Water flows through a horizontal branching pipe as shown
in Fig. P3.109. Determine the pressure at section (3).
Answers
The pressure of the water in section 3 of the horizontal pipe is -945 kPa.
Volumetric flow rate at equilibrium
The volumetric flow rate at equilibrium is given as flows;
mass in = mass out
[tex]Q_1 = Q_2 + Q_3[/tex]
where;
Q1 is the flow rate in section 1Q2 is the flow rate in section 2Q3 is the flow rate in section 3Pressure of water in section 3The pressure of water in section 3 of the horizontal pipe is calculated as follows;
[tex]F_2 + F_3 = F_1\\\\\frac{P_2}{A_2} + \frac{P_3}{A_3} = \frac{P_1}{A_1} \\\\\frac{350\ kPa}{0.02} + \frac{P_3}{0.07} = \frac{400 \ kPa}{0.1}\\\\\\frac{P_3}{0.07} = \frac{400 \ kPa}{0.1} - \frac{350\ kPa}{0.02} \\\\\frac{P_3}{0.07} = 4000 \ kPa/m^2 \ - \ 17,500 \ kPa/m^2\\\\\frac{P_3}{0.07} = -13,500 \ kPa/m^2 \\\\P_3 = 0.07 \ m^2 \ \ \times \ \ -13,500 \ kPa/m^2\\\\P_3 = -945 \ kPa[/tex]
Thus, the pressure of the water in section 3 of the horizontal pipe is -945 kPa.
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