Compression formula

1. Nov 23, 2004

swabbie58

Is there a formula to calculate the amount of force needed to compress air from a low atmosphere to a higher one? In my situation, I have a chamber wherein the pressure is one (1) atm and I ned to push the air from that area into a reservoir where the pressure will increase with each cycle. Am not using a fan, but rather a sort of bellows (one side of the rectangular chamber moves toward an opposite side with hoses to the reservoir). As the reservoir pressure mounts, how can I calculate for the growing amount of energy needed to push more air in and raise the atm pressure?

2. Nov 23, 2004

Clausius2

It is necessary to have more information about the process on your chamber. Is is adiabatic (i.e fast) or it has times of heat relaxing?. If it is adiabatic, you can obtain the energy needed (work):

$$W=\Delta U=m c_v \Delta T$$ where m is the mass contained into the chamber in one cycle.

Adiabatic process: $$W= m c_v T_o \Big(\frac{T_f}{T_o}-1\Big)= m c_v T_o \Big(\Big(\frac{P_f}{P_o}\Big)^\frac{\gamma-1}{\gamma}-1\Big)$$

where T_o is the temperature at which you start the process of compression,
c_v=714 J/kg
gamma=1.4
and if you define your compression ratio as: $$r=\frac{P_f}{P_o}$$ then:

$$W= m c_v T_o (r^{\frac{\gamma-1}{\gamma}}-1)$$ is the work per cycle (Joules). Surely you will have to multiply this by some kind mechanic efficiency of your device.

HINT: the less $$T_o$$ the less $$W$$. Do you understand the meaning of cooling before compressing air?

3. Nov 23, 2004

swabbie58

While it is adiabatic in that it uses a very fast cycle rate, the air being drawn into the chamber for the subsequent cycle comes from outside the system. So, while the temperature of the first-cycle air will rise, that heat will be carried with it down the line to the reservoir. There it will build up as each cycle of air is forced into the reservoir.

4. Nov 23, 2004

Clausius2

So you agree with the formulation posted or not?

5. Nov 23, 2004

swabbie58

Sorry--my computer had to be taken down to install a new graphics card. Frankly, I don't know enough to know whether I "agree" with it or not! That's why I am posting the question on the forum. I'll have to work through it and see what happens.