Solution Manual for an Introduction to Equilibrium Thermodynamics

57 Pages·1972·1.458 MB·English

by  Bernard Morrill (Auth.)

SOLUTION MANUAL FOR
An Introduction to
Equilibrium Thermodynamics
Bernard Morrill
Professor of Mechanical Engineering
Swarthmore College
Pergamon Press Inc.
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Copyright^ 1972. Pergamon Press Inc.
Library of Congress Catalog Card No. 73-173824
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08 017184 2
1
Chapter 1
1-1 1-5
t = TOO C t. = OvC w kx2
l
s
t = 212°P t = 32°F k = 200
s ±
!00 x 2 = 400 in-lbf
1°C 1.8°P
33.33 ft-lbf
t°P = (32 + 1.8°C)
1 = 0.0428 BTU
1-2
1-6
pV = n PT
W = 4k x2
e
Y »
=
P where
k1+k2
ke " k1 k2 "k1k2
40x60
k - = 24 Ibf/in
e k +k 40+60
1 2
1-3
W = - i x 24 x (i)2 =
JR
k =
N.
.25 ft-:
W = -T2 = 0
k =
3.21 x 10~4 BTU
k = 5.662x10"24 ft lbf/part. °R
1-4
P = 190.99 lbf/ft2
¥ =
k = 40
= C
W = A r\ - 45 in-lbf
W = p(v - V) = 190.99(5-3)
-3.75 ft-lbf 2 1
= 381.97 ft lbf
W = - 3.75/778 = - .00482 BTU
W = 0.491 BTU
2
1-8 1-9 cont.
For a constant temperature process
c) z = J" y cos x dx + sin x dy
pV = C
From initial condition dz = y cos x dx + sin x dy
C = 200 x 144 x 3 = 86400 ft lbf
Then
cos x = cos x
dz is exact
W = C In
1-10
V = 86400 x In 5/3 = 86400 x 0.5108
u = A ( RT + pB)
W = 44135 ft lbf
du = A(Rdt + Bdp)
= 56.73 BTU
1-9
2 1-11
a) z = 3xy + 4x
AU = AQ - AW
dz = (3y + 8x)dx + 3x dy
AW = p (V - V. ) constant pressure
If exact then 9 process
AW = 5.553 BTU
3 = 3
AQ = - 8.5 BTU
dz is exact
AU = - 8.5 + 5.553
2 3
b) z = x + xy + yx
= - 2.947 BTU
dz = (2x + 3y + 3yx2)dx m = 0.22<
+ (3x + x5)dy
Au =
Au = - 12.85 BTU/lbm
2 2
3 + 3x = 3 + 3x
dz is exact
continued
3
11--1122 1-14 cont.
11 HHPP == ^^ QQ °°°° == 4422..4422 BBTTUU//mmiinn = .375 x .171 x (425 - 70) + 40
¥ = - 62.76 BTU
ddVV == ddQQ -- dd¥¥
ddUU ddQQ__ dd¥¥
ddtt ddtt ddtt 11--1155
ccAATT == AAuu
== -- 3300.. ++ 4422..44 yy
mm cc __ JJ^^44 __ ..112244 BBTTUU//mmiinn°°RR
yy
HH == 1122..44 BBTTUU//mmiinn
MM ~~ 117711 ~~ **^^22^^ llbbmm//mmiinn
111---111333
1-16
WWW === ^^^pppdddVVV
u = (j(x x,x)
1f 2 3
FFFooorrr cccooonnnssstttaaannnttt eeennneeerrrgggyyy ppprrroooccceeessssss,,, TTT === CCC
TTThhheeennn
"-12, DX1 +^D J C2+%D X3
pppVVV === CCC
Extensive property conjugate to
vvv
, v v
¥¥¥ === CCC JJJ ^^^ lllnnn^^^---
===CCC
^ax ' xi
vvv 111 1
111
PPP,,, --- --- SSS^^^HHHPPP222*** === *** »»»°°° PPP***
QX ' ^ X2
2
„„„ PPP111VVV111 222888555222000 xxx 111 ___ ... •?S_ 1 „
000 === 777777888 ~~~ 777777888 333 666'''666555
^ x - *3
3
WWW === 333666...666555 IIInnn 222...
11--1177
WWW === 444000...222666 BBBTTTUUU
GGiivveenn ppVV == CC
¥¥¥ === 333111333333000 fffttt lllbbbsss RRTT
QQQ === WWW === 444000...222666 BBBTTTUUU
TThheenn II
1-14
OOTT rr""11 -- ""ll ((..))
RT 53.3 x 530 iD*
-A V = AU -
= JB Oy 4kT - ,A Q
continued ccoonnttiinnuueedd
4
1-17 cont. 1-19 cont.
starting with (a) x 10
RT
(b) V_ = 3.715 ft3
V = —
P
then (a) becomes T
(a) T = 602.2°R
2
T p
R 1-20
For adiabatic process
(b)
r 2
A U = J p dV
where V1
i/r
-r
p = CV
( 2 -r
Au dV
1-18
T2 = T1
= 530
bXJ = - 94.08 BTU
= 530 (.235)-286
1-21
= 530 (4.25)*286
Along 1-3 path
T = 801.71
2
Pi7/ "
Au = c T = .171(801.7-530)
?
along 2-3 path
Au = 46.46 BTU/lbm
PV = P3V3
22
V
1-19 Divide 1st by 2nd
1/2T
continued continued
5
1-21 cont. 1-22 cont.
AU = C(T-T) = .171(1069,3-534.6)
2 3 y 3 2
AU = 91.43 BTU
2 5
/3
2*3 J dv = P(V-V)
2 3 2
2.5
= 36.63 BTU
V, = .0466 ft 2W3
3
Q =^ U +AW = 91.43 + 36.63
2 3 2 3 2 3
128.06 BTU
= 2897 psia 2Q3
0
3W1
1-22
3Aui =3*Qi = WV
P = ^ 15
2
= .171 (405.3 - 1069.3)
P = 39.58 psia
2
5^ = _ 113.54 BTU
AQ = - 113.54
3 1
534.6°R
Wnet = ^W = ~22-11 + 36.63 + 0.
= 14.52 BTU
1*U2 = l /- 1W2 = CV*T
2 Q = X Q = 0. + 128.06 - 113.54
net
= .171 (534.6 - 405.3) = 14.52 BTU
AU = 22.11 BTU
1 2
&W =-22.11 BTU 1-23
1 2
W
in
_ ^3V3 _ 39.58 x 144 x 10
3 ~ R " 53.3 W . = H-52 BTU
net
T, = 1069.3°R
Qin = 2Q3 = 1 2 8 , 06
3
_ 14.52
x 100
" 128.06
continued
continued
6
1-23 cont. 1-26 cont.
FFoorr llaarrggeerr eennggiinnee
1-24
For smaller engine
^= ( i 3) l o o - ( i- f g) 100
c
^ = 22.73$
c Vl/2
If ¥ = W then
Co.OuP.r . - ^ - _5 10 T 0
T 660 510
H L 2
C.O.P. = 3.40
TT == 22 TT -- TT
11--2255
LL wwnneett xx22 552200
II cc QQ.. '' TT.. '' 11446600
iinn 11 1-27
\ = .6438 ¥hen T„ = 1000°R and T = 500°R
c T
¥ net. = KI cQ . m h 1000-500 , /
I L" 1000 ^1°0-5O/
O
^in ~ ^out \ c ^in . ? T -2 T
t, H lL 1000 „„,
1 s= 2 T-T = 1500X 1 00 = 6 6'7^
v U ) Qm. = ^Qo ut. =100 Wn TL
Qin " 1. - .6438 = 280'77
1-28
¥ , = Q. - Q .= 280.77 - 100
net in out
a) constant temperature process
C
¥ . = 180.77 BTU
net
P = V
v
(¥) = c/ " f = c m^
1-26 T
v 1
1
In general from equation 1.15.11 V2
V2 (¥) = m RT In —
T 1
¥ork = JQ^ - |Q ^mR^-T^lity-
2
b) constant pressure process
rv2
continued ( w ) = pi Jvi dv = pi(VV
P
7
1-28 cont. 1-29 cont.
oO
cc)) AAddiiaabbaattiicc process 11
5 | e 3 * dt
w = - v. RRll
ppVV == CC 0
so
T2 dV C Ll-2 tt
W = |- RC £ RC
0
W = - ^C joules
H =o - 1- r v 2 vi
ZZ 33
111---333000 AAA---^^^ HHHaaa -
rr " a "' " V " 1
11--2299 ii I—11 —. i 1 -
_ J • /
MMoo ||77\\
• li Mi i ,,,|||WWW === 000 sssiiinnnccceee MMM iiisss
* Pl'r * 222 ccoonnssttaanntt
By Kirchhoff• s Law
2W = ^0 HOdM
^ (idt + Ri - V = 0 5
c 7 FFoorr ccoonnssttaanntt HH
Differentiate to D.E.
ddTT
1 . di ddMM == -- CCHH ~~
n 0
c 1 + R dt = TT
Then
di 1 .
A
0r d t+R C1 = ° 2 dT
4-Y
m = - CH
This is a seperable equation 2 3 0 Q
T
dT
di 1 W = -4^CH2J
flt 0 2 5 0 0
r
+ rc dt - 2T T2
Q
ln 1 = - RC i T
t + K 2 1
e~ = 4 0CH0 T "
i = K ? 2 o
£ VHo
at t = 0 w
2W3 T
V Q
or
V - R— 1 " R
- 1- t
i =x e 2W3 =^0H0M0
RC amps
R
(volts x amps) dt
U
continued
8
Chapter 2
2-1 2-6
P(1,1) = 1/6 x 1/6 = 1/36 VA = 4; V = 3; V = 2
B c
P. = exp [-o< - ft V] i = A,B,C
2-2 i
No. Heads P(A) P(A) x (A) exp[«K ] = 2 exp[-jSV]
i
0 1/8 0
1 3/8 3/8
2 3/8 6/8 C<= ln(e±p[-4p] + exp[-3^] + exp[-2y3])
3 1/8 3/8
£ P(A)x(A) = 12/8
3
-4exp[-4(j]-3exp[-3£]-2exp[-2£]
- =2.5
<X> = 2 Pi X (V = 1 ,5
exp[-4£]+exp[-3jb]+exp[-2£t]
i=0
LLeett xx == eexxpp[[--^^33 ]]
2-3 11..55 xx44 ++ 00..55 xx55 -- 00..55 xx22 == 00
U =-k ln P =-k ln 1/8 3 x2 + x - 1 =0
5
U= k In 8 = 2.0794 k x = - ~ - lj\l'l+y\2 = 0.434, -0.768
UUssiinngg tthhee ppoossiittiivvee rroooott
2-4
exp[-/3] = 0.434
Tl=-k ln P =-k In 3/8
y3= 0.834
U.= k In 8/3 = 0.9808 k
c(= ln(exp[-4x0.834] + exp[-3x0.834]
+ exp[-2x0.834])
2'5 3
c<= ln(exp[-3.336] + exp[-2.502]
S =-k £ p. in p.
+ exp[-1.668])
i=0
- - k r-ln-^ln^ln^ln-1 ©<= ln (0.3061)
- l-8in8^ 8^ 8^ 8J
©(= - 1.184
S = k(jln8 + |ln|) = 1.2555 k
P = exp[-0^] expf-^]
±
exp[-0(} = exp[L184] = 3.267
continued