lovely video , i noticed a mistake in the conversion of the internal diameter of the suction pipe , D=12'' =1' not 1.5' used in calculating friction loss , fL/D(v^2/2g), minor loss is actually 1' ,thanks
Why do you use 144 to convert the pressure to feet? I need to do the same calculation but for diesel which has a different density than water. Would 144 still apply?
Yes it would because this is to convert sq. inch to sq. ft (i.e. 144 sq in = 1 sq ft). Perhaps it would be easier if you just used the specific gravity of the diesel rather than going 'back' to using the density of diesel. It might save you a couple steps.
Hi Kenneth, I noticed the calculation does not include the head required to pump the liquid after the pump. How to count the entire head required for this system ?
Hi Kay, when we are evaluating the net positive suction head available in a pump system, we are only concerned with the hydraulics that exist between the suction side storage and the suction flange of the pump. Other analyses we consider the pressure downstream (or what comes "after" the pump), but not with NPSHa. Does that answer your question?
@@kennethwlamb Hi Kenneth, thank you for your reply! Do you have the video to count the "head after the pump" ? As in my case I need to consider the head create from all the outlets (let's say ststic pressure: 50000 Pa each from 10 outlets), 90 deg bending as well as the piple L/D. I suppose I need this to ensure the pump HQ curve exceed "head after pump".
@@WOOTYNGFENG If you are curious about how to find the residual pressure at any point after the pump then you can use my videos on analyzing a pressurized pipeline. ru-vid.com/group/PLKbpg9Z0_Xwn52vueUU_ZIFFsqD4mdhj5 The most relevant videos on this playlist are the Conservation of Energy Applied, and Residual Pressure. Hopefully, those are helpful to get what you need.
It requires more information than knowing that the NPSHr is 32-ft. You must also consider how much more NPSHa there is than the NPSHr. For example if your pump's NPSHr is 32, but your system's NPSHa is 30, then this would not be a good pump to choose (or you should consider revising the elevation of your pump, or the storage tank, to ensure there is enough NPSHa).
Your explanation is breathtakingly good, thanks for that. But, I only have a doubt in the first term calculation. Why is considered the height at the pump(1140) instead of the height at the reservoir(1150)? because it is supposed that the atmospheric pressure mainly acts on the top of the reservoir tank water.
NPSHa is considered at the pump elevation because that is the location where the risk of cavitation exists. There is no risk of cavitation in the reservoir.
It is a very fantastic explanation. But I do have one question regarding the NPSHa formula. From time to time I saw people using another formula to find the NPSHa, NPSHA = Ha +/- Hs - Hf + Hv - Hvp, they have the term Hv (velocity head) to be positive and I really don't understand why. Would you mind explaining the difference between your formula and this one? Thank you very much!
The derivation of the NPSHa equation is from conservation of energy which has + Hv. After doing some research there is no particular reason that I can see for omitting it other than including it makes NPSHa less conservative. The authors of Pumping Station Design, 3rd Edition state, "There is no term for velocity head in [the NPSHa Equation] because velocity head is part of the absolute dynamic head." However, I just attended a webinar from the hydraulics institute where they are including the Hv term. The Hydraulics Institute is the final word, in general.
If your system have many different pipe diameters and roughnesses then you simply add another pipeline term to your hp equation. Example: let's say you have three pipelines. One is 24-inch, another 20-inch, and a third 18-inch. Then you would have three terms added together in the system curve to account for the energy loss in each pipeline.
@@slengoslengaw8510 The NPSHa calculation does not consider anything that is on the discharge side of the pump. Therefore, I do not see how the system being a closed loop or open loop would change the calculation. Is there a sample system you have in mind? One small difference in the NPSHa calculation can occur when the source of water on the suction side has a starting pressure that is greater than atmospheric pressure. For example the pump is pulling water from a pressurized storage tank. In this situation it is not just the elevation difference (ztank - zpump) but the elevation of the tank plus the tank pressure (then subtract the pump elevation)..
Great video, thank you so much. I think the velocity used in calculating the losses is wrong, because you found the velocity by using the flow rate ((after pumping)), I found that usually the suction velocity is considered 0.6 to 1.5 m/s Please correct me if I am wrong
On the Operation of a pump, the flow through the system constantly varies as the pump drains the suction reservoir and fills the discharge reservoir. Therefore you only need to use the maximum flowreate for the NPSHa calculation. In this example I only use the "pump design flow rate" for the sake of simplicity (because that is how the problems are set up in licensing exam questions in the US ).
The suction pipe is sized with a different criteria. Typical suction pipes have a velocity between 10-12 Ft/s while the discharge pipeline is sized for another criteria using a lower max velocity or unit headloss. The suction and discharge pipelines are typically different for that reason.
@@ATTRUONGNGOC the equation covered here applies to a booster pump situation as well. In this example the water level feeding into the pump is higher than the pump suction flange. Is that the case with your booster pump as well?
@@kennethwlamb I have an old pump and an old piece of pipe. I installed a booster pump and the 2" pipe is 18.5M high and 675M long. My pump is 2M higher than the old pipe.
So, answering this question might require a couple new videos ;-) but I'll give it a try... For a potable water system you start by computing the water demand and the required water storage for your pumping system. Once you know the water storage you must determine how long you want to take to fill the storage tank. Most systems do not spend all day (e.g. 24 hours to fill the storage tank). Pick a time that allows the tank to be filled outside the peak power usage time for your electricity provider. Once you have the volume, divide it by the time and you now have the "design flow rate" of your system. There are other complicating factors so you are motivating me to make more videos.
@@kennethwlamb Thank you for your answer! But what I was thinking of is how can you determine the pump suction flow rate if you have a filled tank that is supplying it. I mean with time, the level in the tank will change right? and therefore, your flow rate will too. Add to that, how can you calculate flow through a nozzle to the pump?
I believe that pump suction flow rate will depend on your tank level and gravity effect. Correct me if I am wrong. And I am not certain how can I find flow rate at pump suction.
@@vadkaa5053 for the NPSHa calculation the velocity is not determined by gravity because the pump is adding a significant amount of energy to the fluid and is drawing water from the tank at a faster rate than gravity alone. Therefore you still need to create the system curve and select a pump to obtain a pump curve to find the operating flowrate from the intersection of these two curves. I have another video explaining system curves and pump curves. You are correct that the water level in the suction tank has an impact on the flowrate. When the tank is higher, the velocity is higher at the pump suction flange than when the water level is lower and the tank is almost empty. For the NPSHa calculation we focus on the time just before the suction tank empties because that is when the risk of cavitation is highest. Does that help answer your question better?
@@kennethwlamb It is helpful thank you very much. But there is one question left, does that mean that flow rate at pump suction and discharge is similar?
Believe it or not, it would be the same process because the NPSHa would be computed at the first pump. You could compute NPSHa for each subsequent pump but it likely would not be useful as the most critical is the first pump in the series.
@@kennethwlamb sir NPSHa for first pump in series I calculate as patm+ (water level minus impeller eye level) minus vapour pressure. Here first pump takes water from source ie dam or reservoirs. Now for second pump which is in series here NPSHa= residual head left behind at impeller eye of second pump minus vapour pressure. Here I doesn't take atmospheric pressure. I think I m on right path.
@@greentechguru362 Yes I think you are on the right path. Perhaps I would have explained it as: When considering the NPSHa for the second pump you must replace the potential energy calculation (Res elevation - Pump Eye) with the static head between the first and second pump (Dynamic Head - Pump Eye). You could say that the Dynamic Head at pump 2 is the residual pressure at pump 2. Therefore we are in agreement.
@@salemrachdi8704 If the suction tank is pressurized it is becuase you (the operations engineer) are pressurizing it ... intentionally. Therefore it isn't something that you compute but it is a given. If your objective is to add pressure tothe suction tank to improve the NPSHa in the system, then you could solve for the pressure by using the steps covered in this video (assuming there is no additional pressure applied to the tank), then add the pressure needed to meet the NPSHa requirement. However, I can not think of a scenario where you will solve for the pressure in tank 1 as an objective by itself.
@@kennethwlamb Thanks. Some costumer ask to close the tank from the top and basically for concrete tanks and provide a vent. I will consider these tanks as opened tanks because there is a vent on tanks top.
It just means that the tank is above the pump inlet. You can set the elevation of the pump at any elevation *below* the storage. If the pump elevation is above the storage then the pump will need to be primed before operation (this is less ideal but possible as well). Does that help answer your question?
@@kennethwlamb if the pump and the storage tank have same elevation, Does the height of the water inside the storage tank considered? for example water inside tank is 1m..Does Zr-Zp equal to 1m?
@@muhdimran5389 You still consider the elevation of the tank even if it is equal to the elevation of the pump inlet because that is simply good bookkeeping of your computations. All this would mean is that the static pressure normally contributed by the tank is '0'.
how to determine NPSH 3% or what we say NPSHR and what should be the NPSH margin ratio for pumps watch this ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-L_kAv0wpST8.html
So sad! what would have otherwise been just a great video but because of an error in the conversion of inches to feet I spent an hour trying to figure this out Being someone without any type of background in this area it's really frustrating to see that videos like this are still published with errors. For heaven sakes please just take a few extra minutes to review your work before publishing it. you tell me you have students that make errors and then you do it yourself and make a similar mistake.
