Compressed Air Engine Calculations

AI Thread Summary
The discussion revolves around designing a compressed air engine and the calculations needed to determine its operational duration using a 500L tank at 8 bar pressure. The user seeks clarification on whether this pressure is sufficient to push the piston down and how to calculate the engine's runtime. Key points include the need to establish the supply pressure and the realization that torque decreases as the engine operates. The calculations presented involve determining force, acceleration, and torque, but there is uncertainty about their accuracy and relevance to the engine's performance. Understanding the relationship between acceleration and pressure requirements is emphasized as crucial for the project's success.
karan_s
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Hi,I am designing a compressed air engine for a project. A little stuck with calculations.

I need to work out how long the engine will run at its top speed using a air pressure which is supplied from a compressed air tank.Compressed air pushed down a piston which is connected to a camshaft. As the piston reaches bottom, a slot opens allowing the air in the chamber to escape and for the piston to come back up again.A second piston will eventually be added which helps the first piston come back up using the air that is released.

For now I am just doing calculations for the first piston.Weight of go kart including driver and tank: 200kg

Compressed air pressure pushing piston down: 8 bar

Max engine speed: 1500 RPM - 17 MPH max driving speed

Piston stroke: 60mm

Piston bore: 62mm

Crankshaft radius: 19mm

Weight of piston: 100g

Min acceleration to top speed: 6 secondsCould someone please explain how long this engine would run on a 500L tank of compressed air? This is assuming that the air is supplied at a constant pressure.

I have attempted it below but do not think its correct and am not sure what to to next.

1500 rpm = ω = 157 rad/s

α = ω/t , t being the time needed to reach the max rpm.

α = 157/6 = 26.16 rad/ sec^2

I = 1/2(200)(0.019)^2 = 0.361 Kgm^2

τ = I x α = 0.361*26.16 = 0.94437 Nm

τ = F x r
0.94437 = F x 0.019 So, F = 49.703NThanks in advance
 
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Welcome to PF.

We'd need to know the pressure of the tank.
 
russ_watters said:
Welcome to PF.

We'd need to know the pressure of the tank.

I was going to have a tank with pressure 8 bar. But again am not sure if this is enough push the piston down. The tank pressure can be changed at this stage.
 
karan_s said:
I was going to have a tank with pressure 8 bar. But again am not sure if this is enough push the piston down. The tank pressure can be changed at this stage.
Well, now you have two problems:
1. You haven't established the supply pressure you need.
2. If the supply pressure is equal to the tank pressure, the instant you turn the machine on the torque starts to drop.

...Though I see your question doesn't match the math you did. To find the power required to maintain the top speed, I'd research real gokarts. It is a very difficult thing to calculate.
 
russ_watters said:
Well, now you have two problems:
1. You haven't established the supply pressure you need.
2. If the supply pressure is equal to the tank pressure, the instant you turn the machine on the torque starts to drop.

...Though I see your question doesn't match the math you did. To find the power required to maintain the top speed, I'd research real gokarts. It is a very difficult thing to calculate.

That makes sense.

I also did the calculation shown below to find the force. Was thinking I could get the required supply pressure from this and then choose a tank with appropriate pressure. Is this also incorrect?

f = 1500/60 = 25 Frequency in revolutions per second
w = f*2*pi = 157 Angular frequency in radians per second
x = 0.06 Stroke in meters
a = w^2*(x/2) = 739.47 Acceleration in meters per sec^2
M = 0.1 Mass of piston in kilograms
F = Ma = 73.947 Force in Newtons
g = 9.8 Acceleration due to gravity
P = F*g = 725.420 Equivalent Force
 
Well, you're probably right that the acceleration is more important to the pressure requirement than the top speed, so for that purpose that may be valid.
 
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