Pumping water out of a spherical tank, calculating work, calculus 2 tutorial

Pumping water out of a spherical tank, calculating work, calculus 2 tutorial
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[@jeremytq]

You're my hero after studying all day. Why are professors so good at avoiding important explanations? Thank you very much for your thorough solution!

[@isaact.477]

You connect the concept and intuition to the written mathematics so well. Great video!

[@matthewshaabo5967]

best teacher ever, this dude teaches better than any calculus teacher I had before

[@jarifahmed977]

2023 and still helps students like me. Thanks for literally existing, dude...❤

[@Panda-ch3lz]

Love the way you explain things !! super thorough

[@kroganoperator]

You explained it very patiently and with passion. I liked and subbed, thank you very much.

[@johnbarron4265]

A shortcut to solving this problem is to locate the center of mass of the water both before and after pumping it. The geometry of the tank is simple enough to do this. Beforehand, its center is at the geometric center of the tank, y=0. After pumping, all of the water has been lifted to an elevation of y=3+1=4 m. The mass of water pumped equals volume times density. Volume is 4/3*pi*3^3=36*pi m^3. Mass is 1000*36*pi=36000*pi kg. Weight=9.8*36000*pi=352800*pi N. Work equals force times distance. 352800*pi*4=1411200*pi=4.43*10^6 J. This approach neglects viscous losses and changes in kinetic energy that would likely occur during pumping. These losses may be small in comparison to the gravitational component of the work, so that the pumping process is nearly 100% efficient. For example, a 5000 W pump takes 887 seconds to do 4.43*10^6 J of work. The volumetric flowrate required would be 0.128 m^3/s. Suppose the outlet pipe has a diameter of 0.3 m. All of the water must at some point flow through a cross sectional area of pi*(0.3/2)^2=0.0707 m^2. The required flow speed is thus 0.128/0.0707=1.80 m/s. The total kinetic energy attained by the water during pumping would be (1/2)*(36000*pi)*1.80^2=1.84*10^5 J. Thus the total work done would be 4.43*10^6+1.84*10^5=4.62*10^6 J. Here 96% of the work is used to lift the water, and 4% is used to accelerate it as it flows through the outlet pipe. A more powerful pump will draw higher flowrates, so that the kinetic energy accounts for a larger percentage of the total work done on the water. A smaller outlet pipe will also increase flow speeds, which has a quadratic effect on the kinetic energy attained. Even this analysis neglects the nonuniform velocity profile of the water, which depends on how turbulent the flow is. A kinetic energy coefficient is often used to account for the various velocity profiles; however, these details are beyond the scope of calculus II curriculum and their study is reserved for a course in fluid mechanics.

[@lorostotos5647]

maybe we can work both ways to locate the center of mass,assuming its this we dont know?

[@gary3ward]

I noticed that also and used it to check my answer using integration. The center of mass based calculation was definitely easier. Usually he loses me with his twenty step solutions.

[@marstruth1578]

John Barron I was disappointed that he stopped at the top of the sphere. He took the height of the 1 meter spigot into account but then neglected to calculate the work necessary to force the water through the spigot.

[@tabatuby]

But now you are getting into center of mass, and lets say that it was a sphere half full of water, then you would need to integrate to just find the center of mass. But regardless your method is very clever, props

[@m0zzar353]

dont sound like no shortcute 😅

[@user-mr5ze2vi3c]

Thank you!!

[@camdenjohnson2988]

Very clear and well done video. It was very helpful. Thank you!

[@michaelrislingnb306]

thank you bprp.. youre a great help and you always brighten my day with your positive attitude. I remember I was so frustrated trying to solve this integral and so I watched a video of you doing it and like halfway through the solution you paused and looked at the board and said "okay... this looks...cool..." and I thought it was so funny and it made me less frustrated. Wish u all the best always

[@blackpenredpen]

Thank guy. Best wishes to you as well 😃

[@mn-yv2ur]

OMG. You are amazing. I love your explanation. I have final next week. Now I've learned a lot. Tanks again for being so helpful.😀😊🤗

[@remingtonrimple7357]

Thanks for the video! Your explanation was really helpful :)

[@SeanEMusic]

Great video, got a test in two hours. Thanks much!

[@AR-fh4qu]

just cuz i saw a comment from 2019 we can start a trend. This is greatly helpful in 2022. Thanks :)

[@InfernoPoisn]

It's 2019 and this video is still super helpful! Thanks a bunch c:

[@usamabintahir5285]

2021*

[@jessiemullett5719]

2023 and still helpful!

[@chastaine813]

I subscribed after months of watching your videos. You're a very effective teacher.

[@blackpenredpen]

+Chastaine eniatsahC Thank you, I am glad that you find my videos helpful!

[@FuzeGamer11]

Great video! I appreciate the quality audio from the mic

[@alisonmalekian8811]

thank you!! literally saved my life

[@kazeyomi4131]

You are my life saver!! thx for the decent explanation!

[@Wolfstar26]

very awesome work, it really helped me understand

[@rockyboy5545]

The way you explain makes perfect sense unlike my professor who think I was born Einstein and should know these thing.

[@AshrafAli-qs6ep]

Extremely helpful. Thank you so much.

[@blackpenredpen]

Ashraf Ali you're welcome

[@thesulliedone9787]

In case anyone will answer four years later, could you set displacement to equal y+1 then integrate from 0 to 6 instead?

[@arfanm9440]

I have exam in 3 days and this really helps me! Better than my professor’s explanations!

[@derekfrostbeard6419]

As a math tutor, I find it more helpful to explain the expression for r differently. The equation for that slice is x^2+z^2=r^2. Lower case r is a variable and is NOT always 3. My students often set the radius as 3 which winds up with them finding the water pumped out of a cylinder instead of a sphere. The equation for the sphere is x^2+y^2+z^2=R^2. Lower case r is the radius of the slice that we need to find, and upper case R is the radius of the sphere. This means lower case r is a variable while upper case R is a constant. That means we have x^2+z^2=r^2 and x^2+y^2+z^2=9. We can substitute x^2 and z^2 in the second equation for r^2. So now we have r^2+y^2=9. This can be solved for r giving r=(9-y^2)^1/2.

[@ivyknudsen7423]

thank you very helpful ! you're very concise!

[@afilmbyPac]

Great video. Thanks a lot.

[@Rockstar1376]

Thanks your vid helped a lot!

[@chriscamilo8334]

thanks a lot. Very clear example!!!

[@MichaelMagill1990]

I like your enthusiasm. Subbed.

[@blackpenredpen]

Thanks

[@linaaa2535]

Thank you so much!

[@daniyard2012]

Why can't we derive radius from ratio of triangles within sphere?

[@creativeclaude5304]

If we reference the area of a cylinder in the sphere we wouldn't obtain a more exact approximation right? Because we are assuming that y-value our radius stays the same till the bottom of the cylinder. But we will still be missing some area covered on the sides of the cylinder because we made that assumption right? Please correct me if I am wrong but everything else seems accurate. Thank you.

[@soccerstudut]

In this case the easiest solution is to simply use the equation for the volume of the sphere to calculate the volume of the water, then multiply by 1000*9.8*4 and you get the same answer. The reason you multiply by 4 is that the center of the water mass in the sphere is obviously the center of the sphere. If you consider all the mass to be concentrated at that point then you simply need to raise/pump that mass up 4 meters to get it to leave the spout. This method will work for any problem, but if your professor wants you to show work of the method using integrals you need to know how to do that as well. It is a great way to check your answer though.

[@kaitlynleffler4019]

Awesome, thank you!!

[@intelligentpress7598]

Thanks a lot! I finally got it)

[@sanalchhabra7901]

Can you please tell this question is from which book?

[@AllanPoeLover]

麥克風為何不用懸掛的方式呢 ?

[@cefalopodo5717]

excellently explained

[@mryup6100]

Very helpful man!

[@dabsraee7431]

There is one more zeo at the end, isn't it?

[@natekim8518]

You should consider getting a lapel mic. It looks much more professional and your audio levels won't fluctuate based on the distance between the mic.

[@karangarg4631]

I think he chooses to hold the mic because it feels more natural.

[@me89315]

I love you my dude thank you

[@carlosrr9097]

Gran explicación que sugiere otro método, en mi caso lo corroboré con la esfera fuera del origen en (0,3) y los limites de integración de 0 a 6 (cero a seis), el valor de la distancia es de 7-y ; el resultado es de 1411200 x PI , lo que significa en términos cerrados de 4.43 MJ. Saludos desde México

[@liliawongg]

Thank you

[@py6342]

Actually, we can just assume the water as a point mass, 4/3 pi r^3 density, located at the origin. The height is 4m. We will have the same ans.

[@Mr.Jack_learning]

why isn't the range of the integral from-3 to 4

[@IkeSan]

I am going to subscribe man you are great.

[@blackpenredpen]

Kev San LOL! You are great too!!!

[@Awawaeaawawawawawawa]

what if we drew the coordinate plane below

[@jmscshipp]

Thank youu!

[@pipertripp]

So when I integrated from 0 to 3 and then multiplied the result by 2 I get 3186.5 kJ. If I integrate over [-3,3] I get the same answer as you. So what's the issue here? Do we need to consider a different distance if we change the bounds of integration? I'm not sure what else would be different. Any idea what is causing the discrepancy?

[@gagnant4]

It takes less work to pump out the top half than it does to do the bottom half. The bottom half has to move a greater distance.If you integrate the bottom half from -3,0 and add that to your 0,3 answer you should get the same as him.

[@dinaraturagulova4634]

Love yaaaa! Thanks!!!!

[@znhait]

This is one of the problems that makes calc II much harder than calc I. A mechanical engineering student or a physics major will find the set up to be easier because this is a fluid mechanics problem-albeit calc II is likely a prerequisite for fluid mechanics.

[@user-up8or2tr1e]

please take my firstborn you have saved my calculus marks.

[@sabi_jaha9389]

You are awesome

[@zhou4168]

Good video

[@rosanaferreira4296]

You integrated from the bottom of the tank (y=-3) to the top of the tank (y=3). How about the spout, which is one meter above the top of the spherical tank?

[@blackpenredpen]

There's no water there.

[@rosanaferreira4296]

Thanks, I agree, but the water that is in the tank still has to move that extra 1 meter distance, right? Did we take that into account?

[@maxsongreen1910]

+Rosana Ferreira That was taken into account when he included the expression (4-y), which represents the distance that any slice has to travel to reach the top of the spout. Notice: The slice located at y= - 2 has to travel [4 - (- 2)] = 6 meters which is accurate. The limits of integration do not have anything to do with the distance we are moving the slices, they only represent the interval over which we are *counting* and *adding up* our slices. Since our slices are only representing the interval where there is water, we integrate over the interval [-3,3].

[@xrisku]

can someone explain why the distance is 4-y and not 7-y?

[@XDmofasa]

I believe it is because of what he chose to be his x axis. Since his x axis is in the center its 4-y and he integrates from -3 to 3. If you chose your x axis to be the very bottom, then you would choose 7-y and 0 to 6. Hopefully I'm right, sometimes I'm wrong.

[@RunninAD]

Subbed, thank you.

[@blackpenredpen]

+Jared Shorten Cool!

[@adnanyousaf4049]

a spherical tank with radius 5m is partially filled with water ,5 meters deep in middle .How much work is required to pump all the water out through a hole at the top of the tank? (use the fact that the mass density of water is 1000kg/m^3 and the acceleration due to gravity is 9.8m/s^2) solve this qs

[@erick_fernandez_78910]

Can height be 7-y then set up integral from 0 to 7?

[@znhait]

You can shift your coordinate system around, where the bottom of the sphere is the point (0,0). But you’d need to adjust other parts of the problem accordingly.

[@ryoyamamoto4894]

Quarantine and I'm learning calculus thru this wonderful yt teacher more than my real uni prof. Thanks a lot

[@vmdvaldez]

nice video

[@imtiazhassan6242]

If you don’t mind I have a question. Water being a liquid attracts itself for the the surface tension and also attracts the wall of the container (due to adhesive and cohesive forces),right? Of course these forces are very tiny. But still becomes non negligible if present in a sufficient volume. Am I correct if I say the calculated work is the work required to pull the water considering it an assembly of tiny water pieces with no forces in between?

[@DoctrinaMathVideos]

You are correct there other forces and dynamics involved. These later discussed in a more advanced class such as fluid dynamics but at least we have the basics to work from.

[@imtiazhassan6242]

@@DoctrinaMathVideos Thanks for relpying. I am in 12th grade now and I am trying to improve my understanding of physical situations. Identifying these subtle differences help a lot specially when textbook of my level doesn’t say these.I just wanted to confirm my assumptions. I truly thank you from my heart.

[@DoctrinaMathVideos]

@@imtiazhassan6242 It's great that you are curious about these applications. Never stop thinking and wondering. Curious minds development into intellectual minds.

[@derrickcastillo8077]

Love the video but why don’t we integrate from -3 to 4?

[@williamperezhernandez7331]

The explanation in video is not clear about limits of integration. Best is to separate problem into two parts. 1. Work to raise all water from -3 to 3m. Here the sectional area varies as a circle. 2. The tube (cylinder) of length 1m has constant area. So no integral is needed. Work is just (length)(volume of water)(density)(gravity). Therefore 4m - y is actually 3m - y for calculation of Work of part 1, plus 1m times the integral for volume of water x 98000 as Work of part 2, giving the total Work.

[@KoobyYin]

Why is it not bound from -3 to 4? doesnt it need to go out of the pipe?

[@umang1737]

thats what i was thinking too

[@mryup6100]

We are taking slices from -3 to 3. So the "last slice" would only travel a distance of 1 (4-y) -> (4-3), which the 4 takes into account the pipe.

[@Prxwler]

Why when you were integrating you set it from -3 up to 3 and not up to 4?

[@danielrutherford9456]

Because -3 to 3 are the bounds of the sphere. You’re just taking those volumes up an additional 1 meter while pumping.

[@MrKA1961]

Why use integration when you can calculate it almost instantly? You must rise a sphere of water 4 meters higher than its actual position. Work necessary for doing that is W=m*g*h=(mass of the water)*9.8*4 The mass of the water is (the volume of the sphere)*(density of water)=4*pi*3^3/3*1000 so, you get finally: W=(4*pi*3^3/3)*1000*9.8*4 which gives the same result.

[@jairogen90]

that´s not fun, is it?xD

[@diegor3194]

Great video. I'm confused as to why the distance is traveled is (4 - y). The water is being pumped from the bottom of the tank all the way up and out the spout. Shouldn't it be (7 - y) ?

[@kokokoko-sj2ex]

he must delete (4-y) when integral for all the layers

[@DoctrinaMathVideos]

The distance from the x-axis to the element piece is y. The total distance to get all of the way to the top is 4, so therefore the difference is 4-y. This all depends on where you place your axes and element piece but the solution will be the same.

[@JensenPlaysMC]

but isnt this calculating work to overcome gravity? not taking account the distance the water not directly below the pump as it needs to move some horizontal conponent. in pretty sure this will affect result

[@AshFeatherRocks]

Technically to move something horizontally, where theres no horizontal force working against you, you can use any force no matter how small and it will eventually get wherever you want it to, so the minimum force is infinitesimal, and there's no contribution to the total work. If you said you wanted the water to move to the centre in some time, like 1s, then you could calculate the extra work needed.

[@blakeelzinga1168]

I love you bro

[@jupiteraniel4851]

I love 😍

[@CountryJukebox56]

isn't the distance traveled by the layer 7 - y? what if the layer is at the bottom?

[@CountryJukebox56]

+John Williams Nvm I figured it out!

[@chriswass1176]

+John Williams i also got 7-y. why is it 4 - y. isn't that just half of the water?

[@daanishrasheed]

+Chris Wass In case you still need help, if you use 7-y then you would have to change the bounds to 0 to 6 instead of -3 to 3

[@daniel6678]

+Daanish Rasheed My calculator still gives a different answer doing it that way, though. Conceptually it makes sense, but I get 1108353.88819 .

[@iceverything2000]

+Daniel extra account did anyone figure this out yet?

[@user-sf7qz5kg3b]

Isn't the volume gonna be in m^3 ? It should be in liters/dm^3 for the density to work. ?

[@wraithbackup4286]

very pedagogical

[@kokokoko-sj2ex]

u must delete (4-y) when integral

[@yaxeenrahman]

Why it's -3 to 3!! isn't it would be -3 to 4? cause you took the 1-meter height as well for the distance too @blackpenredpen

[@danielrutherford9456]

You’re only integrating where there is water. The extra 1m of distance starts out empty.

[@omarmaya1263]

I tried to take a shortcut and evaluate the integral from (0,3) and just multiply by 2, but it did not work lol Remember to evaluate from (-3,3) otherwise you will spend an hour wondering why you aren't getting the same answer

[@jordanthrailkill4792]

Wouldn't the Y distance be (6-Y)??

[@MikeySureHanded]

+Jordan Thrailkill , The distance traveled is 4-y because y can include the whole y*axis (negative numbers). So for example, when the level of water is at y=-2, you get 4-(-2) = 6.

[@JensenPlaysMC]

no as you take the 0,0 to be the centre of the sphere

[@gabrielhendricksen6213]

I also found another way of solving this specific problem. I usually set the lowest point of the tank as my origin, so the new equation of the circle is (x)^2+(y-3)^2=9, in which we solve for x to get r in the equation pi*(r)^2*h. I then said that my distance was the very top of the pipe minus y [(7-y)]. My limits of integration would then be from 0 to 6. All in all my integral would look something like this --> integral from 0 to 6 of (7-y) (9.8)(1000)pi(9-((y-3)^2)) dy , which also gave me 4.43 x 10^6 J. I just wanted to share my method in case there was anyone out there that was taught where the bottom of the tank is the origin like I did and was confused.

[@dantefox6556]

Gabriel Hendricksen thanks! I was really confused for a mite there lol

[@averydrago]

Amazing video, but at 1:52, a CD is more like a washer than a cylinder

[@DoctrinaMathVideos]

Think of as a flat cylinder with an infinitesimal height. Remember each piece is acting as an element (representative piece).

[@jiezheng9454]

Nice shirt man!

[@blackpenredpen]

Jeremy Zheng thanks!

[@DanielPerez-ws6ts]

Is this Calc. 3 stuff?

[@blackpenredpen]

Calc 2 actually.

[@DanielPerez-ws6ts]

@@blackpenredpen Do you have any Calc. 3 problems on video?

[@jikaikas]

Poggers Just what i need

[@chritophergaafele8922]

4.43 Mega Joules

[@justinbeal3920]

is that a devil fruit

[@matthewwesterhoff6470]

The dy technically be written as a "delta y" because the disk is the element piece. You should also mention how to go from the finite sum to the continuous sum (integral setup) but otherwise good job.

[@christianwimmer415]

You need no Integration for this example: W=V*rho*g*h W...work, V...Volume, rho...density, g..acceleration of gravity, h...average height V=36*Pi m^3, rho=1000kg/m^3, g=9,81m/s^2, h=4m Therefore W=4,43*10^6 J

[@comma_thingy]

I imagine finding the average height gets harder when the shape of the tank changes however?

[@christianwimmer415]

Yes, you are right. But you shouldn't shoot with a big calibres at little birds!

[@maryamelamiri5514]

omg

[@elliberger6079]

Bad Boy

[@user-vs6en1du7e]

Who is from 2023