FE Fluid Mechanics Review Session 2022

[@TrucNguyen-xk3ud]

Thank you Mark for all of your videos! I just passed my FE exam after several failed attempts. This time I studied with you and passed it. Your videos are incredible, I appreciate how you put them together like this.

[@sulimanhamed2751]

What did you do in addition to the video? Lots of practice problems? Glad to hear you passed

[@keiajackson7446]

Good morning, were there alot of conceptual questions?

[@ThatWasNew]

Thanks Mark for your help incredible, I have passed FE

[@jasonlewis460]

what similar questions came up on your exam?

[@shamanaveed9329u]

In question# 8 specific gravity SG =γ/γw we can directly solve this.SG*γw=γ put value 13.6*9810=γ

[@arslankhalid760]

Thanks Mark, I have a quick workaround for problem #10, instead of calculating slant distance for "Yc" and "Ycp" we can directly calculate the vertical distances "hc" and "hcp". Hc will be 50cm, however little caution needs to be taken when calculating "Ixc" and Area "A" of the gate, instead of using 16cm, use 8cm as a vertical dimension of the gate to calculate "Ixc" and "A". Just wanted to put this method as a comment as it reduces time to calculate vertical distance to the centre of pressure.

[@curtisreidy4028]

Mark, just wanted to say a quick thank you for putting these videos together. I watched them a few days prior to my exam and I am happy to say I passed. THANK YOU!

[@kautz00100]

For the Pitot tube question you could also do v=sqrt(2gh) and get the same velocity without finding what the pressures are.

[@mateusbernardodeoliveira7641]

To find velocity on Question #3, a faster route, although it does not explain all the concepts you did while solving, is to equate the difference in height between the two tubes to (V^2)/2g, and then solve for V. edit: Mark did explain it right after solving the first way.

[@Cdwilliams1999]

Thank you Mark! Thanks to your helpful videos I was able to pass the FE on my third attempt! P.s. Your jokes are the best!

[@amanuelzemedhun8768]

Hello Mark! Thank you for these videos! In question 11, why did you make P2 zero? Why not plug in the atmospheric pressure itself?

[@Paul-oo6uu]

It may be too late for your question at this point. The reason is the 100kPa at point A is gage pressure. Gage pressure 0 is atmospheric pressure. The atmospheric pressure would need to be added if the pressure at A was absolute pressure.

[@hulb1670]

Thanks Mark. I am using this to study for the PE and it has been very helpful.

[@MarkMattsonPE]

So how do you spell gage/gauge? USGS and the Steel Deck Institute uses gage. However, ASTM and others use gauge. I normally associate gage with stream flow gages and steel deck, and gauge with meters used to measure a pressure. Merriam-Webster indicates gauge is preferred. Both are in the FE Reference Handbook. Thoughts?

[@nabinshrestha6594]

Hi Mark, In question number 7, why Pvinegar=5386N/m^2?

[@jace3254]

for question 11 why did you use 0kpa for Pb instead of using 101.3 kpa for atm pressure?

[@unmaroolable]

gage pressure is measured with respect to the atmospheric pressure. Overall whenever the flow is exposed to atmosphere the pressure at that point is 0.

[@lay6787]

Dear Mark, Im taking this video from korea and it's really wonderful! Just a question. From the 1:17:48, you explained about the applicable b and h if it is not the square type gate. So the h value will be derived from the shown value, e.g.16 cm as per problem and the b will be the value from page direction? You mentioned 12cm in y direction and 20cm in x direction and I suppose Ixc will be then 20 x 12^3/12 but the video says the other way. Appreciate if you can clarify. Thank you so much!

[@MarkMattsonPE]

Hey Lay! thanks for the question from the other side of the globe :) You are totally right. It should be 20 × 12^3 / 12 if the 12 is in the direction of y and 20 in the direction of x. I flipped it when I was trying to add some clarification :( Thanks so much for pointing this out! Hopefully, the comment and response helps clarify it for you and others.

[@lay6787]

@@MarkMattsonPE Thanks for swift response! I really appreciate your lecture, always happy to see your smiley face on the right bottom of screen with stunning jokes from time to time :)

[@lay6787]

Dear Mark, I passed it thanks to you! (2months of studying with 2 hours/weekday+5hours on weekends) Could you please advise me on how to pass in PE Geotech in general? e.g.) List of books to study with, Online lecture will help, right?, sequence of study, preparation duration and etc… Thank you!

[@MarkMattsonPE]

Congrats! For the PE, I used the CERM almost exclusively and felt super prepared.

[@Fleur874]

Hi Mark, thank you for the FE review sessions! I find them extremely valuable! For Fluid Mechanics Problem 10, can you go over how you calculated Ixc ?

[@MarkMattsonPE]

Ixc is the moment of inertia of the plate about its centroid. So you take the base (or width into the page) times the height cubed (length of plate) divided by 12. I hope that helps clarify. I did make a mistake when talking in the video. See the comment string from @Lay also.

[@DatNguyen-wi3fs]

Incredible as always Mark. Appreciate the video

[@ricandres1000]

Good morning Mark I have a question on problem # 7 When you were getting Resultant 2 why did you do the average of the two pressure and why not use only the 5386 like when you got resultant 1 using 3973 only?

[@MarkMattsonPE]

You actually do the average of the top and bottom pressures for each area... only for a triangle, the top pressure is zero, so you take (0+3973)/2, versus (3973+5386)/2 for the trapezoid. Does that help?

[@ricandres1000]

Yes, it helped Thank you very much. Question will you do a review for environmental?

[@02AJO]

Thank you so much for the answers for these concepts. But can I ask a doubt about the question 3 about pitot tube. Since there is a height difference between the two tubes will the equation used in the video still be valid. Since his equation does not account for the additional potential energy due to the height difference Δh.

[@jossyurael9042]

question No.6 needs to be corrected. We should assume that the Vat is closed. otherwise, the atmospheric pressure will be added.

[@crawsabi]

For problem 3, why is it that we use .15m for the height of Ps instead of .10m? would we use .10m if the problem didn't have the first pipe sticking out on the left?

[@tundratales449]

Hi Mark, Thank you for the lecture. On question 5, you multiplied the height of the raft (7.5"/12") by 6' by 8' feet to get the volume of wood in the raft, but since the raft is made of 7.5x7.5 timbers, there has to be some space between the timbers since 7.5 inches does not evenly divide into 6 feet. Only 9 of these timbers can fit into a 6 foot span, unless of course you are using a partial timber (0.6 the width of the others), which kind of defeats the purpose of telling the width dimension of the timbers, right?

[@danielj.berrios960]

Question 12: the problem ask for the horizontal Thrust force ( forces acting in the X axis ) why is it that the answer is 2,828 Lb acting vertically ?

[@MarkMattsonPE]

The z direction is vertical and ignored in this problem. The horizontal thrust is the resultant thrust in the x-y plane.

[@dilipsaha94]

Can you explain why, in problem 12, the impulse momentum principle doesn't come into play? Why can we set Fx & Fy to 0? Doesn't our momentum change in both x & y directions because our V1 and V2 are in different directions? Why isn't it "P1A1 - P2A2*cos(α) + Fx = Qp(V2*cos(α) - V1)" and "Fy + P2A2*sin(α) = Qp(V2*sin(α) - 0)" ?

[@MarkMattsonPE]

The distribution lines are are usually sized to keep the velocities relatively low in the pipes. Adding in the mass flow due to a velocity would increase the forces slightly. I probably could have defined the problem more fully by asking for the thrust due to the static pressure.

[@dilipsaha94]

@@MarkMattsonPE Thanks for the reply. And thank you for this wonderful video series, you really do a good job of hammering home key concepts.

[@danielj.berrios960]

Question 5: W= γ*V, But the problems states that the density ρ=45 lb/ft3. why are you using density as Specific weight γ ?? Thanks Matt. Keep the jokes up man!

[@MarkMattsonPE]

You're right... gamma is unit weight and includes gravity, whereas density or rho is mass per volume. The problem should say the "unit weight of the saturated wood is 45 pcf..." Thanks!

[@LeaderZ2000]

Grateful you asked that question, Daniel. Just spent almost 10 minutes racking my brain on why gravity wasn't used, but you and Mark cleared it all up here.

[@shamanaveed9329u]

Me to was confused, thanks for help.

[@carultch]

@@LeaderZ2000 Everybody brought gravity to the party in that example. Both the surrounding water and our raft, are both affected by the same gravitational field, and buoyancy comes in proportion to the weight of the water. We'd get the same answer, regardless of what planet the problem takes place upon. So whether we use specific weight, or density, we'd get the same answer, as long as we were consistent.

[@Sxciroq]

For question 3, if we know that v^2/2g represents the change in height, then why can't we use the .2m and set it equal to that equation to find our velocity? Wouldn't it be faster?

[@usmitapokhrel9764]

Thank you Mark.

[@engraghad8663]

Hi Mark! Finally I paaaaaaseeeed the FE civil exam from the second try 😍😍 I am soooo happy and want to thank you for all your videos I watched it several times Also I want to recommend the book of “ Islam “ which contains 800 questions it was also amazing and let me understand many additional things that not included in the yellow book So it was great knowledge from your videos, the yellow book, and Islam book after completing it I passed the exam 😍😍😍

[@nikhildumbre4076]

In question 7, while calculating R1, what does 0.4 represent and from where did you get that

[@MarkMattsonPE]

I believe that's the width of the vat. Please see the previous problem for the diagram showing the width.

[@carultch]

@@MarkMattsonPE How do you know whether the given ratio in that problem, is a volumetric ratio or a mass ratio? Is it convention to give the ratio by volume if not otherwise specified?

[@be1062]

Thanks, Mark!!

[@DRM0087]

I am taking the NCEES FE Mechanical exam, would this this video still be helpful for me? Even though it's for NCEES FE Civil exam.

[@Bri.Charles]

Yea it would still be good!

[@eng_sam4850]

q1) how did u get 1.25 ft for area?

[@trellispowell1332]

1.25=1ft & 3inches

[@raulc02]

Tricky first question... open channel flow using the continuity equation...

[@alban97]

I used Manning's equation and got 7.5

[@MarkMattsonPE]

A correction factor (in this case, 0.9) is often to adjust average velocity and area measurements. For example, see this sample lesson/lab on finding flow in a small stream www.nps.gov/common/uploads/teachers/lessonplans/Climate%20Science%20in%20Focus%20Field%20Trip%20-%20Measuring%20Stream%20Flow.pdf

[@AnailyCarbonellAlvarez]

@@alban97 when I used Manning equation , i got 5.56 cfs. PLease could you tell me how many is S in this equation. For me S got 1.22

[@linnacyzia3003]

Where in the link I can find V1.3

[@5313030]

Thank you for these videos.

[@DatNguyen-wi3fs]

55:49 can you explain why Pv = 0 for mercury again Mark ?

[@MarkMattsonPE]

It's not that there's no vapor pressure, it's just really really small and can be ignored.

[@DatNguyen-wi3fs]

@@MarkMattsonPE roger

[@nadheeralgburi5265]

First question 1.3 Should be 1.025 not 1.25

[@TheColonelJman]

1.25 is correct. 1 foot, 3 inches or 1 + 3/12 = 1.25

[@musicbyverve]

I don't understand why you used Q = Av to solve question #1 when the problem statement defines it as a channel. We can see it is open. Open-channel flow is manning's equation?

[@Godloves24]

Great video.

[@noahbatterson2884]

question 1 is annoying. Why would you make a problem using open channel flow and then not use the open channel equation for flow rate???

[@thepiratebey]

10:10 my name is ish