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1 Grand Valley State University MTH302 – Differential Equations and Linear Algebra Prof. Aboufadel Section A December 6, 2005 Computer Project #2: Finding the Leak By Dan Schwarz 2 Differential equations are very important to scientists and mathematicians because they provide a means of modeling dynamic systems. If the rate of change in one or more characteristics of a system is known, a differential equation can be developed. The solution to a differential equation is a function that describes the characteristic of interest as it changes with time. In this sample solution, a differential equation is formed to describe the rate of change of salt in a tank of water that has a leak in it. The tank of interest contains an unknown amount of salty water at time t = 0. The values of the initial volume of liquid and the initial amount of salt in the tank are symbolized by L gal and Q lb respectively. Throughout the observation period, water is being pumped into the tank at a rate of 4.8 gal/min with a concentration of 0.34 lb/gal. The water in the tank is also being pumped out of the tank at a rate of 4.8 gal/min. In order to obtain enough information to solve for the unknown rate of leakage, symbolized by r gal/min, the salt concentration of the water is sampled at time t = 3 min and t = 5 min. At time t = 3 min the salt concentration of the water leaving the tank is 0.2645 lb/gal and at time t = 5 min the concentration is 0.2910 lb/gal. The first step in the process of solving the unknown values for the tank system, is to create a differential equation. Equation 1 is a first order differential equation that describes the tank system. dx 0.34lb 4.8gal xlb 4.8 r gal gal min L rt gal (1) dt min Note that the first term in the equation is the rate of salt entering the tank and the second term is the rate of the salt exiting the tank. It is already apparent that the known concentrations at t = 0, t = 3 and t = 5 will come in handy since the concentration of the liquid exiting the tank consists of unknowns only. Solving the differential equation with the intial value, Q, produces equation 2. 15 24 1 15 24 1 Q L L rt 5 r rt Q L L rt 5 r xt 17 17 L rt 50 50 24 1 24 1 (2) 50 50 L 5 r L L 5 r Since it is known that Q = 0.2 lb/gal * L gal, a substitution is made to derive equation 3. 24 1 24 1 0.14L L rt 5 r rt 0.14L L rt 5 r xt 17 17 L rt 24 1 24 1 (3) 50 50 L 5 r L 5 r At time t = 3 equation 3 is rewritten as equation 4. 24 1 24 1 0.14L L r 3 5 r r 0.42L L r 3 5 r x3 17 51 L r 24 1 24 1 (4) 50 50 L 5 r L 5 r 3 Since it is known that x(3) is also equal to 0.2645*(L-r3), equation 4 is used to create equation 5. 24 1 24 1 0.14L L r 3 5 r r 0.42L L r 3 5 r x3 0.2645L r 3 17 51 L r 24 1 24 1 (5) 50 50 L 5 r L 5 r Solving equation 5 for L results in equation 6. 3e 0.128647868r r L (6) 1 e 0.128647868r Substituting equation 6 back into equation 3 yields equation 7. 24 1 24 1 3e 0.128647868r r 5 r 3e 0.128647868r r 5 r 0.42e 1 e 0.128647868r rt r 0.128647868r 1 e 0.128647868r rt rt 0.14 (7) xt 51 e 0.128647868r r 17 rt 50 1 e 0.128647868r 50 24 1 24 1 r 5 r r 5 r 0.128647868r 0.128647868r 3e 3e 1 e 0.128647868r 1 e 0.128647868r 1 e 0.128647868r At time t = 5 equation 7 is rewritten as equation 8. 24 1 24 1 3e 0.128647868r r 5 r 3e 0.128647868r r 5 r 0.42e 0.128647868r 1 e 0.128647868r r 5 r r 0.7 1 e 0.128647868r r 5 (8) 51 e 0.128647868r r x5 17 r 50 1 e 0.128647868r 10 24 1 24 1 3e 0.128647868r r 5 r 3e 0.128647868r r 5 r 1 e 0.128647868r 1 e 0.128647868r 1 e 0.128647868r Since it is known that x(5) is also equal to 0.2910*(L-r5), equation 6 is substituted for the term L into equation 8 to create equation 9. 3e 0.128647868r r x5 0.291 1 e 0.128647868r r 5 (9) Setting equation 8 and equation 9 equal and then solving for the concentration at t = 5 (0.2910 lb/gal) produces equation 9. 24 1 24 1 3e 0.128647868r r 5 r 3e 0.128647868r r 5 r 0.42e 1 e 0.128647868r r 5 r 0.128647868r r 0.7 1 e 0.128647868r r5 51 e 0.128647868r r 17 r 50 1 e 0.128647868r 10 24 1 24 1 r 5 r 5 r 0.128647868r 3e 3e 0.128647868r r 1 e 0.128647868r 1 e 0.128647868r 1 e 0.128647868r (9) 0.291 3e 0.128647868r r 1 e 0.128647868r r 5 Since Maple will not solve equation 9 numerically for r, a graphical approach must be taken. The “plot” function is used to create a graph for the function of r on the righthand side of 4 equation 9. The resultant plot in figure 1 shows that the value for r is 0.45 gal/min when the concentration is 0.291 lb/gal. Figure 1 : The value of r is approximately 0.4465 but is rounded to 0.45 gal/min for the subsequent calculations. Now that r is known to be 0.45 gal/min, 0.45 is substituted into equation 5 for r to derive equation 10. 0.14L L 1.35 0.189L L 1.35 10.667 10.667 x3 0.2645L 1.35 17 L 0.459 (10) 50 L10.667 L10.667 Once again maple does not solve for L and a graphical approach is taken. Solving equation 10 for the concentration at t = 3 (0.2645 lb/gal) results in equation 11. 0.14L L 1.35 0.189L L 1.35 10.667 10.667 17 L 0.459 0.2645 50 L 10.667 L10.667 (11) L 1.35 Again the “plot” function is used to create a graph for the function of L on the righthand side of equation 11. The resultant plot in figure 2 shows that the value for L is 24 gal when the concentration is 0.2645 lb/gal. Figure 2 : The value of L is 24 gallons for the given concentration of 0.2645 lb/gal. 5 Now that L is known, Q is easily derive using the relation x(0) = Q = 0.2*L as shown in equation 12. x0 Q 0.224 0.45 0 a lg ebra Q 4.8lb (12) As a result of the preceding process, the unknown values were found to be: r = 0.45 gal/min = leak rate L = 24 lb/gal = initial concentration Q = 4.8 lb. = initial amount of salt Normally, differential equations are not this difficult to solve since more information can be derived through experimentation. However, setting up a differential equation can be done in the same manner, no matter what kinds of values are given. A numeric version of the solution is created by substituting the appropriate values into equation 2. 15 24 1 15 24 1 4.8 24 24 0.45 t 5 0.45 0.45 t 4.8 24 24 0.45 t 5 0.45 24 0.45 t 50 50 xt 17 17 24 1 24 1 50 50 24 5 0.45 24 24 5 0.45

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posted: | 10/5/2012 |

language: | English |

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