A cylindrical tank is being filled with water. The tank is initially empty but t

Question

A cylindrical tank is being filled with water. The tank is initially empty but then water begins to flow into it at a rate of 82 .0O0 kg/min. There is a small hole of radius r = 0.8000 cm at the bottom of the tank where water can escape. Because the flow rate of water leaving the hole is initially at a slower rate than water entering the tank, the water level rises. The average velocity of water leaving through the hole at the bottom of the tank is not constant and is a function of the height of the water level in the tank: v = Squareroot 2 gh where g is the gravitational constant of Earth, 9.81 m/s^2. Additionally, the density of water at this temperature is 1.000 times 10^3 kg/m^3. What is the maximum height of the water level in the tank, if it is initially empty? h = If the radius of the tank is R = 0.9000 m, how will it take the tank to reach 90.0% of this water level? t =

Explanation / Answer

For solving this problem we need to do a volumetric flow balance on the tank, as follows:

qin - qout = A*(dh(t) / dt)

Where we know that:

qin = 82 kg/min * (1m3 / 1000kg) = 0.082 m3/min = 0.0014 m3/sec

qout = vA = sqrt(2gh) * (0.016m)2 * pi / 4

qout = sqrt (19.62h) * 0.000201 m2

Substituting values:

0.0014 m3/sec - sqrt (19.62h) * 0.000201 m2 = A*(dh(t) / dt)

Maximum will be when derivative or change is 0:

sqrt (19.62h) * 0.000201 m2 = 0.0014 m3/sec

sqrt (19.62h) = 6.97

19.62 h = 48.51

h = 2.47 m

Now, knowing diameter of tank:

A = (1.8 m)2 * pi / 4 = 2.544 m2

0.0014 m3/sec - sqrt (19.62h) * 0.000201 m2 = 2.544*(dh(t) / dt)

Separating terms:

0.00055 - 0.00035 sqrt(h) = (dh(t) / dt)

dt = dh(t) / (0.00055 - 0.00035 sqrt(h))

Solving from 0 to 90 percent which is 2.223 m:

t = 18167.8 s = 5 hours

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