Conservation of Linear Momentum in Fluid Mechanics

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MEC516/BME516 Fluid Mechanics, Chapter 3 Control Volume Analysis, Part 5: Derivation of the linear momentum equation for a control volume using Reynolds Transport Theorem. A numerical example is presented that involves calculating the flow-induced forces on a stationary vane that deflects a flow of water.
Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
#fluidmechanics #fluiddynamics

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26 дек 2020

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Комментарии : 21

@FluidMatters Год назад

All the videos for this introductory Fluid Mechanics course are available at: www.drdavidnaylor.net/

@cornerbloke8861 11 месяцев назад

Best explanation I've seen in RU-vid, thank you so much :)

@FluidMatters 11 месяцев назад

Thanks for the kind words. Glad to hear the video was helpful. Best of luck with your studies.

@cornerbloke8861 11 месяцев назад

@@FluidMatters thank you :)

@joaogarcia9855 19 дней назад

I was wondering if, rather than taking the control volume such that V2 is perpendicular to the normal area vector, we took a rectangular box with no θ inclined edge. We would have a dot product of V2 . A = cosθ and hence a Rx = ρV²A (cos²θ -1). Which would mean the reaction force is dependent on the chosen control surface. What am I missing? :( Thanks for the video, it really helps me out!

@FluidMatters 17 дней назад

The force is independent of the control volume you select. You are missing that the mass flow rate must to be calculated using the velocity normal to the control surface. See the explanation of mass flow rate here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iffMXHFueSo.htmlsi=RAaBqj6FX4djHgfM&t=321

@joaogarcia9855 17 дней назад

@@FluidMatters Uhm, it makes sense then. Thanks a lot for the response!

@FluidMatters 16 дней назад

Btw, that we an insightful question. I bet this will help other students with the same issue.

@abdullahashraf8249 2 месяца назад

what da dog doing

@NE0NTOKYO 5 месяцев назад

Hi, I have fluid mechanics final exam tomorrow and I have a question. Do we have to assume the direction of the forces (Fx & Fy)? Or is there any fixed conditions that I have to know to determine the direction of the forces?

@FluidMatters 5 месяцев назад

It's just like in statics. You start by assuming a direction for each force. (I suggest using the positive x- and y-directions.)Then do your analysis. If the force turns out to be a negative number, then your assumed direction is wrong. Flip the arrow on your final answer and lose the negative sign.

@NE0NTOKYO 5 месяцев назад

@@FluidMatters Thank you so much! My lecturer didn't explain the fact that it's just the same as statics so the whole class thought that there's a specific direction for the forces. Your videos helped me a lot. I will try my best tomorrow. Thank you, Mr. Fluid!

@TheeGai Год назад

So if we use the Bernoulli equation to verify that the velocity is the same, do we assume that the elevation difference is zero?

@FluidMatters Год назад

Yes. If you assume the elevation difference is zero, then you are assuming gravity effects are negligible (or gravity acts into the page). Alternately, if you know the elevation change, you can calculate a corrected outlet velocity, and do a more accurate force calculation. (We have our students do that in a lab experiment at TMU). The velocity correction is usually small, btw. I hope this helps. Thanks for the excellent question!

@TheeGai Год назад

@@FluidMatters thank you

@TheeGai Год назад

Another question, why are you considering pressure to be atmospheric and when shouldn't we consider to be atmospheric, when we cut through a control volume?

@FluidMatters Год назад

@@TheeGai The vane surrounded by air. So, we're assuming it's not say, in a tank. But if it was in a pressurized tank, the inlet and outlet pressure different from p_atm, but the same. So, the magnitude of the assumed pressure doesn't matter. The other question: You have to account for inlet and outlet pressure differences when the flow is though a pipe, for example. I hope that helps.

@saddamhecine9024 Год назад

thank you can you show link the book

@FluidMatters Год назад

www.mheducation.com/highered/product/fluid-mechanics-white/M9781260258318.html

@r2k314 11 месяцев назад

A nagging problem I can't resolve is how the "linear momentum" terms have units of MA as you would expect but the fluid is not accelerating, and nevertheless allows us to compute forces.

@FluidMatters 11 месяцев назад

Momentum is a vector quantity. So, acceleration can be either a change in speed (scalar) or a change in direction. In this water jet example, the water jet comes in with no y-momentum and is accelerated in the y-direction. I hope that helps.