Solution Manual for Thermodynamics: An Engineering Approach 10th Edition by Cengel | All 18 Chapters Covered

1-1




Chapter 1 INTRODUCTIO 4t 4t


N AND BASIC CONCEPTS
4t 4t 4t




Thermodynamics


1-1 C4tClassical4tthermodynamics4tis4tbased4ton4texperimental4tobservations4twhereas4tstatistical4tthermody
namics4tis4tbased4ton4tthe4taverage4tbehavior4tof4tlarge4tgroups4tof4tparticles.


1-2 C4 t On4ta4tdownhill4troad4tthe4tpotential4tenergy4tof4tthe4tbicyclist4tis4tbeing4tconverted4tto4tkinetic4tene
rgy,4tand4 t thus4tthe4tbicyclist4tpicks4tup4tspeed.4 t There4tis4tno4tcreation4tof4tenergy,4tand4tthus4tno4tviolati
on4tof4tthe4tconservation4tof4tenergy4tprinciple.


1-3 C4tThere4tis4tno4ttruth4tto4this4tclaim.4tIt4tviolates4tthe4tsecond4tlaw4tof4tthermodynamics.




Mass,4tForce,4tand4tUnits


1-4 C4tPound-mass4tlbm4tis4tthe4tmass4tunit4tin4tEnglish4tsystem4twhereas4tpound-
force4tlbf4tis4tthe4tforce4tunit.4 t One4tpound-
force4tis4tthe4tforce4trequired4tto4taccelerate4ta4tmass4tof4t32.1744tlbm4tby4t14tft/s2.4tIn4tother4twords,4tthe4tw
eight4 t of4ta4t1-lbm4tmass4tat4tsea4tlevel4tis4t14tlbf.

, 1-2
1-5 C4tKg-mass4tis4tthe4tmass4tunit4tin4tthe4tSI4tsystem4twhereas4tkg-force4tis4ta4tforce4tunit.4t1-kg-
force4tis4tthe4tforce4trequired4tto4taccelerate4ta4t1-
kg4tmass4tby4t9.8074tm/s2.4tIn4tother4twords,4tthe4tweight4tof4t1-kg4tmass4tat4tsea4tlevel4tis4t14tkg-force.


1-6 C4tThere4tis4tno4tacceleration,4tthus4tthe4tnet4tforce4tis4tzero4tin4tboth4tcases.




1-7 A4tplastic4ttank4tis4tfilled4twith4twater.4tThe4tweight4tof4tthe4tcombined4tsystem4tis4tto4tbe4tdetermined.
Assumptions4tThe4tdensity4tof4twater4tis4tconstant4tthroughout.
Properties4tThe4tdensity4tof4twater4tis4tgiven4tto4tbe4t4t=4t10004tkg/m3.
Analysis4tThe4tmass4tof4tthe4twater4tin4tthe4ttank4tand4tthe4ttotal4tmass4tare mtank4t=4t34tkg
V4t=0.24t
mw4t=V4t=(10004tkg/m3)(0.24tm3)4t=4t2004tkg 3
m 4tH2
mtotal4t=4tmw4t+4tmtank4t=4t2004t+4t34t=4t2034tkg
O
Thus,
24 t  14tN 
W4 t 4tmg4t4t(2034tkg)(9.814tm/s
1 4t4tm/s 4t4t4t19914tN
t )4t
4t4kg

 24

, 1-3


1-8 The4 t interior4 t dimensions4 t of4 t a4 t room4 t are4 t given.4 t The4 t mass4 t and4 t weight4 t of4 t the4 t air4 t in4 t the
4 t room4 t are4 t to4 t be4tdetermined.

Assumptions4tThe4tdensity4tof4tair4tis4tconstant4tthroughout4tthe4troom.
Properties4tThe4tdensity4tof4tair4tis4tgiven4tto4tbe4t4t=4t1.164tkg/m3.
Analysis4tThe4tmass4tof4tthe4tair4tin4tthe4troom4tis ROOM4tA
IR
m4t4tV4 t 4t(1.164tkg/m4t 4t)(64t4t64t4t84tm4t 4t)4t4t334.14tkg
3 3

3
Thus, 6X6X84tm
 14tN 
W4 t 4tmg4t 4t(334.14tkg)(9.81 )44tt14tkg4t4tm/s4t4t244tt 4t32774tN
24 t

4tm/s
 




1-9 The4 t variation4 t of4 t gravitational4 t acceleration4 t above4 t the4 t sea4 t level4 t is4 t given4 t as4 t a4 t function
4 t of4 t altitude.4 t The4theight4tat4twhich4tthe4tweight4tof4ta4tbody4twill4tdecrease4tby4t1%4tis4tto4tbe4tdetermin
ed. z
Analysis4 t The4tweight4tof4ta4tbody4tat4tthe4televation4tz4tcan4tbe4texpressed4tas
W4t4tmg4t4tm(9.8074t4t3.324t4t10
64tz)4tIn4tour4tcase,


W4t4t0.99Ws4 t 4t0.99mgs4 t 4t0.99(m)(9.807)
Substituting, 0
6
0.99(9.81)4t4t(9.814t3.324t104t 4tz)4t4tz4t4t29,5394tm
Sea4tlevel




1-10 E4tAn4tastronaut4ttook4this4tscales4twith4thim4tto4tspace.4tIt4tis4tto4tbe4 t determined4thow4tmuch4the4twill
4tweigh4ton4 t the4tspring4tand4tbeam4tscales4tin4tspace.
Analysis4t(a)4tA4tspring4tscale4tmeasures4tweight,4twhich4tis4tthe4tlocal4tgravitational4tforce4tapplied4ton4ta4tbody:
24 t  14tlbf 
4t 4 t )4t
W4 t 4tmg4t 4t(1504tlbm)(5.484tft/s 24t4t
32.24tlbm4t4tft/s4t 4 t 4t25.54tlbf
 
(b)4tA4tbeam4tscale4tcompares4tmasses4tand4tthus4tis4tnot4taffected4tby4tthe4tvariations4tin4tgravitational4tacc
eleration.4tThe4tbeam4tscale4twill4tread4twhat4tit4treads4ton4tearth,
W4t 4t1504tlbf




1-11 The4tacceleration4tof4tan4taircraft4tis4tgiven4tin4tg’s.4tThe4tnet4tupward4tforce4tacting4ton4ta4tman4tin4tth
e4taircraft4tis4 t to4tbe4tdetermined.
Analysis4tFrom4tthe4tNewton's4tsecond4tlaw,4tthe4tforce4tapplied4tis
24 t  14tN 
F4 t 4tma4t4tm(64tg)4t4t(904tkg)(64t4t9. )4t4t14tkg4t4tm/s4t4t244tt4t52974tN
814tm/s
 

, 1-4


1-12 [Also4 t solved4 t by4 t EES4 t on4 t enclosed4 t CD]4 t A4 t rock4 t is4 t thrown4 t upward4 t with4 t a4 t specifi
ed4 t force.4 t The4 t acceleration4tof4tthe4trock4tis4tto4tbe4tdetermined.
Analysis4tThe4tweight4tof4tthe4trock4tis
 14tN 
24 t
 4t4t
W4 t 4tmg4t 4t(54tkg)(9.79 ) 14tkg4t4tm/s4t24 t4t48.954tN
44tt

4tm/s
 
Then4tthe4tnet4tforce4tthat4tacts4ton4tthe4trock4tis
Fnet4 t 4t Fup4 t 4tFdown4 t 4t1504t4t48.954t4t101.054tN
From4tthe4tNewton's4tsecond4tlaw,4tthe4tacceleration4tof4tthe4trock4tb
Stone
ecomes
F4 t 101.054t4tN4t4t14tkg4t4tm/s4t 4 t 4t
2
a4t4 t 4t 4t20.24tm/s4t2
m 54tkg  14tN 




1-13 EES4 t Problem4 t 1-
124 t is4 t reconsidered.4 t The4 t entire4 t EES4 t solution4 t is4 t to4 t be4 t printed4 t out,4 t including4 t the4tnumer
ical4tresults4twith4tproper4tunits.
Analysis4tThe4tproblem4tis4tsolved4tusing4tEES,4tand4tthe4tsolution4tis4tgiven4tbelow.

W=m*g"[N]"
m=5"[kg]"
g=9.79"[m/s^2]"

"The4tforce4tbalance4ton4tthe4trock4tyields4tthe4tnet4tforce4tacting4ton4tthe4tr
ock4tas"4tF_net4t=4tF_up4t-4tF_down"[N]"
F_up=150"[N]"
F_down=W"[N]"

"The4tacceleration4tof4tthe4trock4tis4tdetermined4tfrom4tNewton's4tsecond4tlaw."
F_net=a*m

"To4tRun4tthe4tprogram,4tpress4tF24tor4tclick4ton4tthe4tcalculator4ticon4tfrom4tthe4tCalculat

e4tmenu"4tSOLUTION
a=20.214t[m/s^2]4tF_do
wn=48.954t[N]4tF_net=
101.14t[N]4tF_up=1504t[
N]
g=9.794t[m/s^2
]4tm=54t[kg]4tW
=48.954t[N]