## Time to Drain

The figure shown is a cross sectional view of a circular water tank with a 4 ft diameter that discharges water freely into the atmosphere. If the output nozzle has a 5 in^2 cross-sectional area, what is the time (in seconds) required to drain the water line from 6 ft to 2 ft? 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.

$$$A_t=\frac{\pi (4ft)^{2}}{4}=12.566\:ft^2$$$

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.

$$$\Delta t=\frac{2(12.57ft^2/(5\div 12\div 12ft^2))}{\sqrt{2(32.2ft/s^2)}}((6ft)^{0.5}-(2ft)^{0.5})$$$

$$$\Delta t=\frac{2(12.57/0.0347)}{\sqrt{64.4ft/s^2}}(2.45\sqrt{ft}-1.41\sqrt{ft})$$$

$$$\Delta t=\frac{723.8}{8.02\frac{\sqrt{ft}}{s}}(1.04\sqrt{ft})=93.9s=94\:sec$$$

94 sec