Tank 1 initially contains 50 gals of water with 10 oz of salt in it, while Tank

Question

Tank 1 initially contains 50 gals of water with 10 oz of salt in it, while Tank 2 initially contains 20 gals of water with 15 oz of salt in it. Water containing 2 oz/gal of salt flows into Tank 1 at a rate of 5 gal/min and the well-stirred mixture flows from Tank 1 into Tank 2 at the same rate of 5 gal/min. The solution in Tank 2 flows out to the ground at a rate of 5 gal/min. If x1(t) and x2(t) represent the number of ounces of salt in Tank 1 and Tank 2, respectively, set up but do not solve an initial value problem describing this system.

Explanation / Answer

Note that the net amount of water flowing in and out of both tank 1 and tank2 is 0. Let S1 be the salt in tank 1 and S2 be the salt in tank 2. S1(0) = 10 oz S2(0) = 15 oz 2oz/gal flows into tank 1 at 5 gal/min, so 2 oz/gal * 5 gal/min = 10 oz/min As the 50 gal mixture flows out at 5 gal/min, then salt leaves at S1/50 gal * 5 gal/min = S1/10 /min. In tank 2, salt flows in at S1/10 /min. It flows out at S2/20 gal * 5 gal/min = S2/4 /min. Thus, measuring S1 and S2 in ounces, and the rate of change in change/minute. dS1/dt = 10 - S1/10 dS2/dt = S1/10 - S2/4 Again, S1(0) = 10 S2(0) = 15 As a check, note that at equilibrium, 10 - S1/10 = 0 implies S1 = 100, so there are 100 oz in the 50 gal., or 2 oz/gal. This equals the concentration of salt entering. S1/10-S2/4 = 0 implies, as s1 = 100 100/10 - S2/4 = 0 10S2 = 400 S2 = 40 The concentration of salt is 40 oz/20 gal = 2oz/gal, again the concentration of salt entering.

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