[PDF] FE Mechanical Practice Problems Problems

102736_3femepp_mar_2019.pdf

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9.7 kgof neonis stored ina rigidtankat threetimes

atmospheric pressureanda temperature of70 ?

C. 30kJ

of heatis added totheneon. Whatis mostnearlythe final temperatureofthe neon? (A) 70 ? C (B) 73 ? C (C) 74 ? C (D) 77 ? C

10.A deviceexpen ds130kJof energy whilepressuriz-

ing 10kg ofwater initiallyat 17 ?

C. Theisentropic

efficiency ofthe deviceis 50%.Inefficien ciesare repre- sented bya heatloss fromthe devicecasing. Whatis most nearlythe finaltemperature ofthe water? (A) 18 ? C (B) 19 ? C (C) 20 ? C (D) 21 ? C

11.An adiabaticpump receives1.5 kg/sof 15kPa

water anddischa rgesitat15 MPa.The specificvolume of theenter ingwateris0.001055 m 3 /kg. Considert he water tobe incompressible. Theisentropicefficiencyof the pumpis 0.82.Velocity andelevation changesare insignificant. Thewater doesnot increasein tempera- ture significantly.Mostnearly, whatis theminimum electrical powerrequired todrive thepum p? (A) 13kW (B) 20kW (C) 23kW (D) 30kW

12.An engineoperates ata constanttempe ratureof

90
?

C. Througha reversible process,theengine's work

output is5.3 kJ,and theheat lossis 4.7kJ. Whatis most nearlythe changein entropyduring theprocess? (A) 0.013kJ/K (B) 0.014kJ/K (C) 0.015kJ/K (D) 0.016kJ/K

13.In aheat treatingproce ss,a 2kgmetalpart (spe-

cific heat=0.5kJ/kg?K) initiallyat 800 ?

C isquenched

in atank containing200 kgof waterinitially at20 ? C.

What ismost nearlythe totalentropy changeof the

process immediatelyafter quenching? (A) 2.3kJ/K(decrease) (B) 0.65kJ/K (decrease) (C) 0.90kJ/K (increase) (D) 1.4kJ/K (increase)SOLUTIONS

1.The generatorsare drivenby hydraulic turbines.

Since theprocess isadiab atic,and sincevelocityand elevation changesare insignificant, alloftheterms in the steadyflow energyequation dropout exceptwork and enthalpy.Thet urbinewo rkperunitmass is w turbine ¼h i ?h e

In theabsence ofcompressed watertables (giving

enthalpies ofsubc ooledandcompressedwater), thetur- bine workmustbe calculatedfrom morebasic princi- ples. Enthalpyis definedas h=p?+u, sothe turbine work perunit massis w pump ¼h i ?h e ¼p i ? i ?p e ? e Since thewateris incompressible, thespecific volumeis unchanged. Thet urbineworkper unitmassis w pump ¼? i ð p i ?p e Þ

The powergenerated byturbines is

_

W¼_m?

i ð p i ?p e Þ

¼900

Mg s??

0:001m

3 kg?? ð

900 kPa?200 kPaÞ

¼630 MW

The answeris (B).

2.

The firstlaw is

Q?W¼DU

Qis theheat transfer intothesyste macross thesystem boundary.Wis theenergy transferredacrossthe system boundary tothesurrou ndingsin theformofwork. DUis the changein energystored withinthe systemin the form ofinternal energy.This expressionshows thatthe energy transferredacrossthe systemboundary comes from achange instored energy.

The answeris (B).

3.

The energyequation isbased onthe firstlaw of

thermodynamics.

The answeris (C).

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FEM ECHANICALPRACTICEPROBLEMS

Thermodynamics

12.The changein entropyis

DS out ¼ Q out T reservoir ¼ 4:7kJ 90
?

Cþ273

?

¼0:013 kJ=K

The answeris (A).

13.

Calculate thefinal temperature,T

f . ð mcDTÞ metal

þðmcDTÞ

water ¼0 ð

2kgÞ0:5

kJ kg?K?? ð 800
? C?T f Þ

þð200 kgÞ4:18

kJ kg ?K?? ð 20 ? C?T f Þ ¼0 T f

¼20:93

?

Cð21

? CÞ

For asolid orliquid ,the totalentropyis

DS¼mcln

T 2 T 1

So, consideringasyste mconsisting ofthemetalpart

and thewater inthe quenching tank, DS metal

¼mcln

T 2 T 1

¼ð2kgÞ0:5

kJ kg ?K?? ln21 ?

Cþ273

? 800
?

Cþ273

?

¼?1:295 kJ=K

DS water

¼mcln

T 2 T 1

¼ð200 kgÞ4:18

kJ kg ?K?? ln21 ?

Cþ273

? 20 ?

Cþ273

?

¼2:655 kJ=K

DS total

¼DS

metal

þDS

water

¼?1:295

kJ

Kþ2:655kJ

K

¼1:36 kJ=Kð1:4kJ=KÞ

The answeris (D).

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LAWS OFTH ERMODYNAMICS

14-5

Thermodynamics

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46

Manufacturability,Quality, andReliability

PRACTICE PROBLEMS

1.A shaftwith aninterfe rencefit hasamaximum

diameter of3 cmand anominal diameterof 2.990cm.

The upperand lowerdeviations ofthe shaftare

0.005 cmand 0.003cm, respectively.What ismost

nearly theminimum shaft diameter? (A) 2.990cm (B) 2.992cm (C) 2.993cm (D) 2.995cm

2.A shaftwith aclearance fithas anomin aldiame terof

12 mm.The lowerdeviation andupper deviationare

0.036 mmand 0.028mm, respectively. Whatis most

nearly themaximum nominal sizeoftheshaft? (A) 12.028mm (B) 12.032mm (C) 12.036mm (D) 12.064mm

3.A shafthasa nominal diam eterof15mm. Theshaft

diameterisspecif ied withatolerancerang eof14.9 50 mm to 15.027mm.What ismostne arlythe tole ranceof the shaft? (A) 0.015mm (B) 0.023mm (C) 0.050mm (D) 0.073mm

4.A hollowaluminum cylinder ispressedovera hollow

brass cylinderasshown. Bothcylin dersare 5cm long. The interferenceis0. 010cm. Theaveragecoefficient of friction duringassemblyis 0.25.The pressure onthe cylinders is37 MPa.

2.5 cm5 cm7.5 cm

aluminum alloy,

E = 70 kN/mm

2 ,

ν = 0.33

brass, E = 100 kN/mm
2 ,

ν = 0.36

What ismost nearlythe initial axialdisassembl yforce required toseparate thetwocylinders? (A) 57kN (B) 61kN (C) 65kN (D) 73kN PPI * www.ppi2pass.com

Mechanical

Design/Analysis

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SOLUTIONS

1.The fundamentaldeviation, ?

F , isthe smallerof the upper andlower deviations,which is0.003 cm. d min

¼dþ?

F

¼2:990 cmþ0:003 cm

¼2:993 cm

The answeris (C).

2.

For aclearance fit,the fundamentaldeviation, ?

F ,is the upperdeviation ,? u , whichis0.028 mm.

The maximumnominal sizeof theshaft is

d max

¼dþ?

F

¼12 mmþ0:028 mm

¼12:028 mm

The answeris (A).

3.

The upperdeviation is

? u

¼15:027 mm?15 mm¼0:027 mm

The lowerdeviation is

? l

¼15 mm?14:950 mm¼0:050 mm

The shafttolerance is

D d

¼j?

u ?? l j

¼j0:027 mm?0:050 mmj

¼0:023 mm

The answeris (B).4.

The initialforce necessaryto disassemble thetwo

cylinders isthe sameas themaximum assemblyforce . F max

¼2pr

shaft ?pl interface ¼

2p2:5cmðÞ0:25ðÞ37 MPaðÞ5cmðÞ1000

Pa MPa?? 100
cm m?? 2

¼72:65 kNð73 kNÞ

The answeris (D).

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FEM ECHANICALPRACTICEPROBLEMS

Mechanical

Design/Analysis