## Drainage Time

The figure shown is a cross sectional view of a circular water tank with a 2 m diameter that discharges water freely into the atmosphere. If the time required to drain the water line from 1 m to 0.5 m is 1 minute, what is the output nozzle’s cross-sectional area in square cm? Assume no minor losses in discharge.

##
__
__**Hint**

**Hint**

The time required to drain a circular tank:

$$$\Delta t=\frac{2(A_t/A_0)}{\sqrt{2g}}(h_{1}^{0.5}-h_{2}^{0.5})$$$

where
$$A_t$$
is the tank’s cross-sectional area,
$$A_0$$
is the cross-sectional area of fluid flow,
$$g$$
is the acceleration due to gravity,
$$h_1$$
is the initial fluid line height, and
$$h_2$$
is the final fluid line height.

##
__
__**Hint 2**

**Hint 2**

$$$A_t=\frac{\pi D_{t}^{2}}{4}$$$

where
$$D_t$$
is the circular tank’s diameter.

First, let’s find the tank’s cross-sectional area:

$$$A_t=\frac{\pi D_{t}^{2}}{4}$$$

where
$$D_t$$
is the circular tank’s diameter.

$$$A_t=\frac{\pi (2m)^{2}}{4}=3.14\:m^2$$$

The time required to drain a circular tank:

$$$\Delta t=\frac{2(A_t/A_0)}{\sqrt{2g}}(h_{1}^{0.5}-h_{2}^{0.5})$$$

where
$$A_t$$
is the tank’s cross-sectional area,
$$A_0$$
is the cross-sectional area of fluid flow,
$$g$$
is the acceleration due to gravity,
$$h_1$$
is the initial fluid line height, and
$$h_2$$
is the final fluid line height.

$$$1min\cdot \frac{60sec}{1min}=\frac{2(3.14m^2/A_0)}{\sqrt{2(9.8m/s^2)}}((1m)^{0.5}-(0.5m)^{0.5})$$$

Solving for the nozzle’s cross-sectional area:

$$$60sec=\frac{2(3.14m^2/A_0)}{\sqrt{19.6m/s^2}}(1\sqrt{m}-0.707\sqrt{m})$$$

$$$60s=\frac{2(3.14m^2/A_0)}{4.43\frac{\sqrt{m}}{s}}(0.293\sqrt{m})$$$

$$$\frac{60s(4.43s^{-1})}{(0.293)}=\frac{6.28m^2}{A_0}$$$

$$$A_0=\frac{6.28m^2}{907.17}=0.0069m^2=69\:cm^2$$$

$$$69\:cm^2$$$