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FACULTY NOTES
The LTAs and Spinoffs are designed so that each professor can implement them in a way that is consistent with his/her teaching style and course objectives. This may range from using the materials as out-of-class projects with minimal in-class guidance to doing most of the work in class. The LTAs and Spinoffs are amenable to small group cooperative work and typically benefit from the use of some learning technology. Since the objective of the LTAs and Spinoffs is to support the specific academic goals you have set for your students, the Faculty Notes are not intended to be prescriptive. The purpose of the Faculty Notes is to provide information that assists you to take full advantage of the LTAs and Spinoffs. This includes suggestions for instruction as well as answers for the exercises.
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FACULTY NOTES
LTA 11
Helium Usage at Kennedy Space Center
Solutions
- First use Boyle’s Law to determine the volume at 3500 psia.
P 1 V 1 = P 2 V 2 (14.7psia)(70,000 scf) = (3500 psia)V 2 294 ft^3 = V 2
Since we know that each railcar holds 1050 ft^3 ,
1050 ft 3 294 ft^3 per day
= 3.57 days
Thus, the gaseous helium in one railcar will last 3.57 days (or they will use approximately 2 railcars of helium per week).
- Here we use Boyle’s Law once more to determine the volume at 3500 psia.
P 1 V 1 = P 2 V 2 (14.7 psia)(1,000,000 scf) = (3500 psia)V 2 4200 ft^3 = V 2
KSC will use 4200 ft^3 of gaseous helium from a railcar. Each railcar holds 1050 ft^3 , so
4200 ft 3 1050 ft 3 per railcar
= 4 railcars
- There are several steps/conversions we have to perform.
a) First convert all temperatures to the Kelvin scale. C = 5/9(F – 32) K = C + 273 ⇒ C = 5/9(70 – 32) ⇒ K = – 271.5 + 273 ⇒ C = 21.1° C ⇒ K = 1. and K = C + 273 ⇒ K = 21.1 + 273 ⇒ K = 294. So 70° F = 294.1 K and – 271.5° C = 1.5 K
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Mathematical Aside
Direct and Inverse Proportions
Solutions
Questions
k = 0.4 for each temperature/volume pair.
The shape of the graph is a straight line.
- Since the volume divided by the temperature equals the constant 0.4 for each pair of data, the equation can be written as follows: V = 0.4T or
V
T
- k = 140 for each volume/pressure pair.
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Pressure approaches zero as volume becomes very large.
- Since the product of the variables volume and pressure equals the constant 140 for each pair of data, the equation can be written as follows: V ⋅ P = 140 or P =
V
Exercises
- a) direct proportion b) k = 200 c) P = 200T
- a) inverse proportion b) k = 200 c) M =
T
- a) direct proportion b) k = 3 c) y = 3x
- a) inverse proportion b) k = 12 c) y = 12/x
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Mathematical Aside
Unit Analysis
Solutions
Exercises
- a) 22.36 mph b) 1.17 hours
- a) 44.1 psia b) 303.85 kPA
- 14,524 ft^2
