Evaporation An outdoor decorative pond in the shape of a hemispherical tank is t

Question

Evaporation An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R = 10 ft, that water is pumped in at a rate of pi ft^3/min, and that the tank is initially empty. See Figure 3.2.6. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k = 0.01. (a) The rate of change dV/dt of the volume of the water at time t is a net rate. Use this net rate to determine a differential equation for the height h of the water at time t. The volume of the water shown in the figure is V = pi Rh^2 - 1/3 pi h^, where R = 10. Express the area of the surface of the water A = pi ^2 in terms of h. (b) Solve the differential equation in part (a). Graph the solution. (c) If there were no evaporation, how long would it take the tank to fill. (d) With evaporation, what is the depth of the water at the time found in part(c)? Will the tank ever be filled? Prove your assertion.

Explanation / Answer

we have the volume equation as ; V= Rh2-1/3h3

Area, A= dv/dh= 2Rh-h2

A= 20h2-h2= 19 h2 ft2

here A= r2

r2 =19 h2

h2= r2/19= 0.053r2

we get , h= 0.23 r

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