solution manual intoduction to engineeriing termodynamics

SOLUTIONS MA N UAL
P

, 2-1
CHAPTER 2 P




ThePcorrespondencePbetweenPthePproblemPsetPinPthisPfirstPeditionPversusPthePproble
mPsetPinPtheP5'thPeditionPtext.PProblemsPthatParePnewParePmarkedPnewPandPthosePthatP
arePonlyPslightlyPalteredParePmarkedPasPmodifiedP(mod).

New Old New Old New Old
1 1 21 new 41 new
2 2 22 18 42 32
3 new 23 19 43 34
4 3 24 15 44 35
5 5 25 new
6 6 26 20
7 7 27 21
8 new 28 new
9 9 29 new
10 10 30 23
11 11 31 24
12 12 32 new
13 new 33 new
14 new 34 26
15 new 35 27
16 13 36 new
17 new 37 28mod
18 14 38 29
19 new 39 30
20 16 40 31

,2-2
2.1 TheP“standard”PaccelerationP(atPseaPlevelPandP45°Platitude)PduePtoPgravityPisP
9.80665Pm/s2.PWhatPisPthePforcePneededPtoPholdPaPmassPofP2PkgPatPrestPinPthisP
gravitationalPfieldP?PHowPmuchPmassPcanPaPforcePofP1PNPsupportP?

Solution:
maP=P0P=PPFP=PFP-Pmg F
FP =PmgP=P2PP9.80665P=P19.613PN g
FP=Pmg => m
mP=PF/gP=P1P/P9.80665P=P0.102Pkg



2.2 APmodelPcarProllsPdownPanPinclinePwithPaPslopePsoPthePgravitationalP“pull”PinPtheP
directionPofPmotionPisPonePthirdPofPthePstandardPgravitationalPforceP(seePProblemP
2.1).PIfPthePcarPhasPaPmassPofP 0.45PkgPfindPthePacceleration.

Solution:


maP=P PFP=PmgP/P3P
aP=PmgP/P3mP=Pg/3
=P9.80665P/P3P=P3.27Pm/s2 g




ThisPaccelerationPdoesPnotPdependPonPthePmassPofPthePmodelPcar.



2.3 APforcePofP125PNPisPappliedPtoPaPmassPofP12PkgPinPadditionPtoPthePstandardPgravit
ation.PIfPthePdirectionPofPthePforcePisPverticalPupPfindPthePaccelerationPofPthePmass.

Solution: x

FupP =PmaP=PFP–Pmg F
aP=P(PFP–PmgP)P/PmP=PFP/PmP–Pg
=P125/12P –P9.807 g
m
=P0.61Pms-2



2.4 APcarPdrivesPatP60Pkm/hPandPisPbroughtPtoPaPfullPstopPwithPconstantPdecelerationPinP5Ps
econds.PIfPthePtotalPcarPandPdriverPmassPisP1075PkgPfindPthePnecessaryPforce.

, 2-3
Solution:

AccelerationPisPthePtimePratePofPchangePofPvelocity.
dV 60PP1000
aP= = =P3.33Pm/s2
dt 3600PP5
maP=PPFP;
FnetP=PmaP=P1075PP3.333P=P3583PN


2.5 AP1200-
kgPcarPmovingPatP20Pkm/hPisPacceleratedPatPaPconstantPratePofP4Pm/s2PupPtoPaPspeedPo
fP75Pkm/h.PWhatParePthePforcePandPtotalPtimePrequired?

Solution:
V
dVP V => tP=P (75PP20)P1000 =P3.82Psec
aP=P =P =P
dt t a 3600PP5

FP=PmaP=P1200PP4P=P4800PN

2.6 APsteelPplatePofP950PkgPacceleratesPfromPrestPwithP3Pm/s2PforPaPperiodPofP10s.PWhatPf
orcePisPneededPandPwhatPisPthePfinalPvelocity?

Solution:

ConstantPaccelerationPcanPbePintegratedPtoPgetPvelocity.
dVP
aP= =>P PdVP=PPaPdt => VP=PaPt
dt
VP=PaPtP=P3PP10P=P30Pm/s => VP=P30Pm/s


FP
FP=PmaP=P950PP3P=P2850


P N