Question

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 observed that the water level in the tank rises steadily by four inches per minute. based on this information, determine the mass flowrate (lbm/min) of water out the bottom of the tank at pipe 3. report your answer to the nearest whole number. pipe 1: diameter = 1.3 inch, velocity = 4.8 ft/sec; pipe 2: diameter = 2.25 inch, velocity = 4.4 ft/sec; the density of water = 62.25 lbm/ft³; pipe 3: diameter = 2.6 inch, velocity = ?????; pipe 4: volumetric flowrate = 2.0 ft³/min

Explanation:

Step1: Recall mass - flow rate formula

The mass - flow rate formula is $\dot{m}=
ho A v$, where $\dot{m}$ is the mass - flow rate, $
ho$ is the density, $A$ is the cross - sectional area of the pipe, and $v$ is the velocity of the fluid. The cross - sectional area of a circular pipe is $A = \frac{\pi}{4}d^{2}$, where $d$ is the diameter of the pipe.

Step2: Calculate cross - sectional area for Pipe 1

Given $d_1 = 1.3$ inches. First, convert inches to feet: $d_1=1.3/12$ feet. Then $A_1=\frac{\pi}{4}(\frac{1.3}{12})^{2}$ square feet.

Step3: Calculate mass - flow rate for Pipe 1

Given $
ho = 62.25\ lb_m/ft^{3}$ and $v_1 = 4.8\ ft/s$. Using $\dot{m}_1=
ho A_1v_1$, substitute the values of $
ho$, $A_1$, and $v_1$:
\[

$$\begin{align*} A_1&=\frac{\pi}{4}(\frac{1.3}{12})^{2}\approx\frac{\pi}{4}\times(\ 0.1083)^{2}\approx\frac{\pi}{4}\times0.01173\\ &\approx0.00922\ ft^{2}\\ \dot{m}_1&=62.25\times0.00922\times4.8\\ &=62.25\times0.044256\\ &\approx2.75\ lb_m/s \end{align*}$$

\]
Convert to $lb_m/min$: $\dot{m}_1 = 2.75\times60=165\ lb_m/min$

Step4: Calculate cross - sectional area for Pipe 2

Given $d_2 = 2.25$ inches. Convert to feet: $d_2 = 2.25/12$ feet. Then $A_2=\frac{\pi}{4}(\frac{2.25}{12})^{2}$ square feet.
\[

$$\begin{align*} A_2&=\frac{\pi}{4}(\frac{2.25}{12})^{2}\approx\frac{\pi}{4}\times(0.1875)^{2}\approx\frac{\pi}{4}\times0.03516\\ &\approx0.0276\ ft^{2} \end{align*}$$

\]

Step5: Calculate mass - flow rate for Pipe 2

Given $v_2 = 4.4\ ft/s$ and $
ho = 62.25\ lb_m/ft^{3}$. Using $\dot{m}_2=
ho A_2v_2$:
\[

$$\begin{align*} \dot{m}_2&=62.25\times0.0276\times4.4\\ &=62.25\times0.12144\\ &\approx7.56\ lb_m/s \end{align*}$$

\]
Convert to $lb_m/min$: $\dot{m}_2=7.56\times60 = 454\ lb_m/min$

Step6: Calculate total mass - flow rate out

The total mass - flow rate out of the tank at the bottom is $\dot{m}_{out}=\dot{m}_1+\dot{m}_2$.
$\dot{m}_{out}=165 + 454=619\ lb_m/min$

Answer:

The mass - flow rate of water out the bottom of the tank is $619\ lb_m/min$