(*^ ::[ Information = "This is a Mathematica Notebook file. It contains ASCII text, and can be transferred by email, ftp, or other text-file transfer utility. It should be read or edited using a copy of Mathematica or MathReader. If you received this as email, use your mail application or copy/paste to save everything from the line containing (*^ down to the line containing ^*) into a plain text file. On some systems you may have to give the file a name ending with ".ma" to allow Mathematica to recognize it as a Notebook. The line below identifies what version of Mathematica created this file, but it can be opened using any other version as well."; FrontEndVersion = "NeXT Mathematica Notebook Front End Version 2.2"; NeXTStandardFontEncoding; fontset = title, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e8, 24, "Times"; ; fontset = subtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e6, 18, "Times"; ; fontset = subsubtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, italic, L1, e6, 14, "Times"; ; fontset = section, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, grayBox, M22, bold, L1, a20, 18, "Times"; ; fontset = subsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, blackBox, M19, bold, L1, a15, 14, "Times"; ; fontset = subsubsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, whiteBox, M18, bold, L1, a12, 12, "Times"; ; fontset = text, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = smalltext, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 10, "Times"; ; fontset = input, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeInput, M42, N23, bold, L1, 12, "Courier"; ; fontset = output, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L-5, 12, "Courier"; ; fontset = message, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = print, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = info, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = postscript, PostScript, formatAsPostScript, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeGraphics, M7, l34, w282, h287, L1, 12, "Courier"; ; fontset = name, inactive, noPageBreakInGroup, nohscroll, preserveAspect, M7, italic, B65535, L1, 10, "Times"; ; fontset = header, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, italic, L1, 12, "Times"; ; fontset = leftheader, 12; fontset = footer, inactive, nohscroll, noKeepOnOnePage, preserveAspect, center, M7, italic, L1, 12, "Times"; ; fontset = leftfooter, 12; fontset = help, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = clipboard, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = completions, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12, "Courier"; ; fontset = special1, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special2, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special3, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special4, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special5, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; paletteColors = 128; showRuler; automaticGrouping; magnification = 150; currentKernel; ] :[font = title; inactive; preserveAspect; startGroup] Issues involved in Torricelli Model of tank leaking fluid :[font = section; inactive; preserveAspect; startGroup] BRIEF ABSTRACT :[font = subsection; inactive; preserveAspect; endGroup] Several slit configurations in the side of a cylindrical can are considered to develop specific "leaking" patterns using Torricelli's Law. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect; endGroup] 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. :[font = section; inactive; Cclosed; preserveAspect; startGroup] STATEMENT OF PROBLEM :[font = subsection; inactive; preserveAspect] 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. :[font = subsection; inactive; preserveAspect] For small holes, the constant of proportionality is about 0.6. :[font = subsection; inactive; preserveAspect] We consider three situations. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Situation 1 :[font = subsubsection; inactive; preserveAspect; endGroup] 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? :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Situation 2 :[font = subsubsection; inactive; preserveAspect] 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. :[font = subsubsection; inactive; preserveAspect] How long will it take for the tank to empty? :[font = subsubsection; inactive; preserveAspect; endGroup] Compare your solutions in Situation 1 and Situation 2. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Situation 3 :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] 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? :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsection; inactive; preserveAspect; endGroup] Torricelli's Law, first order, separable, differential equation. :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; preserveAspect] ISSUES RELATED TO THE PROBLEM :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] Knowledge of solving simple first order, separable differential equation. :[font = subsection; inactive; preserveAspect] Time allotment - time management :[font = subsection; inactive; preserveAspect] Expectations :[font = subsection; inactive; preserveAspect] Future payoffs :[font = subsection; inactive; preserveAspect] Extensions :[font = subsection; inactive; preserveAspect; endGroup] References and Sources :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION(S) :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Situation 1 :[font = subsubsection; inactive; preserveAspect] 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, :[font = subsubsection; inactive; preserveAspect] 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) . :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] We solve this initial value problem, with y(0) = 8: :[font = input; Cclosed; preserveAspect; startGroup] sol = DSolve[{y'[t]== -4.8Sqrt[y[t]]/ (144*9 Pi), y[0]==8}, y[t],t] :[font = output; output; inactive; preserveAspect; height = 44; endGroup; endGroup] {{y[t] -> (-2*2^(1/2) - (0.001851851851851852*t)/Pi)^2}, {y[t] -> (2*2^(1/2) - (0.001851851851851852*t)/Pi)^2}} ;[o] 0.00185185 t 2 {{y[t] -> (-2 Sqrt[2] - ------------) }, Pi 0.00185185 t 2 {y[t] -> (2 Sqrt[2] - ------------) }} Pi :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Since y decreases, the second solution is the one we want: :[font = input; Cclosed; preserveAspect; startGroup] s[t_] = y[t]/.sol[[2]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] (2*2^(1/2) - (0.001851851851851852*t)/Pi)^2 ;[o] 0.00185185 t 2 (2 Sqrt[2] - ------------) Pi :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] To find when the tank is empty, we solve: :[font = input; Cclosed; preserveAspect; startGroup] solt = Solve[s[t]==0,t] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {{t -> 4798.313573211023}, {t -> 4798.313573211049}} ;[o] {{t -> 4798.31}, {t -> 4798.31}} :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] We examine a plot of the graph of s(t). :[font = input; preserveAspect; endGroup] sPlot = Plot[s[t],{t,0,5000}] :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Therefore the tank is empty in 1.33 hrs. :[font = input; Cclosed; preserveAspect; startGroup] 4798.31/3600 :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 1.332863888888889 ;[o] 1.33286 :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Situation 2 :[font = subsubsection; inactive; preserveAspect] Fix a certain time, and suppose the height of the water in the tank is y feet deep at that time. :[font = subsubsection; inactive; preserveAspect] 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. :[font = subsubsection; inactive; preserveAspect] 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. :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] To get the total change in volume through the entire slit, we integrate the last expression with respect to x from 0 to y: :[font = input; Cclosed; preserveAspect; startGroup] Integrate [-3/5*Sqrt[2*32*(y-x)]/120,{x,0,y}] ;[s] 3:0,0;9,1;12,2;45,-1; 3:1,10,8,Courier,1,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Courier,1,12,0,0,0; :[font = output; output; inactive; preserveAspect; endGroup; endGroup] (-2*y^(3/2))/75 ;[o] 3/2 -2 y ------- 75 :[font = subsubsection; inactive; preserveAspect] 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 :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] And we solve the initial value problem. :[font = input; Cclosed; preserveAspect; startGroup] sol1 = DSolve[{y'[t]==-2/75 y[t]^(3/2)/(9Pi),y[0]==8}, y[t],t] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {{y[t] -> (-1/(2*2^(1/2)) + t/(675*Pi))^(-2)}, {y[t] -> (1/(2*2^(1/2)) + t/(675*Pi))^(-2)}} ;[o] -1 t -2 {{y[t] -> (--------- + ------) }, 2 Sqrt[2] 675 Pi 1 t -2 {y[t] -> (--------- + ------) }} 2 Sqrt[2] 675 Pi :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] We plot both solutions so we can determine which is valid: :[font = input; preserveAspect] s1[t_] = y[t]/.sol1[[1]]; s2[t_] = y[t]/.sol1[[2]]; :[font = input; preserveAspect] Plot[s1[t],{t,0,5000}] :[font = input; preserveAspect; endGroup] s2Plot = Plot[s2[t],{t,0,5000}, PlotStyle->Thickness[.01]] :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] 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. :[font = input; Cclosed; preserveAspect; startGroup] Solve[s2[t]==0,t] :[font = output; output; inactive; preserveAspect; endGroup] {} ;[o] {} :[font = input; preserveAspect; endGroup] Plot[s2,{t,0,10000}] :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Indeed, using the Limit command we see the asymptotic behaviour. :[font = input; Cclosed; preserveAspect; startGroup] Limit[s2,t->Infinity] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 0 ;[o] 0 :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] We now examine the plots from Situation 1 (light line) and Situation 2 (dark line) to compare them. :[font = input; preserveAspect] Show[sPlot,s2Plot] :[font = subsubsection; inactive; preserveAspect; endGroup] 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. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Situation 3 :[font = subsubsection; inactive; preserveAspect] There is no practical answer to this problem. :[font = subsubsection; inactive; preserveAspect] The average pressure and the total area of the hole both decrease, so the rate of change of volume (and height) must necessarily decrease. :[font = subsubsection; inactive; preserveAspect] We can demonstrate this mathematically as follows. :[font = subsubsection; inactive; preserveAspect] 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. :[font = subsubsection; inactive; preserveAspect] Then as in Situation 2 we have: :[font = subsubsection; inactive; preserveAspect] 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. :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] Therefore we must have w(x) = 0, which is the only function that allows a constant rate of drainage, that is, none. :[font = section; inactive; preserveAspect; endGroup] ISSUES IN SOLUTION ^*)