Reciprocating pumps

Why it matters

Reciprocating (piston and plunger) pumps deliver small flows at high pressure with high efficiency. Examples are hydraulic test rigs, boiler feed for small plants, high-pressure washers, metering and dosing pumps, and the plunger elements of diesel fuel-injection pumps. Their pulsating flow brings in acceleration head, separation and air vessels, which do not appear in centrifugal pumps and are favourite GATE topics.

Key ideas

• Positive displacement. Each stroke traps and pushes out a fixed volume, so discharge depends on speed and swept volume, almost independent of delivery pressure. Never run one against a closed delivery valve: pressure rises until something fails, so a relief valve is fitted.
• Working cycle. During the suction stroke the piston moves out, pressure in the cylinder falls below atmospheric, the suction valve opens and liquid enters. During the delivery stroke the piston moves in, the delivery valve opens and liquid is pushed into the delivery pipe.
• Single- vs double-acting. A single-acting pump delivers once per revolution. A double-acting pump delivers on both strokes, giving nearly twice the flow (less the piston-rod area on one side) and a smoother delivery. Multi-cylinder (duplex, triplex) pumps smooth the flow further.
• Slip. Valves close late and leak, so actual discharge is less than theoretical. Slip = (Q_th − Q_act)/Q_th, and the coefficient of discharge C_d = Q_act/Q_th. Negative slip (Q_act > Q_th) can occur with a long delivery pipe, high speed and a short suction pipe: the inertia of the moving liquid column pushes the delivery valve open before the suction stroke ends.
• Indicator diagram. Pressure head in the cylinder plotted against stroke. The ideal diagram is a rectangle between −h_s (suction) and +h_d (delivery), relative to atmosphere. Its area is proportional to the work done per cycle.
• Acceleration head. The piston moves with simple harmonic motion (for a long connecting rod), so the liquid in each pipe accelerates and decelerates every stroke. The pressure head needed is h_a = (l/g)·(A/a)·ω²·r·cos θ:
  • maximum (= (l/g)(A/a)ω²r) at the start and end of each stroke, zero at mid-stroke;
  • at the start of suction it lowers the cylinder pressure; at the end of suction it raises it.
  • It tilts the ideal indicator diagram into a parallelogram, but does not change its area (no net work).
• Friction head in the pipes varies as the square of instantaneous velocity: zero at the ends of the stroke, maximum at mid-stroke. It adds parabolic bulges to the diagram and increases work. The mean friction head over a stroke is 2/3 of the maximum.
• Separation (cavitation). If the absolute pressure in the cylinder at the start of the suction stroke falls to the liquid's vapour pressure — in practice taken as about 2.5 m of water absolute for water (check the problem data) — dissolved air and vapour come out and the liquid column separates from the piston. This limits pump speed, suction-pipe length and suction lift. At the end of the delivery stroke, separation can also occur in the delivery pipe if the delivery head is small.
• Air vessels. Closed chambers containing air fitted close to the cylinder on the suction and delivery sides. They absorb the flow fluctuations, so the liquid in most of each pipe flows at a nearly uniform mean velocity. Benefits:
  • acceleration head is limited to the short pipe between vessel and cylinder, so the pump can run faster or with a longer suction pipe without separation;
  • friction work is reduced (by 84.8 % for a single-acting and 39.2 % for a double-acting pump);
  • delivery is nearly steady.
• Comparison with centrifugal pumps. Reciprocating pumps give high pressure at low flow with high efficiency, can be self-priming and can handle viscous liquids. However, they are heavier, have pulsating flow, more wearing parts (valves, packings) and higher maintenance. Centrifugal pumps suit large flows at moderate heads.

Formulas

Q_th = A·L·N/60 (single-acting) Q_th = (2A − a_r)·L·N/60 ≈ 2·A·L·N/60 (double-acting) Q_th theoretical discharge (m³/s); A piston area (m²); a_r piston-rod area (m²); L stroke (m); N speed (rpm).

C_d = Q_act/Q_th Slip = 1 − C_d

P = ρ·g·Q_th·(h_s + h_d) (single-acting, ideal) Power (W); h_s suction head, h_d delivery head (m). Use 2A − a_r in Q_th for double-acting; add 2/3 of the maximum friction heads when friction is included.

h_a = (l/g)·(A/a)·ω²·r·cos θ Acceleration head (m); l pipe length (m), a pipe area (m²), ω = 2πN/60 (rad/s), r crank radius = L/2 (m), θ crank angle from inner dead centre.

h_f,max = (4·f′·l / (2g·d))·[(A/a)·ω·r]² Maximum friction head at mid-stroke (m); f′ Fanning friction coefficient, d pipe diameter (m). With the Darcy factor f, replace 4f′ by f.

H_atm − h_s − h_as ≥ H_sep No separation at the start of suction; H_atm about 10.3 m of water, H_sep the separation head (absolute).

Worked examples

Example 1 (standard) — discharge, slip and power. Given: single-acting pump, plunger diameter 150 mm, stroke 300 mm, N = 50 rpm, suction head 4 m, delivery head 16 m, actual discharge 4 L/s. Find C_d, slip and theoretical power.

1. A = π × 0.15²/4 = 0.017671 m².
2. Q_th = A·L·N/60 = 0.017671 × 0.3 × 50/60 = 4.418 × 10⁻³ m³/s.
3. C_d = 0.004/0.004418 = 0.905; slip = 1 − 0.905 = 0.095 = 9.5 %.
4. P = ρgQ_th(h_s + h_d) = 1000 × 9.81 × 0.004418 × 20 = 866.8 W. Answer: C_d ≈ 0.905, slip ≈ 9.5 %, theoretical power ≈ 0.87 kW.

Example 2 (GATE level) — maximum speed without separation. Given: single-acting pump, plunger 150 mm diameter, stroke 300 mm; suction pipe 6 m long, 75 mm diameter; suction lift 3 m; atmospheric head 10.3 m of water; separation occurs at 2.5 m of water absolute. Find the maximum speed with no air vessel.

1. Area ratio A/a = (150/75)² = 4; crank radius r = 0.15 m.
2. Separation is most likely at the start of suction (θ = 0), where h_as is maximum and reduces cylinder pressure.
3. Condition: 10.3 − 3 − h_as ≥ 2.5 → h_as ≤ 4.8 m.
4. h_as = (l/g)(A/a)ω²r = (6/9.81) × 4 × 0.15 × ω² = 0.36697·ω².
5. ω² ≤ 4.8/0.36697 = 13.08 → ω ≤ 3.617 rad/s.
6. N = 60ω/(2π) = 34.5 rpm. Answer: N_max ≈ 34.5 rpm. An air vessel near the cylinder would allow a much higher speed, because the accelerating column would then be only the short pipe between the vessel and the cylinder.

Common mistakes

• Forgetting to divide N by 60, or doubling Q for a single-acting pump.
• Using the ratio of diameters instead of areas in A/a (it is (D/d)²).
• Checking separation at mid-stroke — acceleration head is maximum at the ends, zero at mid-stroke.
• Thinking acceleration head changes the work per cycle; it does not (only friction does).
• Confusing work (J) with power (W): ρgQH is a rate.
• Treating the separation head as gauge pressure — it is absolute.

For GATE ME

Expect: theoretical discharge, slip and C_d; power for single- or double-acting pumps; acceleration head at a given crank angle; maximum speed or suction-pipe length to avoid separation; work saved by air vessels; and conceptual questions on negative slip and indicator diagrams. Practise sketching the indicator diagram with acceleration and friction effects.

Quick check

1. A single-acting pump has a 0.1 m bore, 0.2 m stroke and runs at 60 rpm. What is Q_th in m³/min?
2. At what point in the stroke is acceleration head maximum?
3. Why can negative slip occur?
4. A pump delivers 0.02 m³/s against 500 kPa. What hydraulic power does it supply?
5. Name two benefits of an air vessel. Answers: 1. 0.007854 × 0.2 × 60 = 0.0942 m³/min. 2. At the beginning and end of each stroke. 3. Inertia of a long delivery column opens the delivery valve before the suction stroke ends. 4. 500 000 × 0.02 = 10 kW. 5. Allows higher speed without separation; reduces friction work; smooths delivery (any two).