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

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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
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

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## [Solved] TEST BANK FOR An Introduction to Equilibrium Thermodynamics By Bernard Morrill (Auth.)

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• Submitted On 10 Feb, 2022 12:19:44
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...
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