Fluid Mechanics Lesson 08F: Pipe Flow Minor Losses
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- Опубликовано: 9 сен 2024
- Fluid Mechanics Lesson Series - Lesson 08F: Pipe Flow Minor Losses
In this 15-minute video, Professor Cimbala shows how to account for so-called minor losses in pipe flows, by using either equivalent length or minor loss coefficient. He shows values of the coefficient for elbows, inlets, outlets, and other minor losses. Finally, he does two example problems to illustrate how to apply minor losses to pipe flow systems.
This video incorporates material from Section 8-6 of the Fluid Mechanics textbook by Cengel and Cimbala.
An Excel file listing of all the videos in this series can be found at
www.me.psu.edu... .
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Dr. John M. Cimbala is Professor of Mechanical Engineering at Penn State. He is an educator, textbook author, Christian author, husband, father, and grandfather. He also created and maintains a website for helping people grow in their faith called Christian Faith Grower at www.christianf... His RUclips channel is at / @johncimbala
Good afternoon Professor John Cimbala, I wanted to bring to your attention that the correct answer for Welec is 3.502 Watts, not 2.05 Watts as you previously stated. I have confirmed my answers using ESS and Wolfram Alpha. Previously, we required 2.085 Watts to overcome minor head losses of 0.112 m, but now the head losses are much larger, specifically 3.35, so a much greater power is required to counteract these head losses.
I also checked my answers and units in EES and get my same answer of 2.05 W. I did discover that my exponent on the volume flow rate should be -5 instead of -3. That was just a "typo" and the final answer remains the same. If you go through my algebra carefully you should get my answer.
I think I understand this now and I am going to try to express it correctly in the hopes that someone will correct or affirm. The kinetic energy correction factor should not be used with the V^2 /2g terms when calculating losses with the exception of the resistance coefficient K for an outlet since the kinetic energy is dissipated within the control volume. I remember the concept of alpha from college, but I haven't used it since, a consequence of a decade in management...
I'm working on an inlet layout for a water storage tank in terms of analyzing predicted mixing performance relative to the use of an active mixer. The tank in question will fill with a pressure sustaining control valve from another tank at a higher elevation. The "waste" kinetic energy will be used to reduce or eliminate the need to add energy using a mechanical mixer. Of course, we are also dealing with a small and seasonally dependent thermal variation between the incoming water jet and the ambient water temperature, even with an otherwise uniform temperature distribution within the tank, so buoyant forces come into play. Interestingly, buoyant forces negatively affect mixing performance regardless of the vector orientation relative to the velocity. Also interestingly, the higher alpha for laminar flow does not compensate for the lack of eddy currents inherent in turbulent flow.
I am not sure I follow your statement about alpha. If you are talking about minor losses, then you are correct - minor losses are based on V^2/2g without the alpha. Same thing is true with major losses since they are also based on V^2/2g without the alpha. However, in the kinetic energy terms in the head form of the energy equation, you DO need to include the alpha. I am 99.99% confident that the equations in my book are correct.
Good day sir.
Thank you for all you do.
Please, how did you get the values for the sharp inlet and sharp outlet.
Outlet is easy - ALL the kinetic energy of the flow is wasted, eventually dissipated as heat. Thus the outlet minor loss coefficient is alpha. The inlet is not so simple and those values came from experiments (not my experiments, but from references).
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