Question

numeric: 5 points
the circular tank shown below is filled from the top by two faucets, with simultaneous drainage out the bottom via two separate drain pipes. information on the dimensions and/or flowrate at each location is given in the table. for the conditions indicated, it is known that the water - level in the tank does not change, such that the overall output from the drains at the bottom of the tank is exactly equal to the overall input from the two faucets.
given these conditions, use the information provided in the table below to determine the exit velocity of the water (ft/sec) at drain pipe #3. report your answer to two decimal places.
(note: for question #8, you will be asked to upload a file showing the steps you took and the calculations you performed to arrive at this answer. you may wish to download the template in ms word ↓ or pdf ↓ format before you complete question #7, so that you can document your work as you go. it is the responsibility of the student to ensure the submitted work is neat, legible, and that all required elements are visible. a partially completed sample, using the course template, is available here ↓)

pipe 1	diameter = 1.6 inch	velocity = 5.7 ft/sec
pipe 2	diameter = 2.4 inch	velocity = 4.9 ft/sec
pipe 3	diameter = 0.9 inch	velocity = ?????
pipe 4	volumetric flowrate = 8.8 ft³/min

Explanation:

Step1: Calculate the volumetric flow - rate of Pipe 1

The cross - sectional area of a pipe $A=\pi(\frac{d}{2})^2$, where $d$ is the diameter. For Pipe 1, $d_1 = 1.6$ inches. First convert inches to feet: $d_1=\frac{1.6}{12}$ feet. Then $A_1=\pi(\frac{\frac{1.6}{12}}{2})^2=\pi(\frac{1.6}{24})^2$ square feet. The velocity of Pipe 1 is $v_1 = 5.7$ ft/sec. The volumetric flow - rate of Pipe 1, $Q_1=A_1v_1=\pi(\frac{1.6}{24})^2\times5.7$ $ft^3/sec$.

Step2: Calculate the volumetric flow - rate of Pipe 2

For Pipe 2, $d_2 = 2.4$ inches. Convert to feet: $d_2=\frac{2.4}{12}$ feet. $A_2=\pi(\frac{\frac{2.4}{12}}{2})^2=\pi(\frac{2.4}{24})^2$ square feet. The velocity of Pipe 2 is $v_2 = 4.9$ ft/sec. The volumetric flow - rate of Pipe 2, $Q_2=A_2v_2=\pi(\frac{2.4}{24})^2\times4.9$ $ft^3/sec$.

Step3: Calculate the total volumetric inflow

The volumetric flow - rate of Pipe 4 is $Q_4 = 8.8$ $ft^3/min$. Convert it to $ft^3/sec$: $Q_4=\frac{8.8}{60}$ $ft^3/sec$. The total volumetric inflow $Q_{in}=Q_1 + Q_2+Q_4$.

Step4: Calculate the cross - sectional area of Pipe 3

For Pipe 3, $d_3 = 0.9$ inches. Convert to feet: $d_3=\frac{0.9}{12}$ feet. $A_3=\pi(\frac{\frac{0.9}{12}}{2})^2=\pi(\frac{0.9}{24})^2$ square feet.

Step5: Calculate the exit velocity of Pipe 3

Since the overall output from the drains at the bottom of the tank is equal to the overall input from the two faucets and Pipe 4, and the output from Pipe 3 is part of the total output. Using the formula $Q = Av$, we can find the velocity $v_3$ of Pipe 3. $v_3=\frac{Q_{in}}{A_3}$.

First, calculate $Q_1$:
\[

$$\begin{align*} A_1&=\pi(\frac{1.6/12}{2})^2=\pi\times(\frac{1.6}{24})^2\approx3.14\times(\frac{1.6}{24})^2\approx3.14\times\frac{2.56}{576}\approx0.014 \end{align*}$$

\]
$Q_1 = A_1v_1=0.014\times5.7 = 0.0798$ $ft^3/sec$.

Next, calculate $Q_2$:
\[

$$\begin{align*} A_2&=\pi(\frac{2.4/12}{2})^2=\pi\times(\frac{2.4}{24})^2\approx3.14\times(\frac{2.4}{24})^2=3.14\times\frac{5.76}{576}\approx0.0314 \end{align*}$$

\]
$Q_2 = A_2v_2=0.0314\times4.9 = 0.1539$ $ft^3/sec$.

$Q_4=\frac{8.8}{60}\approx0.1467$ $ft^3/sec$.

$Q_{in}=Q_1 + Q_2+Q_4=0.0798 + 0.1539+0.1467=0.3804$ $ft^3/sec$.

Then, calculate $A_3$:
\[

$$\begin{align*} A_3&=\pi(\frac{0.9/12}{2})^2=\pi\times(\frac{0.9}{24})^2\approx3.14\times\frac{0.81}{576}\approx0.0044 \end{align*}$$

\]

Finally, calculate $v_3$:
$v_3=\frac{Q_{in}}{A_3}=\frac{0.3804}{0.0044}\approx86.45$ ft/sec.

Answer:

$86.45$ ft/sec