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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?

Respuesta :

we are given

The pump can generate a pressure of 2.3x107 pa

so,

[tex]p=2.3*10^7[/tex]

a radius of 0.041 m

so,

[tex]r=0.041m[/tex]

a stroke lasting 29 s, the piston moves 0.57 m

so,

[tex]h=0.57m[/tex]

[tex]t=29s[/tex]

Firstly, we will find area

[tex]A=\pi r^2[/tex]

[tex]A=\pi (0.041)^2[/tex]

[tex]A=0.00528[/tex]

now, we can use power formula

[tex]P=\frac{W}{t}[/tex]

F=pA

W=Fh

so, W=pAh

we can plug it

[tex]P=\frac{p*A*h}{t}[/tex]

now, we can plug values

[tex]P=\frac{2.3*10^7*0.00528*0.57}{29}[/tex]

now, we can simplify it

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

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