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Q:Hydrostatic Pressure vs. Energy Conservation Equation
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[QUOTE="erobz, post: 6832213, member: 700856"] My belief is the contradiction is hidden in the assumption of the problem statement, that the flow in the side branch is 0. If you apply COE (using what I shown above) without invoking the "no flow" condition in the branch the following is the system of equations: $$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 - \sum_{1 \to 2} h_{L_m}$$ $$ \frac{P_2}{\gamma} = \frac{P_1}{\gamma} + z_1 - \sum_{1 \to 2} h_{L_{b}}$$ That system reduces to the established result: $$ \sum_{1 \to 2} h_{L_m} = \sum_{1 \to 2} h_{L_b}$$ $$ k_{m} \frac{Q_{m}^2}{A_{m}^2 2g } = k_{b} \frac{Q_{b}^2}{A_{b}^2 2g }$$ Let the total flow into the node be known and equal to ##Q##. It follows that: $$ k_{m} \frac{(Q - Q_b)^2}{A_{m}^2} = k_{b} \frac{Q_{b}^2}{A_{b}^2}$$ $$ \implies Q^2 - 2 Q Q_b + \left( 1 - \left( \frac{A_m}{A_b} \right)^2 \frac{k_b}{k_m} \right) Q_b^2 = 0 $$ I get the following solution: $$ Q_b = Q \left( \frac{1+ \frac{A_m}{A_b} \sqrt{ \frac{k_b}{k_m} } }{ 1 - \left( \frac{A_m}{A_b} \right)^2 \frac{k_b}{k_m} } \right)$$ So if you take ##k_m, A_m > 0## you shouldn't find ## Q_b \to 0 ##, unless ##Q = 0##, or in the limit as ## k_b \to \infty, A_b \to 0 ##. [/QUOTE]
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Q:Hydrostatic Pressure vs. Energy Conservation Equation
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