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Fluid Mechanics · Chapter 10 of 10 · Intermediate

Turbomachinery: Pumps and Turbines

Pumps add energy to a flow; turbines take it out. This closing chapter sizes a pump's head and power, finds where it actually runs, and scales it to a new speed with the affinity laws.

01

Readiness check

This capstone applies the energy equation to machines. Tick only what you can do closed-notes.

  • Recall the energy equation with pump head (Chapter 5).
  • Compute hydraulic power ρgQH.
  • Use efficiency to relate input and output power.
  • Recall head loss from Chapter 8.
  • Work with ratios and powers (squares, cubes).
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview pump head and power in Chapter 5.
3 or more weak itemsRevisit the energy equation and head loss first.
02

The core idea

A pump adds head H to a flow at hydraulic power ρgQH; its brake power is that divided by efficiency. It runs where its curve meets the system curve, and its performance scales with speed by the affinity laws.

Ẇwater = ρgQHẆbrake = ρgQH/ηQ ∝ N, H ∝ N², P ∝ N³

Turbomachines exchange energy between a rotating rotor and a fluid: pumps and fans add it, turbines extract it. A pump's useful output is the hydraulic power ρgQH; dividing by efficiency gives the shaft (brake) power the motor must supply. A pump does not run at one fixed point: it settles where its head-versus-flow curve crosses the system's resistance curve. Change the speed and the whole curve shifts, predictably, by the affinity laws.

The skill works when: you compute hydraulic power, divide by efficiency for brake power, and scale with the affinity laws.
The skill breaks down when: efficiency is omitted, or head is scaled linearly with speed instead of as the square.
The concept. A pump runs where its head-flow curve crosses the system curve (static lift plus friction). That intersection, the operating point, sets the actual flow and head.
03

The skills, taught in order

Turbomachinery applies the energy equation to spinning machines. Five skills cover the machines, pump power, the operating point, the affinity laws, and selection.

10.1 Pumps and turbines

Turbomachines transfer energy between a rotor and a fluid. Pumps, fans, and compressors add energy (raising head or pressure); turbines extract it (producing shaft work). Centrifugal machines handle high head at modest flow; axial machines move large flows at low head.

10.2 Pump head and power

The pump adds a head H, giving useful (hydraulic) power Ẇwater = ρgQH. The shaft or brake power is larger by the efficiency: Ẇbrake = ρgQH/ηpump. For a turbine the relation inverts: output power = ηturbine ρgQH.

10.3 Pump and system curves

A pump's head falls as flow rises (its characteristic curve). The system demands head that rises with flow (static lift plus friction hL ∝ Q²). The pump runs at their intersection, the operating point; nothing else is stable.

10.4 The affinity laws

For a given pump at different speeds N, the performance scales predictably.

QuantityScaling with speed
Flow rate Q∝ N
Head H∝ N²
Power P∝ N³

So a small speed increase raises power steeply, the cube law again. The same laws scale across geometrically similar pumps of different diameter.

10.5 Selection, specific speed, and cavitation

Specific speed groups Q, H, and N into a dimensionless number that points to the best machine type. As at a pump inlet, the local pressure must stay above the vapor pressure (adequate net positive suction head) or the pump cavitates, the link back to fluid properties in Chapter 2.

Engineering connection: water-supply and HVAC pumps, hydroelectric and steam turbines, fans, and jet engines are all selected and scaled with these relations, closing the loop from properties to working machines.

04

Worked example 1: pump power and efficiency

A pump delivers 0.03 m³/s of water (ρ = 998 kg/m³) at a head of 45 m, with a pump efficiency of 80%. Find the hydraulic power and the brake (shaft) power.

Figure 1. The hydraulic power delivered to the water is ρgQH. The motor must supply more, the brake power, because the pump is only 80% efficient.
  1. ProblemFind the hydraulic and brake power for the pump in Figure 1.
  2. Given / findQ = 0.03 m³/s, H = 45 m, ρ = 998 kg/m³, η = 0.80. Find Ẇwater and Ẇbrake.
  3. AssumptionsSteady operation, the 45 m is the net head added, efficiency is the pump's overall value.
  4. ModelHydraulic power from ρgQH; brake power by dividing by efficiency.
  5. EquationsẆwater = ρgQH Ẇbrake = Ẇwater/η
  6. SolveẆwater = 998 × 9.81 × 0.03 × 45 = 13.2 kW. Ẇbrake = 13.2/0.80 = 16.5 kW.
  7. CheckThe brake power exceeds the hydraulic power, as efficiency demands; the 3.3 kW difference is lost to friction and turbulence in the pump. A motor would be sized at 16.5 kW or the next standard size up.
  8. ConclusionHydraulic power is what reaches the fluid; brake power is what the motor draws. Efficiency is the gap, and it sets the energy bill.
Result. Hydraulic power 13.2 kW; brake power 16.5 kW (at 80% efficiency).
05

Worked example 2: speeding the pump up

The same pump runs at 1450 rev/min (Q = 0.03 m³/s, H = 45 m, brake power 16.5 kW). It is sped up to 1750 rev/min. Use the affinity laws to find the new flow, head, and power.

Figure 2. Raising the speed lifts the whole pump curve. Flow scales with speed, head with its square, and power with its cube, so a 21% speed rise needs 76% more power.
  1. ProblemFind Q, H, and power at 1750 rev/min for the pump in Figure 2.
  2. Given / findN₁ = 1450, N₂ = 1750 rev/min; at N₁: Q = 0.03 m³/s, H = 45 m, P = 16.5 kW. Find the new values.
  3. AssumptionsSame pump and fluid, geometrically unchanged, efficiency roughly constant over the speed change.
  4. ModelApply the affinity laws: Q ∝ N, H ∝ N², P ∝ N³.
  5. EquationsQ₂ = Q₁(N₂/N₁) H₂ = H₁(N₂/N₁)² P₂ = P₁(N₂/N₁)³
  6. SolveThe speed ratio is 1750/1450 = 1.207. Q₂ = 0.03 × 1.207 = 0.0362 m³/s. H₂ = 45 × 1.207² = 45 × 1.457 = 65.5 m. P₂ = 16.5 × 1.207³ = 16.5 × 1.759 = 29.0 kW.
  7. CheckA 21% speed increase raised the flow 21%, the head 46%, and the power 76%, exactly the 1, 2, 3 powers of the affinity laws. The steep power rise is why oversized pumps are throttled or speed-controlled, not just run faster.
  8. ConclusionThe affinity laws scale a known operating point to any speed. The cube law on power makes variable-speed drives a major energy-saving when demand drops, the practical close of the course.
Result. At 1750 rev/min: Q = 0.0362 m³/s, H = 65.5 m, power 29.0 kW.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Ignoring efficiencyMotor undersized"Is this hydraulic or brake power?"Brake power = ρgQH/η, larger than hydraulic.
Scaling head linearly with speedHead wrong at the new speed"Does H go as N or N²?"Head scales as N²; power as N³.
Picking a flow off the pump curve aloneOperating point wrong"Where does the system curve cross?"The flow is set by the pump-and-system intersection.
Ignoring suction (cavitation)Pump cavitates and loses head"Is inlet pressure above vapor pressure?"Ensure adequate net positive suction head.
07

Practice ladder

Level 1 · Direct skill

A pump moves 0.05 m³/s of water against 30 m of head. Find the hydraulic power (ρ = 998 kg/m³).

Show answer

Ẇwater = ρgQH = 998 × 9.81 × 0.05 × 30 = 14.7 kW.

Level 2 · Mixed concept

The Worked Example 1 pump's flow must be cut to 0.024 m³/s by reducing speed. By the affinity laws, what new speed and head result?

Show answer

Q ∝ N, so N₂ = 1450 × (0.024/0.03) = 1160 rev/min. H₂ = 45 × (0.024/0.03)² = 45 × 0.64 = 28.8 m. Slowing the pump drops the head as the square of the flow ratio.

Level 3 · Independent problem

A turbine takes 2 m³/s of water under a net head of 40 m at 90% efficiency. Find its output power.

Show answer

Output = η ρgQH = 0.90 × 998 × 9.81 × 2 × 40 = 705 kW. For a turbine, efficiency multiplies (output is less than the available hydraulic power), the reverse of a pump.

Level 4 · Transfer to real engineering

Find a real pump or turbine (a building water pump, a hydro plant, a cooling fan). Estimate its head, flow, and power, and discuss how speed or size would change them.

What good work looks like

Head, flow, and power estimated with ρgQH and efficiency, and an affinity-law argument for how a speed or diameter change would scale them.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I divided by efficiency for brake power, not multiplied."
"Give me five speed changes; I will scale Q, H, and P by the affinity laws."
"Compute the pump power." Applying ρgQH/η yourself is the skill.
"What is the new head?" Using the N² scaling is the point.

Portfolio task

Analyse one pump (or turbine): compute its hydraulic and brake power with efficiency, locate or estimate its operating point, and scale it to a new speed with the affinity laws.

Must include: ρgQH and efficiency, an operating-point note, and an affinity-law speed scaling.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the pump hydraulic and brake power.

Ẇwater = ρgQH; Ẇbrake = ρgQH/η.

2. What sets a pump's actual flow?

The operating point, where the pump curve meets the system curve.

3. State the affinity laws.

Q ∝ N, H ∝ N², P ∝ N³ for a given pump.

4. How does a turbine's power relate to ρgQH?

Output = η ρgQH; efficiency multiplies, unlike a pump where it divides.

5. What causes a pump to cavitate?

Inlet pressure dropping to the vapor pressure (too little net positive suction head).

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the pump power and affinity scaling from a blank page.
+3 daysOne power and one affinity-law problem.
+7 daysConnect the operating point back to head loss in Chapter 8.
+30 daysReview the whole course from the Fluid Mechanics hub.
10

Textbook mapping

ItemMapping
Primary sourceÇengel and Cimbala, Fluid Mechanics: Fundamentals and Applications, Chapter 14 (Turbomachinery)
Cross-referenceWhite, Ch. 11 · Munson, Ch. 12
Core topics10.1 Pumps and turbines · 10.2 Pump power · 10.3 Operating point · 10.4 Affinity laws · 10.5 Selection and cavitation
Engineering connectionWater and HVAC pumps, hydro and steam turbines, and fans.
Read nextYou have completed the Fluid Mechanics course. Return to the course hub or continue to Heat Transfer.