Consider two interconnected tanks as shown in the figure above. Tank 1 initial c
ID: 2865496 • Letter: C
Question
Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains 30 L (liters) of water and 480 g of salt, while tank 2 initially contains 20 L of water and 485 g of salt. Water containing 25 g/L of salt is poured into tank1 at a rate of 3 L/min while the mixture flowing into tank 2 contains a salt concentration of 10 g/L of salt and is flowing at the rate of 3 L/min. The two connecting tubes have a flow rate of 7 L/min from tank 1 to tank 2; and of 4 L/min from tank 2 back to tank 1. Tank 2 is drained at the rate of 6 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 salt as the tank as a whole. (This is not completely realistic, but as in real physics, we are going to work with the approximate, rather than exact description. The 'real' equations of physics are often too complicated to even write down precisely, much less solve.)
How does the water in each tank change over time?
Let p(t) and q(t) be the amount of salt in g at time t in tanks 1 and 2 respectively. Write differential equations for p and q. (As usual, use the symbols p and q rather than p(t) and q(t).)
p'=
q'=
Given the initial values:
[p(0)] = [ ]
[q(0)] = [ ]
Explanation / Answer
Set up the mass-balance equations for each of the tanks. That is, account for the salt being added and removed from each tank due to the different flows. Note that the net fluid flows into and out of each tank are zero, so the fluid volume in each tank is constant.
For tank 1,
dp(t)/dt = (25 gm/L)*(3 L/min) - p(t)*(7 L/min)/(30 L) + q(t)*(4 L/min)/(20 L)
and for tank 2:
dq(t)/dt = (10 gm/L)*(3 L/min) + p(t)*(7 L/min)/(30 L) - q(t)*(4 L/min)/(20 L) - q(t)*(6 L/min)/(20 L)
where p(t) and q(t) are the amounts of salt (in grams) in tanks 1 and 2, respectively, at time t.
Simplifying:
p' = dp/dt = 75 gm/min - (7/30)*p/min +(1/5)*q/min
q' = dq/dt = 30 gm/min +(7/30)*p/min - (1/2)*q/min
The equilibrium (steady state) solution is when dp/dt = dq/dt = 0, s:
0 = 75gm - (7/30)*(p_ss) + (1/5)*(q_ss)
0 = 30gm + (7/30)*(p_ss) - (1/2)*(q_ss)
Add these two equations together to get:
0 = 100 -(3/10)*(q_ss)
q_ss = 1000/3 gm
Plug this back into one of the above equations to solve for p_ss:
0 = 75gm - (7/30)*(p_ss) + (1/5)*(1000/3)
p_ss = 4250/7