Consider two interconnected tanks as shown in the figure abo

Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains of water and 305 g of white tank 2 initially contains 40 L of water and 445 g Water containing 15 g/L of is poured into tank at a rate of 1.5 L/min white the mixture following into tanks a concentration of 40 of and is flowing at the rate of 3.5 L/min. The two connecting tubes have a flow rate of 3.5 L/min from tank 1 to tank 2, and of 2 L/min from tank 2 back to tank 1. Tank 2 is drained at the rate of 5 L/min You may assume that the solutions in each tank are thoroughly mixed so that the concentration of the mixture leaving any tank along any of the tubes has the same concentration of as the tank as a whole. (This is not completely realistic, but as in real physics we are going to work with the appropriate, rather than exact description. The real equations of physics are often too complicated to even write down much less solve) How does the water as each tank change over time?

Solution

Inital Volume of Tank 1 = 60L
Initial Volume of Tank 2 = 40L

First, you have to note the inflow and outflow of water for each tank:
Tank 1: In= 3.5 L/min Out= 3.5 L/min
Tank 2: In= 7 L/minOut= 7 L/min ---->(2L/min + 5L/min)

Now let p(t) = amount of salt, in grams, in Tank 1 as function of time per min
andq(t) = amount of salt, in grams, in Tank 2 as function of time per min
where at time = 0 --> p(0) = 305 grams of salt
q(0) = 445 grams of salt

Thus the derivative is:
p\'(t) = the rate of change with respect to time of the amount of salt in Tank 1
q\'(t) = the rate of change with respect to time of the amount of salt in Tank 2

p\'(t) = (Incoming flow in Tank 1)(concentration fo salt in flow)
+ (Incoming flow from Tank 2)(Concentration of salt in Tank 2)
- (Outgoing flow from Tank 1)(concentration in Tank 1)

p\'(t) = (Incoming flow)(concentration of salt in flow)
+ (Incoming flow from Tank2)[p(t)/(Volume of water in Tank 2)]
- (Outgoing flow from Tank 1)[q(t)/(Volume of water in Tank 1)]

p\'(t) = (15g/L)(1.5L/min) + (2L/min)[p(t)/40L] - (3.5L/min)[q(t)/60L]

p\'(t) = 10 + (2/40)p(t) - (35/600)q(t) g/min
p\'(t) = 10 + (1/20)p(t) - 7/120 q(t) g/min


q\'(t) = (Incoming flow)(Concentration of salt in flow)
+ (incoming flow from Tank 1)[p(t)/(Volume of water in Tank 1)]
- (Outgoing flow from Tank2)[q(t)/(Volume of water in Tank 2)]

q\'(t) = (40L)(3.5L/min) + (3.5L/min)[p(t)/60L] - (7.5L/min)[q(t)/40L]

q\'(t) = 11.42857 + (0.058333)p(t) - (0.1875)q(t) g/min


 Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains of water and 305 g of white tank 2 initially contains 40 L of water and

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