Help Needed: Calculating Coal Delivery Rate with 60kW Motor

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    Coal Motor Rate
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The discussion focuses on calculating the coal delivery rate using a 60kW motor, where only 90% of the power is available for work. The formula derived involves using potential energy considerations, specifically the equation m = (Pt) / (gh), where P is the effective power, t is time, g is the acceleration due to gravity, and h is the height of the hopper (19.5m). The efficiency of the motor must be factored in, leading to the effective power being 54kW (60kW * 0.90).

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steelcap
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Hey everyone, just wondering if anyone can help me out with a homework question I'm having trouble with. I've been stuck on it for about 45 minutes now, any help will be appreciated. Thanks.

"A conveyor belt is driven by a motor rated at 60kW. Only 90% of the rated power is available for useful work. The coal is to be raised and deposited into a hopper 19.5m above. How many kilograms of coal can be delivered by this motor per minute?"

- Matt
 
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If you use t = 60s, then find the mass through work/potential energy considerations.

[tex]P = \frac {W}{t}[/tex]

[tex]W = mgh[/tex]

[tex]P = \frac {mgh}{t}[/tex]

[tex]m = \frac {Pt}{gh}[/tex]
 
Dont forget to forget to multiply the power by the motor's efficiency!
 

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