Solved Exam Problem: Conservation Linear Momentum

MEC516/BME516 Fluid Mechanics I, Chapter 3: This is a sample solved problem from Fluid Mechanics Final Exam (2015). The problem requires the calculation of the force needed to hold a water jet nozzle in place using the conservation of linear momentum
All of the course videos and a copy (pdf) of this fluid mechanics presentation can be downloaded at:
www.drdavidnaylor.net
Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
#fluidmechanics #fluiddynamics

[@PeymanSayyadi]

Thanks for sharing these videos. I work in the industry and I wish the students know how useful it is to master these concepts.

[@FluidMatters]

Thanks! Glad to hear the videos are helpful

[@Szhihihihihihi]

Thanks !

[@FluidMatters]

Glad to hear the video was helpful! Best of luck with your studies.

[@mrsengineering3663]

Peace be upon you sir... I need the name of the source for the Fluid Mechanics book and the number of copies. Thank you

[@FluidMatters]

The textbook (White) is given in the video description. That's the best I can do.

[@jondory8134]

Nice job... Related to the FBD, do you need to include a Moment, M, along with the Fx & Fy? (assume a direction...?)

[@FluidMatters]

This analysis is based on conservation of LINEAR momentum. So the are the forces needed to redirect the jet. No moment.

[@beholdayman6995]

why the different notations for x components and y components, those components being F(y) = m(v2 - v1) and F(x) = m(x2 - x1)

[@FluidMatters]

You must mean F(x)=m_dot(u2-u1). This is because F is a vector. u is the x-component of velocity (and v is the y-component).

[@pavandevang551]

sir can u please explain me how to take the direction of force

[@FluidMatters]

As in your Statics course: You start by assuming a direction for the force (which may or may not be correct). You then do the analysis using the assumed direction. If the force comes out positive then the assumed direction was correct. If it turns out to be negative, then the direction of the force needs to be flipped (180 degrees). I hope that helps.

[@DhushanSuresh]

SIR, IN THE QUESTION, ATMOSPHEARIC PRESSURE GIVEN AS 100 KPA SO HOW CAN YOU TAKE P2 GAUGE = 0?

[@FluidMatters]

Gauge pressure is defined as the pressure relative to local atmospheric pressure. So P_2=0 gauge. A Bourdon pressure gauge at the outlet would read zero. In these types of momentum problems you have to work in gauge pressure, for reasons that are explained in this video: kzbin.info/www/bejne/ppyWn3mEhregfLc

[@DhushanSuresh]

@@FluidMatters thank you for the explanation sir.

[@DhushanSuresh]

@@FluidMatters but as p2 we are using straight 200kpa in that why we are not using relative gauge pressure

[@FluidMatters]

@@DhushanSuresh I think you mean p_1. The problem states that "the gauge pressure is p1=200 kPa". So, the absolute pressure at point 1 would be 300 kPa.

[@88_shubhamkumar27]

@Fluid matters sir can I convert gauge pressure into absolute pressure and then use the equation?

[@mrsengineering3663]

Pleas sir I want this book pdf for fluid mechaincs

[@FluidMatters]

Sorry. I can't do that.

[@mrsengineering3663]

Peace be upon you sir... I need the name of the source for the Fluid Mechanics book and the number of copies. Thank you

[@SakeKaama]

Why is the momentum force in the opposite direction as the pressure force? Seems like they would both be acting in the same direction, no?

[@FluidMatters]

The pressure force acts upward on the nozzle, as shown in the Free Body Diagram. So, the component of F_y (the forced needed to hold the nozzle in place) due to pressure alone must be downward. In contrast, the fluid momentum in the vertical direction increases through the nozzle. This change in momentum requires a force (F_y) in the upward direction to hold the nozzle in place. (Think of trying to hold a jet engine in place.)

[@ghbbllk4095]

life saver.finally know why😁