TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.)

SOLUTION MANUAL FOR
An Introduction to
Equilibrium Thermodynamics
Bernard Morrill
Professor of Mechanical Engineering
Swarthmore College
Pergamon Press Inc.
New York • Toronto • Oxford • Sydney • Braunschweig
1
1-1
t = TOO C
s
t = 212°P
s
Chapter 1
1-5
t. = Ov l C
t± = 32°F
t°P = (32 + 1.8°C)
1-2
pV = n PT
Y = »
P
1-3
k =
JR
N.
k =
k = 5.662x10"24 ft lbf/part. °R
k = 200
!00 x 2 = 400 in-lbf
33.33 ft-lbf
1 = 0.0428 BTU
1-6
W = 4k x2
e
where
k1+ k 2
ke " k1 k2 "k1k 2
k -40x60
e k1 +k2 40+60
W = - i x 24 x (i)2 =
W = -T2 = 0 .25 ft-:
= 24 Ibf/in
3.21 x 10~4 BTU
= C
1°C 1.8°P
w kx2
1-4
¥ =
k = 40
W = - 45 in-lbf A r\
-3.75 ft-lbf
W = - 3.75/778 = - .00482 BTU
W = 0.491 BTU
W = p(v2 - V1) = 190.99(5-3)
= 381.97 ft lbf
190.99 lbf/ft2 P =
2
1-8
For a constant temperature process
pV = C
From initial condition
C = 200 x 144 x 3 = 86400 ft lbf
Then
W = C In
V = 86400 x In 5/3 = 86400 x 0.5108
W = 44135 ft lbf
1-9 cont.
c) z = J" y cos x dx + sin x dy
dz = y cos x dx + sin x dy
1-10
u = A ( RT + pB)
du = A(Rdt + Bdp)
1-11
AU = AQ - AW
AW = p (V9 - V. ) constant pressure
process
AW = 5.553 BTU
AQ = - 8.5 BTU
AU = - 8.5 + 5.553
= - 2.947 BTU
Au =
Au = - 12.85 BTU/lbm
continued
cos x = cos x
dz is exact
= 56.73 BTU
1-9
2
a) z = 3xy + 4x
dz = (3y + 8x)dx + 3x dy
If exact then
3 = 3
dz is exact
b) z = x2 + xy + yx 3
dz = (2x + 3y + 3yx2)dx
+ (3x + x5)dy
3 + 3x2 = 3 + 3x2
dz is exact
m = 0.22<
3
1-14 cont.
= .375 x .171 x (425 - 70) + 40
¥ = - 62.76 BTU
1-12
1 HP = ^ Q ° ° = 42.42 BTU/min
dV = dQ - d¥
dU dQ_ d¥
dt dt dt
= - 30. + 42.4
H = 12.4 BTU/min
1-15
cyAT = Au
m cy _ J^4 _ .124 BTU/min°R
M ~ 171 ~ *^2^ lbm/min
1-13
W = ^pdV
For constant energy process, T = C
Then
pV = C
v
¥ = C J ^=Cln^-
v1 1
P, - - S^HP2* = * » ° P*
„ P1V 1 28520 x 1 _ .
0 = 778 ~ 778 36' 65
W = 36.65 In 2.
W = 40.26 BTU
¥ = 31330 ft lbs
Q = W = 40.26 BTU
1-16
u = (j(x1 fx2,x3)
" - 1 2 , DX1 + ^ D J C 2 + % D X 3
Extensive property conjugate to
, v v
^ax1 ' xi
Q X 2 ' ^ X2
•?S_ 1 „
^ x 3 - *3
1-17
Given pV = C
RT
Then I
OT r"1 - "l (.)
continued
1-14
RT 53.3 x 530 iD*
-A V = A U -
= JB O y 4kT - ,A Q
continued
4
T p
R
where i/r
(b)
T2 = 801.71
Au = c ? T = .171(801.7-530)
Au = 46.46 BTU/lbm
1-19
continued
1/2T
1-20
For adiabatic process
r 2
A U = J p dV
p = CV
V1
-r
Au ( 2 -r dV
1-21
Along 1-3 path
Pi7 / "
along 2-3 path
P2
V
2 = P3V3
Divide 1st by 2nd V
1-18
T2 = T1
= 530
= 530 (.235)-286
= 530 (4.25)*286
bXJ = - 94.08 BTU
1-19 cont.
x 10
(b) V_ = 3.715 ft3
T
(a) T2 = 602.2°R
1-17 cont.
starting with (a)
RT
V = —
P
then (a) becomes
continued