Question

Water flows through a valve at the rate of 1000 lbm/s. The pressure just upstream of the valve is 90 psi and the pressure drop across the valve is 50 psi. The inside diameters of the valve inlet and outlet pipes are 12 and 24 in. The flow occurs in a horizontal plane. Determine the loss in available energy units across the valve.

Answer 1

Answer:

$E_{loss}=5660\frac{ft\cdot lb}{slug}$

Explanation:

The loss in avaliable energy associated with the incompressible, steady flow is given by:

$E_{loss}=\frac{P_1-P_2}{\rho}+\frac{V_1^2-V_2^2}{2}$

According to the law of conservation of mass:

$V=\frac{\dot{m}}{\rho A}\\V=\frac{\dot{m}}{\rho \pi r^2}\\V=\frac{\dot{m}}{\rho\frac{\pi d^2}{4}}$

$\dot{m}$ is the rate at which the water flows. Replacing the volumes in the enegy loss formula:

$E_{loss}=\frac{P_1-P_2}{\rho}+\frac{(\frac{\dot{m}}{\rho\frac{\pi d_1^2}{4}})^2-(\frac{\dot{m}}{\rho \frac{\pi d_2^2}{4}})^2}{2}\\E_{loss}=\frac{P_1-P_2}{\rho}+\frac{1}{2}(\frac{4\dot{m}}{\rho \pi})^2(\frac{1}{d_1^4}-\frac{1}{d_2^4}})\\E_loss=\frac{P_1-P_2}{\rho}+\frac{1}{2}(\frac{4\dot{m}}{\rho \pi})^2(\frac{1}{d_1^4}-\frac{1}{d_2^4}})$

Replacing the given values, we obtain this result:

$E_{loss}=5660\frac{ft\cdot lb}{slug}$