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Assignment #1
8/18/04
PROBLEM 1.1
PROBLEM STATEMENT:
Determine the SI values for the following quantities:
6
ft, 200 lbm, 70
o
F, 25 lbf/ft
2
, and 1 atm.
GIVEN:
6 ft, 200 lbm, 70
o
F, 25 lbf/ft
2
, and 1 atm.
FIND:
SI values for each of the given quantities
ASSUMPTIONS:
None.
GOVERNING RELATIONS:
1.
1 ft = 0.3048 m
2.
1 lbm = 0.45359 kg
3.
T(
o
F)= [1.8*T(
o
C)] + 32
4.
1 lbf/ft
2
=47.88 Pa = 47.88 N/m
2
5.
1 atm = 1.01325 bar = 101325 N/m
2
PROPERTY DATA:
Not applicable
QUANTITATIVE SOLUTION:
6 ft = 6 ft * 0.3048 m/ft = 1.8288 m
200 lbm = 200 lbm * 0.45359 kg/lbm = 90.72 kg
70
o
F = 1.8 * T(
o
C) + 32
(70 – 32)/1.8 = 21.11
o
C
25 lbf/ft
2
= 25 lbf/ft
2
* 47.88 (N/m
2 )
/ (lbf/ft
2
) = 1197 N/m
2
1 atm = 1.01325 bar = 101325 N/m
2
DISCUSSION OF RESULTS:
Selfexplanatory
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View Full DocumentAssignment #1
8/18/04
PROBLEM 1.3
PROBLEM STATEMENT:
The "British gravitational" system of units takes
length, time, and force as the base units (foot, second and pound force,
respectively.)
Mass is a derived unit called a "slug"; one slug, when
accelerated at 1 ft/s
2
,
produces a force of 1 lbf. Define the relation between
the slug and the length, time, and force units above and determine the value
of the conversion factor g
c
for this unit system.
GIVEN:
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 Spring '08
 DEINERT

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