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

$P=\frac{W}{t}$

F=pA

W=Fh

so, W=pAh

we can plug it

$P=\frac{p*A*h}{t}$

now, we can plug values

$P=\frac{2.3*10^7*0.00528*0.57}{29}$

now, we can simplify it

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