Guys,
I'm struggling with a realistic approach to calculate die flowrate for a vacuum relief valve of a stripping column (sour water).
#1
First approach was to consider the amount of steam inside the column to be equal to the volume of the column. It is assumed that steam supply stops but water is still fed to the column. The (cold) water will immediately (within seconds) condense the steam. The vacuum breaker would need to have an extremely high (not practical) capacity.
#2
Second approach was to take the interlocking into account that stops the water supply if the pressure drops below a certain value. The vacuum breaker would then only be calculated for a further cooling (and condensing) of the steam due to heat loss (through the insulation) to the atmosphere. Assumed conservative heat transfer coefficient is 12 W/m²K. By doing this the capacity of the vacuum breaker would around 40 times lower than for apporach #1.
But....is it ok to do it like this? Interlocking might fail. And furthermore there is still liquid holdup inside the column. How to deal with this?
#3
Third approach. Just an idea so far: Similar than approach #1. The cold water will condense the steam inside the column. BUT it is assumed that the steel (some tons) of the column is hot. So the assumed dT/dt is heat_cap, water * delta T * flowrate, water / (heat_cap, steel * mass, steel)
Is this a realistic approach? Or wouldn't the heat inside the steel help at all because the heat transfer (and condensing) of the steam would be much much faster?
#4
Any other realistic approaches you would recommend?
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Please have a look the sketch. If seen in the past that the PRV and VRV are both connected to the low pressure steam line which is connected to the column (no valves in between).
What do you think? Is it ok to place PRV and VRV there?
For PRV I would say yes because high pressure inside the column can only result from high pressure in the steam line.
For VRV I am not so sure. But to be honest I have problems of imagination here. For low pressure (p2 < 0 barg) inside the column is the pressure p1 the same as p2? Or is it possible to reach the design pressure of -0.5 barg inside the column while the pressure p1 is higher? My best guess here is that p1 is p2 plus liquid level.
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Thanks for your help!
Jon