Pressure-Time relation for polytropic thermodynamic process

Relation between pressure and time in polytropic thermodynamic process

Poll closed Apr 29, 2014.

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3. Derivation of Milton-Beychok's model.

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1. Apr 14, 2014

vishwanaya

Dear Friends,

I carried out an experiment of sudden release of oxygen (open nozzle) from an oxygen cylinder used for medical college and hospitals. I found that pressure drops quite rapidly and cylinder surface cools from outside such that water droplets accumulate on its surface. This experiment is carried out in an oxygen factory in Goa India. Now I do want to know what is theoretical relation between pressure and time in such a case.

Can anyone show me the steps to arrive at such an equation?

Vishwajeet, Goa-India.

2. Apr 14, 2014

Tobychev

You can probably describe this with the ideal gas law: PV/nk_B = T, since the cylinder volume is fixed, the decreasing pressure P means the temperature T in the cylinder must decrease also.

3. Apr 14, 2014

vishwanaya

ideal gas equation pV = nRT = n^2 k_B T. Therefore, variation of pressure and square of mole quantity should decrease so that your suggestions stands true. Please go ahead to discuss any further clarification for my help.
Thanks for quick response Tobychev.

Last edited: Apr 14, 2014
4. Apr 14, 2014

Q_Goest

Hi vishwanaya. Welcome to the board. I went through this a few years back here. Basically, you need to apply the first law of thermodynamics to the vessel such that:
dU = Qin - Hout

Do this in small time steps to determine the change in internal energy of the oxygen cylinder over time. The internal energy and pressure equate to a temperature. Read through the other thread and see if you have any questions.

Best regards.

5. Apr 14, 2014

Staff: Mentor

Hi Guys,

In my judgement, to get the pressure as a function of time, you also need to characterize the pressure drop vs flow rate relationship for the nozzle, and combine this with Q_Goest's recommendation.

Chet

6. Apr 16, 2014

vishwanaya

Thanks Q_Goest.
Relationship between P and T is not explicitly given there.
Can you help me in deriving such an equation?
Vishwajeet

7. Apr 16, 2014

curioso77

In my opinion, you must take into account the fluid mechanics involved in the process. If you try to relate pressure with time, you need to solve the pressure drop as transitory effect of fluid flow. This is a complex problem that could have more dependencies that pressure ratio. With enough pressure ratio, you can reach supersonic flow, or at least a mach number that justifies include air compresibility in the equations. If you want to be more accurate, you need include heat transfer also. The probe that your system is not adiabatic is the behavior that you are reporting in the exterior. As a first approach, you can assume incompresibility and adiabatic system, and solve Bernoulli's equation over time in order to have mass as a function of time inside the cilinder, pressure vs time and temperature vs time applying the equation for ideal gas. You can use a numeric method like euler looking for a solution of the differential equation.
I hope that this hel you.

8. Apr 17, 2014

vishwanaya

Thanks Curioso77.
While using Bernoulli's equation for Gravitational head mgh_in will get cancelled with mgh_out.
Under kinetic head, 1/2mv^2, m will be equal in and out. v_out will be 1 mach as max speed that can be attained by a shock wave. speed v_in should we assume equal to what?
Under pressure head, P_out will be assumed 1 atm in SI units. P_in will drop with release of the gas.
To calculate P_o and P_in ratio, should I use adiabatic equation (P_in V_in)^1.4 in stead of isothermal equal?
Regards,
Vishwajeet

9. Apr 18, 2014

curioso77

vishwanaya:
v_in=0
I recommend you to read about isentropic flow in order to have a better understanding of the whole phenomenom. For example, you can read the chapter 4: Isentropic Flow, from "The dynamics and thermodynamics of compressible fluid flow", by Ascher Shapiro, a classic book. At least at a first approach, it should be work.
regards

10. Apr 24, 2014

vishwanaya

release of O2 gas

Last edited: Apr 24, 2014