Question

1. the tank shown in the accompanying figure is being filled by pipes 1 and 2. if the water level is to remain constant, what is the volumetric flow rate of water leaving the tank at pipe 3? what is the average velocity of the water leaving the tank? pipe 1: ( d_1 = 1 ) in. ( v_1 = 2 ) ft/s pipe 2: ( d_2 = 1.75 ) in. ( v_2 = 1.5 ) ft/s pipe 3: ( d_3 = 1.5 ) in. ( v_3 =? )

Explanation:

Step1: Calculate the volumetric flow rate of pipe 1

The formula for volumetric flow rate \(Q = A\times V\), where \(A=\frac{\pi d^{2}}{4}\). For pipe 1, \(d_1 = 1\space in=\frac{1}{12}\space ft\), \(V_1 = 2\space ft/s\). Then \(A_1=\frac{\pi(\frac{1}{12})^{2}}{4}\space ft^{2}\), and \(Q_1 = A_1\times V_1=\frac{\pi(\frac{1}{12})^{2}}{4}\times2\space ft^{3}/s\approx0.011\space ft^{3}/s\)

Step2: Calculate the volumetric flow rate of pipe 2

For pipe 2, \(d_2 = 1.75\space in=\frac{1.75}{12}\space ft\), \(V_2 = 1.5\space ft/s\). Then \(A_2=\frac{\pi(\frac{1.75}{12})^{2}}{4}\space ft^{2}\), and \(Q_2=A_2\times V_2=\frac{\pi(\frac{1.75}{12})^{2}}{4}\times1.5\space ft^{3}/s\approx0.027\space ft^{3}/s\)

Step3: Determine the volumetric flow rate of pipe 3

Since the water level is constant, by the principle of conservation of mass (continuity equation for incompressible flow \(Q_{in}=Q_{out}\)), \(Q_3=Q_1 + Q_2\). So \(Q_3=(0.011 + 0.027)\space ft^{3}/s=0.038\space ft^{3}/s\)

Step4: Calculate the average velocity of water in pipe 3

For pipe 3, \(d_3 = 1.5\space in=\frac{1.5}{12}\space ft\), \(A_3=\frac{\pi(\frac{1.5}{12})^{2}}{4}\space ft^{2}\). Using \(V_3=\frac{Q_3}{A_3}\), \(V_3=\frac{0.038}{\frac{\pi(\frac{1.5}{12})^{2}}{4}}\space ft/s\approx4.0\space ft/s\)

Answer:

The volumetric flow rate of water leaving the tank at pipe 3 is approximately \(0.038\space ft^{3}/s\), and the average velocity of the water leaving the tank is approximately \(4.0\space ft/s\)