A right circular conical tank loses water out of a hole at its bottom. Part A: D

Question

A right circular conical tank loses water out of a hole at its bottom.

Part A:
Determine a differential Equation for the height of the water h at time t. The radius of the hole is 2 inches, g= 32 ft/s^2 and the friction factor is .6
Dv/dt = -cA(sub h)(2gh)^1/2
h of tank = 20 ft
radius of top of tank = 8 ft

Part b Find the time required to empty the tank

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Explanation / Answer

Use toricelli's law: v(t)=0.6*sqrt(2*g*h) The outflow of water: V=a*v*t where V is volume outflow, a is area of hole and t is the short interval of time. the outflow must be equal to the the change in volume (Vc) Vc= -(1/3)*b*h (B is the base area of tank, and h is the change in height, the formula is the one for volume of a cone and the negative sign is there because the volume of water will decrease) since V and Vc are equal: a*v*t= -(1/3)*b*h h/t= -3*(a*v)/b we know from toricelli's law that v=0.6*sqrt(2*g*h) and when t approaches 0, h/t becomes dh/dt dh/dt= -3(a*0.6*sqrt(2*g*h))/b a is pi*r^2, r is 2 inches or 0.16 feet, a=0.0256pi. B area of base of cone==> pi*r^2=64pi g=32ft/s^2, hence sqrt(2*g*h)=sqrt(2*32*h)=8*sqrt(h) hence: dh/dt= (-3*0.0256pi*0.6*8sqrt(h))/64pi dh/dt= -0.0576sqrt(h) is the differential equation of height w.r.t time. check for any algebraic errors, hopefully this is correct.

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