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Part (Based off week 6 workshop content ) Cousider Vrater tank with hole of arca in its Di. through this hole, The tank will drain Tale proportional to the height o...

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Part (Based off week 6 workshop content ) Cousider Vrater tank with hole of arca in its Di. through this hole, The tank will drain Tale proportional to the height oi tlc wuter (as this influcnces the pressure). Torricelli $ Law descrilxuse this proce $Vzgh() ,where:h(t)the height of water in the tank tme in m,the cross-sectional area of the tank in m?,the area of the hole in the hottom of the tank in m? 9.81 m/s? the acceleration gravity:For the following questions, use A = 10 m?,0.1 m?_h(0) 10

Part (Based off week 6 workshop content ) Cousider Vrater tank with hole of arca in its Di. through this hole, The tank will drain Tale proportional to the height oi tlc wuter (as this influcnces the pressure). Torricelli $ Law descrilxuse this proce $ Vzgh() , where: h(t) the height of water in the tank tme in m, the cross-sectional area of the tank in m?, the area of the hole in the hottom of the tank in m? 9.81 m/s? the acceleration gravity: For the following questions, use A = 10 m?, 0.1 m?_ h(0) 10 u_ Solve the ODE using sepmration variables: Use first order Taylor series show' that when Torricelli's law: linearised about the initial height, 2h(o) (h() ~ h(O)) Solve the linearised ODE using the integrating factor method, Plot your solutions to Q1 and Q3 in the same figure: Discuss the accuracy of the linearised ODE solution. Plec se Gnsbe C1 l Luestions



Answers

Torricelli's law An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli's law (see figure). If $h(t)$ is the depth of water in the tank for $t \geq 0$ s, then Torricelli's law implies $h^{\prime}(t)=-k \sqrt{h}$ where $k$ is a constant that includes $g=9.8 \mathrm{m} / \mathrm{s}^{2},$ the radius of the tank, and the radius of the drain. Assume the initial depth of the water is $h(0)=H \mathrm{m}.$ a. Find the solution of the initial value problem. b. Find the solution in the case that $k=0.1$ and $H=0.5 \mathrm{m}$ c. In part (b), how long does it take for the tank to drain? d. Graph the solution in part (b) and check that it is consistent with part (c).

Hello, everyone. Here it is, given the laminar flow, laminar flow are reasonable model for for water level over the time divide upon the eighties Cult oh d to the power for roll into the upon 32 de square mule toe Help given right add by zero It's called toe by not and by except is given Why not exponential off minus de for Rajhi and Toti divided by 32 The square mule Why and plus one is called toe Why and plus HK right here Okay is given by get to the power for roadie upon 32 d square We will for time interval the end plus h here it is given diameter of the two with three mil Limited diameter of the tank is to 15 millimeters Why not 1 m and is called for density is 99 kg per meter Cubed dense viscosity is point gee Robin Newton's second per meter square Etch 12 minute to six minutes. Blot Ailor on object result. So for the about giving the question the complete rights value are to be calculated using above equations. Computerized generated values are as follows and using it the graph can be obtained. Time on the depth, White. But And this is Euller. That's all for it. Yes, right. Thanks for watching it.

In this problem, we have a cylindrical tank that is vented at the top with an outlet at the bottom for liquid to train through. And we know that the depth of the liquid hft and it's area at the surface, which is represented by this red circle A of H are related by a of H times D H D T is equal to minus K times the square root of age, which follows from Torricelli's law. And in this problem K is equal to 0.25 in part A. We want to find the depth of the water at Time T so to do so, we will need thio find a of H, which is easy because this is a cylindrical container or a cylindrical tank. So that means that that the area at the surface of the liquid will be constant as it drains through this tank. So to determine a of age, we know that the radius of the tank is 1 ft, so that means that the area of the liquid at the surface this pie feet squared. So in part a, we have that pi times D H d t. That's equal to minus 0.25 times the square root of H, And then we can solve this using the method of separation of variables. But this differential equation also comes with the initial condition that each of zero is equal to four because the tank is completely filled, which means that the liquid as a depth of 4 ft at time T equals zero. So from here we get that pie. Times D H over the square root of H is equal to minus 0.25 Do you see? We can integrate both sides to get that two pi times. The square root of age is equal to minus 0.25 plus C that that's equal to minus 0.25 times T plus E, and now we can use our initial condition. So this becomes two pi times. The square root of four is equal to minus 0.25 times zero plus c, which means that four pi is equal to see, so we have that to pie times. The square root of age is equal to four pi, minus 0.25 times T From here we can divide through by two pi We have that The square root of H is equal to two minus zero point 0 to 5 over two pi Times team and 0.2 5/2 pi. That's approximately 0.0 3979 And then from here we can take the square of both sides to get that H is approximately the quantity to minus 0.3 0.3979 times t all squared and then in part B. We want to find the time it takes for the tank to drain. So that means that h of t will be equal to zero. So we have that zero is equal to the quantity. Tu minus 0.3979 times t squared. We can take thes square root of both sides to get that zero is equal to Tu minus 0.39 79 times team, which means that two is equal to 0.0 3979 times t Finally, we can divide through by 0.3979 When we get that T is equal to two over 0.39 79 which is approximately 8.4 minutes. This number is actually in seconds. Um, which is 502.6 seconds. But if you work that out, you get that. That is approximately 8.4 minutes, and that completes the problem.

Yourself. Costa number 81. Here. The area of 100 orifice is given as a sequence to 25. Square 25. Mm. Square initial volume of urine Tank we notice equals 25 m Q in total volume of tank. It is equal to 10 m cube praising initial preserve. The notice equals to one mega pascal. So were the other figure. This is the tank. This is the tank. So this is a figure here. This is VT total volume. And now the beauty is this. This is pressure in itself is a peanut. This is present P. This is area No, this is the volume flow. This is pressure p. And this volume is We note we not now playing the Bernoulli question between the liquid surface and the office. We can right be the very little plus we square debated or two is equals two p a t. M divided veru place. We be there to square to whatever do were we 0 to 0 hands from here? We can write. Visit is equals. Two route is square root of two multiplied. B minus ph. D um divided way two divided by zero or we can rewrite it as root over two p development room s p A T M B zero gets now the mass flow rate of the water leaving the tank is, um, hum daughters calls to through a visit which is constant dealt I multiplied way fruit or to p development route. Roy. So this is the way divided multiplied wear and the route to pee. We were Where to? From here we get mass Florid. Mm vehicles two square root over two square out of two. Two p. Rule to Peru multiplied way. Now the rate of change of mass in tank can be written as GM divided. Validity is equals to build developer dlt integration of who will be the or we can write it as DM developer. Deity is equals two road w multiplied way Levi. Do you ever did David W two Whatever deity it can be. Return. It's minus row w multiplied way the we're leave a little bit duty. Note not. We t z equals two with it is equals two. It place we w VW now for I saw thermal flow. We know for a solar thermal flow I saw terminal Hello. We know PD weather room is equals to Artie is equal to p noted midway through, we're we're always equals two densities Goes to mass of you got a volume of here? No, here. A density be into me is equal to p note into we note or from here we can write p s equals two We know to be no to where they will be not from continue to prison from continue to present Continue to continue to question we can write Ow Levi W diverted by duty list I am not a is equal to zero minus row W d v a divided by duty plus and that can be written in square root of two p Throw w uh is equal to zero or the V divided by duty is equals to square Roto Toby divided by oW which can be Lieutenant the square root of to the note We not divided by throw w into multiplayer baby. Now, after separating the variables, we can write it we to the part of one way to multiply where TV is equals to square photo two peanut we note Divided way P W. Multiplied by a duty. So by integrating we get two divided by three. We to the power of three were too. Integrating from zero to be is equals two. Yeah, the square root of two peanut. We note the whatever PW multiplied by a duty 80. 80 80. Yes. So on rewriting we get two divided by three. We to the part of three were to minus zero way to report of two by three Well integrated from the note to be so we took part of three were two minus we know to the part of three Were too is equals two to we know to be part of three were too divided by three multiplied way we do whatever we note to the part of three were two minus one or we can write it as this time is equals two square out of Toby Nor do we not divided by road of you multiplied way 80 So from here we can write, we do whatever we note to the part of three way too is equals two one plus three were too We not purport of t were too multiplied by the square root of to the note We note divided by Ow multiplied with 80. No, we did Very well. We note is equals two one plus 1.5 multiplied by the square root of two peanut divided way ow multiplied by 80. They were They were being note to the part of two way three. This whole to the part of to wait three. Now we know that mass M w can be written as row. Ow, ow! Ow! Multiplied by. We t minus week. Or we can let end the blue is equals. Two ow multiplied where we note multiplied where VT divided Where we note minus we develop, we note or we can write MW is equals two row w multiplied where we note well multiplied way we develop, We note minus bracket closed one plus 1.5 square root two peanut divided Were ow multiplied away. 80 defended me a note to the part of two way three now giving the values of T from 0 to 20 seconds. To get different Mosses, we have t We have 20 seconds to zero missiles to fold over 95 20 is equal to two musicals. 24862 20 is equal to four. We have must 47 30 Similarly, six it then 12, 14, 16. No, no, no, no. Not 38. 2014. Similarly, here 404,600 Double 472 4354 43 45 four Double to zero, 4096 3973 Dr D. Eight 2695 two For you it for now. We will draw the drop. Now we need to plot the graph between T and M W. So let's draw the graph here only. Let's start the graph here only. This is Deidre. We need to draw. So at y axis we have Masum and attack success we have time to So we'll take time from zero live then 15, 20 25 30 35 40 45 similarly in. Why access to be here? 0 1,002,000. 3000. 0 1000. 2000. 3000 4000. 5000 6000. So we need to plot each point and after plotting each point, we have a graph almost straight line, which is a bit inclined. It will be like this So it will be like this. Yes, this is what we are going to get after plotting each point. Thank you so much.

So for this problem, we're told that the rate of change of the volume overtime is equal to two separate equations. So that allows us to set These two are individual equations equal to each other. And so I'm essentially, all we're going to do is use this these two equations right here or these two expressions as as one equation, and we're going to use a separation of variables to solve for H, I'm so right away. We can divide both sides by our area because we know that the area of the hole at the bottom of the tank is not changing. Um And so from here we can go ahead and say that the negative square root of to G H is equal to D H over DT. And so when we use our separation of variables, we yet negative skirt of two g duty is equal to H to the negative 1/2 power D h. And so from here, we can go ahead and integrate each side of our equation. Um, and actually, since this to G is a constant, I'll just put the integral in front of the d t. And I'm going to swap the sides of my equation because I like to have my h on the left side. So when we integrate, we get to H to the 1/2 power is equal to, uh, negative t times the square root of two G. And so when we solve for H, we're going to get H to the 1/2. Power is equal to, uh, negative t times the square root of G over too. Um, and so from here, we'll just square both sides of us, and we can't forget our constant on the right side of her equation over here. So plus C um, and plus C So this time, when we square both sides of the equation, we're going to get h is equal to you. Negative teeth and the square root of G over too. Um, plus scene squared. And so we have our initial condition. Um, h sub zero, I feel do this in green. H subzero is equal to 100 which means that our initial height is going to be 100. And so from here, we know that the height of water in her tank, which is 100 is equal to negative T we said, which is zero, um, plus seen squared. And so when we solve for C, we get C is equal to 10 and that allows us to find our final equation on, which is the height of water in the tank is equal to negative teeth. Times the square root of G over too close, 10 squared.


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