7 replies to this topic

[Pauwl]

Hello All,

I am a junior engineer and I am having trouble with calculating the relieving temperature for a blocked flow psv scenario.

If the normal operating temperature is T1 at Pressure P1.

Then how do I calculate the relieving temperature T2 ant P2, where P2 is the Set point + overpressure.

We had previously used the following equation

T2 = T1 * (P2/P1)

Which is Boyle's/Charles law if I remember correctly, but this is assuming an ideal gas with a constant volume and number of moles which isn't true. I believe the equation was taken out of an older version of GPSA.

If your operating temperature P1 is far smaller than your Set Pressure, then you end up with a huge temperature that is unrealistic.

I believe the we should be using isentropic compression with the following equation

T2 = T1 * (P2/P1)^(k-1/k) Where k = Cp/Cv

And this would determine our relieving temperature T2 at our relieving pressure P2.

Is this correct?

[JoeWong]

First of all, welcome...

Both equations are applicable when they are in gas/vapor at both stages (Normal & relieving condition). If either condition contains condensate, then it will starts to deviate. Both equations may not be suitable.

T2 = T1 * (P2/P1) assuming constant volume and mass, i think for a gas/vapor (compression) involve volume change as pressure increase. Those my opinion is this approximation may not be suitable.

T2 = T1 * (P2/P1)^(k-1/k) Where k = Cp/Cv is an isentropic compression which assumed adiabatic and reversible. In my opinion, bring pressure from normal to relieving condition would be adiabatic and irreversible (real world). Polytropic exponent (n) may be used instead of isentropic exponent. However, it is hard to get this polytropic exponent (n). Those in most cases, we consider n=k and isentropic compression stands.

JoeWong

[eastorca]

WOw, thank you so much. I got the same troblem like Pauwl did. Now it's clear.

[bernath]

Hi Joe, if the fluid contains condensate, how to calculate the relieving temperature? what equation should be employed? Can we use hysys to solve this problem? Please kindly advise.

thanks
regards,
bernath

[paulhorth]

Pauwl,

You have to consider what is causing the rise in pressure in your blocked-in vessel. The equations given above are for the case of a piston compressing a constant mass of gas in a cylinder. However this will probably not be realistic for all cases. Consider these two cases:
(1) Compressor discharge KO drum, blocked outlet, compressor running
In this case, the gas from the compressor will go through a cooler, so while its pressure will rise, its temperature will stay constant. So the gas entering the KO drum willl not rise in temperature. The mass of gas in the drum will not be constant but is higher at the relief pressure than at the normal pressure - it is the volume which is constant. There will be a slight rise as the initial gas trapped in the drum is compressed but I do not think this is significant. So for this case the relieving temperature can be taken as the same as the normal temperature.

If this gas is partly condensed in the cooler, then raising the pressure will cause greater condensation, but the temperature will still stay fixed.

(2) Gas-filled vessel blocked in in a fire
In this case, the mass is constant, the volume is constant, and heat is added, so obviously this is not an adiabatic process. The temperature will rise so that ( z2.T2)/(z1.T1) = P2/P1 where ABSOLUTE temperatures are used. Do not omit the compressibility factor z.
If there is liquid in thie vessel, this liquid will tend to vaporise when heat is added, and the temperature at relieving pressure should be determined using a simulation package.

Paul

[bernath]

Hi Paul,

refer to equation "( z2.T2)/(z1.T1) = P2/P1"

What we know is P1, T1, z1 and P2 (set pressure). There's only one equation with 2 degree of freedom. Thus the equation still can't be solved. We need to find another equation or assumption.

When using hysys, if the vapor and liquid are both present, then I'd say that the bubble point temperature is our relieving temperature. Therefore we just have to set the pressure to be at relieving condition and also set the vapor fraction to zero. The resulting temperature - the bubble point temperature, will be the relieving temperature.

As you have mentioned before, the heat is added, obviously the process is not adiabatic. Deviation will occur if we insist using isentropic exponent instead of polytropic one. However, it is quite hard to calculate polytropic exponent (n). In most cases, we can only assume n=k and isentropic compression seems like a good choice for me.

Please correct me if I'm wrong.

thank you
regards,
bernath

Edited by Art Montemayor, 16 March 2011 - 01:22 PM.

[paulhorth]

Bernath,

You are wrong on all three points.

To solve the simple equation that I provided, just requires a few trials. Put Z2 = 1, solve for T2, then look up the true Z2, then re-solve, until T2 is determined. Two or three trials should be sufficient.
This is how most engineering was done before computers were invented.

In the case of a vessel containing vapour and liquid, the liquid is already at its bubble point, that is, liquid and vapour are in equilibrium. When you add heat, the vapour and liquid get hotter, so the composition of both phases will change and the bubble point rises. It is not correct to take the liquid composition at normal pressure, then find its bubble point temperature at zero vapour fraction at the relief pressure, because the liquid composition will not be the same at this condition. Some liquid will have vaporised and contributed to the rise in pressure.

Isentropic and polytropic compression are irrelevant for the case of block-in in a fire. This is a constant-volume pressure rise with heat addition. These processes are also irrelevant for the other case that I described, block-in with feed from a compressor. This is a pressure rise due to increasing mass at constant temperature.

Paul

[bernath]

thanks Paul for pointing me again into the right direction.
regards,
bernath