A closed and elevated vertical cylindrical tank with diameter 2.00m contains wat

Question

A closed and elevated vertical cylindrical tank with diameter 2.00m contains water to a depth of 0.900m . A worker accidently pokes a circular hole with diameter 0.0190m 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 x 10^3Pa at the surface of the water. Ignore any effects of viscosity.

A. Just after the hole is made, what is the speed of the water as it emerges from the hole? in m/s

B. What is the ratio of this speed to the efflux speed if the top of the tank is open to the air?

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

D. 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?

Please help I have no idea how to solve it

Explanation / Answer

Use Bernoulli's equation: (pressure at surface of tank)+(density)(g)(height of tank)+(1/2)(density)(velocity of water at top)^2=(pressure at bottom of tank with hole)+(density)(g)(0)+(1/2) (density)(v_2). Now, let's solve for v_2 (by the way, I will use atmospheric pressure as p_2, which is 1.013e5 Pa, and p_1=1.063e5, since gauge pressure is just the amount of pressure exceeding atmospheric pressure):

(1.063e5)+(1,000)(g)(.9)+(1/2)(1,000) (so small, can be considered to be zero)^2=(1.013e5)+(1,000)(g)(0)+(1/2) (1,000)(v_2)^2.

v_2=5.06 m/s

(b) If we want to determine v_2 assuming p_1=1.013e5, just substitute this value for 1.063e5. You should get 3.96 m/s. So, the ratio of this speed to the efflux speed is 3.96/5.06=.78, or 78%.

(c) You need the volume flow rate: dV/dt=(v_2)(A_2), where A_2 is the area of the hole in the bottom: pi(0.0190/2)^2. So,

dV/dt=(5.06)(0.9 e-4)=4.54 e-4 m^3/s

Next, we need to determine the volume of the tank, and to determine that, we need the equation for the volume of a cylinder, which is:

V=pi(.9)^2*.9=2.28 m^3

Next, compare these ratios: 4.54 e-4 m^3/s=2.28/t. Now, solve for t:

t=5.03 e3 sec

(d) You need to find dV/dt for the slower velocity now: dV/dt=(0.9e-4)(3.96)=3.564e-4

Compare these ratios: 3.546e-4=2.28/t

t=6.44 e3 sec

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