Question

A log splitter uses a pump with hydraulic oil to push a piston, attached to which is a chisel. The pump can generate a pressure of 2.3x107 pa in the hydraulic oil, and the piston has a radius of 0.041 m. In a stroke lasting 29 s, the piston moves 0.57 m. What is the power needed to operate the log splitter's pump?

Answer 1

we are given

The pump can generate a pressure of 2.3x107 pa

so,

`p=2.3*10^7`

a radius of 0.041 m

so,

`r=0.041m`

a stroke lasting 29 s, the piston moves 0.57 m

so,

`h=0.57m`

`t=29s`

Firstly, we will find area

`A=\pi r^2`

`A=\pi (0.041)^2`

`A=0.00528`

now, we can use power formula

`P=(W)/(t)`

F=pA

W=Fh

so, W=pAh

we can plug it

`P=(p*A*h)/(t)`

now, we can plug values

`P=(2.3*10^7*0.00528*0.57)/(29)`

now, we can simplify it

Power=2386.92414 watt.........Answer