Suppose that a large mixing tank initially holds 700 gallons of water in which 5

Question

Suppose that a large mixing tank initially holds 700 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 2 Ib/gal, determine a differential equation for the amount of salt A(t) in the tank at time t > 0. (Use A for A(t). stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 2 lb/gal, determine a differential equation for the amount of salt A(t) in the tank at time t>0. (Use A for A(t).) dA dt Ib/min

Explanation / Answer

Let A(t) be the amount of salt in the tank at time t. Then, the amount of salt in the tank at t=0 is 50 pounds, so the initial condiiton is:

A(0) = 50

Now, salt is flowing into the tank at 3 gal/min at a concentration of 2 pounds per gallon. Therefore, it is entering at (3 gal/min)x(2 lb/gal) = 6 lb/min. This is the first term in the following differential equation.

dA/dt = 6 - [2A/(700+t)]

But, brine is flowing out of the tank as well; this is the last term subtracted from the differential equation.. At any time, t, the tank holds (700 + t) gallons of fluid because a net increase of 1 gallon per minute is happening. 2A/(700+t) represents the pounds of salt leaving the tank at any time, t.

or

dA/dt = ( salt input rate ) - ( salt output rate )

        = ( slat inflow rate ) ( concentration) - ( salt output rate ) ( concentration )

slat inflow rate = 3 gal / min , salt output rate = 2 gal/min

        = ( 3 ) ( 2 ) - 2 [ A(t) / ( 700 + (3-2) t ]

        = 6 - [2A/(700+t)]

so dA/dt = 6 - [2A/(700+t)] lb/min

Note : Volume of the solution at the time of t , is given by

        V(t) = initial volume + t ( change in the volume at the ' t ' th unit of time )

              = 700 + t ( 3 - 2 )

            = 700 + t

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