Issues involved in Torricelli Model of tank leaking fluid BRIEF ABSTRACT Several slit configurations in the side of a cylindrical can are considered to develop specific "leaking" patterns using Torricelli's Law. GENERAL INFORMATION FileName: TANKLEAK Full title: Issues involved in Torricelli Model of tank leaking fluid Developer: Bart Goddard, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA, and Brian Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Brian Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@nextwork.rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering. STATEMENT OF PROBLEM Torricelli's Law states that the velocity of water exiting through a hole is proportional to the square root of 2 g y, where g is acceleration due to gravity and y is the depth of the water above the hole. For small holes, the constant of proportionality is about 0.6. We consider three situations. Situation 1 A cylindrical tank of radius 3 feet and height 8 feet is full of water. A hole of area 1 square inch is punched in the bottom. How long will it take for the tank to empty? Situation 2 Instead of the hole, a slit 1/10 of an inch wide is cut in the side of the tank, running vertically from top to bottom. How long will it take for the tank to empty? Compare your solutions in Situation 1 and Situation 2. Situation 3 Can we design a vertical slit of varying width for this tank which causes the level of water in the tank to drain at a constant rate? KEYWORDS Torricelli's Law, first order, separable, differential equation. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Knowledge of solving simple first order, separable differential equation. Time allotment - time management Expectations Future payoffs Extensions References and Sources POSSIBLE SOLUTION(S) Situation 1 If the tank is y feet full, then the volume of water is given by V = Pi 3^2 y = 9 Pi y, so dV/dt = 9 Pi dy/dt. On the other hand, the change in volume with respect to time is the area of the hole times the velocity of the water squirting through the hole, therefore, dV/dt = -(area)(velocity) = -1/ 144 ft ^2 (0.6) Sqrt(2 32 y) ft/sec or dy/dt = dV/dt/ (9 Pi) = -1/144 (0.6) 8 Sqrt(y) / (9 Pi) = -4.8 Sqrt(y)/ (144*9 Pi) . We solve this initial value problem, with y(0) = 8: sol = DSolve[{y'[t]== -4.8Sqrt[y[t]]/ (144*9 Pi), y[0]==8}, y[t],t] 0.00185185 t 2 {{y[t] -> (-2 Sqrt[2] - ------------) }, Pi 0.00185185 t 2 {y[t] -> (2 Sqrt[2] - ------------) }} Pi Since y decreases, the second solution is the one we want: s[t_] = y[t]/.sol[[2]] 0.00185185 t 2 (2 Sqrt[2] - ------------) Pi To find when the tank is empty, we solve: solt = Solve[s[t]==0,t] {{t -> 4798.31}, {t -> 4798.31}} We examine a plot of the graph of s(t). sPlot = Plot[s[t],{t,0,5000}] Therefore the tank is empty in 1.33 hrs. 4798.31/3600 1.33286 Situation 2 Fix a certain time, and suppose the height of the water in the tank is y feet deep at that time. Let x be a variable which varies from 0 to y, and consider a small piece of the slit dx high. That is, consider a hole at height x of area dx*1/(120) square feet. Then the change in volume through that hole is ±0.6 * Sqrt( 2*32*(y-x)) /120 dx, where y - x is the height of the water above the hole. To get the total change in volume through the entire slit, we integrate the last expression with respect to x from 0 to y: Integrate [-3/5*Sqrt[2*32*(y-x)]/120,{x,0,y}] 3/2 -2 y ------- 75 That is, dV/dt = -2/75 y ^(3/2). As in Situation 1, we replace dV/dt with 9Pi dy/dt to get the initial value problem: dy/dt = -2/75 y ^(3/2)/(9 Pi), y(0) = 8 And we solve the initial value problem. sol1 = DSolve[{y'[t]==-2/75 y[t]^(3/2)/(9Pi),y[0]==8}, y[t],t] -1 t -2 {{y[t] -> (--------- + ------) }, 2 Sqrt[2] 675 Pi 1 t -2 {y[t] -> (--------- + ------) }} 2 Sqrt[2] 675 Pi We plot both solutions so we can determine which is valid: s1[t_] = y[t]/.sol1[[1]]; s2[t_] = y[t]/.sol1[[2]]; Plot[s1[t],{t,0,5000}] s2Plot = Plot[s2[t],{t,0,5000}, PlotStyle->Thickness[.01]] Since the height of the liquid, y, must decrease, we choose s2(t) and we notice that s2(t) is asymptotic to y = 0, but never yields a solution to s2(t) = 0. Solve[s2[t]==0,t] {} Plot[s2,{t,0,10000}] Indeed, using the Limit command we see the asymptotic behaviour. Limit[s2,t->Infinity] 0 We now examine the plots from Situation 1 (light line) and Situation 2 (dark line) to compare them. Show[sPlot,s2Plot] It appears that in Situation 2 the liquid drops faster at first but then slows down its flow when we are near the bottom of the tank. Situation 3 There is no practical answer to this problem. The average pressure and the total area of the hole both decrease, so the rate of change of volume (and height) must necessarily decrease. We can demonstrate this mathematically as follows. The slit in Situation 2 was 1/120 ft wide. We replace this number by an unknown function w(y) which tells the width of the slit at height y. Then as in Situation 2 we have: The change in volume though that hole is ±0.6 * Sqrt( 2*32*(y-x))w(x) dx, where y-x is the height of the water above the hole. To get the total change in volume through the entire slit, we integrate the last expression with respect to x from 0 to y, but since w(x) is unkown, we suppose that F(x) is an anti-derivative of the integrand, so Integrate [-3/5*Sqrt[2*32*(y-x)]/120,{x,0,y}] = F(y) - F(0), That is, dV/dt = 9 Pi dy/dt = F(y) - F(0). We want dy/dt to be constant, so we set dy/dt = (F(y) - F(0)) / (9 Pi) =k. Now if we differentiate with respect to t, we have (F'(y) dy/dt ) / (9 Pi) = 0, but since dy/dt = k, we have F'(y) = 0. Since y varies from 0 to 8, and F(x) is an antiderivative for ±0.6 * Sqrt( 2*32*(y-x))w(x), we conclude that the latter equals zero. Therefore we must have w(x) = 0, which is the only function that allows a constant rate of drainage, that is, none. ISSUES IN SOLUTION