Energy to transfer last half of compressed air between equal tanks

[genergy]

When I have a full tank of compressed air that I need to move to another tank will it require the same amount of energy that it took to compress it to move it?
2 -- 318 cubic feet (equal volume) compressed air tanks are side by side.
One is full at 312 psi and the other is at atmospheric pressure.
I know that if I open up a valve between the 2 that I am able to transfer half of the compressed air and lose some energy to friction/heat.
How much energy will it take me to move the last half of compressed air over?

312 psi / 2 = 156 psi x .9 = 140 psi (assuming 10% lost in friction/heat)

How much energy to compress/move the 318 cubic feet of air at 140 psi over to the other tank?

Will cooling help to reduce the energy loss?

http://gravitybuoyancy.com/g_energy/Conserving_Air_Compressor_Energy.html

 

[haruspex]

When I have a full tank of compressed air that I need to move to another tank will it require the same amount of energy that it took to compress it to move it?
2 -- 318 cubic feet (equal volume) compressed air tanks are side by side.
One is full at 312 psi and the other is at atmospheric pressure.
I know that if I open up a valve between the 2 that I am able to transfer half of the compressed air and lose some energy to friction/heat.

Initially it will transfer rather less than half. This is because the decompressing tank will cool and the other will warm, so equilibrium pressure will be reached before half of the gas has been transferred. You'll get 50-50 when the tanks return to ambient temperature.

How much energy will it take me to move the last half of compressed air over?

Depends how much of a hurry you're in. As above, the receiving tank will warm and increase resistance to further transfer.

312 psi / 2 = 156 psi x .9 = 140 psi (assuming 10% lost in friction/heat)

Judging from that calculation, the 312psi is additional to atmospheric. But I don't understand the 0.9. There's no loss of air, so if the initial and final temperatures are the same it should be 156.

How much energy to compress/move the 318 cubic feet of air at 140 psi over to the other tank?

To compress a volume V0 at pressure P0 to vol V1 at P1, constant temperature (i.e. slowly) would take energy P0*V0*ln(V0/V1) = P0*V0*ln(P1/P0)
So if you have two tanks vol V, pressure Pm = (Ph+Pa)/2, and you want to end up with one at Ph and the other at Pa (atmospheric), the energy required is V(Ph ln(Ph/Pm) + Pa ln(Pa/Pm)).
All pressures are absolute here.
With your numbers, that's 318*(326 ln (326/170) + 14 ln(14/170))*144 = 8*10⁶ foot-pound-weight. That seems rather a lot, so maybe I have something wrong.

Will cooling help to reduce the energy loss?

Allowing to cool to ambient, yes. Any additional cooling will come at an energy cost, almost surely exceeding what you save.