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2-way control valve pressure loss

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mfqd

Mechanical
Jul 3, 2009
60
Hi,

In previous posts in the "Valves" section, i discussed the use of 2-way and 3-way control valves, for plate heat exchanger arregements.

One big question that occurred me was: when we use 2-way control valves, that are constantly modulating, how can we calculate the pressure loss to help in the dimensioning of the pump?

Thanks!
 
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Chicken or egg?

Use the pump curve data to determine the range of capacity coefficients required for the control valve. Typically process provides equipment data to mechanical then provides control data to CSE. Mechanical selects pumps, CSE selects control valves. It is good if someone in process looks at the system before things get bought.
 
Thanks for the reply.
I understood your point, but it didn't help me in nothing...
 
Let's assume that you are using the control valve on temperature control to bypass flow around the heat exchanger; and you selected a three-way diverting valve. In this case the control valve can take about the same pressure drop as the exchanger in the bypass mode. For a diverting valve the Cv is somewhat difference depending for each outlet port but not by a significant number. However, you likely want minimum drop when going through the exchanger. Thus you likely select a line sized three way valve. The throttling action will adjust the flow around or through the exhanger based upon the temperature in this example. You use the available Cv data for the selected control valve in the through exchanger route for the pressure drop calculation.
 
I think that we would need more information on the two-way control valve arrangement. Is the control valve only trimming the temperature as a bypass? What percentage flow would be bypassed? What is the ratio of pressure drop taken by the control valve and the exhanger. If the control valve is a very large butterfly or ball you can bypass most flow. If the valve takes a high percentage of the parallel drop then you bypass a smaller percentage of the flow. You have to evaluate the parallel exchanger and control valve together. This sort of piping system design is often done by the process engineering group.
 
I find a sketch really helps. Essentially, you have to determine the pressure drops in the system.

Let's say you have a pump, a control valve, a heat exchanger and a long run of pipe going to atmospheric pressure (say a pond so it's at atmospheric pressure). If it's an existing system then you have a pump curve and know the discharge pressure. The inlet pressure to the control valve is the pump discharge pressure (if needed, you can correct for piping and/or instruments like an orifice plate). Working back from the pond you start with 0 pressure. You figure out the line losses through your piping for length and/or elevation, that gives you the outlet pressure on the heat exchanger. The data sheet for the heat exchanger gives you the dP across the exchanger (corrected for flow as needed) which is the outlet pressure on the control valve. Call that the normal case.

You then repeat it for maximum flow and minimum flow conditions.

A control valve does not take a certain pressure drop (unless it's fully open), the system imposes the pressure drop on the valve and the valve opens/closes as needed. The dP across a valve, if it's not fully open, is the same regardless of the control valve size for the same process conditions. This may seem minor but it's a critical difference.

If this is a new system, you typically allow a certain dP for the control valve in your sizing calculations for the pump, 33% of the dynamic losses is a typical value. You go through the same exercise as above to come up with a required pressure out of the pump for the pump data sheet.

Why 33%? It's to allow you to handle changes in flow. Let's say you had a piping system taking 100 psi in dynamic losses and you allow your control valve to take 15 psi when you sized it. Let's say Operations needs to increase the flow by 10%. The line losses increase by about 21% (the square of 110%) to get that additional flow or 21 psi. That extra pressure drop has to come from the control valve and it won't have it because you only allowed 15 psi when you sized it. In effect, Operations will be able to go somewhat above 105% before your valve is wide open and the system is at maximum capacity.
 
Thank you all!

I think i understood your explanations correctly!
Also, i've searched more about the subject and found some texts about "Valves authority". Here, i found what i was searching for and after i red all your replies, i confirmed what was mentioned by you: TF2K and JLSeagull.

In fact, for calculations in new systems, it's advised to use a valve ratio between 20% and 50%. The reason why is this, you already gave it.

To conclude this, please confirm me just one more thing. From what you've said, the pressure drop is imposed by the control system. This means that the pressure drop that the valve can create, is the imposed value or can we fall in a situation that it produces a bigger pressure drop and the pump may not have sufficient head?

Thanks
 
The pump would only get in the situation if the control valve was wide open. Basically, you've reached the limit of your system at that point.
 
TD2K,

I think i understand you because is logical.

I explain better my doubt:

1) is obvious that the bigger the blowrate through a 2-way valve, bigger it will be the pressure drop (in a quadratic proportion)

2) but in a situation that de consumer is almost satisfied, the valve will be almost closed. This will increase de velocity of the water through the valve

Doesn't this creates also a big pressure drop?
 
Not for the two examples you gave.

Take the example of water coming from a constant pressure system into a open pond. The pressure drop across the valve is going to be constant. You can open the valve (increasing the flow rate) and the pressure drop isn't going to change. If you close the valve, the same is true. That's a simplistic case but it's the system that sets the pressure drop for the valve, not its opening.

If I want more flow to a valve, I likely will have more pressure drop through the upstream piping leading to a lower inlet pressure to the valve. Similarly for the outlet, I need more pressure to get the flow through the downstream piping to its destination. The combination of a reduced dP for the valve and higher flow rate means the valve has to open more to pass the required flow (a valve is essentially a variable orifice as indicated by its Cv versus % opening curve).
 
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