Water flows through the 100-mm-diameter pipe at 0.025 m³/s, Fig. 5–19a. If the pressure at A is 225 kPa, determine the pressure at C, and draw the energy and hydraulic grade lines from A to D. Neglect friction losses and take γw = 9.81 kN/m³.

Question

Water flows through the 100-mm-diameter pipe at 0.025 m³/s, Fig. 5–19a. If the pressure at A is 225 kPa, determine the pressure at C, and draw the energy and hydraulic grade lines from A to D. Neglect friction losses and take [latex] \gamma_w=9.81 kN / m ^3 [/latex].

Accepted Answer

Fluid Description. We have steady flow, and we will assume water is an ideal fluid.

Bernoulli Equation. The average velocity of flow through the pipe is

[latex] \begin{gathered} Q=V A ; \quad 0.025 m ^3 / s =V\left[\pi(0.05 m )^2 ight] \ V=\frac{10}{\pi} m / s \end{gathered} [/latex]

Since the pipe has a constant diameter over its length, this velocity must remain constant in order to satisfy the continuity equation.
The pressure at A and B is the same, since segment AB is horizontal and we have neglected the effects of viscous friction. We can find the pressure at C (and D) by applying the Bernoulli equation at points B and C, which lie on the same streamline. With the gravitational datum through AB, noting that [latex] V_B=V_C=V [/latex] we have

[latex] \begin{gathered} \frac{p_B}{\gamma_w}+\frac{V_B^2}{2 g}+z_B=\frac{p_C}{\gamma}+\frac{V_C^2}{2 g}+z_C \ \frac{225\left(10^3 ight) N / m ^2}{9.81\left(10^3 ight) N / m ^3}+\frac{\left(\frac{10}{\pi} m / s ight)^2}{2\left(9.81 m / s ^2 ight)}+0= \ \frac{p_C}{9.81\left(10^3 ight) N / m ^3}+\frac{\left(\frac{10}{\pi} m / s ight)^2}{2\left(9.81 m / s ^2 ight)}+(4 m ) \sin 30^{\circ} \ p_C=p_D=205.38\left(10^3 ight) Pa =205 kPa \end{gathered} [/latex]

The pressure at B (225 kPa) is higher than the pressure at C (205 kPa) because the pressure at B has to do work to lift the water to C.

EGL and HGL. The total head remains constant, because there are no friction losses. This head can be determined from the conditions at any point along the pipe. Using point B, we have

[latex] \begin{aligned} H & =\frac{p_B}{\gamma}+\frac{V_B^2}{2 g}+z_B=\frac{225\left(10^3 ight) N / m ^2}{9.81\left(10^3 ight) N / m ^3}+\frac{\left(\frac{10}{\pi} m / s ight)^2}{2\left(9.81 m / s ^2 ight)}+0 \ & =23.452 m =23.5 m \end{aligned} [/latex]

The EGL is located as shown in Fig. 5–19b. The location of the HGL along AB is

[latex] \frac{p_B}{\gamma}+z_B=\frac{225\left(10^3 ight) N / m ^2}{9.81\left(10^3 ight) N / m ^3}+0=22.936 m =22.9 m [/latex]

or along CD it is

[latex] \frac{p_C}{\gamma}+z_C=\frac{205.38\left(10^3 ight) N / m ^2}{9.81\left(10^3 ight) N / m ^3}+(4 m ) \sin 30^{\circ}=22.936 m =22.9 m [/latex]

Along BC the elevation head will increase, and the pressure head will correspondingly decrease [latex] \left(p_C / \gamma The velocity throughout the pipe is constant, and so the velocity head is always

[latex] \frac{V^2}{2 g}=\frac{\left(\frac{10}{\pi} m / s ight)^2}{2\left(9.81 m / s ^2 ight)}=0.516 m [/latex]

Question 5.7: Water flows through the 100-mm-diameter pipe at 0.025 m³/s, ...

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