Fluid Mechanics · Chapter 14
Fluid Mechanics · Chapter 14

Turbomachinery

Turbomachines trade energy between a fluid and a rotating shaft: pumps add it, turbines take it away. This chapter classifies the machines, quantifies pump performance through net head, power, efficiency, and the cavitation-limiting NPSH, and uses dimensional analysis — the affinity laws and specific speed — to scale designs and pick the right machine for a job.

Chapter Objectives

  • Distinguish energy-adding and energy-extracting machines using a consistent work sign.
  • Construct control-volume energy and angular-momentum balances.
  • Determine an operating point from supplied pump and system curves.
  • Calculate power, efficiency, affinity scaling, and available NPSH.
  • Check cavitation margin, similarity limits, uncertainty, and manufacturer-data dependence.
Section 14–1

Classifications & Terminology

Turbomachines exchange energy between a fluid and a rotating shaft. The split is fundamental: pumps add energy to a fluid; turbines extract energy from it.

Pumps vs. Turbines

A pump (or fan, blower, compressor for gases) does work on the fluid, raising its pressure or head. A turbine takes energy out of the fluid, converting flow energy into shaft work. The same general analysis — the angular momentum equation of Chapter 6 — governs both.

Physical meaning

A rotating blade changes fluid angular momentum. With \(x\) along the machine flow and positive shaft work defined into the fluid, pumps have positive fluid work and turbines have negative fluid work. Declaring this convention prevents a correct magnitude from receiving the wrong physical label.

Angular momentum and Euler work
\[ \tau_{\text{on fluid}}=\dot m(r_2V_{\theta2}-r_1V_{\theta1}), \qquad w_s=U_2V_{\theta2}-U_1V_{\theta1} \]

Energy-Adding Machine Geometries

By the direction the fluid leaves the rotor: centrifugal (radial) machines fling fluid outward and develop high head at modest flow; axial machines push fluid along the shaft axis and move large flow at low head; mixed-flow machines fall between.

Section 14–2

Pumps

Net Head

The increase in useful head (Bernoulli head) the pump delivers to the fluid, from inlet to outlet. For a liquid pump it equals the rise in the energy grade line across the pump.

Net Head
\[ H = \left(\frac{P}{\rho g} + \frac{V^2}{2g} + z\right)_{\text{out}} - \left(\frac{P}{\rho g} + \frac{V^2}{2g} + z\right)_{\text{in}} \]

Water Horsepower, Brake Horsepower & Efficiency

The useful power delivered to the fluid is the water horsepower; the shaft power that must be supplied is the brake horsepower. Their ratio is the pump efficiency, which falls short of one because of friction, leakage, and recirculation losses.

Power & Efficiency
\[ \dot{W}_{\text{water}} = \rho g \dot{V} H, \qquad \eta_{\text{pump}} = \frac{\dot{W}_{\text{water}}}{\text{bhp}} \]

Performance Curve & Operating Point

A pump’s net head falls as flow rate rises — its characteristic curve. Where this crosses the system demand curve is the operating point. The best efficiency point (BEP) is the flow rate at which the pump runs most efficiently, and the target for sizing.

Cavitation & Net Positive Suction Head

If local absolute pressure approaches the liquid vapor pressure at its operating temperature, vapor cavities may form and collapse. Available NPSH is a system calculation using absolute pressure; required NPSH comes only from manufacturer test data at the operating point. Their difference is not automatically an adequate design margin.

Problem-solving strategy

  1. Draw the machine control volume and declare elevation, flow, pressure reference, and positive shaft-work direction.
  2. Use continuity and the mechanical-energy balance to determine required or delivered head.
  3. Intersect supplied pump and system curves rather than selecting flow from either curve alone.
  4. Apply efficiencies to the correct direction: divide fluid power by pump efficiency and multiply hydraulic input by turbine efficiency.
  5. Compute available NPSH from absolute states and compare with supplied manufacturer required NPSH plus the project margin.
EXAMPLE 1Shaft power for a transfer pump
A pump transfers water steadily at \(Q=0.0350\ \text{m}^3/\text{s}\) while adding \(H=24.0\ \text{m}\) of head. Given \(\rho=998\ \text{kg}/\text{m}^3\) and efficiency \(0.78\), find shaft input.

Recognition strategy

Head added, flow, density, and pump efficiency directly identify hydraulic power followed by an efficiency correction for shaft input.

Given information

  • Supplied \(Q=0.0350\ \text{m}^3/\text{s}\), \(H=24.0\ \text{m}\), \(\rho=998\ \text{kg}/\text{m}^3\), \(g=9.81\ \text{m}/\text{s}^2\), \(\eta_p=0.78\).
  • Density is supplied for liquid water at \(20^\circ\text{C}\); pressure reference is gauge for the head difference.

Model and interpretation

Use a steady incompressible control volume around the pump, flow positive inlet-to-outlet, and shaft work into fluid positive. Given head already includes pressure, velocity, and elevation changes; equations are exact within the stated one-dimensional model.

Step-by-step solution
1. Establish mass flow. \(\dot m=\rho Q=(998)(0.0350)=34.9\ \text{kg}/\text{s}\), satisfying steady inlet-outlet mass conservation.
2. Compute fluid power. \(\dot W_{\text{fluid}}=\rho gQH=(998)(9.81)(0.0350)(24.0)=8.22\ \text{kW}\).
3. Apply pump efficiency. Since \(\eta_p=\dot W_{\text{fluid}}/\dot W_{\text{shaft,in}}\), shaft input is \(8.22/0.78=10.5\ \text{kW}\).
4. Quantify loss. The modeled internal loss is \(10.5-8.22=2.32\ \text{kW}\), positive as required by irreversibility.

Verification

Units: \(\rho gQH\) reduces to \(\text{kg}\!\cdot\!\text{m}^2/\text{s}^3=\text{W}\), and efficiency carries no units.

Signs and direction: Positive head and shaft input mean the pump adds energy to the fluid; shaft input exceeds useful fluid power for \(\eta_p<1\).

Plausibility: Multiplying the \(10.5\ \text{kW}\) shaft result by \(0.78\) recovers \(8.22\ \text{kW}\); incompressibility is appropriate and Mach, Froude, friction-factor, and NPSH checks are not inferable from these givens.

Final result: Required shaft input is \(\dot W_{\text{shaft,in}}=10.5\ \text{kW}\), with \(2.32\ \text{kW}\) modeled loss.

Engineering interpretation

The computed power is a hydraulic duty, not a motor selection; drive efficiency, service factor, transients, electrical rating, and the actual pump curve still require vendor review.

Available Net Positive Suction Head
\[ \text{NPSH} = \left(\frac{P}{\rho g} + \frac{V^2}{2g}\right)_{\text{inlet}} - \frac{P_v}{\rho g} \]
Section 14–3

Pump Scaling (Affinity) Laws

Dimensionless Pump Parameters

Dimensional analysis collapses pump performance into dimensionless groups: a capacity coefficient, a head coefficient, and a power coefficient. Two geometrically similar pumps with matching capacity coefficients are dynamically similar, so their other coefficients match too.

Affinity Laws (one pump, speed change; or geometrically similar pumps)
\[ \frac{\dot{V}_2}{\dot{V}_1} = \frac{n_2 D_2^3}{n_1 D_1^3}, \quad \frac{H_2}{H_1} = \frac{n_2^2 D_2^2}{n_1^2 D_1^2}, \quad \frac{\dot{W}_2}{\dot{W}_1} = \frac{\rho_2 n_2^3 D_2^5}{\rho_1 n_1^3 D_1^5} \]

Pump Specific Speed

A dimensionless group, formed from speed, flow rate, and head, that characterizes pump type independent of size. Low specific speed indicates a centrifugal machine (high head, low flow); high specific speed indicates an axial machine (low head, high flow).

Pump Specific Speed
\[ N_{Sp} = \frac{n\,\dot{V}^{1/2}}{(gH)^{3/4}} \]

Affinity laws are approximate similarity relations: geometry, fluid regime, and efficiency must remain sufficiently similar. They do not replace a manufacturer curve, and large speed or diameter changes can invalidate the estimate.

EXAMPLE 2Operating point from supplied curves
For water at \(20^\circ\text{C}\), a vendor supplies \(H_p=32.0-800Q^2\) and the system model is \(H_s=12.0+1200Q^2\), with \(H\) in metres and \(Q\) in \(\text{m}^3/\text{s}\). Find operation.

Recognition strategy

A supplied pump curve and system curve signal an intersection problem; neither shutoff head nor free-delivery flow is the operating point.

Given information

  • Vendor-given \(H_p=32.0-800Q^2\), system-given \(H_s=12.0+1200Q^2\), and supplied \(\eta_p=0.74\) at the resulting point.
  • Use supplied \(\rho=998\ \text{kg}/\text{m}^3\) at \(20^\circ\text{C}\); curve heads use a consistent gauge-pressure datum.

Model and interpretation

A steady incompressible system control volume uses positive flow downstream and positive pump work into water. Curve fits are approximate supplied data, not invented universal correlations.

Step-by-step solution
1. Enforce compatible operation. Set \(H_p=H_s\), because one steady flow must satisfy both machine head and system demand.
2. Solve flow. \(32.0-800Q^2=12.0+1200Q^2\) gives \(Q^2=0.0100\), hence \(Q=0.100\ \text{m}^3/\text{s}\).
3. Recover head. Either curve gives \(H=32.0-800(0.100)^2=24.0\ \text{m}\), confirming the intersection.
4. Determine shaft demand. \(\dot W_{\text{in}}=\rho gQH/\eta_p=(998)(9.81)(0.100)(24.0)/0.74=31.7\ \text{kW}\).

Verification

Units: Coefficients carry the implied units stated with the supplied curve, both sides produce metres, and the power expression produces watts.

Signs and direction: The positive root is selected for downstream delivery; positive pump head and shaft input are consistent with an energy-adding machine.

Plausibility: The operating head lies between static head \(12.0\ \text{m}\) and shutoff head \(32.0\ \text{m}\), and substitution into both independent curve fits returns \(24.0\ \text{m}\).

Final result: \(Q=0.100\ \text{m}^3/\text{s}\), \(H=24.0\ \text{m}\), and estimated shaft input is \(31.7\ \text{kW}\).

Engineering interpretation

Selection also requires the vendor’s efficiency island, allowable operating region, motor rating, transient envelope, and NPSH-required curve; one algebraic intersection is not certification.

Section 14–4

Turbines

Impulse vs. Reaction Turbines

An impulse turbine (e.g. Pelton wheel) converts the flow to a high-speed jet that strikes buckets at atmospheric pressure — suited to high heads and low flow. A reaction turbine (Francis, Kaplan) runs full of water with a pressure drop across the runner — suited to low-to-moderate heads and large flow.

Impulse and reaction classifications describe where pressure changes occur, not efficiency guarantees. Site head, flow range, speed, cavitation behavior, sediment, controls, and generator requirements govern professional selection.

Turbine Efficiency
\[ \eta_{\text{turbine}} = \frac{\text{bhp (shaft output)}}{\rho g \dot{V} H} \]

Selection by Head

Pelton wheels serve very high heads (hundreds of meters); Francis turbines the broad middle range; Kaplan (axial, propeller) turbines low heads with large flow, as in run-of-river plants. The angular momentum (Euler turbine) equation links runner torque to the change in the fluid’s tangential momentum.

EXAMPLE 3Available NPSH audit
At a pump suction flange, supplied absolute pressure is \(82.0\ \text{kPa}\), speed is \(2.50\ \text{m}/\text{s}\), and water vapor pressure is \(3.17\ \text{kPa}\) at \(25^\circ\text{C}\). Find NPSH available and compare with supplied NPSH required \(5.20\ \text{m}\).

Recognition strategy

Absolute suction pressure, vapor pressure, and manufacturer-required NPSH identify a cavitation-margin calculation at the stated operating point.

Given information

  • Supplied \(p_{\text{in,abs}}=82.0\ \text{kPa}\), \(p_v=3.17\ \text{kPa}\), \(V=2.50\ \text{m}/\text{s}\), \(\rho=997\ \text{kg}/\text{m}^3\), and NPSH required \(=5.20\ \text{m}\).
  • Properties and vapor pressure are supplied at \(25^\circ\text{C}\); \(g=9.81\ \text{m}/\text{s}^2\).

Model and interpretation

Take a steady incompressible suction-flange control section, \(x\) into the pump, and use absolute pressures. The standard flange NPSH equation is approximate because local impeller-eye minima are represented by vendor test data.

Step-by-step solution
1. Form pressure margin. \(p_{\text{in,abs}}-p_v=82.0-3.17=78.83\ \text{kPa}\), which must remain on an absolute basis.
2. Convert pressure to head. \(78{,}830/[(997)(9.81)]=8.06\ \text{m}\) of liquid.
3. Add velocity head. \(V^2/(2g)=2.50^2/[2(9.81)]=0.319\ \text{m}\), so NPSH available is \(8.38\ \text{m}\).
4. Compare supplied requirement. The simple numerical margin is \(8.38-5.20=3.18\ \text{m}\), with available exceeding required.

Verification

Units: Pressure divided by \(\rho g\) and velocity squared divided by \(2g\) each produce metres, matching the vendor NPSH unit.

Signs and direction: Vapor-pressure head is subtracted because it reduces liquid-pressure margin; velocity head is positive in the flange stagnation-head definition.

Plausibility: Absolute inlet pressure exceeds vapor pressure substantially, and independently summing \(8.06+0.319\) reproduces \(8.38\ \text{m}\); no claim about an adequate project margin follows without owner/vendor criteria.

Final result: NPSH available is \(8.38\ \text{m}\); it exceeds supplied NPSH required by \(3.18\ \text{m}\).

Engineering interpretation

The arithmetic comparison is favorable, but temperature excursions, barometric uncertainty, suction losses, transient level, dissolved gas, wear, and the mandated margin need vendor and project review.

Section 14–5

Turbine Scaling Laws

Turbine Specific Speed

As with pumps, a dimensionless specific speed — formed here from speed, power, and head — identifies the best turbine type for a site, independent of size. Low values point to impulse (Pelton) machines; high values to axial (Kaplan) machines.

Turbine Specific Speed
\[ N_{St} = \frac{n\,\dot{W}^{1/2}}{\rho^{1/2}(gH)^{5/4}} \]

Model Testing

Large turbines are validated by testing small geometrically similar models at matched dimensionless conditions, then scaling the results. A small efficiency gain predicted on a scale model translates into large energy savings on a full-size machine, justifying extensive model programs.

Instructor Note

Two ideas carry this chapter. First, the affinity laws — flow \(\propto n\), head \(\propto n^2\), power \(\propto n^3\) — are worth memorizing cold; they answer most "what if we change the speed" questions instantly. Second, specific speed picks the machine: compute it and you know whether the job wants a centrifugal or axial design before drawing a single blade.

Chapter recap

  • Pumps add head; turbines extract it, and the declared shaft-work sign controls interpretation.
  • Operating point is the intersection of supplied machine and system curves.
  • Affinity laws require similarity, while cavitation checks require absolute pressure and temperature-specific vapor pressure.

Before moving on

You should be able to balance head and power, apply efficiency in the correct direction, solve a curve intersection, estimate similarity scaling, and audit available NPSH against traceable vendor data.