Baixe Termodinâmica e outras Notas de estudo em PDF para Engenharia Mecânica, somente na Docsity! — CHAPTER ONE
"GETTING STARTED:
INTRODUCTORY
CONCEPTS AND
DEFINITIONS
PROBLEM Li
mass enters when .
j : mass extts when exvoust
take valve 1s ogen Velho te open
hot sugfnces aiston exerts force ongas
, VAZ COMPrESSIM, gas
surroundinas Pads Con om eiatê á
durma expansum
hot-surtuces iuteract
vitta suvroundingS aj im
hat
Aves porq
snafh torque
3 transmits force
PROBLEM I.4
| 1 A control volume encisses the valve.
o! vive and turkine.
Bt e Sicam enters ar) and exits at O .
omiis o A torqueis transmaitled +erough tha
E a Antabing Shajh. o .
* lyarm surfaces of the turbine interact
with the turrovndrngos.
toithia tha control volume, Shaw flows acrost vhs valve andthmdugi ja lmrbine lolades,
When tha generator is includad wi tia control volvma,
. Slam enter at O and exite at BD.
luarm turfaces of tha turbina aus tia generator
interact uti tha, Surround .
+ Eleciroa current fiowo from tha generator,
Node Had 4a trantwlhd drque does Vet crose tha boondary of
Aba ateegas cmol volume
p PROBLEM 1.57
A control volume enclesças tha engine-
devem pomp.
* later endere at O) and aut at E)
* Aw fer Combushos of tha on-board fuel
enters; and comburhim gases exit.
+ Carma surfeces of ABS no intevraet
with tus surroundings,
Goitia tha pump, à piston ix kept ui moon witun a counder own
to combushiom of tha on-looard fuel. The prston motor 3 Narmassed
te pump tra Jigurd, The amount of fuel wibhiu the system olecreases wi fume.
loben +ha Lose and mosale ama metudad, a Wsh-sprnd adam
joel emits the auetendad cobol Uoiuna ab tis noggle ext,
I-4
PROBLEM Llo
system boundary
* tuo phases are present (liquid and gas).
- not à pure substance because composition
is different in each phase.
system boundury
- three phases are present ( Solel, liquid and qu),
“not a pure subtunce because composition ef
gas phdses is dt erent tan at ol Pre solid
amá liquid phases.
The system 15 a pure substance,
AlMhough the liguid is Vaporized,
te system reméius fixed tu chem-
ical composition qnd is chemica ly
homoganesus.
The system 15 à pure substance, Alhough
+ Ave phases change, Lhe. system remains of fixed
k chemical composttina and is chemically
homogencous.
The System 18 net à pure subo stance
du g te process since the com-
: osition of +ue qas phase changes
db o water exaporates into the doa.
E
l
|
|
Once all of the water evaporates,
|
1
the gas phase comes ta equilibrium
and'+He composition becomes
= esses el homogencous. At Hhis point
phase am be trealed os a gure Ps
1-5
ACE.
PROBLEM (to
- = = ! tt Abe
101b Ejs Tma=(t0 beto a pftre
| GENOA
E
=Qu3 o, e Eae
Fara
PROBLEM LÊ
gra» [5 EN 000 NI || legum/5*
3 A m/s UN
Fgrav* AS kn
PROBLEM LIZ
2a . Fyeay [4580] lg s>
é m=10 leg (a) Goal det (E) OT
, a7s m/s* Ilocal
, cb) Ymassis unchanged. m
LN
Farav = 95 N Fray > ma «(10 kata m/ç2) Uaumisa
= 98IN Faray
PROBLEM 113
«tolb (6) — Ega «(2 to) 32.2 Ib-£Hs2
Greca * Tm “Vick à be
= 30.4 tHs% Sogal
(b) massisuncanged am
o - + | Ibt
Egas * MI e (10 lb)(02.2 É HEM Ei
= 0 PR
PROBLEM LIB
aa
a & For a linear spring , Fpri ing = ELMO, where
& AX à the spring extension. Since Fprias =
w Bray = mg » we have Kiax)=mg. Sina
; m and kK are independent of Aecatron,
Fgmy tha oca accelerahom of Gravity à
propor tona titho deftetim. Thus
mars,
Qmars . (ARmers Ortlbin Ta = Eee me
Torto (AN) eoth e-zalin
art = 1% gasáo a —
meent Prooa CAN 2 (Ay) pson = Ge (Om) edi
Rent (ax) esrth
SAY o agrin)= 0.049in
32174
PROBLEM 114
Decelevatum ocuws from 5 milh to vestiu 015. The average
osceler atum megnitudo is
la | . Ar. (Imlnco) 5280 || th
aEo 0.15 Imi |la600s
="73.33 Hh/s”
or, in g's
(al -[/73.33 a) - 2.28 a's deceleratim
ing 32.2 Hist J lin 9's)
Thus, +he magnitude of the average force applied is
. < (601 & Hs
lia m lala (Go b93.33 to) 32.2 Wo: fifsà
=/139 bt dia
PROBLEM 1.20%
a.= 5 m/s% (up)
g=48 m/s?
Fg rav
app” Fgm Md
Fapp 2=ma+rmg -m tara
= (Akg)(S+ 2.80) 28 pre
| kg-m/32
= AGA NL Papo
PROBLEM 1,21 *€
Fapp= 45 Ih£
é m=3S
9=32.2 Bis
Fgrav
app” Fara :
Fogo Form
m
ma
as
a age
(as Hb
(EE Er lb Hjs?
Il Tb
Fogo - ma
m
- 32.2 Hs
= 424 €t/s* (downward) e Al
PROBLEM 1.2 2
Fgrav 4:30:5 Hs”
From Table AE : dd =
oras É MG E nHg
=(0,5 Ibmol (18.02 ES TeD(so.s
= 8.53% Ibf
Mo. (0,5)(18.02)
Save E (0.145)
18.02 Ib/Ibmol
3] Lib
Sara
Parar
<br Ib/gã Save
PROGLEM 1.23 o
; =sa Fopnce =m Ispace Espace Feortla
4 t E .m aa
q ; carta dem ) q space Teor
a =. 3 £
Eme MN Ra F êpce (quant « (42 (BE! det)
weight Ispucé b ms? F.
a8 m/s2 =68.61 Ne earth,
Tea
I-to
ProgLEM 1.2.4
q La.si- (33 vab/adz mist,
4 ; T=
tom -— where 2 o km.
-"
"
me
— —
+ É
CA PA E = = 22
Suer weight io Wzmg, the percent Clhangeii weight landins Co
% changez Lhe 3 Jem)
mw
= | Cagimd - [caem (33x05%/6º) (0% )] Je £ | soe? ce
o [ (B.8- GaxWM)] “GIF
of
PROBLEM 1.25
les 30H
[Sua 02 bip
7 G= 324 Hs”
go uohere Vis Me
Fra
The mass of walker is mz
Thus
volume of Je spherical tam
V= Em? (Em(3o = Ii3ixto Se?
and the mass is
a = 3
m= gV=(024 Bo ttstuos) A
=706K0ºlb cm
The weight is
º e Hs VE
= mg =(T.Obxtp (3215) ae a
G vav
= 7.04 X10º Ib£ Egray
PROBLEM 1,29
o) Ar à temperature 0$ 240%, He Specified pressuve of 125Mpa falls
between The tabe values of 10 amd LS MPa. To determine Jhe specific
volume correspondiug to 1.25 MPa, we think of the slope of a
straight live joming "se adjaceut +able stades, as follows:
Y =0,.2275 ms
Z2.
(UV p=1,25)
uv, 18 :P
3
meg JY Vs DINB3 my leg
Ay ; ' P, MPa
to 125 AS
similar triang les :
v- 01463 o.2215-0.I483 . 025 a
| slopel * Te-tãas "sas P VeomBst e (0. 2275-0148
=0./879 mêla (a)
(1) Ma pressure of 1SMpa re quer specific volume et oisss mk falls
between tue tule values of 240 and 280%. To delevmine le temperature
correspording to +he ques specific volume , we Hinle fi He sibpe 58 a
shroag ul Vime jon ing ge adjaceut table states, às follows:
280 (0.1621,280) ç
Slope = T-2Zã4o . zpO-Z4o ”
Pe ASSS-4B3 1627-1483
> T=240+ ad] Ho
[res (to)
d
WOT ciyBs,2do)
= 260C Ch)
q
Mas dC 47 0 mYg
(€) Indhis cause, he specrfiêd pressure falls between He tuble values of Lo
and 1.5 MPa and the specified temperature fuls between the tale values
of 200 and 240%. Thus, double interpolatin is reguived.
. Hi z20%€, tha specific Volume od eaci pressure is Simply He average
over the infervol : zoo 4.2245
4 O +.Z2
at 1.0 MPa 220%; E > = o 21615 wº/kg
1325+,
at 15 MPa, 220% = 1325+ 1483
z = 0.404 Veg
. Thus, with the same approach as iu Ca)
W-OtãoY O.21675 - o. 40% - PN os
LS- 14 > 5-6 > UV =otoy “(85 )to.erers toy)
= 0/5567 fra co
t-Id
PROBLEM |. 30
ca) At a temperature of 120ºC , tra specifred pasauwre of 54 Ibf/in2
faiis beto the table values of 50 and 60 lbf/in? To determine.
the specific volume correspondina to 54 Wflint, tua thank of
tha slope of à straight Aine j raina tua adjacent a bs, states,
as follows:
1,
3
seo v=sga ff
1 4a.
50 = Te sp Py tbflin?
Similar Arrangles:
|stope) = N=88a1 = IHo- 5.881 = vz 6.841 + &-(auo-s89)= corr
co-s4 69-50 Rota
to) At o pressure cf 60 Hoffin?, the given speafic volume, ef
catrttih falis betuwcen the table values of 120 and ISO F. To
determine the temperature corres pondens To tia given spesifi
voloma , we thninn of tha slope ota eirarsht Line jormima Te
asjacent table states, às fottows:
140 (et, 140)
sbpe= T-izo - 140-120 “age
o Pe comia Guz- 8.8]
E
8.982. 8.841 Je
» panot [SE )
zo )
AL 1 Ls =74ºF
s8 54 co é! 5, gtêilo
() Tn this case | the specified prasruve fails between ha table
values of Go and bo lbf/in? and tha spewfved temperature
falls bekweem the table values of 100 and 20ºF, Thus, double
unterpe lah 3 required.
e A UOSF, due specific volume at each prsture JE empty tia
average ouartia intervol:
as Sof ao*F; va quo det = 6.973 1º/th
ut) ? 3
at colá, np; v> ssh = EM
. Then, unth tus Same approach 09 Un tas
V-SH€ CMS yo gar t Etr sm)
c-sg 6o- sv
= Cos AY ato
t-15
PRogLEM Lspk
m=5kg From the pressure - Specific votume relation
é 13 Na !
( E pu = const, pis LATE
N 1 4, = [Ab =|-—— 3/k
[x] A (555) (ora mthg)
paibar ,U,=0,2 m/kg = 05810 mYbg
P,:0.25 bar — MawmazgIOos mi Mo
1.2 -
14 1
0.8 +
Eos]
04 a
d Te
02d z
od et
o 0.5 1 15 2 25 3
V-m
PROBLEM 1.32.
e wmm=l db From Te. pressure - volume relation
+
+; pV" = const. PAR 209º
Í EP VE (&) v = (55) 1044)
£
Fut . 3 na; Vocz HH
p,=20 Ibflin) Vi= 10H ne; Vi IS AE? Va
p= 100 lb£/in? naLz; Ei aoo ns €&————— 0 ————+&
nete, Voz 3.168. 44º
100
80 4
cabo
2255
nara
suvo
60
p-Ibfinz
40
20 +
PROBLEM 135"
Thermo dgmanic cycle!
t-a: pVa constant
= B= ibar, Va lmi, Voz 0.200
3.1 V= constant
rr, pV= constant. The constant cam be
Fr process
usas data at stater:
P v= constani
=» Nr
= (bar )C im) = E bar-m?
Accordinglys on à pressure - Volume plot process I-2 is
deserihed b
+ lbar. ms
Tal
Tm portiuslas, ben Voz olm3 , p= bar,
PE:
Tue therredynamic cycle +apes the forma
2-3 peconstat, VW >V Lexpansion), NV = Omo
ewlvated
Pa=Sbars = constant
5 O Vs 0.2m LL > É Bys Sber
Va= tm
+ V= constant
p
(ue) 3
2
PVE constant
! As lbar
vw Im3
opta to 14
2 oa 04 os ob of og oq Lo
(wi)
1-19
PrRogLem 1.36%
Pam = VOL 33 PA,
Prlitm 5891
q=13.54 Ya
9:4.81 mist
P-Ram
2 [1028
tg
tNjm
VkPa
$3
= (io4,O = 101,33) ka
. Pleg-mifs*
(13.59 9/ems)c7 Si mis?)
EN
vm?
108 cm
=002m = 2Zemgç L
(b) Page * P 7 Futm = (04.0 = [01335 2.67 LA Foge
PROBLEM 1.37
PE
(0) F,vac = Pe 9 L
! ! - db Hp ha LIbg HZ
= (ua, af Y32.2 EM qa A TD
4 ( ps X $ tz l= tb -M/6?] | tuq jut
Po vaç Melia. Ho = Odair dbgfive
Pato é nes bela t
4=3.a Hs? R,abs * Em” Rvac b
- 3
$ç MAM lol o = 14.5 OU = (429 Tão Brabs
PROBLEM 1.38%
+27) Pb * Bbm > vacdum
Tank, Bum AS efa Proc = Ram” Pabs
AB-dO=SEkhe Pg
Pas E OM bar =40 L Pa
I-zo
PROGLEM 1.34
- 1b$
Fête e |APopge] = 8,41
9=320 FH/sz
v:o.oou Hb Sho? Lsuã 62.344 1b/gº
[Bope | a(e2.30u à lb Y32.0 fix E)
32,2 Ib-Ss2] | 144 uz
3 0.39H4 Ibf/ivt LPyago
(decreases)
PROBLEM L4O
Pagto efe L=20cm =60,2m
Tau 6 Foge" Sug gh
T =(13.59 Téo o.2m)
E=20 tm
A |Jogea rode Um? | Ilkgmis2
] 26, ot N/m? =0,26bb bar wi
Tgnge p= LM bars Mercury (g = 19/54 9/emê) -
Gage a 959.81 mjs2 bs,6 Gurrt Gogesa
= LOl+Ze6b=L21b bar e
Pos, = lc gt Fraga A
=L2T66 HLH=267b bar e
PROBLEM LH
Bim E Su Lug
AIR «(3.54 SL )(9.81 me) GES)m
Praç*02 bar ess
Vig | 10 em IN Ibar
Pet 750 vom Hg 1034 im || egemjstl io? Njm?
Su; 213.59 9/em3 2) bar
=4.81 z -
359.8! mis Pass * Bem Prue
= Ibw-02bw=08barç Fabs
t- 21
PROGLEM 1.485
L/x= Sin 6
L
x sine
oe XL
inclined manometer
gives greater resolution
PROGLEM |
» daçS sp=e9d
«(62.2 m 32.2 EX 8)
Libg
32.2 Ib-fils;
gr
teu in?
g=b22 Ice unter =2.89 Ibglivê à
9=32.2 Bjs
1-2%
PROBLEM 1.47
Consider Jue variation of pressure owing to fra effect of
gravita. br an element Am a motion less
gas or ligurd, due foras acting are tia forces ef
pressure on tua upper and lower surface! and tha argh
of tha au atem Jose
Z+HAz
Zz [=]
* Í Toma
2-0 O= plerazjAÃ tmg — pl A
hero m= pp V
da a Vraaz]
= Lad)
=
Rearcam e
Y A (ptrsaz)— pexj=(422) 4
or
plsor)- Pl) dg.
As au
TA Fe Amit ao AZ» O. TUOs gives
de. 2 Ge)
3z Ar
. Bi .
(4) Atmosphere. TE wa C/p tuhere Cs 72 435(mVkg (kPa = Ta gas No,
is insovfed us Es (8), ue hmm
RR
= -4P» dinbo plo = eps RE + ink
cl
om
p= Kemp(- 42)
lonen 200, P=h (latm), giving p= Remp(-92/c).
Tnsertim erewn Valado Gises pomatm when 21 taum
o | IM - (conssz)
[Se] = ep
dia
This telakonshmp is siown ui tra Accompargins plot.
Gestmfs2) (x em)
Tay SE
pés exp (
|-25
PROBLEM 1.47 (Conta)
08
= 08 s
É =.
- 0.4 1 =
=
02.
0 T DA
0 2 4 6 8 19
z(km)
(o) Ocean. Lething 2É-Z ) denote dont, Eq. (K) reads
dp. 4
d2
v-
Than, AStumina Ar constant
paG) ak
o = (294
' lim 2=6, pe fp (lama) quina ps (2)2 +B.
Tasertiny trsum valiis aives Pi atm atuam Zé tm
Cast mist) EC) ida] atm
p= bot 0-4 S6 x1073 m3Tkg [Tum || poisaspio” Mm ameno]
=p+ios Z
Ts celakincoup is shewn x the acto vepamenina plot.
200
N
160
120
p (atm)
so
40 A
PROBLEM 1.5
Using Eq. 1.22
T(F) = 1.8 T(c)+ 32
(o Tltej=21
Ter) =(Le)(z)+32 =69.8
T(R) = T(E)+ 45267 = 529.47
(b) TELE) = 40
T(r)=(h8)M-40 )+32=-40
T(R) =40 + 459.67 = 414.67
ce) T(C)= 500
T(e) = (1.8Hs00)+32 = A3A
TOR) =Ã0A +4459.67 = 13ULT7
cet) TCC)= O
Tltp)cÔLECO +52= 3A
T(ºR) = 3A + 959.67 = UM
te) Tlf) =100
T(ºF) = C1.8HX100)+ 32 =212
TC R)= 2121945967 =67.67
(4) T(Q=-273145
T(ej= (1.8 -273.15)+32 =-459.67
TOR) =-45SA6THUSIGI=O
PROBLEM 1528
Usmg Eq. 1.27
<a)
e)
e)
(43
eo
G)
TÊc) =
TCE) - 32
T.8 18
= T(E) 178
-õ
tise)= 68
(0 = SE -1276 = RO
T(K = 20+27318- 223.15
(0) = HO
T(ºe)= e «278 =-40
T
Tte)= -40+ 2735 = 23315
T(ºp)= Seo
e)= SOS 17.782 AGO
Tt e) - 17.:*8 ae
T(r) =abo+ 273155 633.15
TCE) O
me fa -l78 = 1278
T(K)= “ABAS: 255.37
T(er) = Sa
= BUM 17:28 = 100
T(c) TB
TO: 100 +27: 373.15
T(CE)= —48747
T(C)= — 48% -1A78= -27315
.B
T(x) = O
!-30
ProBCLEM 1.53
Tel TCC) = [nte +agais] -[m, (CO+2741S] = Talk)- Tick)
Tal CP) TF) El Tel CP) 450] TCA) ESET] E TCRI = TR)
PROBLEM 1.54
The expression Tor resistance is
R=Reexpl (+ 40] 3
where Roz 2.70 and To=30k 25) |
Since R= 0.31. ab T=422K 1] À
Bis (az)emp Lo Cais - Lo] E |
Solvihg for 2 1
. In *(.31/2,2) é 44
e: Ti dy + t286AK
4iz 3o 054 me
Thus 01 mm TEA
. 1 250 300 350 400 450 500
R=2.2 exp |r2084 (L- 45] Ta
PROBLEM 1.55%
GIVEN DATA: From the data ot T=32ºE
TRE) R(O) 5139 =Re[t+a(a-5)] > Ro= 51.370
At TEAG'F, R=S/.72. Thus -
Testl 32 51.39 3 s
Tes2 146 5172 staz = BL8I[i+ (IDT x ss a
Finally, R= SIA [+ 34USS HO? (T-33)]
RE BoLitaCr1] * Ras326 + dotre x lo TT
Let To=32º E T=406.A1(R-SL.326) T
TIRESIMTO) = UF «———
PROBLEM 1.5to
Using Eg.1.22, TIP) =I.8 TCC) +32, For T(P)= TC) =
A A
T=:187+32
4
Solving 4 e-40 Foca T
ink: Tl(k)= TÍC)+2ZIZIS = -VOJZTI3IS = 233. Be Tt)
TR: TERETUEASGOT = -dOrESIbIE MIT TER
1-31
PROBLEM 2.1
Known: Am automobile of knovn mass acelerates from a given
veloci to another.
Eron: Determine tira inrbal kinetre eneray and the change ra
Kmeho tnerqu.
ScHampTic ACIvEsM DATA:
ma 200 9
mf= 50 km/h.
No = (00 em/h
ASSOMPT.ONS: (1) Tha autommobrle is ts cloçed SyStemm. (2) The veloutes
ond tinche energies are felative to tia, Aoad.
ANALESIE! The initiot kinefic energa Casturmplio 2) is
Ke, bw +
El im/am | |ixT pin
2 so em VI TT) | Ha
z aU eo( e) [ie | to%Mem Tea)
KE,
= usP4 KI “1
Tua Cla vi Junetre enerys va
Ktkeço Su LE V]
e E]
- à (200) [geo (se | | ie] | El
= 34% KJ a %KEacks
PROBLEM 2,2
KMOwWN: An ooject of knowa mass is located at a specified elevakica
relative tote surface of the earth.
FIND: Determine gravitational potential energy of lhe object.
SCHEMATIC É GAUEN DATA: =400 kg
sa Mm
ASSUMPTIDNS: (1) The Object IT 92418 vis
isa closefcustem. (2) The ac -
celeration o gravity is Constomf. 2.30
LA Á (e L
ANALISIS: The gruvitational potential enerau ig
PE = ma z
UN LRN (ts
= (400lkg (9:18 a Xas libelo MN TEN TA |
=92 kJ «
A
PROBLEM 7.3 *
KNOWN: An object of known wetgkt undergoes à specified change ru its
kinetic energy while its potential energy inereases .
FIND: Delêrmne the final velocity. Rs 3=32a0/8“
SCHEMATIC É GIVEN DATA: s , vça40 Sy/s
ASSUMPTION: The object is à closed system. “object AKE = -S00 4k-bF
AnáLysis: The change tn linekic energu is Foray 2 GO be=m3
O axe a Sontri ENT m= feto ja go |= 100 lb
Thus solving for N7 and insertihs values and à unit comiension Factor
. 2 ACE
GS [om |
- - ) A bs
2z(- 500 St-Ibf [A E.d/s . (40 Hs)?
(100 1b) 4 4
= 3598 Bs Na
1. The analysis makes no use of he Information related to potential energy.
2-2
PROBLEM 2.4
KNDWA: | A bricla of Knogon volume and densiky expenênces a given
decrease. ua grawiratinal potential energy. V=2.5X3.Sxlo vê
FIND: Determine the change im elevation. pras
SLHEMATIC É GIVEN DATA:
LEY sro by
| 9=32.0 M/8*
ASSLMETLONS : (1) The brickis a closed Ape =- Soo fHIbf
system. (2) The aceeleration E Gravity z
às constant. (3) The density of thê. body
às Uniform throughout.
ANALYSIS: (e? ;
V=(0S SB) Es =0.030384 = 5=5517
Based on assumplion (3)
ma=gV =(noblm Vo 030381) = 3.6S1b
The change in pobential energy and the elevation are related by
Ape = mgy4%
Thus, solving for 4z .
AP (-s00 &.ibe) 32.2 W.fh/sa
SE = 9 * Guião Mis | IbE
D =-138 He dz
I The negative sign denotes a decrease, in elevaton -
2-3
PROBLEM 2.8
Knowa: Anobject of known mass moves with a given velocity.
Einp; Determine (o) the final velocity for a given change iu kinetic
enevad »andíb) the change iu elevation for a given change im
potential energy.
. Pes melho
DeHEMATIC À GIVEN DATA: CM wtotis
ASsUMETIONS: (1) The objectis a closed g=32.0 Hs?
System.(2) The acceleration of gravity z (a) AXE = -100 H-tbg
is constant. | (by 4pE = 100 Pt- bg
ANALYSIS: Ca) The change in kinehic energy is related to the initial and
final velocities ly
Ares dm bu]
Thus, selving fer +he Final velocity Vi
2 4kE
Va = TA
sv,
Tnserhiug values and converhng units accordimgly
der . | (2)C100 H-IbE) [32.2 Mo-Skls? 2
Vs (1) | ribe |! Os
D =54,67 Hs « Nz
fb) The change in potential energy is related fo the change tu
elevation ly
APE = mg AZ
Thus, He change in elevakon is
. APE
Az a m
2 Moo He tbe) 32.2 Wo. fHs*
(1 IbY 32.0 Hs2) vibe
o = 100.6 HH « Az
1. The velscity decreases, as expected.
2. The elevation increases , as expected.
2-6
PROBLEM dA
KNOWN: Am object 0€ known mass accelerates from a. given initial
velocity to a given final velocity due do He action of a cesultant force.
EiND: Determine the work done by the resultant force.
ET &Iv : 28 me à bg
a v
h N5=200m]s
Rá Vizsoomis
ASSUMETLONS: 4) The object is à closed system. (2) The resultant force.
is the onty interaction betucen the object ab its surroundings.
ANAIYSLS: By assumphon(2), the work of +he resultant force must equal
Te change iw Kinetic energy. Thus, using Eg 2.6
O work = dm
a 4 (aka) (So0t-2.00* IN
Ikgm/s2
=aio ES work,
E
102 Nm
t. The increase in lunekic energy of the Object is the result of
energy +ransferred to it by EM
e work of the resultawt force. —
PROBLEM Ao
Known: Anobject of known mass undergoes a change of leinebic energy
due to the action of a. resultuut force. Te final velôciy and the work.
done by the force are given,
END: Determine the initial velocity. 22223 m=º300lb
é | V5=200 Hs
SCHEMATIC É GIVEN DATA; FO ENA work done by
fone dor todos = 1008ty
(Odd Cera
ASSUMPTIQNS: (1) The object is a closed system. (2) There is no change in
elevatia. (3) The resultaut force ie the only interaction between the
object ondits sunromadiags.
AnAN Sis : By assumphon (3), the work. of He. resultaut force must equal
The change iu kmetic energy, “Thus, using Eq. 2.6
work = bm (V5*-vr2)
Solving for VT; inserhing values and converking units
2W 2
Vis a tz
tur
2 (00 Blu) | 778 A-ib$ [e Mo. fls? 2 pé
+ 100
(B001b) | Ba | VE 5
= 23300 7 ç>
er
o VN, = 152.0 se Nr
!. The ihcrease m velocity reflects the increase in livetic energy of
the object as a result of energy Iransterred te ih by +he usork
6F He vesultaut force.
PROBLEM 2.12
ENDwa: Am object of Enown mass moves along a stroigut Ive worth a
Known velocity.
End: Determine He e otutiônal speel sta WU wheel whose votatronal
kineho energy is equal ta mag nitude to Ye, objects Incar kúetic energy.
SCHBMATIC É GAVEN DATA: KEo; * KE py
AE i
f===- meto db TAN
| Losv-10Hls ED
4 ,
= object + r=-150 b-H
AssompTON : (1) The Objeck and the Flyuheel are both closed sstems,
ANMNySiS: The kmekHo enerqu of the object is
ke. dmy?
ob í (1) 2 13 UE
«= 4 (Golb(ivo 32 l5za Wenos| * 1853 HH ibf
For the fuwbeel (See the soluhtewu to rroblem 2.11 )
Ko ET?
ee to « [2KEtu . [esses H-ibf) [32.2 tb fhjs?
T (so lh. 42) pot
= 25.82 4
Tn terms of RPM
+) vevi | tos =. 24 ve :
- (25. sz ) | E | mat 246.b revimin Q
aii
PROBLEM 213
+ Two objects fall freely under le infueuce ef gravity feom restand
He same initiod elevaton.
FIND: Show that Abe nitudes 6€ He velocitie are equal at the momeat
just before They stilo fe earth. 8
SCHEMATIC É GIV : 7 A o,
| vºo
z
E fa
Ze pl Ed 6 cp Tur
s:(1) Mn object in free fall is a closed system 2.) The acceleration
of GTONUM AS constawt. (3) There, is no affect 0€ au resistance . (4) The only
forde octing is trai dus do gravity.
ANAaNAISA For an object falling freely under fhe infuence of gravity + ER
Ogg les imivà +mg(z-2,)=0
For N,=O aud E,=0
w. mn? = Mgz,
Thus We 2 9z
Súnce the Ema! velocity doesn't depend on mass, both objects
vil have identical velocittes at fhe moment just before Shey
strike Ihe car.
PROBLEM Z. It
KNOWN: An object of known mass is projected upward from the surface
0f the carta with à lenewn initial velocity. The only force acting on the
object is the force 0€ Gravity.
Em: Plot the velocity of the object versus elevation and dekermine He
elevation when its velDeity reaches zero.
To 13 W20
SCHEMATIC É GV. ETR
ês 9=31.5 Hs?
AssUMETIONS: (1) The object is a closed v7=200 Hs
system. (2) The occelevratiow of Gravity is 4 -meBolb
Cônstaut (3) The only force aching on Fiz ES
objectis the force o gravity. q ———
ANALNSIS: Since he only force oching ou the loody is He force cf Quavi
E 210 aqplies. Tuas the velocity and eleva are related do da E
conditima
o
im Ceniga = Ly tea
Tuus, solving for NT
Ex om for
N” a/Nç-agã inecms of E
When V5=O, Z,1S
ONO 200 Roz
29 (gaste)
Plotting the above velat onship
ted
2 =bmaizHtç EE
200 4
160 1 -
<<.
120 À
ve (fis)
80 +
40 |
T T T T T T
0 100 200 300 400 500 600 700
=(t)
a-i3
PROBLEM 2.17
enown: A box sides doum a pamp. Themass of tin box and ctg
veload amd the. top of tm Lam p oa Eram. The Aa jeometry vs alto
Speatfiod.
FioD* (4) Tx jue oba ! feccdtim , deternana tha velocity oftia hex
at Aa ba à 4 and the clum wu trafo que potentras
certa ta box. Ch) Determuva qua Clanges ui kinchrc aud
potes Aus qr box when frrehm Lã acting and +ou Veto city
ato +ta bate qto dam à msm: Compor uatia dado aaqulth of part la),
seHemaTIC É CiVT DATA:
m= 25 1b
st = ato
AsgsomproNa: |. The loox is the etored system. 2. Tn part(a), fecchon
is negligibie. 3. Tha acceloratia ef gravo 3a.oftist,
ANALTSIS: (4) Tr lue absence 0f any resultont force acting on tira
system, including fricham, E4.2.3 veduces to
dm (NEN?) + mg (20-25) =0 68)
Sovrna,
Vez Wizg (2-2e)
2 7 Ve
=) (24%) +2 (22 É) (SH) = (8;
As shown bu Eq (Rr), tu ineo and potenhal euergy Cumget [a
ta same ma gua tuda bro dra opposite A Sima. Thus
Uh
APE= mg(Ze-20)= Qsw (32.0 (ct É)
= 12427 E-lbf 4 APE
and ARG = + 4 hE.
th) luhether thure à frchon or not; tha potential eue ni que
case vê Ae tame ao dejermined va qurt (a): cIZAT pr DE. TF Vez 4HIS,
à bei ; i ade?
tira €lhanas wi kincht Curso vo smaller in mag ni d : nn
ave = dm[WiNH] sp (a) - (287) | aonde ame |= 218 Hlbf
4xE
Reterrina again to E.29, the decreaes wi potenhal €nar ui URL
se Cam dor ccounted dor wi Jerma of wicrease um kuatic ensrgs
opta box and cuark t+ overmme Lfrichm,
a-tb
PROBLEM zag
KNdWN: Asystem of known mass anda given mitialvelocity experi-
ences à constant deceleratiow due to the action of a resultaut force and
comes to rest.
END: Determine the lengkh cf time lhe force is applied and +he work.
SCHEMATIC É GIVEN DATA :
vs, m=5kg
Fe v) Nicd0 mis
+ wo
x ax =—amjs>
LU tt dt dC td A
AssuUMPTIONS: (1) The system is closed. (2) The horizontal deceleration is
constant.
ANAINSIS: To trhd 4he time, use the fuct that the acceleration is
constant, as follows
av =
4: q > am = asdt
Va tz
Siav= | at
Y É
o
or NEN, = Qx st
Thus - o
at= DE COMO sos. a
Ay E amis?)
The work of Me force Fi is found feom Eq. 2.
oO
Work = LmpÃEwo) —
IN VkT
| kgmis? 102 N-m
6) =-4 ki work.
=- Ss ka)(40) E
1. The negative denotes energy transfer out of lhe system,
PROBLEM qd
Enouon: An object of known mass,
fone for a speufred time, experience a constant agelera bon
from o Eron initial Velocrta,
FiND: Deteruune tu work of fha Agultuut force.
açted on ba a resuHant
scHtMaTic A GIVE DATA!
? /Z Sw 4: 0 2
esto) az fts Lo. x
No sm-WSS
1 Tha objecf tia closed system. 2. Metrom is lusrigontas,
ASSUMPTIONS:
um potenhtas ementa. s. Tha
soma system ex periencos NO Cum
horgomtat acceleratim «3 coustant
ANAL TESES: Artigas Eq24 wite astumphm 2, tha mork % ha
force Rd gistn hy va
Vgorte m o LV, = MO)
Tha finas vetocihy Vo Câm be detarmnes irma astonjh'm 3
dV = à (constant)
dx
Tums dve ad E mm=attito)
o
or q att)
=026 (05s)=18 H/s
be
mes uork= 4 (ao) (19 83 | AE, |
-ASSR Elbç < toorie (6 log)
Blu ;
= (1558 Bebe) [araacta | = 471 to iria
PROBLEM 7.72. (Contd)
(b) Letim N range fem Oto TOwmalh, Aus Ae 00 tgp trair
plots” co boo developed,
40
30 4
—— Wiotal
— 1 e Wr
E —- Wa
5 20 +
z
5
a 1
a
10 4
a
1 AT
a so
AE TT
044 FATITITIT IO IDO
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
Velocity (min)
Wa par frnndia plote 4d up fo amous SEua/h | fia
poser sagured + ovarmuma Aotlina Aosistunce vi mena
sig fi cmd Hm tia poa to ouer coma dvas, “a
lucia specdo, fo cdyas sf los comas damanant becunse pj
5 peedo, 3 .
Va VE dava ua Ta Jim prossima Pre Va. Sina tos
quo E ouartma +hair aff Se ve td 6 Ana
onane from toe fuar stoved em looard tua velecha,
baço pasa elyrvins bus gu sapo alla Sijuficno e fbot om
Lab emsurmphm.
arai
PROBLEM 2.23
Known: Mensured data for pressuve versus volume duving +he compressica
of a refrigeraut ave given.
ElND: (ay Determine n for a fit of the data by pV"= constant. Cb) Use fhe
result of part (a) to evaluate the woric dône during the compression.
(e) Evaluate the porte ustug graphical er mumerical integ vretum of
+he data, (d) compare au clisuss paris Cb) aud (e).
SCHEMA É GIVEN DATA:
Data Point p (bf/in?) V (Gn)
1 112 13.0
2 131 1.0
3 157 9.0
4 197 70
5 270 5.0
6 424 3,0
assumprtons: |. The refrigeraut in the piston-cylinder assembly form a
closed system. 2. The pressure values provided appromimate” Hhe
pressure of the piston face.
AN MYSIS;
cay One approach to fiud mn. is fo begiu with pV "= constant. Taking He
º log o Toth sides ok Vis equation P mt. Metag
Log p+ n £egV = Loge
bo ow Log p = (-n)logv tloge
Tuus, (-n) covresponds lo Ha slope of a plot gp vs. logv. Us
Da spreadsheet program to obtaiu to. Plor a + aqua es bes
Sil qurve:
2.7
2.6 º .
25 Prom the wurve fit
241 Cn) =-04688]
2234 8
F 22 “ acDIOBBI Mm
21 Thus
0.40887
1 py = tonstant
1.8
0.4 0.6 0.8 1 1.2
log(V)
a-àa
PROBLEM 2.23 CCont'id)
(b) Using He results ok gorilas and Ha procedure of Example 2.1, jhe work. is
Vz
. BNE cp V
We dave Ee
Cum bene (3.0 03) - (uD)(13.0) [1 | Blu
o U- DaoLEBND 12 mf | 78 Hibe
=- 0,21b3 Su 4 w
o «o 4 graphical evaluatim of he wore involves a plot of He tabulated
data and & smook urve druwn Hurough Ha data points :
450 =
400 |
350
«300
E 250
E
2450
100
50
0
0 5 10 15
dd
Vin.)
Each elemental rectungle in Hae plot contribules He following Jo Ho
area umber Ve curve:
be Vos ins )jtRI 1 Su Is ot6
(to SÉ NO. iu pls EE) * 5-3 56 x10'! oh
The numbor of vetangles is approximately 40, Hus
Wa(moLi)(S.356xt07“)=- 0.2148 Blu NM
(Ay The results oblalued in parts Cb) awd Cc) gre iu g00A agreement. Each
should be consideredia plausibleestimale for He reasons presented ou
age 40 iu tre discussion oF atual expansum amd conpressioa
processes.
|. The software IT could be used to oblaiu He lensk Squaves Curve fit by
rogramm tas the equatisas for curve Lithing. His custer to use &
spreadsheeF program in Huis Mshance, however.
2. The outy measured cata are We tabulated data paints shown as
filled cirdles , The smooHh cuyve does not necessavily represent He
actual pressure af Yhe piston fare for tha corespondai volume.
a-a3
PROBLEM 7.2 5*
Known: A known amount of gas undargoes à constaut- pressure process
in a piston- culinder assembly besiuning af a speertied Specifi.
volune. The Dorle is Enovsnd Gun tag p pecific
guyp: Determine the final volume.
SCHEM é Ge TA! Pl | ya
wa [pdl=-ISKT
vis 0,2 mig 7 o Je
ES we=-lsty 77
O LO pesar 7
Assumpr ins: (1) The gas is a cloged system. Vo
(2) Pressure is cmstaut duvins Ha process.
ANMNSS: Usmg Eg.2.17
Va
w= S pdv = pl(Ve-V)
= Pp(Y4- mu)
Solving for V> and mserting values
Va & É tm,
CISET) ! bay 10ºN-m
É (Sbar) TES ear! + (02Skg( 0: Mk)
= 002 mº4 Va
2-2
PROBLEM Z. 26
KNONAS Ogg a piston-aqlindew assembly undevgoes a process for thich
pv b3 = constant he vb is bene y '
Finn: Determine the final volume aud pressure.
2
ACHEMATIC É GIVEN DATA: p
Va
Wa Jpar- -4S Blu
v as HW?
AssuMeTIONS.: (1) The Oq is q closed system .(2) The process is
Pelytroçio.
Anasis: To determine V,, substitute He pN velatim into Eq. 2.17
and integrate va vv?
a a constNay = 2 MN
wi = d pax 4 (Sm ont. 25
Now, Const = PV"? With +his expression , amd solving for A
30 (=-3)W —3
Vo pv! Us Y .
C3)(-4S Bh) vs bel 4 HT?
Co bis po? Bla ||iiu* +ês
= 440
PV,<Ig3S Ha Yz
Now, we can use +he p-V relation to get p
1.3 .3
- V - bev/25 .
PR. - (4) =(30 Tê (RS = 44,85 btw Pa
1, The shaded avea on +ue p-N diagram represents the
work for the polytropic process.
PROBLEM 2.27 *
KNOWN: À gas undergoes a compression process. Pressure and votume
are given at Ie in" Rial and Pitol sines! Pressure and vo lume are
related linearty during he process.
EIND: Defermine the work. E
SCHEMATIC É GIVEN DATA: 8 2
p=ibar,Vi=03 m? ; a
fe=3bar ,V,=0.| m? GAS | Log r
Ir
o ol 0% 03
- mê
ASSUMPTIDNS: (1) The gas is a closed system. (2) The compression is a quasi -
equilibrium process. with a linear relation between pressure and volume,
ANANSIS: Based on the given data, He p-V relation cam be expressed
GA
p=4-10V
where pis in bars und Vis inmê The work is delermined using Fg. 247
. w=[pdV
' ê
'
Inserting He p-V relation and integrating
Vos Olm3
w=[" [4- 10] totiim er lay
V:03 m3 — dear
PO oa V,204 —
=[ev-(2)v']] fiel
vz 03
= fu(o1-03)-S (12-08 |too|
O =40kT W
1 The negative sign for work denotes energy frausfer to -+he
System.
a-a8
PROBLEM 2.30%
KNôwWN: Ar undergoes a polytropic process belween two specified states.
FIND: Determine the work.
Schema É GIVEN DATA:
n=0, lomol
p= zo Wf/in3 vu; siso A/ib
p= GO Ibetin3 v,=3984ib
pv”= constant
ASSUMETIONS: (1) The air is a closed system.(2) The system undergoes a
polytropic process.
AntiMSIS: From te pressure - volume relahon for a polytropic process
pu = pv > pulp”
Solving for n
logtr/rO — Yog(2o/go)
log Cu) loq (328/1,50)
Now, using 45% 2.17 te defermine work. and cora tu. molecular Weight
= [,307
of air from Táble UE
“ár =?
W= fear o m [7 pv e mens 1
. (Gr ju, rm. Pura AV PY
«mf mo ml Bu)
Nov am = (0,2 bmol) (284 1b/Iomol) = 5194 Ho
and o (80 Ibshu2)(3AS 2214) -(2(11,80)] has mf] Bu
W =(SAM o (1- 1,307) | EIS? | [TS Skdbf
o =-308,8 Blur w
L The Negative sign or vsork denotes energy transfer sinto the
3yste M.
&-3!
PROBLEM 2.31
kNouN: Warm aim cools slowly im a piston cylinder assembly from a Enoun
initial volume Jo a Evown Sinal volume. During the process, à
spring exerts a foree on the piston Hat varies Linearly from a
nona intrial value Jo a fulad value of zevo. 3
Emo: Determine Jhe initial ud Linol pressure of He air, and He work.
SCHEMAME É GWEN DATA:
A=0.018mº
v=0.003m?
V -D002m? Pass = 100 kPa
. a
Spring force varies linearly from 900 N when
003 m? to zero when Va = 0.002 mP
ET
ASSUMPTLONS : (1) e ar is a closed system.(2) The process oxurs slowly, so
Here is no aceeleration of Ha piston (3) There is ho frichom hetweck He
piston aud He ey lui der wall.(4) The spring force varjes Vinearty LutHy volume,
ANALYSIS: Tue initial aud ftual pressures of he air are defermu ed roma
free -toody diagrum of Fa pidon,as follows. Thafis, ZF=0,so
Lnitialla : Fspriiso = 9)o0N
Pp = Pam + Brig
- 100 bp, 4 COND | IkPa | .ssokp
(0.018m2)
to? n/m? N
Fiada: Fprim =0 >p,=100 kP h
A.
F,
Va +
Nou, the worlk is determined usina Eg.2.17, W=f páv, but from above
p= Pet + Fspring
Since the spring force varies linear from 4300N to zero as volume ques
+Arom Viz0.803 Mm? to Vo 20.002 M
400 N
spring = Uooor, X V-0.002)
and
wW «Vips Eprina) dv Soo «(ge V-0.002) Tu
, &
y O.b0I] [D.oI8) lie?
1
Vo 0. 062m?
= st IPS + S0000V - Io JAY = (ssgoo)yz
A
v,:0.003 m?
HT
D
3 2
10“ Nim
mia) | mg) = - O,125 ES, w
VPa JON -m OizSbda +
t. The negative sign denotes that the prston does worle on Ha Gir 06 the air
cools. Also, tua cotmosphere aud Yha spring do more ow Ya piston,
2-32
- 0125 kPa-m?
PROBLEM Z. 33.
Known: Air undergoes wo processes in series.
FND: Sketch lhe processes ona p-u diagram and determine +he work,
per umt mass of aur.
SCHEMATIC AND GIVEN DTA.:
pReessi-2: pulPiconst
A=icokfa,v; 20.04 mYhg
Des Vz= 0.02 mY kg
PROCESS 2-3! p=const, ve
ASSUMPT NS: (1) The air is à closed system. (2) Bom processes are quasi-
equilibrium processes.
1.3
ANALISIS : For process l-2, R.=R ) = 24b.23 kPa. Thus, je pur
rágram is
as * Z 3
q Zoo
a
100
o OL O 04
vw - m?/kg
The worke for eada process is determined usting Eq. 2.17
wW=[pavemípdu > Wa (pdu
Thus as
Wa ( “entido. (Pur
mw L 50 = a )
º to co bo XD mt) - Ctooicoomo | to Nim | HT
Q-1.3) Via lioFN-m
=-3.082 KS/kg
W. Va
e = Sp páv = PalVg- ta)
3 No?
= (246.23 LAY 0.04-0.027) E es
=4. 4246 k3/eg
Finally
o Wa - Wiz, Wa 4 | 842 ps Wialm
E
1 The result is postiive, denot mg rod Ha net energg transfer
by worke is from the system to He surroundiugs.
2-33
ProBLEM 2.35”
knows: Aun
around à
ED:
+he
object of krown mass is
puiley
atiichad fo a tepe (uound
and falls at constant speed.
Delternune fina pomem transuultad to tia Pullea and
totatronel spres ef ha pultey.
Sepematic Leves DATA:
ue
m=50b
bos
W=3+&s
F=mg
AssumPrONS: |. Tue object falis at constant Speed. 2. Tue acceleratim of
Yravita wu constant)
ANA LTSIS:
una,
Since
E 2i3
W= FO
= 32.0 friçã
Tha power tranenaies do tr pule com be determuned
= ma Vo
16$
Colo Yao gt (38) ||
(14% 4. EMP oraith |
0.27% hp w
n
VzReo , we have.
to =
o 38ts rRey eos
RO CGldt | 02% eua |) Iuin
= us RPM at
2-3
PROBLEM Z.3b
KNowN: The rotational speed and diameter of a drive shafk pull
are known. The net force aphl red by the beit on the pulley isalso giver,
FIND: Determine +he applied torque and +he power transmitted.
SCHEMATIC É GIVEN DATA:
-SEHEMATIC 4 GIVEN DATA —s]
| [— R=D/2
D=03m
300 RPM
k z E- F, =2000N
Lo eps (net tangential Sorce due
A
to but 'tension)
E F. >F,
ANAmSio: The torque is calculated using He tangoutial force amd
te radius ot which Lt acts
S=ER
= (zoc0 N)J( 23 wo) = 300 Nm « dd
Thus, with Fq 2.20 Me power transmitted is
Wma * Í:0O
Uma] OT fu
bos ||t N-milid! ais
= 42 kWe W
H
!
(00 Nem ( 300 Enf Fo)
2:37
ProgLEM 2.3€
EDU: Operating data ate provided for an elechit motor ar
Stendy state.
ED: Deternuna the elechit power required by the metor and
tia pruor developed ba tha outpul Shuatá - Deternune E La
net power mput teta motor. Also, eterna tha amounk
of energy transtar by electricar «vork and by the Slug dorvns
24 of operam ,
SeHematic À GIVEN DATA :
pes==0==.
fOamp h
uoy
- 4
t
AssuMPTONS |. As Shown inths Schomant., the motor is the cloged sygim.
2. Tha system is at Steady state,
ANDALTSIS: (4) Using E 22!
Wetectrae =- CVolinge) (current) al
| warn fai KW
= — (loves) amp) | TvolF TE» -l100 veria
“9 ati 4 Wejeghic
Using Egirto
Wit = CGequea ngular peloch)
La Lew
= (oz 19:m pos Rev | enem] ="))| eu |)
= LotkW 400 Wa
(6) Nuk Welechic + Wokoft
g = (ewjt (roteW) = -0.03kW q we
(6) Tategrahm, + fina enersy transfy Awovats
2h
, - w i
Weleckic = É Wetecdt = Cottew)(2h) = -22 kwh q Welectric
bo, Guopwo)teh)= anétieih + ——Wistass
Wishga = [ Wiugrdt
o
1.º The nuimus sign is needed because energy is dransttrced to doa
Wa tor elechicathy
2. This vou mapusoenti +iu porhon of th electric pomer rapto Ahad «q
vt obtocved 9 a shoga poser output because fe! effecia witiua tia
mejor such sq eleciricas netishuro aut frich'on.
2-38
PROBLEM 2.4
KNOWN; À Wire suspended vertically is sivelched by an applied force,
FIND: Determine (ay the work done, amd Cl) the Young's modulus.
SCHEMATIC É GIVEN DATA:
datas
ASSUMPTIONS: (1) The wive is a closed
sustem, (2) The moving boundary is
the only Work made (3) The appled
force vártes Linenriy with X. (4) Thechange
“in area À is neglecttd.
— x = 10 (unstvetched)
— x, = 10.01 $5+
ANALYSIS !fThe applied force varies vita y
: F varies from O to
x ALCOrÁINo,
25800 Wf
F = (500 É (x IO M) = 2,5x108 X- 2.5x10€
there x às in Hr ond E de im Ibf,
Thus, apply ing Eq. aiz to evaluate the ore of Fha torce E
Xz
w=-frdx
x;
10.01 4+
=- J (z.sx1002x-2,5x108) dx
0 Ft
= [232408 (30,01 10*)— 2.5 x10º (10.01- 10 3]
212,5 Hibfa W
(b) From Problem zm0, He Youngis modulus € ca be expressed ou
Ko
e = o (ão)
where q =E/A is He normal strecs,aud X, is Fhe unshretched lewgth.
Thus E( ko )
cs El K-Xo
. (2500 bg) 40%)
* Corn?) (o. =10) E
. Young!s
2.5x10" Ibt luz ng
a modulas
ad
PROBLEM 2.42
KNowN: À wire 0€ constaut cvoss- sectional area and à a gue initial
length is stretched. The stress - straiu relation is leno
EIND: Derive au expression for the work done onthe toire as a functum
of Straiu.
SCHEMATIC À (IVEM DATA: 5]
c WWEM D Z x
À-area g qn
G&- normal slress 4 Xo————]
£s “ão (strara)
AssUMpTONS:(1) The Wire is à closed system. (23 Stress and straiu are
related linearty . 43 The cross- “Sectonal? area remeins constant.
ANMMSIS: The voor dome on the wire is gem by E. 2.18
We -J SA dx
Xy
From the given stress - straiu relation
o)
lohere CG is à constant (Young's modulus). From His expression
e dx
de = Ke > dx= x de
Substituting info the work expression
E £
w=- | (cejA (x de) = — cAx [ade
o o
Finally Wa - CAX E voork .
z
e Cap ressior
A-HA
PROBLEM 2. 43
know: A soap Bim ow a wire frame is siretched.
EinD: Delermine the work. done.
SCHEMATIC É GIVEN DATA:
te -s
ASSUMPTIONS.! (1) The Lilmis a closed T=2SxiO “Nom
System. (Z) Tha moving boundary is desu
The only Work mode. (3) The suctace
tenston is constant actina on both =
sides ot the Lim. 3 Ira
ANMNSIS: The work Ás determined usina Rg. 2.14
w=- Staa = - [caga
For constaut surface fension
W=- TA Ax
== (osn Pa (Sem) (ie tom | MT
totem! mem
a -5
D =-2,5 40º Tá vw
I. Thenegatie sign denotes work done on the film. Note Ha small
magnitude of the wovk reguired to stretch He film.
PROBLEM Z.4U
knows: A liquid Film ona we frume is stretched,
FLND: Determine the york done.
SCHEMANEO & GIVEN DATA:
ASSUMPTiONS :(1) The Film is a T
closed system. 23 The moving
bowadary às He only sorte mode. Qezin.
GB) The surface Fenton is constant, I é
acting on bolha sides of the Film. — pax =
ANANSIS, The vosrlo is determined using Eq uu
e xa
wW=-S 2a <- Jo tata
For constaut surtace tension
W=-TZRAXx
.- -4 WE : ' VA
=-(2.5x10 TE) ZCM tm) EM
O =-C53u0P Hds q w
I.The negative sign denotes work dene en jhe film. note He small
magnitude of ha work required Jo streteh the £ilm.
243
PROBLEM 2.17
known: Energy transfer by conduction occuvs at Steady stute
+urough a pane wall.
dimensions, termal conductivity, and
surface temperatures are specified.
cinn: Determine the rate of energy transfer by conduchon,
SCHEMATIC É GIVEN DATA:
|. K=0.038 Blu/h.fºE
TE 1ÓF
ASSUMPTIONS:(1) The wall is af steady state .(2) The temperature
varies linearly Lhrough the wall. (3) The thermal conductivity is uniform
and heat transfer is by conduchom -
ANAINSIS: Using Eg.2.3! together with assumption 2
Mn ar dr , To-N
Qt qro ae Ta
Thus
decr
Inserting values
. 30-71 o
Ás o (0.0318 pese too 6) — º) =
=407 Bulhe Ou
|. The values given ave for a typical insulated frame wall.
alo
PROGLEM 7.4 8
Know: À surfuce of Gwen diameter emits Hhurmal radiatum at a
known vate ab a Specifizd temperature.
ED: Determine the emissivity of the surface and plot Jhe rate
of radiant emission vs. Su temperature,
SLHEMATIC É GWEN DATA: .
CI) The Steaua — GeaisW
Boltamamm law applies.GlThe. iafasa” To * 000
emissiviky is constant. o 3
d=2cm =o.0ozm
T= 567 wimik!
ANAINSAS» Using Eq 2.33
& * EGAT, GE
Setyring for &
e. Be da
SATA Ta? “4
xa)
Us W)
(sem x108 (=0.92m!oso KR”
wk! 4
=0.842. « E
Using the emissivity just calculatel, and varging 7; from o fo 2000k, the
folowing plot can be constructed based em Eq. tw) above
250 +
200 1 /
150 -)
Go tw)
100 4
so 1
0d == —— 7
0 400 800 1200 1600 2000
TLCk)
Nokicu fa ropid growth in de with Th.
247
PROBLEM 2. 49%
KnNdwn: A sphere of known surfaga area, temperature emiisivi
emits mermél radiatia . > TEMP . aMef emissi ty
emp, Determine the rate of thermal emision .
SCHEMATIC É GAEN DATA: q Ra
ASSUME TION: The Stefan - fl E=0b Te ormiquo? ,
Boltzmann law applies, N A=tag 6hu/h 8 0p
To= |000ºF = (4O?R
ANAVÓIS: Using Eq 2.23
: “4
Qe = ECAT,
«(06 Hp my xiS* saia H(Idoo ce)!
= Sb07 Btu/h de
2-48
PRobuem 2.51 (Cont'd)
Thus »— IZb0- SO
= —>—— = 97,19 btu q
(9.535 46 bbT) 1719 Bralh
Now, inserkha values in (4%)
T = 1260 — (347119)(0.5357)
= 208 R Tz
L The form of (kk%) illustrates the analogy between heat
conduction through a composite wall and electric curreut flovo-Hrough
a series of resistances. The temperature difference iu He numerator
is analogous to à veltage difference , and tin vuluo ap Ry ind: are
“termos resislonces amalogous to -he electrical resistamees.
Rs
,
SI
PROBLEM 2. 52
Known: Energy transfer ocours from the inside air to tue nulside air
Hhrough am musulated frame wall.
EnD: Determine the stendy-state heat trausfey rate Hrrough he wall,
ÔCHEMATC É GIVEN DATA:
rnside air outside au
Tj-70ºE To 2 10º F
Acts uu/h Hier hos bBm/h HR
disome non (1) The system is at Sendo To; Tio
stute.(2) Newlows law of cooling applês dx em.
tor heat from jue air to the voalt.
(3) The temperature distribution is
linear Hmrough +ue wall. (4) The tuermal
kk: 0.038 Bhy/h. HR
wall
conductwity sf the wall is uniforra. ex
ANMNSIS: For energy hanster between the inside air awi the wal
Q 3 h,A CT; = Tui) <m
For conduettow Hurough rue voall
: dr Twi "Two (2)
d=KA de = KA (Tie)
' Fially, for energy transfer belweeu the wall aud the outside air
Bo =h A (Tue-Te) CG)
Now, from (1)
— O A +)
T 8: (qa) :
Is
And, From (3) .
, Toe * To + ãc( a) (5)
Com bin g (2), (4), and (5), and nokiug Hat A: 0, QE A
á a A (To = TO)
L A
O (x + ze + ES)
Thsertig values
Ci6o Hi) (530 -450)ºR
,
Q = Í 4 ER + 1 )
E E Bio
(= Pa E 0318 rãs “e b PICTTA
Gto)tBo)
= CAES) = 772.7. Blu/ha à
l The quant in te denominader is called Fe overall reciriance, or “Rvalue !
d-52
PROBLEM 7.53
Known: Á hot surfuce ie coveved uitA musulation, Enevgy Jrauster
occurs from nuter surfuce of the msulatiza and The surround air.
END: Determine the minimum tinickmess of insulatiom to mata
the outer surfuce temperature below & dest red value at steady stude.,
CAE A TA
Tair = 30º
f=/0W/mix
kK=0.08 W/m-K
insulatiom
qurt surface
ASSUMPTIONS: (4) The system is at ate state. (2) Newton's law cf
ecoling Gpplies for heaf hunster from He mulato to the air.
(3) The temperature distributiom Through the iu sulatiow às linear.
(4) Tre thermal conduchvity of the insulatiom is UA tfora,
ANAMSIS: For emergu Irançter bu conduciim Harousk the
nda AS ME CMruis TO) “9
Quz-KA dx * KA E
Lu
=300º€c
Further, tor energy transfer from the insulatiow to He air
ão ta. A (To Tai)
A! sieady ctute, Que As ,5S
KA tia TO E BACGTao
Solving for L.
K CTouee — To)
PAO Tao)
Wit &,=0.08 w/m-x , À 210 w/mix,T «300% avá'T. 230%
the plor below can be Lo Sur , air
e
108
Ê Luva
qm A
1 -bb
q
oY
.ez
2-53
PROBLEM 2.576
nown: A System of Krewn mass Undergoes o proces fer toluceh the
hect transfer and work art Exowr, Tha inihol speufc internet
enem is also kreun,
EroD: Determine tha Anel speupc internal energy
schemaTic £ GIVEN DATA:
scbemáric d GIVEN DATA
O Q=-|000:J
W =-200 87
u,= 300 IS)
ASSoMPTIONT: |. A closed Syçhum of km mass is under convderatim,
2. Changes in Einehr and potenhrs enersy com ba heslectd. 3. The
inibiad and final pato are equiibrivm States,
ANALYSIS! The el wi internet €ners cam ta determined fra
me enera balanee vimg assurephm
sKE + SPÉ + su = Q-w
» o 4AU=Q-W
Then, ida asomphm 3, AUS miW-U,), So
- tus) OW
o
Wi = + [DM
we
= 3005 + (-1000) =(-200)] KT
20 leg
- es
ato eg uz
+ According to tha Sign Comvenhons for Q and WI, Qu negahua
hem Amore is à net heat transfr of energy from pha tyrttm and
WU & wegaive Uhen Iuire Ta net work hrans for efeneras
to Are syelem.
PROBLEM 2.57
knowa: Five kg of sitam undergo an expansion in a prstun -eqndom astemb
from state d do stata 2. During tha proces dhere is a Enoum hear transpor
+o tha Siam and à Enmm cvork tranca of en erjy to Jo sicam by e
paddte trheel. Tio Chomas ui speópre interna tiraram qr team 6s
alto |Emown.
gb: Determune tha amount 4 energy Pranstm by work from ta.
segu to Fa puston during tia prcss,
senemaTiC GIVEN DATA!
o=+s0H À my = 2709.9 k]g
o = 2689.6 KJ/kg
ASSUMPT(UNS: | The Steam is Hu closed system. 2. There ds mo change.
wi fue Einehe or potenhos enerau Leoa state do to sht 2.
AMALTSIS: The net vort can le determined Som an energy
balanes. That aj eta &SsUmphonz
GRE + pré +ou = GW
ara Q- at
The net User 1 tha tum of fa Work assocalia ifia fina
peddlenee! Why Gud + Uoork done mta pistom Weixton:
INT = Wow + Woidon
From His aver informa tim Wu = -18SES , here tia penta is
requirts O becanta (ho quddia what fransfers energy to tis iqstim.
Gllectrng Assu da
US + Upiston .Q - ot
Wprtr = Q— SU— Wa
= Go lt) Wu
- 8043- 58 (cost -z70 DEI - (188)
=+3SD lc 3 a
. Was
here +ha posibve sign indicates tur euersy transfevred Agra Amar Piston
sperm do Ho pistom ao tha sim Supands ariny Tia proces,
2-51
PROBLEM 2. 58
kNDwaL: À system of known mass undergoes huo processes iu series.
FIND: Determine the work and hect transfer for the second process.
S T :
Process Lx: U,=U, = H434 4P/IL pa100 Ibflin?
U,= 105,8 6lu/Ib, Qua, =— 38/.36 6h
Process 2-3; p =p, =bOlbfhn? va 7.828Yb
Uç= 214 Btu/lb
dásUM TIDAS: (1) Aclosed sústem is under consideraton. (2) kúetc
vd potential energy effects can be neglected. (3) There is no work
for process |-2.(4) “he pressure is contam d duvida process 2-3.
ANAINSLS Bg assumphon (4), he work for process 2-3 cam be found
Using Egd!
Vz
Wi = SJ, páV = BG) = pemlvçar)
=(bo RX 2 tb.0a 4.434) 8 E mese
o = 75,21 Me Was
Use He energy balance to find Q;s
AU,, * Qs -W3
With AU zmau
Quis E MlUgUz) +Ws Ca)
To get us, use the energy balance for process |-2
o
MiUs-U,) = Garbo
u, = Bia +U = 581.36 Blu) , 105.8 Bhullh
m Po Ch)
= 815.12 Ghu/lb
Thus, from (%)
23 *C)lM2ig- 8ISUZ) +(75,21)
= 687.8 Blu Gas
a-58
PROBLEM 2.6]
known: A closed system undergoes à process with a lenown hear transfer
rate, and Hue power varies as a specified functim of time.
END: Determine cas the vate of change of system energy ot t «0.bh
and Cb) the change mm system energy after 2h.
ScmEMATIC É GIVEN DIA: Qa-tobw
Gem) . (-Bt oct<Ih
5 pé W [a trih
== “o where tisim hand Wisin kw
ASSUMPTION; The syskem is closed.
AnALNSIS: (A) The Hime rate of change of energy at any Hime tis gen by
dE = à O)
M t=0bh
de = [à - (-84) a(-10 kw) (- 8-06) kw
t=0.6h €-0.eh
o a-52 LW4 dEldr
(b) The change in system energy is oblauned by integrating (4) over Ha
hme period 0f Ah. thatis
t=2h
AE = J (a-wde >
CO O tem €=2h
= Qu E (-8t) dt + J (-8) «e]
tºo t=th
=(-10)2) - [co 4] Vo [ese] z
-20- [C4)] -[0-8)C2-0]
2-8 kwW.h
Thus
t, Att=o.bh, the energy of Me system is decreasing at a rat of 5.2kW
because the rate ef endrgg prons cut by heat êxceeds He vate of
enemy transfer tu by work.
2. The negatwe siqu denojes a net decrease. of energy over the fume
period.
d-bl
PROBLEM 2.62
gv A storage battery develops à known power output as a funchen
“me.
END: Plot he power output and the change 1h energy of the
battery, each ada fun dh, of me. Í
SEHEMATIC É GIVEN DATA:
=== IT Es Wa=12 exp Ético)
Cohere Wisin kw and Eis Hime ins)
Q.=0
AsSSUMPTIDANS: (1) À closed system is under consideration (2) There is
no heat transfer.
ANadsa: The hume rate of change of system energy is
o
E A. E.
És ow E =w
dE = —12 -t/60
TÉ I emp( / )
Integrating from +=0 to deny me E
€
AEs - / iIxexpl-tiogdt = E Lexplt/to) =]
om
= m2lexpl-t/to) -1] ud
() Usihg software to plot Wand dE vs. t rangivg from Oto 2505:
15 0
1 «0 -
1 -20 À
41 1
1 a -30 À
Ê 2º 40d
E 05 1 4 -50 À
0 -
“70 À fa
0 rem -B0 T T T T
O 50 100 150 200 250 O 50 100 150 200 250
t(s) t(s)
(b) Inthe limit as t +00, we have Nx O and dE + -IZkT. The plols show
nat Hrese limits gre approached closely attey alsout 2255. For +mes
greater than 4Mis, no more eleclvric power is developed by Pe lbatrerq
amil He energy remains constant.
&-bl
PROBLEM 2.63%
ENOWN: A gas of lenown mass expands iu a piston- cylinder from a specified
initial stute to « lenown final pressure. The pressure -volume for the
proces is given. Also, the apecific internal energy change is leu un,
EIND: Determine the heat transfer for the process,
SCHEMATIC À GAVEN DATA:
m= 025 kg
p= 8 bar, V= 002 m?
fre 2 bar
pvtãa constaut
au: -SS LI/kg
ASSUMPTIDN:C1) The gas isa closed sgstem. (2) kinetic and potential energy
eftects are negligible. (3) The process is pelghrapic wity pv constant
ANAysIS: The heat transfer cam be found Using au energy balance. First,
fino Je worle using Eg. AT, as follows:
Va Ve
=Todv = Coust
w Jp Í, mst: Ay
Integrating and simphf yr
. PeVa = BY
w= CI=[2)
The volume at Yhe final state is Vi a(P/p) = 0.0635 m?. Thus
= 4Qban)(B0bIswl) — (2)(0.025 toSN/m 1kT
W= G- 42) | ibar Hiofn-m
Ea
ha
V
= los bT
Noto, +o get te heut transfer, begin vita the energy bulunce
NEGA AU = Q-W
Solving for Q and nofrisg that AU = mau
Q=mau +W
=(oaskgK-sS kT/k9) + C lS kT)
D = 295 tT« q
|. The heat transfer is posttive, denoting energy transfer by heat
to ue gas as it expands.
a-63