M Chapter 3, Problem 3/116 Google Chrome Secure https:// /edugen/shared/assignme
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M Chapter 3, Problem 3/116 Google Chrome Secure https:// /edugen/shared/assignment/test/aprint.uni edugen.wileyplus.com Print by: RUBEN PAQUINI Section 03 Homework #4 *Chapter 3, Problem 3/116 An escalator handles a steady load of 34 people per minute in elevating them from the first to the second floor through a vertical rise of 29.9 ft. The average person weighs 125 lb. If the motor which drives the unit delivers 6.2 hp, calculate the mechanical efficiency e of the system. 29.9 Answer: e the tolerance is +/-2% Question Attempts: 0 of 1 used right 2000-2017 John WM & Sons related com hts reservedExplanation / Answer
Average person weight =125 lb
1lb = 0.4536 Kg
Average person mass/weight (in kg) = 125 * 0.4536 = 56.7 kg
Mass of 75 people(m) = 34 * 56.7 = 1927.8 kg
Further 1 feet = 0.308 m
Vertical Rise (h) = 29.9 feet
Vertical Rise (h) = 29.9 * 0.308 = 9.209m
Force required by escalator (F)to lift 34 people = mgh
m is weight of 34 people
h is height to be lifted
g is acceleration due to gravity (9.8 m/s)
F = 1927.8 * 9.8 * 9.209
= 173980.48 N
Power Required by escalator = F/s
s is second
Power required by escalator = 173980.48 /60
= 2899.675 Watt
Motor Power = 6.2 hp
1 hp = 745.7 Watt
Motor Power = 6.2 * 745.7 = 4623.34
Efficiency of Motor = Power Required by escalator/ Motor Power
= 2899.675/ 4623.34
= 0.62718
Efficiency of Motor(in %) = Efficiency of Motor * 100
= 0.62718 * 100 = 62.718 %
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