Question

A closed and elevated vertical cylindrical tank with diameter 1.40 m contains water to a depth of 0.800 m . A worker accidently pokes a circular hole with diameter 0.0140 m in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of 5.00×103Pa at the surface of the water. Ignore any effects of viscosity.

Questions

1. How much time does it take for all the water to drain from the tank?

2. What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

*If your answer for #1 should be given in minutes not secons and also if its 26 or 26.25 is wrong. For number 2 if yur answer is 0.78 is wrong

Answer 1

1) Applying Bernoulli's equation, we get,

P_t/ρ + gh + v_t²/2 = P_b/ρ + 0 + v_b²/2

Since the area of the hole is much smallert than the area at the top, v_t << v_b. So, we can ignore it.

v_b²/2 = gh + (P_t - P_b)/ρ

=> v_b = [2gh + 2(P_t - P_b)/ρ]^1/2

=> v_b = {(2 * 9.81 * 0.800) + [2 * (5 * 10³) / 10³]}^1/2 = 5.07 m/s

So, time taken to empty the tank is,

t = V/Av_b = h/v_b = 0.800 / 5.07 = 0.16 s

2) When the top of the tank is open to air, P_t = P_b. So,

v_b' = (2gh)^1/2 = (2 * 9.81 * 0.800)^1/2 = 3.96 m/s

Time taken to empty the tank, t' = h/v_b' = 0.800 / 3.96 = 0.20 s

So, ratio of time taken, t/t' = 0.16 / 0.20 = 0.80