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Suppose the tank has a vertex angle of 60° and the circular hole has radius 4 inches. Determine the differential equation governing the height h of water. Use c = 0.6 and g = 32 ft/s2. dh dt = Solve the initial value problem that assumes the height of the water is initially 8 feet. h(t) = If the height of the water is initially 8 feet, how long (in minutes) will it take the tank to empty? (Round your answer to two decimal places.)

Answer :

The time it will take the tank to be empty is;

t = 1.26 minutes

The image of the initial tank is missing and so i have attached it.

Let's call the area of the hole be A_h

  • From conservation of energy, velocity at which water leaves the tank is; v = √2gh

Thus, volumetric rate;  = A_h(√2gh)

  • If we consider possible friction and contraction, we differentiate to get;

dV/dt = -cA_h(√2gh)

Now, volume of tank is;

V = ¹/₃πr²h

V' = dv/dt =  (¹/₃πr²h)dh/dt

Thus; -cA_h(√2gh) = (¹/₃πr²h)dh/dt

  • Integrating to get t gives us;

t = (2(√H - √h))/(3c√2g × (r'/r)²)

where;

r' is radius of circular hole = 4 inches = 0.333 ft

h = 0 since the tank has a hole

c = 0.6

g = 32 ft/s²

H = 8 ft

  • Since the vertex has an angle of 60°, then a line of symmetry across it divides the angle into two which will be 30° each. We can use trigonometry to find the current radius r.

Thus; r/H = tan 30°

r = 8 tan 30°

r = 4.6188 ft

Thus;

t = (2(√8 - √0))/(3 × 0.6(√2 × 32) × (0.333/4.6188)²)

t = 75.5757 seconds

Converting to minutes gives; t = 1.26 minutes

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