Theoretical Temperature Rise of Highly Compressed Air

Sounds like Old Faceful to me....

Spoken like an old geyser geezer.

1.4 or 0.4 ? could you make a choice or a correction ?

Temperature will accord to environment by heat transfer from hot air to walls or water this will take time, this process is not to be considered if the pressure change is rapid since heat transfer is a low speed process due to on one side the limited convection coefficients, the limited radiation (if the temperatures are high radiation starts to play a role) and on the other side the thermal difusibility which depends on the wall material and is also limited.

The constant for an diatomic gas(air) is 1.4....it's reduced to .4 in solving the equation...In thermodynamics there are different fundamental processes, namely isobaric, isothermal, isentropic, and polytropic and isochoric .

Here's another example....

QUESTION
What compression ratio Vmax / Vmin will raise the air temperature from 20C to 1000C in an adiabatic process?
the gas is diatomic
ANSWER
For an adiabatic process, with Q = 0, the first law of thermodynamics is:
Change in thermal energy = Work. Compressing a gas adiabatically (W > 0) increases the thermal energy. So an adiabatic compression raises the temperature of a gas.
Pf = Pi(Vi/Vf)^gamma
gamma = 1.40 for a diatomic gas (like air)
Pf = Pi(Vmax/Vmin)^1.40
so to solve..:
(Tf) Vf^(gamma - 1) = (Ti) Vi^(gamma - 1)
(1,273K)Vf^(gamma-1) = (293K)Vi^(gamma -1)
(1,273K) Vf^0.4 = (293K) Vi^0.4
1,273K / 293K = Vmax^0.4/Vmin^0.4
4.3447 = (Vmax^0.4 / Vmin^0.4)
(4.3447)^2.5 = (Vmin^0.4 / Vmax^0.4)^2.5
Vmax/Vmin = 39.346

http://www.physicsforums.com

basic thermodynamics of course, temp equalizes with its surroundings

Halliburton fracturing operation in the Bakken Formation, North Dakota,United States....

http://en.wikipedia.org/wiki/Hydraulic_fracturing

https://www.youtube.com/watch?v=tJ-zfUUjBiQ

https://www.youtube.com/watch?v=8MukcZYpD3M&list=PLPTlepYtxLVeweEaCeMTs6CZ76VqVCBMP

Our stuff is way better! Some of our crew came from Halliburton and will never go back for good reason.

Heres where I am at. Trican Well Service.

So far it's been an above average company for me to work at and that says a lot.

I like the upper picture of the burrito transfer station.

You could also use, entropy production of air at those states to know the Temperatures.

I use a 'rule-of-thumb' formula to get a close approximation of final temperature due to adiabatic compression of air (using absolute values)

T2=r^0.286 x T1

T2 = final temp. T1= start temp (assume 20C) = 293K

r= pressure ration (which is final pressure divided by start pressure)

0.286 comes from (k-1)/k, where k is 1.4.

Thus r = (9500 + 14.7)/14.7 = 647

Hence T2 = 647^0.286 x 293 = 6.37 x 293 = 1,865K or 1,592C …. Which is pretty hot !

In practice this is not likely to happen. I don't know anything about fracking. If using compressed air to pressurize the pipe, I guess the air will come from a tank already at 10,000 psig. On entering the pipe it will cool, and as pressure builds up it will heat up - back to something like the original temperature. It will not heat up as such.

What does heat up, as you know, is the air already in the pipe at ambient temperature when it becomes displaced and compressed. The heat quickly dissipating in the walls of the pipe.

(If it is critical to you, you need to look at post 2 for a different answer, and an explanation of where '17' comes from and why '0.4' is used).

We don't use compress air to test the system. Preferably we don't want any air system but give the volumes of line we have in place its unavoidable.

Mostly I was just curious as to theoretically how hot the air that is trapped in the lines gets when we do a pressure test and spike the line pressure from 0 to 9500+ PSI in less than a second.

What I d know is that when we have a good deal of trapped air in the system once it hit the peak testing pressure we can see several minutes of steadily dropping pressure simply due to the highly heated air cooling down and contracting.

By the slowly dropping pressure numbers during the test it looks like a slow leak but in reality it's not.

Not only the higher the pressure the more gas can water absorb so that you have a combined effect: gas in water solution plus contraction. I would bet that the temperature effect is less than the loss of volume due to solubility.

Solar Eagle and Horace 40 have good answers, but I suspect even if the compression is this rapid, the process will not be entirely adiabatic. Those temperatures will be the upper limit, where in reality, any water mist produced suspended in the air will result in a large error due to formation of steam. Turbulent motion will result in some heat transfer also. The real answer probably lies between the adiabatic result and 80% of that value, just a wild guess.

From the <1s and the rate of fluid injection, it should be possible to calculate the volume of trapped air, and if enough heat would result to weaken the pipe at some location. If a blowout actually results, it should be detectable.