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

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SUMMARY

The discussion focuses on calculating the efficiency and power in a belt drive system, specifically a scenario where a belt drive transmits 56kW from the driver pulley to the driven pulley with a power loss of 2.5kW due to friction. The power supplied by the driver pulley is determined to be 58.5kW, while the efficiency of the belt drive is calculated using the formula %Efficiency = (power output/power input) x 100%. The final efficiency is derived from the relationship between the input and output power values.

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  • Understanding of basic power transmission concepts
  • Familiarity with efficiency calculations in mechanical systems
  • Knowledge of belt drive mechanics
  • Ability to apply algebraic equations to solve engineering problems
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  • Research the principles of mechanical efficiency in belt drives
  • Learn about power loss mechanisms in mechanical systems
  • Explore advanced calculations for multi-pulley systems
  • Study the impact of friction on power transmission efficiency
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Engineering students, mechanical engineers, and anyone involved in the design or analysis of belt drive systems will benefit from this discussion.

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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