Physics 34 Fluid Dynamics (4 of 7) Bernoulli's Equation
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- Опубликовано: 14 окт 2024
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In this video I will show you how to use Bernoulli's equation to find the velocity of water draining out of a tank 2.4m in height.
Next video in this series can be seen at:
• Physics 34 Fluid Dyna...
I struggled with these exact problems in my lectures notes for a long time when I studied Continuum Mechanics and those lecture notes always made things look really confusing and messy, and then professor Biezen explained them in these videos and then they got easy all of a sudden.
Incredible.
I thought physics are very difficult, but after his explanation, seems everything so easy and simple. Thanks a lot!
❤
Bill,
The size is not important and does not change the speed of the flow. The reason we pick a small pipe so that the velocity of the water at the top of the tank is negligible
+Michel van Biezen If one would compare two scenarios, using the same water tank of 2.4m in height but given that the drain hole in the bottom is now twice as large. Would the velocity still be the same?
IF velocity are the same, does that mean that the water will make the same parabola and touch the ground at the same distance from the tank?
The time it would take to drain the tank must have some dependency on the size of the hole.
I find it confusing trying to get an intuition of velocity and flow I guess.
Will nozzle help increasing the velocity V2? how?
U nailed this video great job
@@ericohman the speed actually is : V = (2gh/1-{A/a}^2) ^1/2
In which "A" is the area of the big hole and "a" the área of the smaller hole where water flows trough.
So the speed is different, but when "a" veery small compared with "A" it will make virtually no diference.
When A>>>a then {A/a} ^2 will be very close to zero , and 1- (~0) =1. So the velocity will be just (2gh)^1/2
This equation called " Torricelli's Theorem"
Torricelli's theorem, also called Torricelli's law, Torricelli's principle, or Torricelli's equation, statement that the speed, v, of a liquid flowing under the force of gravity out of an opening in a tank is proportional jointly to the square root of the vertical distance, h, between the liquid surface and the centre
If the tank is closed, then the pressure above the water (P1) may be different from atmospheric pressure. In order to solve that problem P1 must be given.
Don't confuse yourself brother, if its gravity feed system only height matters everything else cancels out.
Thanks, that's was I going to ask
I was struggling with this since a few days,but you are an amazing teacher...Wish I had teachers like you so I'd never missed class😅... Thank you so much
I loved your video professor.
Basically all the initial gravitational potential energy was transformed in kinetic energy because we've ignored the friction and other kind of resistive forces.
My favourite Lecturer😍👏. You Helped me so much Sir on my exams. Thank you very much.
Glad to hear that 🙂
Just wanted to comment and commend the professor for being so very intelligent as well as having such a great grasp of these concepts that he can explain it to people through youtube. Thank you kind sir.
Thank you thank you thank you. Your videos get me through college and I am so grateful.
Glad we were able to help.
I have always said if you can relate math to a real world application it becomes easier to understand and to truly understand the art of math is to teach it. Mr. Biezen you make it easy by relating an analytical function to a real world tactile application. Thank you.
This is amazing. You, sir, added another dimension in accessing physical problem solutions quickly and with amazing explanation. I highly value that.
Can you please help me with this question ? A water storage tank is filled to a height of 16 ft. (a) With what speed will water come out of a valve at the bottom of the tank if friction is negligible? (b) To what height will this water rise if the opening is directed upward? (c) What quantity of water will emerge from the tank in each second? The area of the valve is 1/3 in^2
Hello sir, really appreciate ur efforts, tq so much. I have an question regarding the P2 , final pressure which i think should be the sum of the Patm + Pwater = 1 atm + (1000)(9.81)(2.4).... i guess... pls help. tq sir
It depends where you want to measure P2. If it is just at the exit of the small pipe, it will only be atmospheric pressure. At the bottom of the tank it will be P atm + density x g x h. Inside the pipe it will be between the two.
This was exactly what was on my quiz. Thank you for the great explanation!
me too LOL
was it quiz 9?
Heron fountain
Siphon
Inverted siphon
Bell siphon
All are in a series so how much longer pipe required to raise the temperature for boiling under it's own energy
As the sun goes down in next few years so we need some energy by atmospheric pressure ?
Good one💯..What if opening at point 2 is larger ?? Then what should be velocity at point 1?? Please reply soon
If the opening at 2 is significantly larger then you must use the equation: A1v1 = A2v2
@@MichelvanBiezen But initially we don't know V1 and V2 ..so should we have to use discharge equation i.e Q=AV (at point 2) But now to know Discharge is the question..So I think we need to practically calculate discharge and do further calculations..am I right please reply??OR is der any other approach so that we don't need to calculate discharge practically??
i have watch a lot your video all thought it is not necessary for me but i like the way you explain . so thank you very much for your explain
I have a question. Why is the pressure at point 2 equal to atmospheric pressure? Shouldn't it be the pressure from the weight of water above it minus the atmospheric pressure?
The point 2 is just outside the influence of the pressure due to the height of the column of water and therefore is equal to atmospheric pressure.
Prof with you I don't need text book you make physics easy for me pls keep good work from Vincent your student may you send me more on angular momentum moment of inertia and work done during this motion eg w=TQ
We have many videos on moment of inertia. Did you see them? 🙂
This what I expected, I watched many but it's awesome.
Sir Please provide video for mc cabe thiele method please
Michel van Biezen, Thank You for your explanations. I wish I could comprehend more of the math, but I am working on it.
At V2, shouldn't the pressure be equal to hydrostatic head pressure, meaning that H2 would be 2.4m instead of H1 being 2.4m?
Now listen carefully , if the orfice ( the opening ) of the tanks radius is very very small relative to the upper face of tank then we can't apply velocity of efflux .
I am so sorry but I am really confused, in Bernoulli's equation P is supposed to be fluid pressure at a certain point why do we substitute here by the atomshperic pressures which is the pressure exerted on point 1 or 2?.Thanks in advance
No need to apologize. It is a good question. Once the fluid comes out of the tube (just outside the opening), it is no longer subject to the force created by the weight of the fluid above it and the internal (gauge) pressure quickly drops to zero, so that the pressure at that point is only the atmospheric pressure.
But at the same heights, aren't the pressures the same? at the same height at 2 inside the tank, the pressure will be different, say Patm + h(rho)g, why isn't this the pressure at point 2 too? please explain.
Based on the given condition, the water on top of the tank is only subjected to atmospheric pressure
Dear Sir,
I have a problem which is kind of similar to this one but I have no idea how to solve it.
I have a tank ( of 8 Bar) where 1000 Liter/min air is pumped in and I have a slip of 10mmx1meter which leads the air out to an environment of 1 Bar.
How many liters per min of air it getting out?
THANK you for the help !
Hi Michel, Why pressure at the bottom is 1 ATM + 0 instead of 1atm + row.g.h. due to the depth and 1 atm+ 0 On top
thanks
At the bottom of the tank you would include the rho g h term, but not just outside the hole.
thank you
question - you can't define V1 as zero because of A1V1=A2V2. If V1 = zero, then wouldn't V2 also equal zero (assuming everything else you said was constant)? So shouldn't you plug in A1V1 relationship into Bernouillis?
You would be correct if the size of the hole is significant compared to the diameter of the container. In this example the hole is so small that we can ignore the velocity at the top. (Remember that the velocity term in Bernoulli's equation is (1/2) (density) (v^2) the velocity is squared, so a very small velocity squared is nearly zero)
From the Philippines, thank you sir for this explanation, specially in this time of pandemic that online classes are not enough for our profs to explain.
Welcome to the channel!
such a good channel to learn Unit operations
Isn't this assuming the overall tank volume doesn't really change ...becuase as the water empties, the height changes, therefore the water exiting would slow down correct? I feel like this was assumed in the process.
You are correct. The assumption is that it is a very large tank and the level does not drop in any appreciable manner.
@@MichelvanBiezen thank you. Vidoes are very accurate and helpful
What if you have the opposite situation, I mean what if you have an empy tank closed, and a hole into the water(opened space). More specific, what if we have a compartiment ship(watertight), and a hole into it, how would you approach this matter. Really looking forward for your answear.
Essentially, you would have the same problem in reverse.
@@MichelvanBiezen , a hole of 0.5m (2) would let the same amount of water as one hole of 1 m(2).... this is my question how would the diameter of the hole would affect the income water in the equation. Thank you for your time here answering my silly questions.
A bigger hole will let more water through. Remember that dV/dt = v A
Thank you for the amazing videos.
At point V2, shouldn't the pressure equal hydrostatic head pressure which should be taking into account the 2.4m at point H2?
Depends on where you consider point 2 to be. There will be a gradual change from the hydrostatic pressure to zero gauge pressure as you move from left to right.
@@MichelvanBiezen Pascal's principle suggests otherwise. Pressure is equal at points on the same elevation. I don't think it's true to say there will be a gradual pressure change as you move from left to right
Would the velocity increase if some one jumped in the tank?
please sir for V1, you assumed that it was zero. But when water moves from away from point 2, the water level at point 2 will reduce at a constant velocity simultaneously, so why did you assume v is 0, if it were then the water level wont reduce. please I don't understand
You are correct in assuming that the level of the water in the tank will slowly decrease. But since the velocity of the lowering water level is very small and the term associated with that in Bernoulli' s equation squares that velocity, a small number squared is a very tiny number which can be ignored.
what if the opening was bigger what would have happened meaning the velocity would have not being zero or what i need your reply over that
i am enjoying your teacher so much but help me with my question
thank you...............................................
The assumption with this problem is that the diameter at the top is large and the hole is small such that the water level drop is insignificant. We ignore viscosity, turbulence, size of the hole, and other factors that would change the actual outcome in a realistic scenario. This example just illustrates the basic principle of Bernoulli. Since the velocity is squared, unless the water level dropped at a significant enough rate it wouldn't make any significant difference.
I've watched several videos on problems very similar to this and I still don't understand why the Pressure of the left and right sides of the equation are equal. The water velocity is faster for the smaller area, and velocity causes a drop in pressure, so shouldn't the pressure on the right side of the equation be lower?
It depends on where you take the pressure measurement. The pressure at the bottome of the tank will be p = density x g x h + Patm The pressure inside the tube will be less, and the pressure just outside the tube will be atmospheric pressure.
@@MichelvanBiezen Thank you, I understand now, I was misunderstanding the inside/outside of the tube thing. It's helpful to understand the bottom of the tank pressure being different than the pressure outside the tube.
Found myself going through a lot of videos, but this one did the trick! Thank you!
Glad it helped!
What if you have 2 or 3 incoming and varying flow rates going to different pipes and meeting at one point?
1. A1V1
2. A2V2
3. A3V3
Is the total output flow rate (total AV) equal to = A1V1 + A2V2 + A3V3?
Yes, because the sum of the volumes per unit time flowing through each of the smaller pipes equals the volume per unit time flowing through the big pipe.
This is also a proof for Torricelli's law, using the Bernoulli equation, right? Thanks for the clear explanation!
Yes, it is.
Hi sir, why is the pressure at point 2 the same as point 1 at atmosphere pressure? Should it not be equal to P = pgh due to the depth of the fluid? Sorry if this is a basic question I couldn't understand that part...
At the point of exit, the only pressure there is atmospheric pressure. Inside the tank (at the bottom) you must indeed include the P = pgh term.
@@MichelvanBiezen Thank u! So if the hole was punched at the bottom of the container, not entering out of a spigot, then you would include the P = pgh. Thanks!
let’s say the tank is sealed and you let the water out the pipe and no air could go back in creative a negative pressure inside the tank. how would you calculate that?
You would have to set up an equation that would calculate the pressure in the top part of the tank with respect to the level of the water at the top and then set up Bernoulli's equation with a varying pressure difference between the top and the bottom.
Sir please tell, how much time will it take to drain all water from that tank? Because pressure will change continuously as the water will start coming out from that tank. All the values will change every second. Please reply
You can find your answer in this playlist: CALCULUS 1 CH 6.5 RELATED RATES
could you please do the same question with an extra hole on the other side of the tank and little higher, please?
Hello Professor, what is the relation of cross sectional area of the bottom opening with the equation. As we know the velocity is inversely proportional to the area.
We assume that the size of the hole is very small compared to the opening at the top. But if that is not the case, then you have to take the speed of the drop in the water level at the top into account. (Also keep in mind that these are the basic principles, the actual speed is affected by a number of other factors, such as the viscosity of the liquid, the friction on the sureface, etc. )
I once read somewhere that if you were to hook a hose up to that lower spout where the water comes out... and then you let the water flow through the hose, then you put your thumb over the end of the hose and try to get the water to squirt farther..... that in fact it would NOT go farther..... Unlike if you put your thumb over a hose connected to your house's Water spigot.... where the water does form a fast stream and goes further.... I have yet to figure out WHY that's the case... Something to do with the city's water supply PRESSURE versus the Atmospheric Pressure of the Open Container in your video.... Hmmmmmmmmmmmmm..... Nice video series Michel.... :D thanks again..
well, I found out that Flowing water in a system is affected BIG TIME by internal FRICTION. The faster the water flows the greater the resistance.. and that lowers pressure.. that is the force that is fighting the internal friction.. WOW... now I have to go check out your videos on Viscosity and other factors........ This quest for knowledge is something else.. :D
Hi Philip. This set of videos explains that in detail: FLUID DYNAMICS 1 HEAD LOSS & RESISTIVE FORCES
what if the tank is small that the cross sectional area of the outlet is big compared to the cross sectional area of the tank, and you cannot assume that velocity at the surface is negligible, is there a way to solve this?
then you need to add the velocity term for the top surface in the equation. (1/2) density x velocity ^2
Hello. I have one question if you can help me with. Is anything changing if we have a sealed body under constant pressure? Am i thinking right that the only difference will be the starting pressure P1 have to be the pressure of the pump that is keeping the body under pressure. And second question, probably more complicated, if this is a sealed body under pressure but without constant pressure from outside (gas bottle for example). How can i determine the outgoing speed and the total time for emptying the body. I guess the pressure will fall over time and the speed will fall overtime. Thanks in advanced.
Hello sir .
Can we use the following formula to know the velocity of the water flow from Nozzle
V = Q / A WHEN Q IS Measured in cubic meters
The area of the nozzle is measured in square meters
I measured the speed of the water
Through these equations the speed was exactly the same
Yes, that is the equation.
2:29 Isn't the height of h1 2.4m + diameter of the tube. Since the reference line is at the bottom of the tube.
We are assuming that the diameter of the tube is insignificant compared to 2.4 m
@@MichelvanBiezen Thank you
I almost died when my teacher explained our class about this equation but thank goodness your video saved us!
what about friction or buoyancy force?, I mean it gives the same value if we worked on the conservation of mechanical energy, but mechanical energy is not conserved here, or is it conserved?
We are ignoring friction. Buoyancy force does not play a factor here.
thank u
Thank you Sir. It's definitely your teaching that got me into med school. Thanks a lot.
Glad to hear it. Congratulations!
i have a bit of an exam tomorrow and honestly this is exactly what i needed
Good luck on your exam
This is called velocity of efflux of fluid. I.eV2
What if the overhead tank is above ground level? Say 35 mtrs..do we consider height of water plus height of the tank above ground?
Only the top of the water level in reference to the height of the hole matters
This velocity will be the same for different dia of opening at the bottom?
In a real world application, no. But in a simplistic representation of the Bernoulli equation, yes.
I am also wondering about the Pipe Diameter, and the effects on the velocity? If a 10" inch Pipe is P2 vs a 6"inch Pipe as P2.....ect, ect to 11/2" Pipe.
I mean obviously,... A smaller diameter pipe would give a Higher velocity,.... While a Larger pipe would give more GPM at a slower fluid velocity.
But which equations should I use to figure the various Flow Rates?
Thank You Joe D.
This problem has ignored the effects of pipe diameter, viscosity, pipe friction, etc.
What if the height of the block the container is on was provided, how come that is not taken into account since it raised above the ground, do we just assume our container is on ground surface?
Just like with potential energy, we can choose the zero height wherever we want.
Michel van Biezen Thank you sir
Could you just add height plus atm together to get same answer. So 2.4 + 4.48 = 6.88 or 22.57 in feet?
The change in height relative to the atmospheric pressure is negligible. That is why we ignore it.
why is the diameter of the exit pipe not a factor in the equation? Doesn't the exit diameter determine the exit velocity?
As long as it is small compared to the diameter of the tank, and we ignore viscosity and friction, it doesn't matter. (This is a simplified version of an otherwise much more complicated problem).
What should be the value of P(density) if there are two immiscible liquids of different densities and what would be the value of h to be taken then?
The term rho*g*h has units of presser. You have to calculate the pressure by adding the pressure contribution of each layer.
Hi Michel van Biezen, A question please.... I operate a pump test tank, Total Height to P1 is 47" (inches) while the Reference Line of P2 is 11" (inches).
Do I need to convert the h1 variable from 36" to 914mm? As in the last step,.... Square Root of 2(9.8m/s Squared) the Gravity is written in S.I. Units.
It is always a good idea to keep all the units in the standard SI units., unless all the units are given in cgs or in Imperial units.
defs gonna donate money to this guy when i get a job
Excellent explanation but I have one question; does the v2 stays the same after some of the water is already poured outside the tank? I mean the hydrostatic pressure plays a role there.
+Erblin Halabaku
As the water level drops in the tank, v2 will diminish proportionally.
saved me on this fluids hw! thank you!!
Glad you found our videos! 🙂
Cancelling rho, is this mean that the example apply to all kind of fluids? That we'll get same V2 for all fluids
Yes, that is correct. (Although we are ignoring head loss, internal friction, etc. )
@@MichelvanBiezen Understand! Thank you Michel!
Will the equation at the end be applicable for all questions of this type?
As long as the conditions are the same. It is better to start with the original equation and work it down to the particular application.
@@MichelvanBiezen thank you
hi we have a water tank rooftop gravity fed 5 feet high 10 feet long and 5 feet wide how much pressure it will create at buttom of the tank with 1 inch pipe
sir what would happen if there are 2 immiscible liquids and 2 holes are there in problem
the best explanation ever heard of, thanks a lot sir
in hydroelectric power station, to find the potential energy, the solution need me to use average height. In this case, it does not use average height. When should we use height and average height?
To find the total energy in the reservoir, taking the average height is correct. But typically the inlet is near the top of the water level and hydroelectric production tends to stop when the water level drops too low.
Dear Sir....May I ask something about Pressure because I am confusing. Let say .. A 5' tall water tank is at high of 15' from the ground. From the lowest side of it to the ground, I set up 2" PVC pipe. what is water velocity at the end side on the ground? Head pressure ( about 8.66 PSI ) can be assumed as Velocity? Please May I know answers or any of your video link that have explained about this ............. with regards..
+Aung Toe
The video here that you are looking at is the exact example that should help you solve this problem. Without a diagram, I cannot be sure if there are any other conditions or restrictions we should be aware of. Note that the reduced pressure in the pipe indicates a higher velocity.
Thanks a lot Sir +Michel van Biezen :) ... Your explanation is very perfect.
Sorry for my little of background knowledge and because of it I miss to catch that point. I will replay and learn it again. Thank you very much........
hi sir.. i love how you explain the formula.... but what if the tank is presurize.... is the result the same..?
If the tank is pressurized then you have to add that to the pressure at point 1.
If the outlet pipe dia is increased,will pressure increases?
In this simplified example, the diameter of the outlet pipe is not relevant as long as it is much smaller than the diameter of the tank. Otherwise you need to refer to the other examples (internal to the pipe) to see how the change in diamter changes the pressure.
At the exit point of the tank, in addition to the atmospheric pressure, shouldn't there be an additional pressure due to the water column?
Inside the tube yes, but not at the exit point.
Good day Sir, may I ask why the velocity is the same? Because from my understanding, the tube will increase the height used in the equation when it is at the bottom , is this correct?
If the tube is very large, then yes, but here we ignore the diameter of the tube in relation to the height of the tank.
Oh I see, thank you so much.
v2 = 6.86m/?.
Hi, what is the unit of '?'. I could not see it clearly. I assume it to be 's' or 'min'. Please help.
m/s (I usually write sec instead of s)
@@MichelvanBiezen Thank you
thank you so much sir, your videos wil help me pass my final. keep up the great work
Thank you sir your lectures are amazing
Thank you. Glad you found our videos. 🙂
why would they have both the same pressure(atmospheric) when both points have different altitudes. I thought pressure changes with altitude
That is a good observation, although the difference would be insignificant. (The same reasoning why we call the velocity of the liquid zero at point 1, which is also not strictly true.)
Well I have a dout. The or pressure at the small opening. Must be equal to the atmospheric pressure + rhogh . But why we have take atmospheric pressure here? Do we take the pressure opposite to the velocity vector of fluid at that particular point?
Just outside the small hole, the pressure will only be atmospheric pressure.
I have a question, why is P1 all the way on top of the liquid?, why isnt P1 at the bottom right corner, right before the exit spout?
+J Ferro The location of P1 is arbitrary. You can put it anywhere you like. But if you put P1 at the top of the liquid, you will know the value of P1.
One question sir, why we have water tanks at height when we are considering pressure as atmospheric at both ends.. We want pressure or velocity at the end.. Will wait for your reply.. Thanks
The difference in atmospheric pressure is typically small compared to the difference in pressure to the difference in water level. Therefore typically it is ignored. If necessary the difference in atmospheric pressure can be added.
Velocity formula is independent of density. If I replace water with other fluid(oil) , velocity will change or not.And in reality is it possible velocity not change?
This just explains Bernoulli's principle, without viscosity, nor pipe friction, etc, the way we initially learn projectile motion without wind resistance. See the videos with viscosity and pipe friction: PHYSICS 34.1 BERNOULLI'S EQUATION & FLOW IN PIPES
I think rather than neglecting v1, you should equate it using flux. VA=VA
That excellent. Thank you sir. But I have a problem : when we use pump at top of the tank to uptake the water, at this case for work done we use the centre of mass of water (as height) .but why we are not use c.g. of water in this case? Please reply me.
The pressure at the bottom of a water tank (or any vessel containing liquid) = density x acceleration due to gravity x depth. So you can see that it only depends on the depth of the liquid.
you're a life saver!
But at the same heights, aren't the pressures the same? at the same height at 2 inside the tank, the pressure will be different, say Patm + h(rho)g, why isn't this the pressure at point 2 too? please explain.
The difference in pressure due to the height of the water is accounted for in the equation. The difference in the air pressure is so small that it is ignored.
Hi, I'm studied electrical but my scoop of work contains some of this stuff i see you did not consider the area of the piping would that not make a difference?.
Only through physics we can see this is PE=KE. I like how universal physics is
Yes, physics is the foundation to a lot of science.
How is point 2 at atmospheric pressure what about all the weight of the water above it?
the point 2 is just outside the hole at the bottom. (No pressure due to water above)
@@MichelvanBiezen oh !
Thanks Prof.
sir i mconfused in one thing . if the point 2 where the water comes out we take pipe instead of opening water. the formula of velocity is same for pipe also or not.. plz sir correct me AS EARLY AS POSSIBLE.IM WAITING UR REPLY SIR... THANKS
At the end of the pipe, the velocity would be the same (ignoring the effects of fluid flow in the pipe and the viscosity of water).
But sir if i want to determine the velocity in the pipe..what will be the formula..
This formula v=(2gh)^1÷2
I think this is used to determine the velocity at orifice or at outlet of the opening.
How can we determine velocity inside the pipe when a tank is connected with a pipe and the fluid is flowing inside the pipe.
How to calculate time taken to discharge water completely from a reservoir
Those types of examples can be found in this playlist: CALCULUS 1 CH 6.5 RELATED RATES
this is so helpful, u make physics easier sir, thanks! all the love :)
This video really help me to solve my hydraulic task from my lecture..thank you sir for your excellent explanation
Glad it helped
What happens if I put an other outlet pipe with the same diameter at the same height? Will the velocity be the same on both outlet pipes as it was with one outlet? Can we apply Bernoullis law in this case?
As long as the diameters of the pipe are small compared to the size of the tank, it will be the same.
Thank you for the quick answer! Then if this is the case there is one more question I have.
The domestic water system's pressure is based on the same principle. We have a water tower (it is the tank in your example) and the height of the water level is supplying the system with its hydrostatic pressure. Now let's say that we connect a shower head to the system (the opening in your example). The shower head has several openings at about the same height. If I hold with my hands half of the nozzles, from the remaining half the water will flow faster. This would mean that the velocity is not the same in case of 1, 2, or more openings. Even though the diameter of the nozzles are much smaller then the diameter of the water tower. What is the explanation for this?
So far we have been ignoring things such as viscosity of fluids in pipes, differences in diameters, etc. Depending on the situation, these can make a profound effect how water flows through pipes. Especially when you talk about a city's water system, house pipes, shower heads etc. Those videos are planned for the future when we have time.
this was very helpful, a lot better than the dumb prof i got
Sir michel van biezen, is pressure in the bernouli equation always take the absolute pressure not gauge pressure right?
Yes, it should be the absolute pressure, however it can also be solved if you use gauge pressure on both sides (the atmospheric pressure will cancel out on both sides).
@@MichelvanBiezen thanks sir.. Appreciate that
Very nice explanation
u r great, wish i can find people can explain things like u in my subjects, thank u so much