Draining a Tank of Fluid | First Order Differential Equation example | Fluid Mechanics

[@jamielinsdell4900]

Great video demonstrating the predictive power of mathematical modelling. An outstanding result! Cheers

[@ciaranmcevoy9875]

Thanks Jamie! Yeah, in all honesty, I couldn't get over how close the prediction was to the actual result. I didn't really have any feel for what to expect, so it was brilliant to see.

[@shauryajeevagan5013]

Simply amazing! High quality and high effort video with a great explanation, thanks!

[@ciaranmcevoy9875]

Thanks pal! Really appreciate the kind words.

[@esaumadzura]

Fantastic illustration!

[@ramanandah.s.4116]

Beautiful explanation. I was expecting the same. Thank you Ciaran

[@ciaranmcevoy9875]

Thank you Ramananda. Glad you enjoyed it. 👍

[@barathykirupakaran9725]

It's an amazing explanation with a clear demonstration. Thank you!

[@ciaranmcevoy9875]

That's so good to hear. Glad you enjoyed the video.

[@ytbasketball101]

I love example you can experiment on in the real world thank you so much! 😁

[@ciaranmcevoy9875]

Thanks Brandon! Glad you liked the video.

[@ramanandah.s.4116]

Thank you . Happy new year in advance. Marry Christmas.

[@ciaranmcevoy9875]

And the same to you Ramananda. Have a brilliant new year! 🍻

[@ramanandah.s.4116]

Can you please share similar videos of mathematical modeling of real-life problems?

[@ciaranmcevoy9875]

Absolutely! I will be bringing more videos out similar to this in the new year. Cheers for your interest pal 🍻

[@rananagy6754]

A great fantastic video ❤ Thank you ❤❤❤❤

[@ciaranmcevoy9875]

Thank you very much Rana.

[@peasant8246]

Finally somebody did an experiment to verify the results of calculation. Everyone else on YT is too lazy!

[@kimbalcalkins6903]

it would be interesting to add a few drops of liquid detergent to eliminate surface tension?

[@jamielinsdell4900]

Would be great if you could demonstrate other principles such as Newton's Law for Cooling etc with a similar approach. Calculus in action!

[@ciaranmcevoy9875]

Yeah! It's great to see principles tested out like that, isn't it. More videos like this are definitely in the pipeline! Just so you know, I post the majority of my updates on what's coming up next on Instagram and Facebook, @ciaranwamcevoy. Feel free to follow me to stay up to date.

[@lucaslaue2000]

Seeing as the bottle will never truly become empty, does that mean that the final height being 0 is relative to the position of the hole? If so does that mean the initial height is the "true" height - the height below the hole?

[@ciaranmcevoy9875]

Hey Lucas. Both your statements are correct, however your desire to involve the "true" height is not necessary. In the majority of fluid dynamics problems you tend to set a datum from which you want the height to be measured from, this is completely up to us. So, for ease, I set the height at which the hole is to be zero, because, liquid below that hole is not of our concern since the flow will stop. So in other words, what you where imagining was setting the datum height to be at the bottom of the bottle. Hope this makes sense 🍻

[@lucaslaue2000]

@@ciaranmcevoy9875 Absolutely! I have been trying to do this experiment myself, and the differential equation I apply does not match up with my recorded data. I noticed the difference between your experiment and mine is scale. The size of my jet hole is 3mm, the total volumetric capacity is closer to 5L. I was wondering if perhaps at that small scale (jet hole) other factors such as water tension play a bigger role, skewing my findings?

[@ciaranmcevoy9875]

Yeah there's always going to be differences in experiment set up etc. But the key thing I try and do is to constrain variables. For instance is your container the same cross sectional area all the way down? This was a key part of the experiment and equations (it's why my jet hole wasn't right at the bottom of the bottle). If it isn't then we'd have to include the rate at which the cross sectional area (CSA) would be changing as well as the height. The velocity of the the liquid out of the jet hole in my example comes from the fact that the CSA of the jet is much smaller than the CSA of the vessel. It needs to be to assume the velocity of sqrt(2*g*h). Another thing to consider is achieving laminar flow, of there's any turbulence at the jet then that'll change the results too I'd say. It's great to hear you having a go. I'd suggest trying to repeat my results and experiment first to see if you get the same result. Then move onto other sizes of vessels etc. I love to hear how you get on 👍👍

[@JulianLewis-q7e]

How would you go about finding the t(1/2) the amount of time that passess till the bottle is emptied to half its original level?

[@rio_agustian_]

Cool explanation, Sir! I have I little bit confusion. The equation h(t) = (√(h0) - 1/2kt)² h(t) = h0 - √(h0)kt + 1/4 k²t². Your equation is good at some interval times, the equation shows that the h will decrease when the time increases. But, the h value blows up since it's quadratic equation. In the real world, the water can't go back to the bottle to increase the h. Could you explain to me why this is happening? Or, am I missing something about the equation? Thank you, amazing content 🤓

[@kimbalcalkins6903]

is it correct that if the tank drains fully in say 300 seconds that it will be 3/4 drained in 150 seconds? This could be tested by marking the cylinder at 1/4 full and see if it drains to that mark in 27.5 seconds ?

[@ashafaghi]

How do you model a spherical shape reservoir?

[@jamielinsdell4900]

Cool. For inspiration check out this little gem I think from the 60s - "Angles: Real life applications of trigonometry classroom videos". Skip to 8:00. The guy is very confident with his calculations. Not sure if it's a real world application but you never know what life can throw at you!

[@ciaranmcevoy9875]

Haha that is a little gem, what a classic math video!

[@soninugroho6299]

Nice experiment and explanation, I want to ask, how if we isolated the top of the bottle, will the water still come out from the jet hole?

[@ciaranmcevoy9875]

Ooo nice thought! That would be the same as if you were to cover the top of a drinking straw and lift it from the liquid. After capping the top of the bottle, there'd be a little dribble out of the jet hole, as the air above the liquid expanded, to ensure that the pressure at the jet hole level was at atmospheric pressure both inside and outside the bottle.

[@soninugroho6299]

@@ciaranmcevoy9875 Thank you for the explanation, could you please give me the solution for equation to calculate time needed to stop draining in the case we isolate the top of the bottle. Thank you Ciaran.

[@anilkumarsharma8901]

Why it's become more advance by negative calculation of Whirlpool🌀 caused by this situation so we found out that unconditional situation of universe🌌🌌 according to which all the natural happening become phenomenal success🏆💪

[@PTNLemay]

Every time I see one of these problems, the formula we end up with is Flow rate over time, either as escape velocity of the water or volume flow over time. But it's never given as a height over time. It feels like this would be a simple alteration to the formula, bu I can't figure it out. The best I can do is isolate h in the formula and I get a height that's dependent on the flow rate. Not on time. What kind of manipulation do I need to do to find a simple h(t) function for a problem like this.

[@franciscopen1681]

Well height is also on the volume' formula

[@jishnup1205]

❤❤❤

[@lucaslaue2000]

Hello Ciaran, I am sorry to continuously bother you. I have been performing this investigation on my own however the predicted time for the experiment to cease simply does not match my findings. In my attempt at modelling the situation, I have used a cylindrical pipe, 10.5cm in diameter. In addition, I made an incision (the jet hole) which is 0.5cm in diameter (5mm). This incision is placed approximately 2cm above the base of the pipe. I fill my cylinder up with water, an equivalence of 5000 grams. Through density conversion, I know that this is the same as 5000mL. By applying basic maths I deduce that the initial height is 57.7cm Therefore the height difference between the crest of the water, and the top of the hole is 57.7 - 2, approximately 55.2 cm. My recorded time to have 0 height is 3 minutes and 43 seconds (223 seconds) However, the equation tells me "t=1463.24" If it is not too much of a bother, could you help me understand why I seem to be so far of the mark? I would appreciate any help a lot!

[@ciaranmcevoy9875]

Hey Lucas! No need to apologise. It's really not a bother at all. I'm interested in your findings. It's the one thing I didn't do is try this with different vessels. So my immediate thought is gave you checked your units? I'd recommend converting to metres since gravitational acceleration (g) is in m/s^2 or converting g to the other units. I did this and typed it in on me phone and got 147.9 seconds. It's of the same magnitude but still quite off your results. Is the pipe clear or opaque? Because the next thing I'd suggest is double checking your measurements specifically the initial height above the hole. Also, though I'm sure you have, be sure to measure the inner diameter of your pipe as the bottle's diameter in the equation. Also I timed the water to reach just above the hole, which is also where I measured my height from. This is because the flow gets that slow that water tension becomes a factor. Hope this helps pal. Let me know how you get on. 🍻

[@lucaslaue2000]

@@ciaranmcevoy9875 Hey Ciaran! Thank you so much for taking the time to help me. Yes as you determined the equation does tell me 147.9 seconds (it certainly does help using correct units) The pipe I used is clear, therefore I can and did measure the initial height. What I did was measure the initial height above the base, and subtracted 2 (the approximate height between the hole and the base). I could assume that I was inaccurate, but that would leave me with at most + or - 2cm (let us say this to be strict) And yes, I did measure internal diameter. Even with the uncertainty of the height between the top of the hole, and the initial height (plus or minus 2). I don't see a way of getting from 3 minutes and 43, to 2 minutes and 23.9 seconds. I will perform my experiment again, and I will simply stop at 147.9 seconds, and from there determine how much height is left over between the hole and the water crest. Ill let you know what I got! Thank you again!

[@ProcesswithPat]

I came to double check my calculations XD I am actually pleasantly surprised that the equation works as well as it does in this case. I suspect that it is because the vessel has a thin wall. I have conducted similar experiments from the perspective of Process Engineering - go check it out: kzbin.info/www/bejne/n3ynlXykepyqp6c. Hope it's ok to post here @Ciaran

[@anilkumarsharma8901]

Show flow path 👣 as equation💻 for💦 every time⌚⌚ and range of pressure😖 Show calculations💻 as Diffenrial equation💻💻💻

[@anilkumarsharma8901]

Show path 👣of flow and📐 time⌚ derivative 😁😁😁😁

[@anilkumarsharma8901]

How volume and📐 angle and📐 time⌚ and pressure are interlocking 😝😝😝😝