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ENGRD 221 – Prof. N. Zabaras
9/3/2007
SOLUTIONS TO HOMEWORK
1
Problem 1:
Given
: Oxygen in a pistoncylinder process undergoes a process where
.
The total work done is known.
1.4
pV = constant
Find
: The final volume and pressure
Assumptions
: (1) O
2
is a closed system (2) process is polytropic.
Analysis
:
We start with
, where W is the work done in a pistoncylinder assembly.
2
1
V
V
W
pdV
=
∫
28 ft
3
35 lbf/in
2
50 Btu
p
1
= 35 lbf/in
2
V
1
= 28 ft
3
pV
1.4
= const.
To determine V
2
, substitute the pV relation into this equation and integrate
2
1
V
.4
.4
21
1.4
V
vv
constant
W
dV
constant
v
−
==
∫
.
4
. But we know, constant = pV
1.4
thus the
1.4
11
constant=P V
Using this expression and solving for V
2
, we get
2
.4
0.4
3.
4
1.4
2
3 1.4
2
0.4 W
(.4)(50 Btu)
778 ft.lbf
1 ft
V
+ V
=
(28 ft )
= 0.2928
P V
(35 lbf/in )(28 ft )
1 Btu
144 in
−
−
=+
V
2
= 21.56 ft
3
Now, we use the pV relation to get P
2
1.4
1.4
2
1
2
2
V
lbf
28
P = P
35
= 50.47 lbf/in
V
in
21.56
⎛⎞
=
⎜⎟
⎝⎠
The shaded portion in the pV diagram shows the work done.
Page 1 of 6
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View Full Document ENGRD 221 – Prof. N. Zabaras
9/3/2007
Problem 2
:
Given
: Dimensions of the swimming pool = 100 m x 50 m and depth varying from 1 m
to 4m, density of water = 998.2 kg/m
3
, atmospheric pressure = 0.98 bar.
Find
: total force at the bottom of the pool and pressure on the floor at the center of the
pool
Analysis
:
Total pressure at the bottom of the pool = weight of water inside the pool
+ downward force exerted by atmosphere on the surface of the water.
F
total
=
F
atm
+ F
grav
To find the weight of water, find the total mass of water inside the pool as follows:
3
3
7
kg
1
998.2
(1)(100)(50)
(3)(100)(50) m
m2
(998.2)(12500)
1.25x10 kg
mV
ρ
⎡⎤
==
+
⎢
⎣
⎥
⎦
Gravitational force exerted by water is given by,
F
grav
75
2
2
3
m1
N1
k
N
mg
1.25x10 kg x 9.81
1.226x10 kN
s1
k
g
m
0
N
=
F
atm
=
52
25
atm
surface
3
10 N/m
1 kN
P
A
(0.98 bar) (100x50)m
4.9x10 kN
1 bar
10 N
Therefore, the
total force acting on the bottom of the pool surface is
F
total
=
1.226 x 10
5
kN + 4.9 x 10
5
kN = 6.126 x 10
5
kN
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This note was uploaded on 07/09/2009 for the course M&AE 326 taught by Professor Psiaki during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 PSIAKI

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