# Air Pressure and flow to power calculations?

is there any easy way, or relatively easy way, to calculate how much power is stored in a certain amount of air with a certain amount of pressure and flow? i know thats VERY vague, so lets say (random number's being chosen...) 30 psi at 100 cfm.

use Bernoulis Equation.
$$PE = P_{atm} + \rho gh + \frac{1}{2} \rho v^2$$

Regards,

hmm...call me an idiot if you wish. but i guess im gonna have to guess at these variables:

Patm...no idea

p = density
g = no idea
h = no idea
v = velocity

thats bad i know..but could you maybe help me fill in the blanks?

Patm is Atomspheric Pressure
g is gravity of course
h is height

FredGarvin
What exactly are you trying to get to using this? I have a sneaky suspicion I know, but I'd rather know for sure.

In hydraulic systems, one simply uses

Power = p * Q
Where:

p = pressure
Q = Volumetric flow rate

You can do that here, but you'll have some pretty decent errors due to the high compressibility of air vs. hydraulic fluid and availability to do work.

Clausius2
Gold Member
infamous_Q said:
is there any easy way, or relatively easy way, to calculate how much power is stored in a certain amount of air with a certain amount of pressure and flow? i know thats VERY vague, so lets say (random number's being chosen...) 30 psi at 100 cfm.

The amount of energy stored by a fluid is its total entalphy:

$$h_t=e+v^2/2+P/\rho=c_pT+v^2/2$$(J/Kg) in the case of an ideal gas.

In order to determine the total content of energy of a gas you need a mechanic variable such us velocity and two thermodynamic variables (P,T). If the flow is at low Mach numbers, it is only needed one thermodynamic variable and one mechanic variable because thermal and mechanical states become discoupled.

thanks guys. hey Fred...what's this sneaky suspicion you have? lol. also..how big would that margin of error be? and i'm assuming pressure is kpa and flow rate is m^3/s....(although i really think i'm wrong with the Q unit)