Assignment #2
4/21/2010
PROBLEM 2.2
PROBLEM STATEMENT:
The official weight of a baseball is 5 oz (0.3125 lbf). What
is the kinetic energy of a 90 mph fastball in ftlbf, kJ, and Btu?
DIAGRAM DEFINING SYSTEM AND PROCESS:
GIVEN:
W
=
0.3125 lbf,
v
=
90 mph
ASSUME:
Standard gravitational acceleration, g=32.17 ft/s
2
FIND:
Kinetic energy (in ft

lbf, kJ, and Btu)
GOVERNING RELATIONS:
Kinetic energy,
2
c
m
KE =
2g
v
.
Weight
c
mg
W
g
=
QUANTITATIVE SOLUTION:
Mass of the ball:
2
c
2
Wg
0.3125lbf 32.17lbm ft / s lbf
m
0.3125lbm
g
32.17ft / s
×
=
=
=
g
g
The kinetic energy is:
a.
(
29
(
29
(
29
2
2
2
2
2
c
c
2
s
m
1
miles
5280 ft
1 hr
KE lbf =
=
0.3125 lbm
90
2g
2g
hr
miles
3600 s
lbm ft
= 2722.5 ft
32.174lbm ft / s lbf
s
= 84.62
ft lbf
×
×
÷
v
g
g
g
g
b.
(
29
(
29
1
1
KE kJ =
KE ft lbf =
84.62
737.56
737.56
= 0.115
kJ
×
×
g
c.
(
29
(
29
1
1
KE Btu =
KE kJ =
0.115
1.055
1.0551
= 0.109
Btu
×
×
DISCUSSION OF RESULTS:
Since it was assumed that the gravitational acceleration was standard (32.17 ft/s
2
),
the mass in lbm has the same numerical value as the weight in lbf. If g were not
standard, this would not be the case, since the value of g
c
(a units conversion
constant) is always 32.17 in USCS units, irrespective of the gravitational force.
v
=
90 mph
W
=
0.3125 lbf
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4/21/2010
PROBLEM 2.4
PROBLEM STATEMENT:
Just before reentry from a 250 mile high orbit, the space
shuttle weighs approximately 250,000 lbf and has a velocity of 17,300 mph.
At this altitude the gravitational acceleration is 29 ft/s
2
. What is the total kinetic
plus potential energy of the shuttle in ftlbf and in Btu?
GIVEN:
h
=
250 mile, W = 250000 lbf,
v
=
17300 mph, g = 29 ft/s
2
FIND:
Total kinetic and potential energy (ft

lbf and Btu)
GOVERNING RELATION:
2
total
c
c
m
mgh
E
= KE+PE =
+
2g
g
v
,
c
mg
W
g
=
QUANTITATIVE SOLUTION:
The kinetic energy of the space shuttle is computed as follows:
(
29
(
29
2
2
c
c
c
2
2
2
2
2
12
Wg
m
1
KE =
=
2g
2g
g
250000 lbf
1
miles
5280 ft
1 hr
=
17300
2
hr
miles
3600 s
29 ft/s
KE = 2.775 10
ft lbf
×
×
×
v
v
g
The potential energy is:
(
29
c
c
c
11
Wg h
mgh
PE =
=
g
g
5280 ft
= 250000 lbf
250 mile
mile
= 3.3 10
ft lbf
×
×
×
g
Thus, the total energy of the shuttle is
(
29
(
29
12
11
total
12
E
= KE+PE = 2.775 10
+ 3.3 10
= 3.11 10
ft lbf
×
×
×
g
which is also equivalent to
(
29
12
total
9
1 Btu
E
= 3.11 10
ft lbf
778.16 ft lbf
= 3.99 10
Btu
×
×
×
g
g
DISCUSSION OF RESULTS:
This is clearly a situation in which kinetic and potential energy are not negligible!
When converted to internal energy during reentry, the air temperature and shuttle
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 Spring '08
 DEINERT
 Thermodynamics, Energy, Work, Problem Statement, QUANTITATIVE solution, DIAGRAM DEFINING

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