A closed and elevated vertical cylindrical tank with diameter 1.90m contains wat

Question

A closed and elevated vertical cylindrical tank with diameter 1.90m contains water to a depth of 0.900m . A worker accidently pokes a circular hole with diameter 0.0150m 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

A closed and elevated vertical cylindrical tank with diameter 1.90m contains water to a depth of 0.900m . A worker accidently pokes a circular hole with diameter 0.0150m 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

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)(.8)+(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(0.0140/2)^2. So,

dV/dt=(5.06)(1.54e-4)=.779e-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(.8)^2*.8=1.6 m^3

Next, compare these ratios: .779e-4 m^3/s=1.6/t. Now, solve for t:

t=2.05e3 sec

(d) You need to find dV/dt for the slower velocity now: dV/dt=(1.54e-4)(3.96)=6.09e-4

Compare these ratios: 6.09e-4=1.6/t

t=2.6e3 sec

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