Question

4-12. The morning inspection of the tank farm finds a leak in the turpentine tank. The leak is repaired. An investigation finds that the leak was 0.1 in. in diameter and 7 ft above the tank bottom. Records show that the turpentine level in the tank was 17.3 ft before the leak occurred and 13.0 ft after the leak was repaired. The tank diameter is 15 ft. Determine (a) the total amount of turpentine spilled, (b) the maximum spill rate, and (c) the total time the leak was active. The density of turpentine at these conditions is 55 lbm/ft3.

Answer 1

Answer:

a) $V=759.8727\ ft^3$

b) $\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}$

c) $t\approx29.541\ s$

Explanation:

Given:

• diameter of hole in the tank, $d'=0.1\ in=\frac{1}{120}\ ft$

• position of the hole form the tank bottom, $h' =7\ ft$

• initial level of turpentine in the tank before the leakage, $h_i=17.3\ ft$

• level of turpentine in the tank after the repair of leakage, $h_f=13\ ft$

• diameter of the tank, $d=15\ ft$

• density of turpentine oil, $\rho=55\ lbm.ft^3$

a)

Now, volume of turpentine spilled:

$V=A.(h_i-h_f)$

where:

$A=$ area of the cross section of the tank's volume

$V=\pi .\frac{d^2}{4} \times(h_i-h_f)$

$V=\pi\times\frac{15^2}{4} \times(17.3-13)$

$V=759.8727\ ft^3$

b)

When the tank was full the liquid level was highest:

so velocity form the height of the hole will be given as:

$v=\sqrt{2g.(h_i-h')}$

$v=\sqrt{2\times 32.12\times (17.3-7) }$

$v=25.722\ ft.s^{-1}$

Now we have the flow rate of the spillage given by:

$\dot V=(\pi.\frac{d'^2}{4}) \times v$

$\dot V=\pi\times \frac{(\frac{1}{120})^2 }{4} \times 25.722$

$\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}$

c)

Total time the leak was active can be calculated as:

$t=\frac{V}{\dot V}$

$t=\frac{759.8727}{25.722}$

$t\approx29.541\ s$