07-Str-B11 · May 2013

Question 3 of 6: Two Pumps in Series and in Parallel on a Long Transmission Main

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations — May 2013 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and one approved Casio or Sharp calculator are permitted. The paper prints six questions of 20 marks each and instructs the candidate to complete any five; where a question has more than one part, the parts carry equal marks. Candidates are urged to state any interpretive assumptions with their answer. All six questions are worked below, because this set is intended as a study resource.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, pipe networks, pump-system curves and quasi-steady reservoir routing; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning's equation, compound and divided channel sections, and the specific-energy and momentum treatment of channel obstructions; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American design practice for transmission mains, minimum service pressures and pump selection; Henderson, F.M., Open Channel Flow (Macmillan) — surges, hydraulic jumps and the unsteady response of a channel to a sudden blockage.

Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form $Q = 0.278\,C\,D^{2.63}S^{0.54}$ with $Q$ in m³/s, $D$ in metres and $S = h_f/L$; it is used exactly as printed. Local losses and velocity head are neglected, as instructed in Note 6, so the hydraulic grade line (HGL) and the energy grade line coincide and "pressure head at a node" means $p/\gamma = \mathrm{HGL} - z$. Water properties are $\rho = 1000$ kg/m³ and $\nu = 1.31\times10^{-6}$ m²/s.

Question 3: Two Pumps in Series and in Parallel on a Long Transmission Main (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.

Given. A pumping station lifts water over an intervening hill and then delivers it downhill to a reservoir that is lower than the source.

Given data — Question 3
Quantity	Symbol	Value
Water surface, Reservoir A	$z_A$	50 m
Water surface, Reservoir C	$z_C$	40 m
Ground elevation at the summit, point B	$z_B$	55 m
Chainage of point B from A	$x_B$	1200 m
Total pipeline length	$L$	6000 m
Pipe diameter	$D$	800 mm = 0.800 m
Hazen-Williams coefficient	$C$	140
Head-discharge curve, each pump	$H$	$20 - 9\,Q^{1.8}$ m, $Q$ in m³/s
Minimum allowable pressure head	$(p/\gamma)_{\min}$	14 m

Find. (a) the delivered discharge with the two pumps in series and with the two pumps in parallel, and (b) whether the pressure head at the summit satisfies the 14 m limit under series operation.

Figure 3-S. Longitudinal profile with the computed hydraulic grade line for series operation. The HGL clears the summit, so flow is maintained, but it clears it by only 6.2 m and therefore violates the 14 m minimum pressure requirement.

Approach. Build the system head curve as static lift plus Hazen-Williams friction, build the combined pump curve for each configuration, intersect the two to find the operating point, and then walk the HGL from the pump discharge flange to the summit to test the pressure limit.

Form the system head curve. The static component is the rise from A to C, which is negative because C lies below A: $$H_{\text{stat}} = z_C - z_A = 40 - 50 = -10\ \text{m}$$ With $K = 0.278(140)(0.800)^{2.63} = 21.642\ \text{m}^3\text{/s}$, friction over the full 6000 m is $h_f = 6000\,(Q/21.642)^{1.852}$, so the total dynamic head the pumps must supply is $$\mathrm{TDH}_{\text{sys}}(Q) = -10 + 6000\left(\frac{Q}{21.642}\right)^{1.852}$$ The 10 m of favourable static drop is a genuine assist, but the pumps are still needed: the summit at 55 m stands above Reservoir A at 50 m, so no gravity flow is possible at all.
Form the two combined pump curves. Pumps in series pass the same discharge and add their heads, whereas pumps in parallel split the discharge and share a common head: $$H_{\text{series}}(Q) = 2\left(20 - 9Q^{1.8}\right) = 40 - 18\,Q^{1.8}$$ $$H_{\text{parallel}}(Q) = 20 - 9\left(\tfrac{Q}{2}\right)^{1.8} = 20 - 2.5846\,Q^{1.8}$$
Locate the series operating point. Setting $H_{\text{series}} = \mathrm{TDH}_{\text{sys}}$ and solving, $$40 - 18\,Q^{1.8} = -10 + 6000\left(\frac{Q}{21.642}\right)^{1.852} \quad\Rightarrow\quad \boxed{Q_{\text{series}} = 1.159\ \text{m}^3\text{/s}}$$ at which the two pumps together deliver $\mathrm{TDH} = 16.54$ m against 26.54 m of pipe friction, the 10 m difference being the favourable static drop.
Locate the parallel operating point. Repeating with the parallel curve, $$20 - 2.5846\,Q^{1.8} = -10 + 6000\left(\frac{Q}{21.642}\right)^{1.852} \quad\Rightarrow\quad \boxed{Q_{\text{parallel}} = 1.161\ \text{m}^3\text{/s}}$$ with $\mathrm{TDH} = 16.62$ m. This completes part (a).
Interpret the near-tie between the two configurations. The two answers differ by less than 0.2 per cent, which is not a coincidence. Equating the two pump curves gives the crossover discharge $$40 - 18Q^{1.8} = 20 - 2.5846Q^{1.8} \quad\Rightarrow\quad Q^{1.8} = \frac{20}{15.415} \quad\Rightarrow\quad Q_{\text{cross}} = 1.156\ \text{m}^3\text{/s}$$ and the system curve happens to cut both pump curves within a few thousandths of that point. Series operation is superior below 1.156 m³/s and parallel operation above it, so this pipeline sits almost exactly on the boundary. The practical reading is that the choice of configuration should be made on other grounds here — standby redundancy, suction conditions, motor sizing — because neither buys any useful extra flow.
Compute the HGL at the summit under series operation. Friction is uniformly distributed along a prismatic pipe, so the loss over the first 1200 m is 20 per cent of the total: $$h_{f,A\to B} = 26.54 \times \frac{1200}{6000} = 5.31\ \text{m}$$ The HGL leaves the pump at $50 + 16.54 = 66.54$ m, so at the summit $$\mathrm{HGL}_B = 66.54 - 5.31 = 61.23\ \text{m}$$
Test the summit pressure against the 14 m limit. Subtracting the ground elevation at B, $$\frac{p}{\gamma}\bigg|_{B} = 61.23 - 55 = \boxed{6.23\ \text{m}}$$ Since $6.23\ \text{m} \lt 14\ \text{m}$, the answer to part (b) is no: the pressure at B falls roughly 7.8 m short of the minimum. The line does remain pressurised and will not break the water column, so flow is sustained, but the design does not meet the stated criterion. Raising the summit pressure requires either more pump head — roughly 8 m more, which two pumps in series on this curve cannot provide at this flow — a larger-diameter first reach to cut the 5.31 m of upstream friction, or lowering the alignment through the hill.

Final results — Question 3
Quantity	Series operation	Parallel operation
Pipeline discharge	1.159 m³/s	1.161 m³/s
Total dynamic head supplied	16.54 m	16.62 m
Total friction loss over 6000 m	26.54 m	26.62 m
HGL at the pump discharge	66.54 m	—
Friction loss, A to B (1200 m)	5.31 m	—
HGL at summit B	61.23 m	—
Pressure head at summit B	6.23 m	—
Meets the 14 m minimum?	No, short by 7.77 m	—

Topic. Pump and system curve matching, series versus parallel pump combination, and the summit pressure check on a transmission main.

Key relations. System curve $\mathrm{TDH} = (z_{\text{out}} - z_{\text{in}}) + h_f(Q)$; series combination adds heads at equal discharge, $H_{\text{ser}} = 2H_1(Q)$; parallel combination adds discharges at equal head, $H_{\text{par}} = H_1(Q/2)$; friction distributes linearly along a prismatic pipe so $h_{f,0\to x} = h_{f,\text{total}}\,(x/L)$; and $p/\gamma = \mathrm{HGL} - z$.

Why this works. The operating point is where the head the pumps can produce equals the head the system demands, because both statements describe the same physical head rise across the station at the same discharge. Whether series or parallel wins depends entirely on the steepness of the system curve: a steep, friction-dominated curve rewards series operation, because doubling the head buys more flow than doubling the capacity at a fixed head, while a flat, lift-dominated curve rewards parallel operation. This pipeline sits almost exactly on the dividing line, which is why the two configurations tie. The summit check is separate from the flow calculation and must always be made, because a pipeline can carry its design flow perfectly well while running at an unacceptable or even negative pressure over an intervening high point.

Common pitfall. Using $z_A - z_C = +10$ m for the static head instead of $z_C - z_A = -10$ m, which inflates the required head by 20 m and badly understates the flow. A second frequent error in part (b) is subtracting the total 26.54 m of friction rather than only the 20 per cent that occurs upstream of the summit, and a third is forgetting that the pump head must be added to the reservoir level, not to the summit level, when starting the HGL march.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 4 and Ch. 6 (pumps, system curves, series and parallel operation); Chin, Water-Resources Engineering, 3rd ed., Section 2.6 (pumps and pump selection).

← Question 2 Next: Question 4 →

Quantity	Symbol	Value
Water surface, Reservoir A	\(z_A\)	50 m
Water surface, Reservoir C	\(z_C\)	40 m
Ground elevation at the summit, point B	\(z_B\)	55 m
Chainage of point B from A	\(x_B\)	1200 m
Total pipeline length	\(L\)	6000 m
Pipe diameter	\(D\)	800 mm = 0.800 m
Hazen-Williams coefficient	\(C\)	140
Head-discharge curve, each pump	\(H\)	\(20 - 9\,Q^{1.8}\) m, \(Q\) in m³/s
Minimum allowable pressure head	\((p/\gamma)_{\min}\)	14 m