# Tank depressurization through a valve

1. Jun 14, 2014

### hybro

Hello guys,

I want to study the temperature evolution of a gas in a vessel from 200 bar to 20 bar. I made the assumption the expansion is isentropic. However I have some trouble to represent the heat exchange between the wall and the gas. I used the energy conservation (transfer from a fluid to another through a wall) but I don't know if it's clearly what I need to use. Can someone help me to define the heat transfer? I found a topic talking about This but nothing was explain.

Another question, when the gas from the vessel at 200 bar flows through a valve and the circuit is at the atmospheric pressure, is there a temperature evolution? I don't think so but if We use the relation PV = nRT the circuit pressure rise so the temperature rise too.

Hybro

2. Jun 17, 2014

### Andrew Mason

If it is isentropic (ΔS = 0) it will have to be adiabatic. In that case there is no heat transfer between the gas and the surroundings.

If the gas expands adiabatically in a cylinder against a piston, for example, the gas does work on the surroundings. So apply the first law: Q = ΔU + W where W is the work done BY the gas, since Q=0 (adiabatic) you have ΔU = - W

You will have to explain this a little better. What is the "circuit" and what do you mean by temperature evolution - just a temperature change? Are you talking about an ideal gas? Are you talking about the gas in the vessel adding to the gas in the circuit? Is the volume of the circuit fixed? If so, does n (number of moles of gas in the circuit) not increase?

AM

3. Jun 19, 2014

### hybro

Yes exactly I am talking about a real gas. I have a valve between a pipe and the vessel. The gas flows from the vessel to the pipe from an initial pressure of 200 bar. The pipe is at atmospheric pressure and ambiant temperature with air composition. I want to know if there is an increase of the gas temperature at the outlet of the valve since the gas is flowing in a pipe of a fixed volume. The pressure drop due to the valve is really low, less than 1 bar.

Hybro

4. Jun 19, 2014

### SteamKing

Staff Emeritus
It's not clear what you mean by 'a pipe of fixed volume'. The pipe will be of fixed volume only if one end is blocked by a closed valve or some other obstruction. If there is no obstruction, the gas will flow through the pipe on its way out of the other end.

5. Jun 20, 2014

### Andrew Mason

Welcome to PF, by the way!

Applying the first law: Q = ΔU + W, since there is no work done on the surroundings and no heat flow into the gas, (Q = W = 0) there is no change in internal energy. Since it is a real gas (which I take to mean that there are material attractive intermolecular forces) the increase in internal intermolecular potential energy comes at the expense of the internal kinetic energy of the gas molecules. So temperature would decrease.

AM