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Hi, when finding F_bc, since it is a vertical force, wouldn't the formula be F_bc = pgV ?

❤ beautiful

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?

This channel is helpful for my research project.

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.

Hello Professor, may i know why F(AB) is not considered in the moment ?

This video is absolutely GOLD for someone like me struggling with fluid mechanics. Glad to find this on TH-cam!!

how is there a moment because of Fbc? not sure how that causes rotation

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!

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?

Does this mean the Bernoulli Equation / Venturi effect can be applied to non ideal fluids as well? (approximately of course)

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.

Great video and explanation

Is it the no slip condition that causes the velocity gradient, professor or the internal shear stresses?

when you have losses, you add them to get to EGL?

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

Thank you sir.

THIS VIDEO IS SO GOOD

All your videos helped a lot. Thank you so much!

I like this series

Is that a plaquette on a hinge?

🌹🌹

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?

great stuff

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

He saved my final exam😭😭😭

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?

best instructor

thank you you are a great teacher

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.

thank you for your great explanation

from 3:40 on, why didn't we use the fluid for W and the wood for Fb?

I came just to check for a concept, then proceed to finish the whole series.

what da dog doing

Where is the vertical force acting on the top of the free body in no 2 prblm 7:55

Continue the videos sir ..... following your videos ... ..waiting to learn from you sir.....

Just wanted to extend my thanks for putting together a great TH-cam 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 🙂

Why isn't d^2u/dy^2 not 0?

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

THANKS A LOT SIR!!!!

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.

Cool video!

Greetings from Turkey, this material helped me a lot. Thanks professor.

hi professor, why are there no videos for content past chapter 5?

Impressive profound experiment and explanation. Additionally, I encountered numerous novel ideas in the comments section here.

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!

Nice! Chapter 2 completed！！Next is the two ptional videos haha.

I can’t stop learning, so take a break, click the like button, and then continue!

why do I have a course called wave physics and fluid mechanics? I think both are hard enough on their own 🥲

Sir where can i get its theory part...