For the inlet and outlet conditions, we can write this as $P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho gh_{1} = P_{2} + \frac{1}{2} \rho v_{2}^{2} + \rho gh_{2}$. Since the tube is horizontal, $h_{1}$ and $h_{2}$ are the same and cancel out. We are left with $P_{1}
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