@42:40, when you use the second bernoulli's equation to solve for P3 (Pstag), would the left hand side be equal to 0(because you corrected the P1 and v1 to equaling zero by placing point 1 outside of the vent). Would there then be two answers for P3?
I was thinking the exact thing. there would have to be two answers, because everything is zero in the second equation when solving for P3 and then in 2 to 3 its equal to all the kinetic energy.
According to my opinion, We cannot apply Bernaulli's equation between point 1or 2 and point 3 because at point 3 there appears the viscosity effect of air. Because Bernaulli equation is applicable only for non viscous fluid. Due to this viscosity effect of air, the air molecules get sticks to the front of the vehicle and its velocity becomes zero at point 3.However we can apply it for viscous fluid also but precaution should be taken that the viscous effect of fluid should not appear between these two points. So we can apply it point 1 and point 2 but we can't apply between point1 and point 3. I have learnt these things from him only. His teaching is so great.
41:51 still don't get it why the P3 is zero if we calculate from equation point 1 to point 3. Seems like Prf. Biddle had a problem at that then he moved to the equation from point 2 to point 3. Can someone explain to me please ?
@1:01:15 , even though its not steady state why do we take LHS = 0 , even in main continuity equation @51.15 why is LHS (time.rate change of B(mass) of sys ) = 0 even before assuming its a steady state . PLEASE , IF ANY KNOWS ANSWER IT.
@@yogeshcgowda2071 Left hand side of eq is how B changes in the system which has a fixed amount of mass. Since we chose B=mass which is always fixed in the system the time rate of change is zero and will always be 0 when we choose the mass as property B.
Didn't understand 3.48 part b. For point 1 to 3 Bernoulli eqn, we are getting Pstag = P3 = 0. But different with Bernoulli eqn from point 2 to point 3. Why?
Sir, what if the area at point 2 from the last question is not triangle, but let say, its a parabola, shall we approach it in the same way as triangle?
I have a doubt professor In the 5th example of control volume can't we solve that question by applying bernaulli's theorem ? if not why ? i will be happy to get ur reply
Dr. Biddle I wanna kiss you on the forehead right now! You're an absolutely wonderful professor. Wish I could study under you, life would be so much more enjoyable.
Man, imagine having a professor that could actually teach. Unfortunately, I don't.
This man is a Godsend.
I cannot agree more.
If only I had a John Biddle... In short, friend ahead. Actually explaining how to approach a problem? This man is a blessing upon my grade and
If only there were more Dr. Biddles in this world... if only.
This helped me a lot on my midterm exam. Thanks!
Great!
Well, z1 can be found using the manometer equation. 48:34
@42:40, when you use the second bernoulli's equation to solve for P3 (Pstag), would the left hand side be equal to 0(because you corrected the P1 and v1 to equaling zero by placing point 1 outside of the vent). Would there then be two answers for P3?
I was thinking the exact thing. there would have to be two answers, because everything is zero in the second equation when solving for P3 and then in 2 to 3 its equal to all the kinetic energy.
According to my opinion, We cannot apply Bernaulli's equation between point 1or 2 and point 3 because at point 3 there appears the viscosity effect of air. Because Bernaulli equation is applicable only for non viscous fluid. Due to this viscosity effect of air, the air molecules get sticks to the front of the vehicle and its velocity becomes zero at point 3.However we can apply it for viscous fluid also but precaution should be taken that the viscous effect of fluid should not appear between these two points. So we can apply it point 1 and point 2 but we can't apply between point1 and point 3.
I have learnt these things from him only. His teaching is so great.
@@RAVIKUMAR-hm7iu thankssss
I do not remember the equations used to solve problems 3.6 and 3.12. Which lecture were those derived in?
Sir , have you taught the equations that are required to solve first two examples in this video.i couldn't find them in lectures.
I think that some lectures were not posted , but those equations are developed in Munson 7th Fluid Mechanics chapter 3
41:51 still don't get it why the P3 is zero if we calculate from equation point 1 to point 3. Seems like Prf. Biddle had a problem at that then he moved to the equation from point 2 to point 3. Can someone explain to me please ?
i guess it's problem 3.46 (not 3.48) in textbook (p.145), that prof. Biddle solves at 25:15.. am i wrong ?
yes
@@locom16deen78 as in yes, it is 3.46. For you chegger's out there, use 3.46.
58:10 why is the dB/dt = 0? isnt it unsteady flow?
@1:01:15 , even though its not steady state why do we take LHS = 0 , even in main continuity equation @51.15 why is LHS (time.rate change of B(mass) of sys ) = 0 even before assuming its a steady state . PLEASE , IF ANY KNOWS ANSWER IT.
LHS is not about steady state. CV part of the equation is related to steady state.
Atakan Okan Thanks
@@yogeshcgowda2071 Left hand side of eq is how B changes in the system which has a fixed amount of mass. Since we chose B=mass which is always fixed in the system the time rate of change is zero and will always be 0 when we choose the mass as property B.
@@ryanredhead1261 Thank you for your time, but I have started my carrier in Computer science 😅 now, and not into mechanical subjects. Thanks 👍🏻
@@ryanredhead1261 omg, u saved my life, thanks so much
Didn't understand 3.48 part b. For point 1 to 3 Bernoulli eqn, we are getting Pstag = P3 = 0. But different with Bernoulli eqn from point 2 to point 3. Why?
for V0
V0= 3*V1*D1² over 4*R0²
At 48:43 . If I use bernoulli's point 1(water surface) to point 3 (Top of fountain) . The γ for the Pa pressure is γ of AIR, right?
also why is this a non steady flow? is that because of the number of outlets? or is it just stated in the problem?
At 39:34 He changed point 1 and also the velocity at point 1. I think I could get the answer without the change so why change the point?
He changed it to just outside the wind tunnel so P1 and V1 can be 0 for the case of it being atmospheric. You would in your case have to solve for P1.
On 1:18:20. Anyone knows why Vr = Vo (1-r/ro)?
It comes from the ratios of the sides in similar triangles. i.e. (Ro-r)/R0 = Vr/Vo, if you look at the flow half way.
Tshepo Molale Is this always the case with a non steady linear flow? This is what is tripping me up
1hr 7mins into this video(the bathtub example) sums up this video beautifully lol
in the wind tunnel problem. what is the value of gamma oil ? also, the answer h?
gamma oil is SG oil * gamma water since SG oil was given
Sir, what if the area at point 2 from the last question is not triangle, but let say, its a parabola, shall we approach it in the same way as triangle?
Im confused as to why he sets R=6-n for the force in the n direction Euler equation. Surely from the diagram we can see R is 5?
For anyone in the future, he is taking it from the known point (Point 1) where R=6 as it is where he already knows the pressure.
Did sir cover chap 6 differential analysis of fluid flow?
For problem 3.36 in this video, is point 2 considered to be at atmospheric pressure?
Yes, because if it's not atmospheric then air would not flow vertically straight
Do mini windtunnels for classrooms exist? 🤔 If not I bet they would be pretty easy to build, using clear, see through plastic
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At 1:05:27 the subtitles and the video start to get out of sync
+Josh Pham Thanks for letting us know. We will work on correcting this issue.
Problem 3.36 Will the pressure at point 2, P 2 = 0 ?
Yes. It is atmospheric, as well as the pressure at the nozzle exit.
@38:58, the bernoulli eq used to find h and p2, the h doesnt go all the way to the top of the tube. then why are we just using h for it?
Because the weight of the air is negligible.
bc the density of air
what is name this dear teacher ?
Guys i need some application of continuty equation
thank you
You're welcome.
I wish Mr. Biddle was my prof. wow.
I have a doubt professor
In the 5th example of control volume can't we solve that question by applying bernaulli's theorem ?
if not why ?
i will be happy to get ur reply
Awesome. thanks. ;)
not bad
Dr. Biddle I wanna kiss you on the forehead right now! You're an absolutely wonderful professor. Wish I could study under you, life would be so much more enjoyable.
this was a pretty bad video because professor did not completely solve the first two and left with doubts with the remaining questions
45:19 hehehehehe
not bad , not good