We have two interconnected tanks. Initially tank 1 contains 30 gallons of water

Question

We have two interconnected tanks. Initially tank 1 contains 30 gallons of water and 25 oz of salt, while tank 2 initially contains 20 gallons of water and 15 oz of salt. Water containing 1 0z/gal of salt flows into tank 1 at 1.5 gal/min. The mixture flows from tank 1 to tank 2 at 3 gal/min. Water containing 3 oz/gal of salt also flows into tank 2 at 1 gal/min. The mixture drains from tank 2 at the rate of 4 gal/min, of which some flows back into tank 1 at 1.5 gal/min.

1) Let Q1(t) and Q 2(t) be the amount of salt in oz in each tank at time t. Write the differential equations and initial conditions that model the flow process.

2) Find the values of Q1 and Q 2 for which the system is in equilibrium. Let these be Qe1 and Q e2

3) Let    x1 = Q1 (t) - Q e1    and    x2 = Q2 (t) - Q e2 , determine an initial value problem for X1 and X2

4) When do tank 1 and tank 2 reach equilibrium?

Explanation / Answer

Ans 1) For Q'1, 1.5 gal/min of water enter Tank1 and it contains 1oz/gal of salt therefore 1.5gal/min * 1oz/gal = 1.5 oz/min of salt enter Tank1,

Similarly, 3gal/min of solution leave Tank1 and it contains Q1(t) / 30 oz/gal of salt therefore 0.1Q1(t) oz/min of salt leave Tank1.

Using similar argument, we have 0.075Q2(t) oz/min of salt entering Tank1.

So, Q'1 = 1.5 - 0.1Q1(t) + 0.075Q2(t) ------- (1)

For Tank2,    Q'2 = 3 - 0.2Q2(t) + 0.1Q1(t)    ------ (2)

The initial conditions for Q1(0) = 25 oz and Q2(0) = 15oz, the general form of the equations is

x'1 = p11(t) x1 + p12(t) x2 + g1(t)

x'2 = p21(t)x1 + p22(t) x2 + g2(t)

Since, g1(t) = 1.5 (not equal to 0) and g2(t) = 3 (not equal to 0), it shows that system is nonhomogeneous system.

Ans 2) At equilibrium, Q'1 = Q'2 = 0, From eqns (1) and (2)

0 = 1.5 - 0.1Q1(t) + 0.075Q2(t)     ----- (3)

0 = 3 - 0.2Q2(t) + 0.1Q1(t)       ------ (4)

From (3), Q1(t) = (1.5 + 0.075Q2(t)) / 0.1 ----- (5)

Plug value of Q1(t) in (4)

   0 = 3 - 0.2Q2(t) + 0.1 [ (1.5 + 0.075Q2(t)) / 0.1]

    0 = 3 - 0.2Q2(t) + 1.5 + 0.075Q2(t)

   0 = 4.5 - 0.125Q2(t)

so, Q2(t) = 4.5 / 0.125 = 8 oz/min = 8oz/min * 4min = 36 oz

Plug value of Q2(t) in (5)

Qe1 = (1.5 + 0.075*8) / 0.1 = 21oz/min * 2min = 42 oz

Ans 3) Using values found in ans2

x1(t) = Q1(t) - Qe1 and    x2(t) = Q2(t) - Qe2.

x1(t) = Q1(t) - 42      and

x2(t) = Q2(t) - 36

therefore, x' = Q'1 = 1.5 - 0.1Q1(t) + 0.075Q2(t)

                                = 1.5 - 0.1 (x1 + 42) + 0.075 (x2 + 36)

                               = 0.1 x1 - 0.075x2

similary for x'2 = Q'2 = 0.1x1 - 0.2x2

When t=0, x1(0) = Q1(0) - 42 = 25 - 42 = -17 and

x2(0) = Q2(0) - 36 = 15 - 36 = -21

So initial conditions are x1(0) = - 17 and x2(0) = -21.             

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.