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_{t}^{2}/2 =
P_{b}/ρ + 0 + v_{b}^{2}/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_{b}^{2}/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 *
10^{3}) / 10^{3}]}^{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

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