CH EN 2800 – Fundamentals of Process Engineering
Spring 200
8
HOMEWORK #1 SOLUTIONS
1.
Book problem 1.2
For these, simply put in the appropriate conversion factors and make sure that the units cancel out.
You can find conversion factors inside the front cover of the book.
(a)
3
m
4
3
m
3
ft
hr
lb
ft
3.281
m
h
1
min
60
g
454
lb
m
min
g
0.04
⋅
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
−
10
.50
1
(b)
day
ft
day
h
24
h
1
s
3600
L
28.32
ft
s
L
2
3
3
6100
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
(c)
kg
s
m
m
ft
3.2808
kg
lb
2.2046
s
3600
24
365
yr
cm
100
m
in
m
0.0254
ft
lb
s
yr
cm
in
6
2
2
m
2
2
m
2
⋅
×
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⋅
⋅
-11
10
1.15
2.
Book problem 1.7
This is somewhat reminiscent of the old "if a train leaves Chicago at 2:00 travels 80 miles an hour
towards Philadelphia…" problem.
Basically, what we have to do is determine how many gallons are
used for each of the two scenarios and compare them. Knowing that the final measurement will be in
gallons will help us set up the problem:
gallons should be "on top" in the equations.
Essentially, we
are converting miles required for the trip to gallons required for the trip, so it should be intuitive that
miles should also be "on top" in the equations.
(
)
gallons
hr
gal
2200
miles
525
hr
miles
1000
4190
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
(
)
gallons
hr
gal
2000
miles
475
hr
miles
1000
4211
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
It appears that it would actually take
21 gallons more
if you follow the boss's plan.

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