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

[nettie2311]

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

 

[L2E]

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!

 

[berkeman]

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...

 

[L2E]

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)

 

[DeeNos]

for sharing this homework guide on calculating efficiency and power in a belt drive. I can provide some insights and guidance on how to approach this problem.

To solve this question, we need to use the equation for efficiency, which is %Efficiency = (power output/power input) x 100%. In this case, the power input is the 56kW transmitted from the driver pulley, and the power output is the power received by the driven pulley after accounting for the 2.5kW lost in belt friction.

1. To calculate the power output, we need to subtract the power lost in belt friction from the power input. So, the power output would be 56kW - 2.5kW = 53.5kW. Therefore, the power supplied by the driver pulley is 53.5kW.

2. To calculate the efficiency, we can plug in the values into the equation: %Efficiency = (53.5kW/56kW) x 100% = 95.54%. This means that the efficiency of the belt drive is 95.54%.

I hope this helps you understand how to approach and solve this problem. Remember to always use the appropriate equations and units when solving scientific problems. Good luck with your homework!