280 psia. The temperature ofthe steam le4ving the calorimeter 1s 284~, and baranetric pressure is

.30 in. Hg abs. The temperature of the steam enterille the calorimeter 1s:(A) 2B4~(B) indetenninate(C) 307or(D) h11 of(E) IlO2or

!!E~ A. P'nIlP handles liq\1id vs.ter at 6Oor and a flow rate of 20 gallons/min.The 1n1et and outlet pressures are 20 psia and 80 psia, respectively.Neglecting changes in kinetic and potential energies, the ~~ ~~ mum powerrequired is:

(A) 5.24 hp(B) 0.10 hp(C) 1.00 hpCD) 0.0.31.1 hp(E) 0.70 hp

~~ A Carnot refrigerator operates between -32or and +96°F. The energyabsorbed from the low-temperature space is JOO Btu/lbm. The net work doneon the working sub3tance is:

(A) 69.1 Btu/lbn(B) 1003.1 Btu/lbm(C) 230.9 Btu/lbn(D) 89.7 Btu/lbm(E) 389.7 Btu/lbn

-2 -.~~ Stem enters a throttling valva at 600 psia and 700~. 'nle nOW' through

the v.a1ft.is adiabatic" with negligible chmges in Jdnetic and potentielenergies. The stea J.eaves the Tal '99 at ~oo psia. The surroundingstemperature is 77~. The irreversibility of the throttl.ing process is:

.(.1) 0.0 Btu/lbm(B) indetenrlnate( C) 0.042 Btu/lbn(D) 3.23 Btu.lJ.bm(E) 22.,;' Btu/lbm

No. 7'~ In a basic ideal RanJd.ne cycle using s't.eem as the worldng fi'Cid, thepressure and temperat1n'e at the inlet of the turbine are .500 psis. and SOODr.The e:xhs.ust pressure .fran the turbine is 1.0 psis.. The themal ef'ficiency0.1" the q-cle is:

(1) 25.5%(B) .31.~(0) 19.(#(D) 3S.JL .

(E) 34.JL

No.8. In an ideal vapor-canpression refrigeration system using Freon-12 asthe worJdng .fluid, the Freon leaving the condenser is a saturated liquidat llOor, and the Freon leaves the evaporator as a saturated vapor at lOor.nle coeff1cient of performance for the refrigeration system is:

.(A) 11.00(B) 11.70(C) 3.S8 "

(D) 10.00(E) S.70

}foe 9. .An ideal Ot.to cycle uses sir as the wor1dng fluid. The compressionrai~o is 7.00 J and tJ1e max:1Jn\Z temperature of the cycJ.e is 1400°&. Thethermal efficiency for the cycle is:

(A) 16.7%(B) 43.3(0) 6l.~CD) lh.Jt(E) 51.1.1$

No. 10. An ideal simple BrQ"ton cy-cle uses air as the wor]d.ng fluid. The maximumand JrlniDttml. temperatures in the cy-cle are 1 aX> Dr and 6Oor, respectively. Thepressure ratio across the compressor and the turbine is 6.00 The thermalefficiencY' for the cy-cle is:

(J..) 51 %(B) 49 %(C) indeterminate(D) 60 %(E) 110 %

-3 -

.~utions to The~o~cs' Probl~s

..The srstera is sho\m below. 40,000 lbm/hr

steam at1 psia arid 9~ quality

0

@0 vater at10l.74Dr

waterat 6OOp

@ condenss.tesaturated liquid101.74 Op'

The heat transfer from the stesm is:

m -9.15 X 105 1bm/hrw

The volmuetric now rate of cooling water 192

(9.15 X 105 ~bm/hr)(0.0l6033 ft3/lbm)(7.h805 gal./!t3).v .(60 Irl.n. /hr )

JNS. (D)

~

!~~ .Appl,ing the First Law to the desu,perheater,

8 8 C 8 8 ):m h + m h .In + D'o h2.2 111 W 23

Substituting the numerical'values,

e>

1.00 lbm/min.Site mn at1.00 psia and lOOODr

.m 0

w -~'water Iat BOOF '

@.steam at60 Rsia and100 F sprht.(100)(1$32.0) + 48.04 =w

.(mw + 100)(1229.8)

(100)(1$32.0 -1229.8)

.

(100)(312.2).ANs. (c)-25.57 lbm/min..

:m .

~

.

(1181.76)

w(1229.8 -lIB.oll)

The mass nov rate of cooling vater is found froa:

-h-~l,. From the Hollier chart, it is .found that

the entering state o! the ate- is in the." 't1«>-phase region; there!ore J the steam

..t.aperature is the saturatim1 temperature" at a preSBm"e o! 280 psia.

hANS. (D).4U~

-

t

~

~

S

.!.9.~L For &t.eadjl" now or an incanpressible !"luid in an isentropic process,1~e power is giwn by:

." ...W .V (p -p ) .(20 gil./min.)(O.l3368 rt3/g8l.)(80 -20)(144)/{33,OOO)

2; 1

JNS. (E)

~

!.Q~ The coefficient of performance of a Carnot refrigerator is:

h28~TL ~.3.311 --TH -TL W

ThI! work done on the refrigerator is:

COP ..5560 .;. 4280

AN5. (D)300/3.3lL 89.7 Btu/lbm'W - .

~

No.6. The property values are found !'rom the CombustionEngineering SteS!!i Tables:

0 @-=-r-.~<1-~._.~.I. 1351.8 Btu/l~ at 600 psia and 700~

'nle exit entropy mq be to'lmd from the Mollier chart or the Steam Tables,sinc:e the First Law yields:

700~Temp.. 6~OFPressure

hCiJ psia" h-1335.9s-1.6163

h --13;1..8 Btu/lbm~ 2

:B]' i].'lterpolation,

T.h2

1351.8s

1363.41.6uO6

2s

2

The irreversibill ty is I

ANS. (E):l -22.S Btu!J.bm°

-s -

0/

~

~Z.!. The enthalpy values at the various.~ts in the RsnJdne cy-cle are :

.8 h. 69.73 Bt1)/lbm at 1 psia, sat 8 liq.T2.

~

h2

,

(!){ I

(7---.69.73 + 1.49 .71.22 Btu/lbm

h s3

1.9781 -1.11921s -Sg4~ -~ ..0.2633(1 -.x) .

1.84554,

Big"h -h -(1 -x )h.r

4 g4 4 ..g"

'!be net worlc is:

."wi -W -h ) -(hturbine P1Jmp 3 ..2 ].

I. A- ft o..~ r\r'\ 1.119 .398.25 -1.49 -.396.76 Bt.u/lbm

-h).(hVnet

w t .~~~~.~ -g~~.~~J -ne .

The be:at added in the boiler is:

396.76

U;9.98ANS. (E).O.JaJ .34.3t

~

"}--~

~b The enthalpy values at the variouspoints in the vapor-compre!sionrefrigeration cycle are: p

h1.

hh

h3 4

qg]~o. 8 J cont' cr.'evaporator is:

The heat absorbed in the

-78.335 -.33.531-h -h% 1 "" 0,..' ~'-~'-% .U4.oUU D't.W.1.~

The work input to the compressor is:

w .h -h .90.853 -78.335net 2 ~

Wnet .12.;'18 Btu/lbm

The coefficient of performance is:

ANS. (c).hl1.80h./12.Sl8 .3.58--

~

The thermal efficiency for the ideal Otto cycle is given by:1:J2. 9~\

1.11-1k-l-1 -0.459

ms. (E)'? .o.51J1 -511.1%

~

No. 10. The thermal efficiency for the ideal Bray"ton cycle i~ given by:

(k-l)/k (1.4-1)/1.41"rJ -1-(-)

rp

1-l-{-)

6.1 -0.600

"r? .O.lJOO .40 % ANS.

~

--

r-,

KIT 'EI.AH HE VIEW

Sam'Ole Questions on Thermodvnsmics

1.

(

The state of a thermodynamic S7S'tem is a1.wqs defined b:r its

A) absolute temperat1Ir8B) processc) propertiesD) temperature and pressureE) avaUabilit,-

2. KathematicalJ.'1', a thermodynaDic propert,. is 'Which of the tollo...:Lng?(A.) a point tmction(B) a. path .t\mction(C) an inexact differential(D) discontinuous(E) an exact di.!'ferential .

What is this ~pera.ture3. The normal boiling point of l1qmd oxy-gen is 90 K.in ~?(A) -330~.(B) -lB3Oa(C) +162~(D) +168~(E) +51B~ .

4.

. .Equations of "tate tor a single-canponent pure substancle can be any otthe foUoidng, ~x~'Pt:(A.) the ideal gas 18."", pv .RT(B) the ideal gas la modified by the canpressibllity factor, pv -ZRT(C) 8DY' relationship interrelating 3 or more state fut1ctions(D) a mathematical expression defining a path between "tate"(E) a mathematical expression interrelatiIlg thermott'namic properties of

the material

5. In the pressure-temperature diagrmn for H2O sho.-n below, what is the nameof the point labeled P?(A) pour point -

(B) isentropic point ~(C) cri tic;l point(D) triple point(E) nomal boiling point

~~0C/)

~C/]C7)

~ pVAPOR

m-iPERAI'URE

A. two-phee mixt'Ure of liquid vater and steam exists at a pressure of 200 psiaand a quality of 0.800 The enthalpy of saturated liquid water is 355.5 Btu/l1and the enthalpy of saturated vapor (steam) is 1198.3 Btu/lbm at 200 psia.What is the specific enthalw of the two-phase mixture?'(,1) 1243.0 Btu/lbm(B) 1553.6 Btu/lbm (D) 524.1 Btu/lbm(c) 1314.1 Btu/lbm (E) 1029.7 Btu/lbm

6.

-2 -

An adiabatic process is ~rined as a process in which:(A) the heat trSD-'ter is zero .(B) the entropy change .is zero(C) the enthalpy change is zero(D) the internal energs- change is zero(E) the vol~ remains constant

7.

8. J. nonf1.ow (closed) system contains 1 lbm o! air, which mSY' be assumed tobeha'Ve .., an ideal gas with Cp -0.240 Bt'U/lbm-~ and Cv -0.l7l Btu/1bm-~.The air temperature is increased by 10~ mile 5 Btu or work is done by" theiu. What is the heat transfer in Btu?(J.) -3.29 Btu(B) -2.60 Btu (D) +7.40 Btu(C) +6.71 Btu (E) none of these

.9.

The F1nt Law of Thermodj"nBnics states that:(A.) he.t energs- cannot be can.pletel.,. transformed into work(B) internal energj" is due to molecular motion(C) heat can only be transferred tran a bocb'" of higher temperature toone of lower temperature -

(D) energy- can be neither created nor destrOY'ed(E) entropy of the universe is increased by irreversible processes

10.

The ma:x:im-um thermal efficiency that can be obtained in an ide al reversibleheat engine operating between lS40~ and 340~ is cJ,osest to:(A.). lO~ '

(B) 6\Yp(0) 7~(D) hot(E) 2~

u. A re.frigerator with a ~r input of 3 hp operates between Oar and lOOar.The ma'T"'tmum theoretical heat transfer rate .from. the cold reservoir isne &rest to:(.1.) t. 7~800 Btu/hr

(B) 13,000 Btu/hr (D) 35,000 Btu/hr(c) 23,000 Btu/hr (E) 43,000 Btu/hr

12. The internal energy' or an ideal gas is:(.A.) a function or temperature and pressure(B) a function or temperature alone(C) constant(D) not changed in an i"entropic proce"s(E) decreased when the gas expan~ through a throttling valve

Which of the following characteristics of &ny' absolute or thermodynamicteMperature .scale is fixed by- the Second Law of Thermodynamics?(A) the ice point temperature(B) the difference betwen the steam point and the ice point temperatures(C) the ratio or ice point to steam point temperatures(D) the nature of the thermometer substance(E) the number of degrees on the scale

3

The rate of heat transfer through a given section of' a un1.form ve1.l for agi Ten temperature di.f"ference is:(A.) directly proportimlal to the coe;fficient of' thermal conduct! vi t7 and'to the thickness ot the vall .

(B) inversely proportional to the c~tficient of thermal conductivity anddire ctly proportional to the thickness of' the vall

(C) directly proportional to the c~tticient of' therm&1. coMuctivit,- andiDversely proportional to the thickness ot the vill

(D) inverselY' proportional to the coefficient of thermal conductivit.v and'to 'the thickness at the vall

(E) independent ot the thicIaless of the vall

Given a vall ha'rlng an inner surface temperature of BOor and the outerS"ar.t ace is exposed. to abient .wind and surraandings at .$O~. The. ~nvecti vaheat trmsfer coefficient for a 1.$ mph wind is about 7 Bt"1J/hr-ft2_~.Neglecting 8%]j"' radiatian losses.. cal~ate the overall heat tranSfercoefficient for the Cmlduction and convection heat transfer. The heat nuxt.hrough the wall'is 2h Bt1J/hr-fta..(A) 0.14 BtWhr-ft 2- or(B) 0.80 Btu/hr-ft2_~(c) J..2.$ Btu/hr-ft2_or(D) 7.J.0 Btu/hr-ft2_or(E) 8.20 Btu/br-ft2-or

"

-4-

1.

2.

3.

(c)

4. CD)

5.

6.

7. (A.)

B. (C).

".Q -w. u

9. ~. (.£. .+6. 7l Btu(D) energy can be neither created nor destrqyed. The First Law of Thermo~acics

is also called the Conservation of Energy Principle.

(B) 60%. The thermal efficien~ of an ideal reversible heat engine (Carnot engine) is given by:

10.

BOo~

u.

22. (B)

13.

(C)

~. (C)

15.

(B)u .(2h)/(BOO

-50°) .0.80 ~~~hr-!t -F

EIT Exmn Renew -~ODmA.\!ICS Section

~ "J."U JJ I 0UTLDl.E

1. Conce'Ct,s and Definitions

a.b.c.

Thermodynamic S7Stem and control TO1mu.eSpecific vol~e (v .VIm) and densit,- ( p .l/v )Pressure; the m.an~eter equation,

6P .p (g/g-)~z

d.

Zeroth Law 01' Therm~muics: ...

When two bodies have equality' of' temperature \lith a third body',the,- have equality o£ temperature \lith each other.

Temperature scales:

e.

(Oa) .(~) + 459.70(~) .(1.8)(OX)

(K ) .(OC) + ~~13.2°(~) .(1.8)(°0) + 32°

Property and state!.2 .Prope~,ies of Pure Substances

a.bcdergh,

S~b-cooled Uquid, saturated,liquid (r), saturated vapor (g), andS'Up3rheated vapor

n~ssure-temperature -diagrmPr~ssure-volu.'!1e diagramTemperature-entrow diagram.Hollier chart (enthalw-entrow diagrmn)Steam TablesTw'o-ph~e region: V' v 1" + oX v r ; h. h.f' + x Ar ;". sf ...oX Sf 0'Ideal gas g g 0

J R -R~ -1.54.5 tt-lb1"/lbm- ~pv-RT

;

Ru

3. The First Law of ThemOdynamics

WJ. 2

QJ. 2

Evaluation of work .tor a closed system:

-a.

Ev'aluation of heat in a reversible process:b. .General First Law equation:c.

2: m (houtle ts e ~!inl~Smi(hi

Q -W +

d. Closed systeD1:

........

-2 -

Ste ady' -!'low

e.

.t.

hi -heg.

Can t1nu1 t,- eq ua ti on :

.--m -!._V/v .! Vp

Throt tUng pro c e s s :

h. The Second Law gf Th~modynamic~

a. Thermal etf.icienc,. ("7) and COP for heat engines and refrigerators

b.

"7 .Wnet/~ ; OOp. ~ /Wnet

Carnot engine and Carnot refrigerator

TH -TL TL7]carnot. T T -T

H H L

Clausius and Ielvi!1-Planck etateIl1.ents of the Second Law

., .roP carnot

c.(Clausius) It is impossible to construct a heat engine that produces

no effect other than the transfer 0: heat from a coOler body to ahotter body'.

(Kelvin-Planck) It is impo-,sible to construct a heat engine whichproduces no effect other than doing worlc on the surroundings andexchang"...ng heat with a single the~al reservoir. A lO~ efficientheat engine is impossible.

d.

e.

t.

General Second La equation:

Entropy- production, for a reversibl~ process

for an irreversible process

for an i.!Ilpossible process

Corollaries of the Second Law.I. No heat engine is more efficient thml a reversible heat engine

(Carnot cycle).n. All reversible heat engines haw the same efficienCY' when

operating betwen the same two teMperature limi t8.III. It is possible to de~.ne a temperature scale (absolute scale)

which is independent of the thermanetric medium.IV. It is impossiOle to achieve a negative temperat".Jre on the absolute

temperature scale.

Ps .0

Ps > 0

Ps ..(. 0

3 -

Closed system:g.-(s

-dQ-+P

T S -S1,) stored2

Stsad3'-flow sys'tem:

h.

-dq-+p'S -s"T s e i

GE!neral irreversibi1i ty equation:

I .TASstored + L meToseoutlets

Stead:'1-i'low system:

1.

oj.

a.b.

c.d.

e.#..g.h.1.

RS"Hne cy-cle&aheat cy-cleRegenerati va cycle ( with open or ~osed feedwater hea ~rs)Vapor-compression refrigeration cy-cleOtto cycle (spark-ignition intemaJ. combustion engine)Die~el c7cleStirling cy-cle (external combustion engine)Brqton C7c1e (gas turbine engine)Brqton cy-cle with:

intercoolingrehe atregenerationisentropic efficienay for compressor and turbine

~=!c~Ht)re~~~is~~0~~H~~rJ]

e5&{

~~fo4

~~0tJ 0~~H ~I c~ge...

~H to)~

C

o)0 .~

c~

2 g.

~H~~~CJ' 0.~ 0.Q

,

0.~ 0.~

~ Ir4

~

~.ft

lr4

E-4

E-4

>N

I ~

I

.('I

IE

-t, (-4

~~IPo ct

~ct

~...>

..~

~ ...mC0ori

~~~~

-7cc~

-~ft

1C

Ig.

~

~.""--'"

C P

4 .

~P

4 _,

P4

g. E

-t io4

.E-t~

1 E-t

.E-cftlE

-c~ ...

~~Po

t'a r4

~ -r4

to-

t

cwto--~0 -~"1

~-c:r-IP

4

~P

4~ .of

~.of

~

~I

I

~cw

r-4

CI

~

rf~

rfg,

~I

I

..ft rof

ft

g,

ft

~ft

g.-c

N

~C

I~~cq

...

~

0D-

ocp

->~

I p

-~rf

Prf

c.. ~I~ =If"'f

..

-~

e...

. Eo4

->()

-..E-tI 0 -...

Eo4I= ~I

N f'f

~

-..Eo4

.-"E

-oI

-rtEo4I

CW

Eo4

-0-c~ -...

fotI«fot

-Q.C)

.::

.cI ~N ~

-..~~

~-~r-.= 0~,

g.~m "(II

-4-

-.5

SrE.AK PO'Q. CYa.ES

~eic'l~~ ~c1_~

TURBnlE@_c- ,.-

c ~Wtrom;RA

~

~ @CONIENSER

@ ~~~;~. ~

Wp"""'"s

-h.w .h .' h, t 3 "-hQH -h 1

"3

2

~hee.t_C1'c1e

s

-h)-I

-h ) + (h .h2

.{h3

s23

-p )~

-h)6c.-h -h

~.L. & 1

~~!rati ve Cy-~--~-

@) ~ ""1..~- W tl

~l(!)

~@

~ ~-

-J@~;~~-~ ~

WPJ.

@ ~~Wp2

5

-h )/(h~.h -h e.2.3

~

"

-6 -

JIR-ST.A!fDARD PO~ CYCLES

Otto Cycle

~e~l Qzc_l~

Tp

v

~tirling Cycle

~

-7 -

REI.ATIONSHIPS FOR AIR-SfANDARD R)'Q CYCLES

p~.-.presS'tlre ratio

Pmin~prv

v t~ -~ -thermal e.t'.t'iciency

qH

vD1&X compression ratio

Vmin

T---~ -temperature ratio

TJdnrT

Otto Cycle

k-l k]) . P r2. 1. V

k-lrv

1-1-(-)

r.v

T .T"'J 1.2

k-lrirvT .T

k-l]) - P r Ir'4 J. 1" Vr -V Iv:. v Iv

v 2.2. 43 1..

r", -T trJ.3

Diese1 Q:9'cle

T.2. .cut-off ratio

T~ ; r c

2.

k-J.r

v-TT

kPa -p1,ryr -v Iv

v 2. a J.a

kP3 .P2. -P1, rV

k-lr

vT .T r

1. c3

kP4. P1, rC

kT .T r

" 1. C

~t~lin~ Cycle

.T

Pa.

P2, ry.TT~ L1.2

T .T .T..3 Hr

vD P /p

3 1..r r

T vrrT p

-8 -

~~j--~~;~-i~2 @caiP. TURBnlE

C~BUS1'ER

/i-;:

-w: c~ 1Vnet

@~4~

I

..-~~

(k-l)/k1-1-(-)

rp

(k-l)/k"'J T -T2. 1

rir~k-l)/kT -T" 3.

,QH0 @-~~I~~~;:;;;:.."'-l CQ! PP.E S 5 OR

,-

!}= -.-w~ I .-0

VAI.VE~EVAPORATOR

hh -h

1. 3COP -qt--wc

~-h -h -h-h~ 1 " 1 3b -h1

W .h -hc 2 1 q .h

B-h

2 3

rp