Transmission main with a control valve: 16-Civ-A5 Hydraulic Engineering, May 2016, Q3

Question 3: Transmission main with a control valve (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Find. (a) the valve coefficient $\tau$; (b) the downstream reservoir level $h_B$; (c) the new steady discharge when $\tau$ is reduced to 0.3.

Given data
Quantity	Value
Pipe length $L$	5,000 m
Hazen–Williams $C$	110
Inner diameter $D$	1,067 mm = 1.067 m
Upstream level $h_A$	105 m
Valve constant $E_s$	0.35 m$^{5/2}$/s
Case (a) discharge / valve loss	1 m$^3$/s / 5 m

Transmission main: upstream reservoir (105 m), 4,000 m of pipe to the valve, 1,000 m to the downstream reservoir.

Approach. The valve law fixes $\tau$ from the (a) operating point. An energy balance between the two reservoir surfaces — friction over the full 5,000 m plus the valve loss — sets the downstream level. Reducing $\tau$ raises the valve loss for a given flow, so (c) is found by balancing pipe friction plus valve loss against the (now fixed) available head.

Valve coefficient (a). The valve law $Q=\tau E_s\sqrt{\Delta h_v}$ rearranges to $\tau=\dfrac{Q}{E_s\sqrt{\Delta h_v}}$. With $Q=1$, $\Delta h_v=5$ m: $$\tau=\frac{1}{0.35\sqrt{5}}=\boxed{1.28}.$$
Pipe friction at $Q=1$ m$^3$/s. Hazen–Williams over the full length: $$h_{f,pipe}=5000\!\left[\frac{1.0}{0.278(110)(1.067)^{2.63}}\right]^{1/0.54}=6.47\text{ m}.$$
Downstream level (b). Energy from surface A to surface B loses pipe friction and valve head: $$h_B=h_A-h_{f,pipe}-\Delta h_v=105-6.47-5.0=\boxed{93.5\ \text{m}}.$$
Discharge when $\tau=0.3$ (c). The available head is unchanged, $h_A-h_B=105-93.5=11.47$ m, and must equal pipe friction plus the new valve loss $\Delta h_v=\left(\dfrac{Q}{\tau E_s}\right)^2=\left(\dfrac{Q}{0.105}\right)^2$: $$h_{f,pipe}(Q)+\left(\frac{Q}{0.105}\right)^2=11.47,\qquad h_{f,pipe}(Q)=5000\!\left[\frac{Q}{0.278(110)(1.067)^{2.63}}\right]^{1/0.54}.$$ Here the tighter valve dominates. Solving iteratively (pipe loss $\approx0.9$ m, valve loss $\approx10.6$ m): $$\boxed{Q\approx0.34\ \text{m}^3/\text{s}}.$$ Throttling the valve from $\tau=1.28$ to $0.3$ cuts the discharge from 1.0 to about 0.34 m$^3$/s.

Final results — Question 3
Quantity	Value
(a) Valve coefficient $\tau$	1.28
(b) Downstream reservoir level $h_B$	93.5 m
(c) Discharge at $\tau=0.3$	0.34 m$^3$/s

Topic. Series pipe–valve system between two reservoirs.

Key relations. Valve law $Q=\tau E_s\sqrt{\Delta h_v}$; energy balance $h_A-h_B=h_{f,pipe}(Q)+\Delta h_v(Q)$; Hazen–Williams for $h_{f,pipe}$.

Why this works. Pipe and valve are in series, so they carry the same discharge and their head losses simply add to consume the reservoir-to-reservoir head difference. The valve loss $\Delta h_v=(Q/\tau E_s)^2$ grows as $1/\tau^2$, so a partly-closed valve ($\tau$ small) becomes the dominant resistance. This 1,067 mm main has modest friction ($\sim$6 m at 1 m$^3$/s), so once the valve is throttled to $\tau=0.3$ the valve controls the flow almost entirely.

Common pitfall. Dropping the pipe friction in part (c) (it is small but not zero), or forgetting that the valve loss uses the full $(Q/\tau E_s)^2$ — not the (a) value of 5 m, which applied only at $Q=1$.

Source. Mays, Water Resources Engineering 2e, §4 (pipelines, valves and system curves).

Question 3 of 6: Transmission main with a control valve

Question 3: Transmission main with a control valve (20 marks)