Unknown2x said:
I'm building a compressed air car that will move a 70 kg person and I'm wondering how far it can go off a 135 PSI tank with 10 gallons. The vehicle should have a mass of at most 80 kg and I'm trying to figure out how to do calculations with a coefficent of friction of 0.5. Can anyone help me figure out how far I can propel this vehicle?
Another way of asking this question is: how much work can a 135 psi, 10 gallon volume of air do at an ambient 1 atmosphere pressure?
The first law of thermodynamics (conservation of energy) says that the amount of work you done by a gas:
Work = \int Fdl = \int (F/A)(A dl) = \int Pdv
is equal to the heat energy added (\Delta Q) - change in internal energy of the gas (PV).
As the gas expands, it will pick up heat from the environment and remain at the ambient temperature. So the change in internal energy of the gas will be zero.
The relationship between PV and temperature is PV = nRT or P=nrT/V, where n is number of moles of gas, R is the gas constant and T is the temperature (in Kelvin).
Work = \int_i^f Pdv = nR\int_i^f \frac{T}{V}dv
At constant temperature, nRT is constant so nRT = P_iV_i. So:
Work = P_iV_i\int_i^f \frac{1}{V}dv = P_iV_i ln\frac{V_f}{V_i}
Now let's do your problem. First, let's convert to MKS. Life is to short to work in psi and gallons. 10 gallon = 37.84 l = .03784 m^3; 135 psi = 9.5 kg/cm^2 = 9.3E5 Pa(N/m^2)
P_i = 9.3E5 Pa
P_f = 1E5Pa
V_i = .03784 m^3
v_f/v_i = P_i/P_f = 9.3
So:
Work = 9.3E5 x .03784 ln(9.3) = 3.52E4 x 2.23 = 7.85E4 J.
That is the amount of work that the compressed air is capable of doing. To work out how far that will move you, with a normal force of 80 kg (800 N) and a coefficient of friction of .5 (why is it so high - don't you have wheels?) you will have about 400 N force of friction, so you will consume 400 J/m. So you could go about 200 m., assuming your car was 100 percent efficient in converting compressed air energy into useable work. But I would first check your friction figure.
AM
