Question 5.17: A concrete tank is 10 m long and 6 m wide, and its sides are...
A concrete tank is 10 m long and 6 m wide, and its sides are vertical. Water enters the tank at the rate of 0.1 m^{3}/s and is discharged from an orifice of area 0.05 m^{2} at its bottom (Fig. 5.39). Initial level of water in the tank from the bottom is 5 m. Find whether the liquid level will start rising, or falling or will remain the same. If the liquid level changes (either rise or fall) then find the value of steady state level to which the liquid will reach. Find also the time taken for the change in the liquid level to be the 60% of its total change. (Take C_{d} = 0.60).
The rise or fall of liquid level at any instant will depend upon the relative magnitudes of instantaneous rate of inflow to and outflow from, the tank. Here, the rate of inflow Q is constant and equals to 0.1 m^{3}/s. Initially, the discharge rate from the orifice
q=0.6 \times 0.05 \sqrt{2 \times 9.81 \times 5}
= 0.297 m ^{3} / s
Since q > Q, the liquid level will start falling. As the liquid level falls, the discharge rate through the orifice decreases, and when it equals to the rate of inflow, the liquid level will neither rise nor fall further. Let H_{s} be this height of steady liquid level from the bottom of the tank.
Then 0.6 \times 0.05 \times \sqrt{2 \times 9.81 \times H_{s}}=0.1
or H_{s}=\frac{(0.1)^{2}}{2 \times 9.81 \times(0.6)^{2} \times(0.05)^{2}}=0.57 m
Total change in the liquid level = (5 – 0.57) = 4.43 m. The liquid will attain a level of 2.34 m (= 5 – 0.6 × 4.43) when the change in the level will be 60% of its final value 4.43.
Consider at any instant t, the height of liquid level in the tank to be h, and let this height fall by dh in a small interval of time d t
The amount of inflow during this time = Q d t = 0.1 d t
and the amount of discharge =0.6 \times 0.05 \sqrt{2 g h} d t
=0.133 \sqrt{h} d t
From continuity,
-A d h=0.133 \sqrt{h} d t-0.1 d t
where A is the area of the tank = 6 × 10 = 60 m ^{2}
Hence, d t=\frac{60 d h}{(0.1-0.133 \sqrt{h})} (5.114)
Let H=0.1-0.133 \sqrt{h}
Then, h=\frac{(0.1-H)^{2}}{0.0177}
d h=-\frac{2(0.1-H)}{0.0177} d H
Substituting the value of dh in Eq. (5.114) and writing H for the denominator, we get,
d t=\frac{-120(0.1-H) d H}{0.0177 H}
=-6780\left(\frac{0.1}{H}-1\right) d H
If T is the time taken for the liquid level to fall from 5 m to 2.34 m, then
\int_{0}^{T} d t=-6780[0.1 \ln (0.1-0.133 \sqrt{h})-(0.1-0.133 \sqrt{h})]_{5}^{2.34}
=-6780\left[0.1 \ln \frac{0.1-0.133 \sqrt{2.34}}{0.1-0.133 \sqrt{5}}+0.133(\sqrt{2.34}-\sqrt{5})\right]
= 5080 s
= 1.41 hours