# Pneumatic Lift Problem

## Homework Statement

A pneumatic lift system is being demonstrated at a sales show. The total load is 70 kg, and the lift piston is 15.2 cm in diameter and has a 20.2 cm stroke. A portable air bottle with an initial pressure of 20 MPa and a temperature of 23 Celsius degrees is to be used as the pneumatic supply. A regulator reduces the pressure from the bottle to the lift system. Neglecting all volume in the lines from the bottle to the piston, determine the number of times the piston can operate per air bottle if the air in the bottle remains at 23 celsius degrees and the volume of the bottle is 0.05 m3

P = F/A
PV = mRT
M = ρ v
V = πr2 h
A = 2πrh + 2πr2

## The Attempt at a Solution

V = π (0.076)2 x 0.202
V = 0.0036m3

A = 2 π 0.076 x 0.202 + 2π(0.076)2
A = 0.13m2

P = 70 x 9.81 / 0.13 + Patm
Pcyl = 106607.61 Pa

mair = 1.225 x 0.0036
mair = 0.0049 kg

Finding mass of air in bottle

PV = mRT
2000x103 x 0.05 = m x 0.287x103 x 296
m bot= 1.18kg

mbot / mair = 240 lifts.
I am unsure if that is the correct step to take, I feel as though i am missing something obvious.

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SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

A pneumatic lift system is being demonstrated at a sales show. The total load is 70 kg, and the lift piston is 15.2 cm in diameter and has a 20.2 cm stroke. A portable air bottle with an initial pressure of 20 MPa and a temperature of 23 Celsius degrees is to be used as the pneumatic supply. A regulator reduces the pressure from the bottle to the lift system. Neglecting all volume in the lines from the bottle to the piston, determine the number of times the piston can operate per air bottle if the air in the bottle remains at 23 celsius degrees and the volume of the bottle is 0.05 m3

## Homework Equations

P = F/A
PV = mRT
M = ρ v
V = πr2 h
A = 2πrh + 2πr2
The surface area of the pneumatic cylinder is not what's important.
It's the area of the piston in the cylinder on which the pressure acts to move it up and down that's important.

## The Attempt at a Solution

V = π (0.076)2 x 0.202
V = 0.0036m3

A = 2 π 0.076 x 0.202 + 2π(0.076)2
A = 0.13m2
Wrong area calculation. See comment above.
P = 70 x 9.81 / 0.13 + Patm
Pcyl = 106607.61 Pa

mair = 1.225 x 0.0036
mair = 0.0049 kg

Finding mass of air in bottle

PV = mRT
2000x103 x 0.05 = m x 0.287x103 x 296
m bot= 1.18kg

mbot / mair = 240 lifts.
I am unsure if that is the correct step to take, I feel as though i am missing something obvious.
Yes, you don't know how a pneumatic cylinder works.

I see thank you for your help.
I calculated the area of the piston to be 0.018m2 which gave me a new value of Absolute pressure as 139.48kPa.
Would I be correct in using the density of the air in the cylinder as 1.225kg/m3 as looking back I do not think this is correct.
I think I have to calculate the temperature inside of the cylinder to be able to us PV=mRT to calculate the mass of air but I am unsure as to how I would go about this, any input would be extremely helpful.

Last edited:
SteamKing
Staff Emeritus
Homework Helper
I see thank you for your help.
I calculated the area of the piston to be 0.018m2 which gave me a new value of Absolute pressure as 139.48kPa.
Would I be correct in using the density of the air in the cylinder as 1.225kg/m3 as looking back I do not think this is correct.
I think I have to calculate the temperature inside of the cylinder to be able to us PV=mRT to calculate the mass of air but I am unsure as to how I would go about this, any input would be extremely helpful.
It takes a certain amount of pressure to support and lift 70 kg with the piston in the cylinder.

The density of air within the cylinder is going to depend on this pressure.

Thank you again for your help, although I cannot think of a way to calculate the density of the air without knowing the mass of it inside of the cylinder.
Again sorry if I am missing something very obvious here.

SteamKing
Staff Emeritus
Homework Helper
Thank you again for your help, although I cannot think of a way to calculate the density of the air without knowing the mass of it inside of the cylinder.
Again sorry if I am missing something very obvious here.
You don't have to necessarily use the properties of the cylinder in order to find the density of the air at a given pressure and temperature.

You have the ideal gas law, PV = nRT (I'm using n for the number of moles here as opposed to m for mass), which can be re-arranged into a relation to give density.

https://en.wikipedia.org/wiki/Ideal_gas_law

I rearranged to ρ = P / RT but would the density not differ when the air is in the cylinder?
Or is the temperature of the air inside of the cylinder the same as the temperature in the tank?

SteamKing
Staff Emeritus