I have been working for 6 years as Rotating Engineer in oil & gas downstream and upstream and this man has refreshed my knowledge so much, easy to understand before we plan to procure the centrifugal pump.
Thanks so much for the kind words. Also especially good for students to hear that this stuff can actually have a real world application, not just homework!
But I think the calculation conversion you made from ft3/s to gal/min is wrong.😂 Correct me I been mistaken. But anyway, you did great explanation and easy to understand. Tq😂@@BrianBernardEngineering
oh no, I think you're right. When looking up numbers on the Moody Diagram, it should be shifted left, one section, which changes the f values from about .015 to about .017. Final plot for the system curve would then be above the orange line, but below the blue line. I didn't catch any mistakes with the process. If you follow this process, but don't make division mistakes like I did, you should still be golden.
If I have already calculated my head losses at a certain flow rate, can I simply divide the head losses by the flowrate^2 to get the Q coefficient? For example and ignoring elevation, if I calculated 200kPa of head loss at 100LPS, can I get the Q coefficient by dividing 200kPa by 100LPS^2?
I don't think I fully understand your question, in particular I'm not sure what you mean by "Q coefficient". But perhaps you may be able to answer your question with unit analysis. What happens when you divide kPa / (L/s)^2? It's going to be a pretty weird unit. But suppose I change this to more ordinary units like Pa, which is (N/m^2) divided by (m^3 / s)^2 I'd then get Ns^2 / m^8. This is probably not headed in the right direction. I can't recommend you an exact path forward, but whatever you try next, first check your guess by multiplying or dividing out the units and see if they give you the right units that you expect your answer to have.
@@BrianBernardEngineering First of all thank you very much for your prompt response to a poorly worded question. Your videos are awesome and have helped me immensely on working out some complex problems. Let me clarify. When we develop the system curve, we end up with an equation that looks like this: h_p = h_static + coefficient * Q^2. In this example you calculate it as 10 + 0.000021977 * Q^2. You developed this equation algebraically buy simplifying and collecting terms, keeping Q unknown. However, in my problem, I have multiple pipe sizes and minor head losses calculated from various equations and graphs and it makes this process quite complicated. What I'm asking is, if I calculated h_p = 200kPa at a flow rate of 100LPS. Can I simply rearrange the equation to get: coefficient = (h_p - h_static)/Q^2 using my values for 200kPa @ 100LPS to get a system coefficient that I can sub back in to the generic formula: h_p = h_static + coefficient * Q^2 which I can than use at varying flow rates to develop the system curve? Hopefully this makes more sense. Thank you!
Oh, I understand now. That would mostly work with a caveat though. The main issue I see with your approach would be friction. If friction is close to constant, then yes, I think your plan would work. If friction is going to be noticeably different between low to high flowrates, then you essentially need 2 coefficients - the minor losses coefficient wouldn't change, but the friction factor coefficient potentially could change, so you might need to do it sort of piecewise, only over small regions of the flowrate, where friction doesn't change much.
Water distribution is carried out with the following requirements: 25 % of the time flow rate is 75 l/s and head in variable-speed operation is 20 m 50 % of the time flow rate is 90 l/s and head in variable-speed operation is 25 m 25 % of the time flow rate is 105 l/s and head in variable-speed operation is 30 m To simplify the calculation, the motor and variable-speed drive efficiencies are assumed to be 100 % The price of electricity is 0.12 €/kWh A Sulzer APP42-150 pump with closed 340 mm impeller whose characteristic curves is used to carry out the pumping task. What are the annual energy consumption and costs of this pump with variable-speed operation? Annual energy consumption: ?? MWh Annual energy costs: Euro ?? How much energy and euros are saved with variable-speed operation compared with the use of throttling valves and fixed-speed operation at 1400 rpm? Saved energy: ?? MWh Cost savings:Euro??
Sorry, but I think the method in this video won't help you with that problem. In your givens, you are already provided with both flowrate and head. This video is used when you only know one of those, in order to find the other. You'll need to find a different video with more information about pump power - and I don't have one of those sorry. Good luck to you.
