- **FloEFD, FloWorks & FloTHERM**
(*https://www.cfd-online.com/Forums/floefd-floworks-flotherm/*)

- - **Gate valve flow simulations...**
(*https://www.cfd-online.com/Forums/floefd-floworks-flotherm/128442-gate-valve-flow-simulations.html*)

Gate valve flow simulations...2 Attachment(s)
Hi all,
I have been working on valve flow simulation project in my company recently and this is probably my third or fourth post in this forum here. "Boris", mostly has been very helpful in sorting my problems out from time to time. And I really thank you for sharing your precious time and knowledge with everyone in this forum. Anyways, let me describe briefly again what I am doing. I need to simulate for ball valve and gate valve flows while they are opening from a fully closed position (0%) to fully open position (100%). I am running steady state simulations for all the cases in a stepwise manner with an interval of 10%, which means I have 10 individual simulations to run for Ball valve flows and 10 more for gate valve flows. With the results, I need to compute and compare mass flow rate, i.e. the total amount of fluid (water, in kg) coming out through the valve in each of the two cases. So far I have pretty much managed to complete ball valve flows and now I have moved on to gate valve flows. While working on it, I have encountered few problems, few doubts I should say that has been confusing me. Fig 1 shows the 3D sketch of my gate valve and the pipe with the boundary conditions that I have imposed at the inlet and outlet lids. I started my simulations with 100% valve position. There seems to be no problem. I also compared the pressure drop with the "pipe only" flow, with the same BCs and the total length of pipe. The "delP" is again compared with the analytical result using "pressure drop online calculator" in the internet. Here are the results: delP-analytical = 0.25 bar delP-pipeOnly = 0.234 bar delP-100%Valve = 0.248 bar mass flow rate at inflow-analytical = 629.55 kg/s mass flow rate at inflow-pipeOnly = 627.755 kg/s mass flow rate at inflow-100%Valve = 626.599 kg/s In my understanding, these results made sense, so I then carried on with my further simulations, for valve position at 90% and so on. Pipe roughness and heat transfer effects are not considered. Here in Fig2, I would like to discuss some of the results and would like to hear from the experts here whether or not its looking alright 1. As the valve is closing, 100%-90%-80%-...and so on, delP is increasing-which is desirable BUT Why is the total pressure at inlet increasing despite the inlet velocity is fixed as a BC for all the cases? If my results make sense, I'm looking for an explanation for this. 2. Or should not the velocity at outlet change as well? Mass flow rate through the inlet and outlet are the same for each of the cases. I was expecting mass flow rates through the outlet to decrease from 100-90-80%....valve positions. Mass flow rate is the final result that I need. I need to plot "mass flow rate-vs-valve position", which should increase almost linearly with increase in valve opening area (valve position). With this observation I am most likely to obtain the same mass flow rate even at exactly the valve position, (rho*OpeningArea*avgVelocity), is it so? 3. To answer this question, I even tried obtaining the mass flow rate at the valve position but could not succeed to do so. Could anyone as well instruct me if there's a way to effectively extract those results in SW Flow simulation? Attachment 27909 Attachment 27910 |

Hi Nikesh,
Your error is in the physics and your boundary condition definition. First let's consider the physics:There are some fundamental laws such as Mass and Energy balance. They are always valid and basically say what is put into a system will come out of the system again. m_in = m_out with m = roh * v * A and A beeing the aera and v the velocity Also if you look at the basic Bernulli you'll see that the total pressure is the sum of all kind of pressures: P_total = P_static + P_dynamic + P_potential dyamic pressure (1/2 * roh * v^2) is basically the pressure that can be built up on a flat plate in the find like the pressure you feel on your chest when driving a motor cylce. The potential pressure (rho * g * h) is due to hight difference. You can write this also as: P_total = P_static + 1/2 * roh * v^2 + rho * g * h Here you can also put the mass flow rate into the bernulli equation if you have an unknown parameter by for example using m = rho * v * A as v = m / (rho * A) and give you: P_total = P_static + 1/2 * m^2 / (rho * A^2) + rho * g * h Ok, now let's consider your case:In your case you can ignore the potential pressure. And since you are dictating the velocity, it will not change at the inlet and with the area of the inlet fix only the density could change but it won't for water in the pressure range you are condsidering (water can be considered as incompressible). That will automatically lead to a fixed mass flow rate and since the mass balance is law you will always get the same mass flow rate no matter what the valve does. For the pressure at the outlet you defined the static pressure and with that only the dynamic pressure could vary but since you fixed the inlet velocity the mass flow rate is fixed and with that at the outlet the velocity is fixed too. That leads to a fully fixed total pressure at the outlet since neither the static pressure, nor the density and velocity of the dynamic pressure changes. The changes you see at the outlet are probably in case you have used "average total pressure" as a goal instead "bulk average total pressure". The difference is av will use the surface averaged value and bulk av will use the mass flow averaged value and therefore also consider variations in the velocity distribution over the outlet surface. That leaves only the static pressure variation at the inlet as a free parameter that can be changed and since that is part of the total pressure you will see a total pressure change at the inlet. If you want to see a mass flow rate-vs-valve position diagram then you should not define the mass flow rate but the pressure. I think I suggested you to use rather total pressure at the inlet and static pressure at the outlet. So please try using total pressure at inlet and static at the outlet and you will get better results. If you want to have an initial mass flow rate at 100% valve opening which you might know from measurements then use it as an initial calculation with a fixed flow rate to find the total pressure to define at the inlet and use this for your next calculations. If you talk to your measurement team how they do such measurements, check with them if their pump regulates the flow rate or the pressure at the inlet. This should make it clear how it is done in reality. Regards, Boris |

Boris,
Thank you very much for your swift reply. This is exactly the kind of explanation that I was looking for and I've been trying to hack into these for quite some time now. Ok, let me now put it the way I've understood so far in general sense. We have mass conservation equation, m_in = m_outand Energy equation (Bernoulli's equation) p_total, in = p_total, outto be satisfied for any flow, where p_total = p_static + p_dynamic + p_potential; In this case p_potential can be neglected which leaves us with:p_total = p_static + p_dynamicAnd incompressible flow condition. Case1-BC 1: My previous case where V_in and p_static,out were fixed and imposed as BC.Firstly, Continuity equation will always be satisfied leading to V_out = V_in.Secondly, p_static,out is fixed and from continuity equation V_out is obtained and is fixed too so is p_dynamic,out, hence p_total,out = fixed valve.I see some p_total,out values changed in my result, this is because I've used "average total pressure" as goals instead of "bulk average total pressure". Ques 1: Is "Bulk average total pressure" what I should use for goals??? Now, in the inlet, p_dynamic,in has been fixed too, hence, leaving only p_static,in to be free which changes (increases) leading to increase in p_total,in too. This increase in static pressure is because the flow has to move from inlet towards downstream since velocity has been imparted at inlet. Then,p_total,in > p_total,outleading to NOT satisfying the energy (Bernoulli's) equation, exactly the problem with my setup in the first case. Ques 2: Did I get it right here? Ques 3: Could this also be the reason why I was getting negative pressure values near the wake of the valves at 60%, 50%, 40% and so on? Furthermore, in each of these cases, pressure seemed to be increasing further downstream until the outlet which was fixed at 1 atm. These did not seem very physical to me, yet I wasn't able to figure out the source of this problem. There was no sign of negative pressure for 100%, 90%, 80% and 70% valve position cases. I forgot to mention this behavior earlier before. Case2-BC 2: As you suggested, p_total,in and p_static,out are fixed and imposed as BC.V_in and V_out are not known, hence calculation moves to the energy equation. Since,p_total,in = (p_total,in + 1/2*rho*V_in^2) = (p_static,out + 1/2*rho*V_out^2) = fixed valve, V_out is obtained. The solver then iterates to obtain optimal V_in and p_static,in values based on above equations and values. BUT then my question again:Ques 4: Since continuity equation has to be satisfied, which, for incompressible flow leads to: V_in = V_out, as inlet and outlet areas are the same. From energy equation, this again leads to p_static,in to be the free variable that can change and increase perhaps. This goes back to the same problem as in Case 1. This is not making sense to me again. These are few things that have been baffling me at the moment. I'm sure there're some misconceptions in my understanding or something that I might have missed. A little bit of more insight into it is highly appreciated, Boris. Thanks, Nikesh. |

1 Attachment(s)
The reason why I didn't (couldn't) use pressure as inlet condition as you suggested earlier is because I do not have enough initial data to proceed with it. Figure 1 shows all the condition that's available to me at present. So based on the height of the reservoir, I have used Torcelli's equation to calculate velocity at the the bottom of the reservoir (the inflow position);
V = sqrt(2*g*h) = 12.82 m/s As the measurements and experiments have not be carried out yet, we are at the moment, unsure of what the flow rate is at the inlet. Like your suggestion in this post, I could perhaps use this velocity calculated in above equation, or mass flow rate for valve at 100% position (or pipe-only flow) to obtain the total pressure at inlet. This inlet total pressure value can then be defined as the inlet boundary condition for the flows with valve at various other positions. Will it make sense? Also, while setting goals, I need to use 'Bulk average values' rather than simply 'average values'. Does that sound reasonable? Thanks, Nikesh |

Hi Nikesh,
Mass conservation is correct, but the energy conservation is not correct. I just mentioned this as it is one of the basic laws in thermodynamics. If I put some heat into a system that is cooled by air then that air is heated and the same amount of heat I had in my hot component was transfered into the air which will leave the system again but hotter. So 10W of a CPU will result in a by 10W hotter air leaving the system if we neglect any radiation. This is called the first law of thermodynamics. But since your simulation does not contain any temperature change or heat source you don't need to consider any equations so I left them aside. Bernoulli (sorry I spelled it wrong the last time) is the equation to calculate changes in pressure due to various influences of geometry or mass flow rate etc. You can find more here: http://en.wikipedia.org/wiki/Bernoul..._flow_equation Answer to Question 1:Yes, this is better as it takes into account for a non uniform velocity distribution over the outlet. Otherwise you will find that total pressure loss compared with experiments can deviate. Imagine a pipe that has a 90° bent and a circular diameter. Due to the 90° bent the flow will separate right at the sharp bent causing a low velocity region at the bent and the core flow is very fast towards the outer wall. See the image at this link where you can see the velocity profile at several sections in the bent: http://www.flowcontrolnetwork.com/ex...%20fig%208.jpg The first profile right after the bent has non parabolic profile and the higher velcity is at the top side of the pipe in the image. Here the mass flow is higher and with that the veloctiy of course and so is the total pressure if you would measure it here. If you use "average" for goals the whole profile is averaged over the cross section area only and can be low if the velocity spike is very high but only in a small area since the rest of the aera has a very low speed and is a much bigger area. Lets say 80% of the area is at 1m/s and only 20% has 3m/s. The average over the surface is then calculated roughly like this if we consider the area to be 10cm^2 in total: v_av = (v_1 * A_1 + v_2 * A_2) / A_total v_av = (1m/s * 8cm^2 + 3m/s * 2cm^2) / 10cm^2 which will give us ~1,4m/s as an average. If you use the "bulk average" you are not just averaging by the surface area but also by the mass flow rate which will much better take the high flow area into account although the area is just small. It would then look like this: v_bulk_av = (v_1 * (rho_1 * v_1 * A_1) + v_2 * (rho_2 * v_2 * A_2)) / m_total where rho_1 * v_1 * A_1 is the mass flow rate m_1 at that part of the flow and we can neglect the density as it is the same for water since it is considered incompressible. And the same for total mass flow rate m_total which is the sum of 1 and 2. v_bulk_av = (1m/s * (1m/s * 8cm^2) + 3m/s * (3m/s * 2cm^2)) / (1m/s * 8cm^2 + 3m/s * 2cm^2) This gives us a v_bulk_av = 1,857 m/s you see the difference? Answer to Question 2:Since the understanding of Bernoulli is Energy conservation is wrong (as described above in the beginning, your understanding of "Not satisfied energy equation" is wrong. As I said above, conservation of energy is one of the laws but in your case it is not relevant. Bernulli only helps to determine the pressure and the mass conservation often helps in that formula in case you have two unknown variables. For example you know the static pressure at the outlet but not the static pressure at the inlet, and you also don't know the dynamic pressure at the outlet as it is not defined. But with the mass conservation we know that the velocity on inlet and outlet are the same and with that we know both dynamic pressures on inlet and outlet etc. One thing missing in the Bernoulli equation is any loss coefficient due to a valve or bent or friction. Any losses due to such an object can be added to the equation as a delta_P because you will have a total pressure loss and that delta is what you usually want to find out in many calculations. Here are some examples of the calculations with various applications and losses from friction to pressure differences due to pumps etc. http://web2.clarkson.edu/projects/su...20Equation.pdf Bernoulli is always constant if you include any losses, so P_total_in is the same as P_total_out if you consider the loss delta_P in the equation such as: P_static_in + P_dyn_in + P_potential_in = P_static_out + P_dyn_out + P_potential_out + delta_P In a pipe network you can have several losses for example from three 90° bents, friction and sometimes you can increase the pressure in case a pump is used as in the link above shown. Answer to Question 3:No, the negative pressure can appear in case you have cavitation but not activated the cavitation option in the fluid section of the general settings. FloEFD will show that there is some negative pressure which of course is not possible in reality as total pressure can only become 0 but not negative. If you activate caviation it will be calculated but in your case it is most likely because you fixed too many parameters and causing the valve to work in a region of the flow where it usually wouldn't in reality in this case. Cavitation can happen but I would say in your case it is due to the wrong definition of the boundary condition. Usually it happend in such wake region where the flow is rapidly expaning due to a sudden cross section area jump. Answer to Question 4:This again is part of the pressure loss that I explained above, here you again missunderstood the energy conservation as beeing Bernoulli. Sorry again for that maybe not so clear definition in my previous post. Your second post:Aha, here Bernoulli can help you with your reservoir :-) If you need P_total for your inlet BC you can calculate it as this: P_total_CFD = P_static_reservoir + P_dyn_reservoir + P_potential_reservoir If you consider the height level 0m beeing at the CFD part of the mode the potential part is rho * g * H or in numbers: P_potential_reservoir = 1000kg/m^3 * 9.81N/kg * 8.500m = 83385 Pa (or N/m^2) Sorry for me using SI units but that's how the world really works ;) Since you always consider the two position in Bernoulli the one is on the surface of the reservoir (1) and the other at the outlet of the reservoir which is the inlet to your CFD simulation (2). The numbers are therefore the marker for the two positions to make the equations a little shorter: This is how you got to your V=sqrt(2*g*h) equation: P_t1 = P_t2 P_t1 = P_s2 + P_dyn2 + P_pot2 P_t1 = P_s2 + 1/2*rho*(v_2)^2 + rho*g*h_2 this you can use for your CFD simulation with P_s2 beeing the atmospheric pressure at the location of the surface of the reservoir and since the reservoir is constant level you can neglect the velocity and therefore the dynamic pressure and will get this: P_t1 = P_s2 + rho*g*h_2 (use in CFD)Now to get to your velocity: If we use now the same pressure splitting for P_t1 we get P_s1 + 1/2*rho*(v_1)^2 + rho*g*h_1 = P_s2 + rho*g*h_2 with the height beeing 0m as the ground level is at this position the potential part is 0 and can be neglected: P_s1 + 1/2*rho*(v_1)^2 = P_s2 + rho*g*h_2 Now since the air pressure at the outle and reservoir are basically not changing for just 8 m height we can set them also equal and they will cancel each other out of the equation, leaving: 1/2*rho*(v_1)^2 = rho*g*h_2 This then transformed to the velocity v_1 which is the only one so we can call it v and the height also beeing the only one so let's call it h. Also the two rho will cancel each other out. So we get: v = sqrt(2*g*h) See, Bernoulli is the best and the mass conservation is it's helper ;):D Oh and you had another question before:To measure pressure and flow rate or velocity at a certain location in the flow you can create a lid inside that section and use the "component controls" to deactivate it for the flow so it simply passes through it. But you can use the surface of that lid just like at inlet and outlet for a surface goal. The values are less accurate due to interpolation and not beeing partial cells like on inlet and outlet or on walls so you will need to increase the mesh a littel more on that lid with a local mesh to compensate if the mesh cells are too large. I hope this solves all your questions, Boris |

Thanks a lot Boris! You solved all my problems with such precise and detailed explanation. However, it did take me a couple of simulations until I could completely grasp few ideas. Yet, now its all crystal clear and I have managed to complete all my simulations. I just need to wait for the experimental results and see how it goes. It might be couple of months more.
Talking about Bernouli's equation as the energy equation was rubbish from my side, I don't know how it came into my mind :p Its the Navier-Stokes equation that govern the flow. Anyways, thank you very much Boris. Sincerely, Nikesh. |

All times are GMT -4. The time now is 16:38. |