Question

7.) Water treated in the City of Tallahassee Water Treatment Facility is pumped to the top of the city water tower at a constant rate of 1000 L/min. The water tower has a maximum capacity of 1 × 107 L. At 5 am Monday morning the tank is filled to 90% of maximum capacity. As the city residents wake up and start their day the rate of withdrawal from the water tower increases: the rate of water consumption at 5 am is 700 L/min and increases linearly with time until 9 am when the rate of consumption is 17,500 L/min. The density of the city water is 1 kg/L Hint: don't forget to define the time coordinate and units homogeneously a. Draw the system described above, label all the variables with values or as unknowns. b. Derive and solve the equation for the volume of water in the tank as a function of time. c. How much water minimum needs to be in the water tower at 5 am to ensure the tower is not empty be fore 9 am

Answer 1

V(t) V(out) (iii) Vin

Rate of water input to the tower $V_{in}$ = 1000L/min

The water inside the tower at 5 am =
x 107 001000-L = 9 × 10-1

Water consumption rate at 5 am = 700L/min

dV (0) ปไ

Rate of water consumption rate $\lambda$ =

17500 700 L/min-70L/mn2

Differential equation corresponding to the system can be written as dt2 Net flow rate is the algebraic sum of input and output rates dV(t) =vin dt dt V(t) _ V(0) = Vint-(-) t--M

Vt) V(0) 1000 x (4 x 60) 700 x 4 x 60 x 70 x (4 x 60)- = 240000-168000-2016000-_1944000し So V(0) must be greater than 1944000L at 5am for the tower to be not empty by 9am

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Hope this clarifies your doubt.

All the best...