Open Systems

[MarkE]

So how much energy in joules did we loose from the 20 ltr supply vessel,and what do we now have in the 750ml cylider in joules of energy?-so as we can calculate the loss in total.
Cheers

The equation predicts the maximum pressure that you should read based on conservation of energy.  If you want to know Joules transferred we just need to identify the gas and then plug the constants into the equation.

[tinman]

Sorry, I now see that this was in August of 2013.  Well, congratulations anyway...better late then never.

I stumbled across that story and assumed it was a current one.

Sorry,

Bill

Lol-all good Bill,as i missed it to lol.

[tinman]

The equation predicts the maximum pressure that you should read based on conservation of energy.  If you want to know Joules transferred we just need to identify the gas and then plug the constants into the equation.

The gas is just ambiant air. I was just wondering how much energy was lost during the transfer from tank to ram. So we start with X amount of energy in the tank,and 0 in the ram. We take some from the tank,and put it in the ram. So lets say the temperatures remain the same,what energy in joules was removed from the tank,and what energy in joules was gained in the ram?

[MarkE]

The gas is just ambiant air. I was just wondering how much energy was lost during the transfer from tank to ram. So we start with X amount of energy in the tank,and 0 in the ram. We take some from the tank,and put it in the ram. So lets say the temperatures remain the same,what energy in joules was removed from the tank,and what energy in joules was gained in the ram?

Gamma for the nitrogen and oxygen which makes up most of the air is 1.4.  So:

P*V^1.4 = K.

Whereas the energy is P*V.

Consequently, for a change in volume from V_START to V_END the remaining energy as a ratio to the starting energy is:

E_END/E_START = P_START*(V_START^1.4/V_END^0.4)/(P_START*V_START)

P_START appears in the numerator and the denominator and cancels, and V_START is raised to the 1.4th power in the numerator and the 1.0th power in the denominator, so simplifying we get the remaining energy percentage as:

E_END/E_START = (V_START/V_END)^0.4 and the percentage energy lost is:

E_{INTERNAL_LOSS}/E_START = (V_END^0.4 - V_START^0.4)/V_END^0.4

I'll have to go back and look for the dimensions to plug in values for the starting and ending volumes.  For the first approximation we would just use the pressure vessel volume for V_START and the sum of the cylinder extension volume plus the pressure vessel volume for V_END.

For 20 liter pressure vessel, and a 750ml ram, the approximate percentage loss to heat on each stroke is:

E_LOSS% = (20.75^0.4 - 20^0.4 )/20.75^0.4 = 1.46%

The initial energy is 120psi * 20 liters =  6895 P/psi * 120 psi * 0.02m³ ~= 16.5kJ.  1.46% * 1.65kJ ~= 242J internal energy loss per stroke.  This is approximate because we are treating the ram cylinder volume as completely void, when it at 1 ATM.  This rough approximation s/b within 5-10% of the actual loss.  If you vent the ram to reset it, then after about 50 operations the energy in the pressure vessel and the pressure should both be about 50% of the starting value.

[tinman]

I'll have to go back and look for the dimensions to plug in values for the starting and ending volumes.  For the first approximation we would just use the pressure vessel volume for V_START and the sum of the cylinder extension volume plus the pressure vessel volume for V_END.

Thanks Mark