Calculating Efficiency and Power in a Belt Drive: A Homework Guide"

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of efficiency and power in a belt drive system. Participants are tasked with determining the power supplied by the driver pulley and the efficiency of the belt drive, considering power losses due to friction.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the power input is 56kW and the power loss is 2.5kW, suggesting that power output can be calculated as power in minus power loss.
  • Another participant reiterates the homework statement and equations, expressing uncertainty about how to approach the problem.
  • A later post indicates a correction, claiming that if 56kW is supplied to the driven pulley, then 56kW is the output, while suggesting that the input power is actually 58.5kW.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the correct input power value, with differing views on whether the output power is indeed 56kW or if the input should be considered as 58.5kW.

Contextual Notes

There are unresolved assumptions regarding the definitions of input and output power, as well as the implications of the stated power loss on the calculations.

nettie2311
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Homework Statement



A belt drive transmits 56kW from the driver pulley to the driven pulley. If 2.5kW is lost in belt friction, calculate:

1. kW's supplied by driver pulley

2. efficiency of the belt drive expressed as a percentage.

Homework Equations



%Efficiency = (power output/power input) x 100%

The Attempt at a Solution





I'm really not sure how to tackle this question, if anyone can help it would be greatly appreciated.

Thanks
 
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power in = 56kW
Power loss = 2.5kW
power out = power in - power loss

for efficiency you can just use your equation.

Try it right now!
 
nettie2311 said:

Homework Statement



A belt drive transmits 56kW from the driver pulley to the driven pulley. If 2.5kW is lost in belt friction, calculate:

1. kW's supplied by driver pulley

2. efficiency of the belt drive expressed as a percentage.

Homework Equations



%Efficiency = (power output/power input) x 100%

The Attempt at a Solution





I'm really not sure how to tackle this question, if anyone can help it would be greatly appreciated.

Thanks

copying...
 
Reading was definitely not my strong point last night. Just want to change what I wrote.

if 56kW was supplied to the driven pulley, 56kW is the output.
the input is unknown (well it's 58.5kW)
 

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