Set up a Differential equation to solve the following: A tank with a surface are

Question

Set up a Differential equation to solve the following:
A tank with a surface area A = 1 ft2 is initially filled to a depth y = 0.55 ft. Water is pumped into the tank at a constant rate, 0.45 ft3/min. Water drains from the tank through a sand filter. The flow rate through the filter is proportional to the water depth, y (by Darcy’s law for groundwater seepage). The outflow rate is 0.18 ft3/min when the depth is y = 1.05 ft. a) What is the equilibrium depth of water in the tank? B) After t = 9 min., what will be the water depth?  

Explanation / Answer

Vin rate=0.45 ft3/min, V0=0.55 ft3

Outflow rate: dVout/dt=ky,

where k=0.18/1.05=0.1714.

Rate of change of volume=In flow- Out-flow

A dy/dt=dVin/dt - dVout/dt

1. dy/dt=0.45-0.1714*y

dy/dt+0.1714y=0.45

(i) From this equation, equilibrium depth is dy/dt=0;

0.1714*y-0.45=0

y=2.635ft

(ii) Solving the differential equation we get,

y=2.63-1.63*e^(-0.1714t), where t is in minutes

so at t=9min,

y=2.28ft.

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