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sleet_krkn [62]
2 years ago
13

An exercise room has six weight-lifting machines that have no motors and seven treadmills, each equipped with a 2.5-hp (shaft ou

tput) motor. The motors operate at an average load factor of 0.7, at which their efficiency is 0.77. During peak evening hours, all 13 pieces of exercising equipment are used continuously, and there are also two people doing light exercises while waiting in line for one piece of the equipment. Assuming the average rate of heat dissipation from people in an exercise room is 740 W, determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions.
Engineering
1 answer:
Diano4ka-milaya [45]2 years ago
3 0

Answer:

25.4 kW

Explanation:

There are 15 people doing exercises, each will dissipate 740 W of heat, so they will disspiate a total of p = 15*740 = 11100 W = 11.1 kW

There are 7 treadmills, each has a 2.5 hp motor (1.86 kW) running at a load factor of 0.7 with an efficiency of 0.77. So their total power would be p = 7*1.86*0.77/0.7 =14.3 kW

So the total heat dissipated would be 11.1 + 14.3 = 25.4 kW.

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49. A solenoid coil with a resistance of 30 ohms and an inductance of 200 milli-henrys, is connected to a 230VAC, 50Hz supply. D
Leya [2.2K]

Answer:

69.59 ohms

Explanation:

Given that

L=200\ mH

F=50\ hZ

R=30\ ohms

For inductance

X_L=2\times \pi\times f \times LX_L\\=2\times \pi \times 50\times 200\times 10^{-3}\\=62.8\ ohm

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8 0
2 years ago
An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30
sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

F = LwY_{avg}

where;

w is the width of the annealed copper

Y_{avg} is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm

Now, determine the average true stress, Y_{avg}, for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

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Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

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F = LwY_{avg}

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The power required in this operation is given by;

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2 years ago
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jonny [76]

Answer:

ALL CAREFULLY ANSWERED CORRECTLY

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