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@comment8767 11 сағат бұрын
I always pass the administrative part.
@MeechMwalo 2 күн бұрын
nice
@craig7878 9 күн бұрын
Hi, when finding F_bc, since it is a vertical force, wouldn't the formula be F_bc = pgV ?
@FluidMatters 8 күн бұрын
no. F_BC is the a fluid weight. F_BC is the upward pressure force on plane surface BC. The entire surface BC is a depth R. The pressure at this depth is rho*g*R, which gets multiplied by the surface area. I hope that helps.
@adhimta3306 9 күн бұрын
❤ beautiful
@aliVimtaj 10 күн бұрын
I've got a question: in the example of discharge flow from a tank, if I put my finger at the discharge point, I see the speed of the discharging fluid is increasing, but Bernulli equation shows the discharge speed is not function of discharge diameter. Is there any explanation for that?
@FluidMatters 10 күн бұрын
Interesting question. I haven't tried it, but I suspect your intuition is not correct. Try drilling a small hole and a larger hole at the same depth near the bottom of large plastic water bottle. My bet is that both water streams follow the same arc and extend the same distance from the bottle i.e., they have the same exit velocity. It would make a nice short video. I'm completely open to be proven wrong! Thanks for the question.
@surprisemoleele8004 11 күн бұрын
This channel is helpful for my research project.
@FluidMatters 10 күн бұрын
Glad it helps. What's the project?
@tw5002 14 күн бұрын
I wouldn't say that the rest of the analysis is easy (at 14:34). Firstly, because the term (r+dr)[dθdz] is missing in the formulation of the mass outflow on the infinitesimal area; if you follow this video only you can't arrive at the correct answer. Secondly, when you isolate the inflows and outflows across this face, you have to use the curious mathematical fact that dr^2 = 0 to get rid of one of the terms. Then, you have to see the odd step of multiplying some terms by r/r so you can factor a 1/r term out. Then, after more simplification, you're left with a term like (1/r)[ρVr+∂(ρVr)/∂r], and you have to recognize that the term in brackets is the same as ∂(rρVr)/∂r by virtue of the product rule of derivatives.
@AungHeainSoe 15 күн бұрын
Hello Professor, may i know why F(AB) is not considered in the moment ?
@FluidMatters 14 күн бұрын
FAB=FH. So, this force is considered, indirectly. When you take moments on the gates, you have to consider the forces of the water on the gate. I hope that helps.
@ishitavardhan7697 15 күн бұрын
This video is absolutely GOLD for someone like me struggling with fluid mechanics. Glad to find this on KZfaq!!
@ryanduthie1311 22 күн бұрын
how is there a moment because of Fbc? not sure how that causes rotation
@FluidMatters 21 күн бұрын
I suggest doing a review of basic statics. Fbc has a moment arm that will cause the gate to rotate CCW. It will act to open the gate.
@joaogarcia9855 23 күн бұрын
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 21 күн бұрын
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: kzfaq.info/get/bejne/n8yWgLt5q9vJg6M.htmlsi=RAaBqj6FX4djHgfM&t=321
@joaogarcia9855 21 күн бұрын
@@FluidMatters Uhm, it makes sense then. Thanks a lot for the response!
@FluidMatters 21 күн бұрын
Btw, that we an insightful question. I bet this will help other students with the same issue.
@user-mf7vr3sj4u 24 күн бұрын
Can you explain why (at 1:05) the pressure forces act inwards as opposed to in the direction of flow / where in Chapter 2 specifically should I look to find out why?
@FluidMatters 21 күн бұрын
p. 61 of the White text: "Pressure creates a force due to the molecules bombarding the surface, and it is normal to the surface." Fluid molecules bouncing off a surface push on the surface. So, the pressure force acts inward and normal to the surface. Also, see the proof of that pressure is a point property in Chapter 2, Fig. 2.1. Equilibrium on a small wedge of fluid. Think also about your experience. A balloon of air submerged in deep in water will get compressed by the water pressure, because the static pressure acts inward. If pressure acted outward, it would expand, which it certainly does not. I hope that helps.
@arundhati.parikh 21 күн бұрын
@@FluidMattersthank you! That makes complete sense for the inlet… however, at the outlet, aren’t the molecules moving away from the surface (in the direction of flow) as opposed to pushing against the surface? maybe I am missing something?
@FluidMatters 21 күн бұрын
@@arundhati.parikh The bulk motion of fluid is not the root cause of static pressure. Maybe this will help: Do you expect the outflow side of elbow to be held onto the pipe at the flange by suction? That's what you are thinking, which I hope you can sense is clearly not what happens.
@Tom-ts5qd 25 күн бұрын
Does this mean the Bernoulli Equation / Venturi effect can be applied to non ideal fluids as well? (approximately of course)
@FluidMatters 25 күн бұрын
Yes. In real (viscous) fluids there are some energy losses. But "Bernoulli" is often still a good approximation for air and water, which both have low viscosity.
@francisfrancis4219 26 күн бұрын
A submerged ball when released does not “rocket out of the water”. It rockets up when in the water, but only to reach the surface.
@FluidMatters 21 күн бұрын
www.tiktok.com/@thelahayfamily/video/6846123958639578374?lang=en
@jacobhardy2313 Ай бұрын
Great video and explanation
@FluidMatters Ай бұрын
Thanks. Glad to hear it was helpful.
@URKAEELIZACOLINPOWELL-ci1hw Ай бұрын
Is it the no slip condition that causes the velocity gradient, professor or the internal shear stresses?
@FluidMatters Ай бұрын
I'm not sure which part of this video you are specifically referring to. Here is a general answer. The fluid "sticks" to the surface i.e., the no slip condition. If the fluid above a stationary surface is moving (because of a pump or fan, for example), the fluid's viscosity will decrease (to zero) as you approach the surface. It's the internal viscous shear forces in the fluid, which opposes fluid motion, that slows the fluid in the near wall region. So, to answer your question, it is both effects. I hope that helps.
@URKAEELIZACOLINPOWELL-ci1hw Ай бұрын
So, hypothetically speaking, even if there doesn't exist 'no-slip' at the bottom surface ( but the fluid sticks to the top surface) there would be a velocity gradient in the fluid, solely because of the internal friction between them. Am I right, professor?
@URKAEELIZACOLINPOWELL-ci1hw Ай бұрын
I meant internal resistance between the fluid layers, alone, sir.
@user-pd5bn6qf5k Ай бұрын
when you have losses, you add them to get to EGL?
@FluidMatters Ай бұрын
Yes. The EGL remains constant only in the ideal friction case. But in the real world you always have energy losses in the flow direction. So, the EGL decreases because of energy losses (due to turbulence, wall viscous friction & pressure losses across a valve.) I hope that helps.
@ballerslounge2855 Ай бұрын
Thanks for this lecture Dr Neylor i have a question though at 11:42 is the height difference not supposed to be (Z3 - Z2) or it doesnt matter
@FluidMatters Ай бұрын
Note that z is measured upward, i.e. Z2>Z3. You need that term to be positive, since you are moving downward into zone of higher pressure. So it must be Z2-Z3 to get a positive number. You might find it easier to work in terms of DeltaZ, where DeltaZ is always a positive value.
@john-hf9dt Ай бұрын
Thank you sir.
@antoniamillaraylizanaraban9380 Ай бұрын
THIS VIDEO IS SO GOOD
@zhenccc Ай бұрын
All your videos helped a lot. Thank you so much!
@javierpicazo2107 Ай бұрын
I like this series
@Kawsusstory Ай бұрын
Is that a plaquette on a hinge?
@Mehdiranjb Ай бұрын
🌹🌹
@TerminallyOnline92 Ай бұрын
At 13:29, shouldn’t it be C1 = -1/(2*mu*b)*dp/dx - U/b and not C1 = -b/(2*mu)*dp/dx - U/b ? Doing the algebra, I feel the b should be in the bottom of the fraction and not the top. Can someone explain please?
@FluidMatters Ай бұрын
Before simplification, we have b^2 in the numerator (from y^2 in the original expression applied at b). So, when you divide by b to isolate C1, you end up with b in the numerator. I hope that helps.
@TerminallyOnline92 Ай бұрын
@@FluidMattersThanks! I didn’t see that
@suleyman7478 Ай бұрын
great stuff
@RGCAD 2 ай бұрын
Great stuff, for the last problem i see we have our origin placed midway between the plates, I would therefore expect the y at the bottom to be y = -h
@FluidMatters 2 ай бұрын
Agree. The axis in the graphic (that I stole from the book publisher) is in the wrong place. But I think it's totally clear in the presentation that y=0 is at the bottom. At some point, a long time ago, I fixed it in the pdf download.
@baaa-ej7xj 2 ай бұрын
He saved my final exam😭😭😭
@comment8767 2 ай бұрын
Point of Confusion for me - The cube as 6 faces with viscous stresses on each face. Therefore, there are 6 x 3 = 18 stresses on the differential element. Are we somehow saying that the stresses on opposite faces are the same, so that it is really only one? Then, all stresses are referred to a central point? What happens to the extra nine stresses?
@comment8767 2 ай бұрын
OK, so you are not claiming that the nine stresses cover the entire differential volume. You are simply positing the stresses on three faces, and then using taylor series to get the stresses on the opposite face. The posited stress plus the Taylor-derived stress provide the stress on the entire cube.
@mohammadfallahzade2110 2 ай бұрын
best instructor
@FluidMatters 2 ай бұрын
Thanks for the kind words. Best of luck with your studies.
@mohammadfallahzade2110 2 ай бұрын
thank you you are a great teacher
@jamboreejackson8309 2 ай бұрын
Hi professor, many thanks for the excellent lectures. I tried the derivation of the equation in cylindrical co-ordinates and I couldn't quite get there so ended up looking at another reference. Should the dA in the outlet term you show at 14:30 actually be (r+dr)dΘdz (rather than rdΘdz) to account for the slightly larger area at the outer radius? This is what the derivation I saw suggested.
@FluidMatters 2 ай бұрын
Good point. Yes, I think you are correct.
@mohammadfallahzade2110 2 ай бұрын
thank you for your great explanation
@spiral-dz1eq 2 ай бұрын
from 3:40 on, why didn't we use the fluid for W and the wood for Fb?
@FluidMatters 2 ай бұрын
As shown in the free body diagram, W is the weight of the WOOD and Fb is the buoyancy force. Thus, Fb is the weight of the WATER displaced by the wood (Archimedes principle). That is why I used the specific gravity of wood for W, and the specific gravity of water for Fb. I hope that helps.
@spiral-dz1eq 2 ай бұрын
@@FluidMatters Thank you!
@andrewmerafuentes6683 2 ай бұрын
I came just to check for a concept, then proceed to finish the whole series.
@FluidMatters 2 ай бұрын
Glad to be able to help. Best of luck with your studies.
@abdullahashraf8249 2 ай бұрын
what da dog doing
@rashedshahariar2748 2 ай бұрын
Where is the vertical force acting on the top of the free body in no 2 prblm 7:55
@FluidMatters 2 ай бұрын
If I'm understanding your question: There is no vertical force on "the top of the free body". The vertical force acts upward and equals the weight of the water, which acts downward.
@rashedshahariar2748 2 ай бұрын
@@FluidMatters if we considered only the upper left quarter of the circle instead of the semi circle then there would have been vertical force acting on the upper part of the quarter circle then why not for the semi circle?
@rashedshahariar2748 2 ай бұрын
At 6:57 inclined forces at the upper portion should have vertical components ..isn't it? Can u pls clarify?
@gani82 2 ай бұрын
Continue the videos sir ..... following your videos ... ..waiting to learn from you sir.....
@FluidMatters 2 ай бұрын
Glad to hear you find them helpful. Good luck with your studies.
@colinwarn4606 3 ай бұрын
Just wanted to extend my thanks for putting together a great KZfaq series on Fluid Mechanics. Been a couple years since I've taken the class, the series has been a great refresher on the theory, in addition to tying it to some very informative experiments that visually cement the concepts 🙂
@FluidMatters 3 ай бұрын
Thanks. Great to hear!
@anthonylorenzo1126 3 ай бұрын
Why isn't d^2u/dy^2 not 0?
@FluidMatters 2 ай бұрын
d2u/dy2=0 means that the velocity profile must be linear i.e. no curvature. There is no basis for this requirement, as you can see from the solution.
@jondory8134 3 ай бұрын
Nice job... Related to the FBD, do you need to include a Moment, M, along with the Fx & Fy? (assume a direction...?)
@FluidMatters 3 ай бұрын
This analysis is based on conservation of LINEAR momentum. So the are the forces needed to redirect the jet. No moment.
@KT-dv8qy 3 ай бұрын
THANKS A LOT SIR!!!!
@FluidMatters 3 ай бұрын
Glad to hear it was helpful. Best of luck with your studies.
@renatosureal 3 ай бұрын
To estimate the impact force of a 5-meter fall from a 90 kg person, we can use the work-energy principle which states that the work done by the impact force is equal to the change in kinetic energy. When a person falls from a height, they gain kinetic energy, which is then absorbed by the impact. The kinetic energy (KE) just before impact is given by: $$ KE = \frac{1}{2} m v^2 $$ Where: - \( m \) is the mass (90 kg in this case), - \( v \) is the velocity at impact. The velocity at impact can be found using the equation for velocity of an object in free fall: $$ v = \sqrt{2gh} $$ Where: - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)), - \( h \) is the height of the fall (5 meters). Substituting the values, we get: $$ v = \sqrt{2 \times 9.81 \, m/s^2 \times 5 \, m} $$ $$ v = \sqrt{98.1 \, m^2/s^2} $$ $$ v \approx 9.9 \, m/s $$ Now we can calculate the kinetic energy: $$ KE = \frac{1}{2} \times 90 \, kg \times (9.9 \, m/s)^2 $$ $$ KE = \frac{1}{2} \times 90 \, kg \times 98.01 \, m^2/s^2 $$ $$ KE = 4410.45 \, J $$ The impact force (F) can then be estimated if we know the distance over which the impact force acts (d), which is the distance over which the person's momentum is brought to zero. This distance will depend on many factors, including how the person lands and the nature of the surface they land on. Assuming a certain stopping distance, we can use the formula: $$ F = \frac{KE}{d} $$ For example, if the person comes to a stop over a distance of 0.5 meters, the impact force would be: $$ F = \frac{4410.45 \, J}{0.5 \, m} $$ $$ F = 8820.9 \, N $$ So the impact force would be approximately **8820.9 Newtons**. Please note that this is a simplified calculation and the actual impact force can vary. For accurate results, especially for safety considerations or engineering applications, a detailed analysis considering all relevant factors would be necessary. Source: Conversation with Bing, 4/30/2024 (1) Impact Force Calculator - Calculate the impact force in a collision. www.gigacalculator.com/calculators/impact-force-calculator.php. (2) Impact Force - The Engineering ToolBox. www.engineeringtoolbox.com/impact-force-d_1780.html. (3) Free Fall Force Calculator Online. calculatorshub.net/physics-calculators/free-fall-force-calculator/. (4) Impact Force Calculator | Calculate Impact Force in Collision .... physicscalc.com/physics/impact-force-calculator/. (5) Impact Energy Calculator | Impact Force. www.omnicalculator.com/physics/impact-energy.
@FluidMatters 3 ай бұрын
ok
@ethanbaldwin1181 3 ай бұрын
Cool video!
@majroi 3 ай бұрын
Greetings from Turkey, this material helped me a lot. Thanks professor.
@ericjung7621 3 ай бұрын
hi professor, why are there no videos for content past chapter 5?
@FluidMatters 3 ай бұрын
That's where the intro fluid mechanics course ends at my university. I don't teach "Fluids II", at least for now.
@Frankchouzpy 3 ай бұрын
Impressive profound experiment and explanation. Additionally, I encountered numerous novel ideas in the comments section here.
@FluidMatters 3 ай бұрын
Thanks for the kind words. A colleague and I ended up writing a paper on this effect.
@Frankchouzpy 3 ай бұрын
These interesting videos make your tutorial more and more attractive. These formulas jump out of the screen and appear in our lives. Thank you for your amazing idea!
@Frankchouzpy 3 ай бұрын
Nice! Chapter 2 completed！！Next is the two ptional videos haha.
@Frankchouzpy 3 ай бұрын
I can’t stop learning, so take a break, click the like button, and then continue!