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A conical flask contains water to height \(H=36.8 \mathrm{mm}\) where the flask diameter is \(D=29.4 \mathrm{mm}\). Water drains out through a smoothly rounded hole of diameter \(d=7.35 \mathrm{mm}\) at the apex of the cone. The flow speed at the exit is approximately \(V=\sqrt{2 g y},\) where \(y\) is the height of the liquid free surface above the hole. A stream of water flows into the top of the flask at constant volume flow rate, \(Q=3.75 \times 10^{-7} \mathrm{m}^{3} / \mathrm{hr}\) Find the volume flow rate from the bottom of the flask. Evaluate the direction and rate of change of water surface level in the flask at this instant.

Short Answer

Expert verified
The volume flow rate out of the flask (\(Q_{out}\)) and the rate of change of the water level can be calculated using the measurements and the equations provided. By comparing \(Q_{in}\) and \(Q_{out}\), you can decide whether the water level is rising, falling, or unchanging, and at what rate in \(m^3/s\).

Step by step solution

01

Determine the radius and area of the hole

First, convert the diameter \(d\) to meters: \(d = 7.35 \times 10^{-3} m\). The radius \(r\) is then \(\frac{d}{2} = 3.675 \times 10^{-3} m\). Using this radius, the area \(A\) of the hole can be calculated using the formula \(A = \pi r^{2}\).
02

Determine the flow speed \(V\) at the hole

To find the flow speed \(V\) at the hole, we use the given equation \(V = \sqrt{2gy}\), where \(g = 9.81 m/s^{2}\) is the gravity constant and \(y = 36.8 \times 10^{-3} m\) is the height of the water surface above the hole.
03

Determine the volume flow rate out of the flask (\(Q_{out}\))

To find the volume flow rate out of the flask \(Q_{out}\), you multiply the flow speed \(V\) by the area of the hole \(A\). The result is the volume flow rate in \(m^3/s\).
04

Determine the volume flow rate into the flask (\(Q_{in}\))

The flow rate of the water into the flask, \(Q_{in}\), is given as \(3.75 \times 10^{-7} \ m^3/hour\). This must be converted to \(m^3/s\) by dividing by \(3600\).
05

Determine the direction and rate of change of water surface level

By comparing \(Q_{in}\) and \(Q_{out}\), you can conclude the rate and direction of change of the water surface. If \(Q_{in} > Q_{out}\), the water surface level is rising with rate equal to the difference \(Q_{in} - Q_{out}\). If \(Q_{in} < Q_{out}\), the water level is falling with a rate equal to the difference \(Q_{out} - Q_{in}\). If \(Q_{in} = Q_{out}\), there is no net change in the water surface level in the flask.

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