E5 · HVAC Equipment and Components

Foundation lesson · 30 minute study session

Use affinity laws only for comparable operating states

A modest speed change can produce a large power change. The relationship depends on the machine geometry, fluid density, efficiency and system curve.

By the end of this lesson

Apply fixed-diameter speed affinity laws.
Distinguish head or pressure scaling from power scaling.
Recognize curve, efficiency, NPSH and compressor limitations.

Fixed geometry and speed change

For the same fan or centrifugal pump at comparable points with fixed diameter and the same fluid density, flow scales with speed, developed pressure or head with speed squared, and shaft power approximately with speed cubed when efficiency remains comparable. Speed must be a ratio in matching units. Use $r=N_2/N_1$ only with the stated fixed geometry and comparable efficiency.

Check the actual system

A friction-dominated system often follows a quadratic loss curve, making speed scaling useful. A substantial static-head term changes how the operating point moves. Filter loading, dampers, valves and changed efficiency can make a simple cube-law estimate inadequate. Use the supplied machine and system curves when they govern.

Do not extend the law blindly

Centrifugal pump laws do not replace a compressor thermodynamic calculation. Compressors require attention to pressure ratio, suction state, efficiency and the stated process. Pump cavitation requires a separate NPSH check: available NPSH must exceed required NPSH with any stated margin, using absolute suction pressure and the fluid vapor pressure.

A decision map for this lesson

Affinity laws describe comparable points

Use the speed relationship for an estimate, then establish the installed operating point and equipment limits.

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Speed changes at fixed diameter and density	At comparable points, $Q\propto N$ , $H\propto N^2$ , and $P_{shaft}\propto N^3$ with the stated efficiency assumption.	Using one linear multiplier for flow, head and power.
Static head is material	Intersect the changed machine curve with the actual system curve.	Assuming installed flow must scale directly with speed.
Input power is requested	Separate hydraulic/refrigerant work, shaft input and motor input efficiencies.	Applying motor efficiency to a quantity already stated as electrical input.
A pump, fan or compressor is selected	Check NPSH, stable operating region or gas-flow reference as applicable to the machine.	Treating an affinity estimate as a complete equipment-selection check.

On a narrow screen, scroll across the table to compare the decision and the error.

Relationships and their limits

Fixed-diameter speed laws

\frac{Q_2}{Q_1}=r;\quad\frac{H_2}{H_1}=r^2;\quad\frac{P_2}{P_1}\approx r^3;\quad r=N_2/N_1

Comparable points; same fluid density and geometry; shaft-power estimate assumes comparable efficiency.

Fan pressure and density

\frac{\Delta p_2}{\Delta p_1}=\frac{\rho_2}{\rho_1}r^2

For comparable fan states with fixed geometry. Density effects must not be omitted when they change.

Symbols and unit basis
Symbol	Meaning	Units or basis
$Q$	Flow on stated basis	gpm or cfm
$H$	Pump head	ft of liquid
$P$	Shaft power	hp
$N$	Rotational speed	rpm
$\rho$	Fluid density	lb/ft³
$r$	Speed ratio	dimensionless
$\Delta p$	Fan pressure rise	matching pressure units

One decision, fully worked

Apply the method.

A centrifugal fan delivers 10,000 cfm at 900 rpm, develops 2.0 in. w.g. and uses 8.0 shaft hp. Diameter and air density are unchanged. At comparable points assume unchanged efficiency. Estimate flow, pressure and shaft power at 1,080 rpm.

Calculate the speed ratio, 1.20.
Multiply flow by the ratio.
Multiply pressure by the ratio squared.
Multiply shaft power by the ratio cubed.

Q_2=10{,}000(1.2)=12{,}000\ \mathrm{cfm}

\Delta p_2=2.0(1.2)^2=2.88\ \mathrm{in.\ w.g.}

P_2=8(1.2)^3=13.824\ \mathrm{hp}

Decision

The comparable-point estimate is 12,000 cfm, 2.88 in. w.g. and about 13.8 shaft hp.

Independent check

A 20% speed increase produces about a 73% power increase under these assumptions, not a 20% increase.

Specialist extension

Distinguish isentropic from mechanical compressor efficiency

Isentropic compressor efficiency compares the ideal isentropic enthalpy rise with the actual refrigerant enthalpy rise at the same inlet and discharge pressure. Mechanical or motor losses belong to additional input boundaries if supplied. Obtain the ideal discharge state from inlet entropy and discharge pressure on the stated refrigerant property source rather than substituting an arbitrary outlet temperature.

A new applied case

Inlet enthalpy is 110 Btu/lb and supplied isentropic discharge enthalpy is 130. Isentropic efficiency is 0.75 and refrigerant flow is 400 lb/h. Ignore other losses.

h_{2a}=110+(130-110)/0.75=136.667\ \mathrm{Btu/lb}

\dot W_{ref}=400(136.667-110)=10{,}666.7\ \mathrm{Btu/h}

Decision. Actual discharge enthalpy is 136.667 and refrigerant work is about 10,666.7 Btu/h.

Cross-check. An efficiency below one makes actual work greater than the ideal 8,000 Btu/h for this specified comparison.

Unit basis. h is Btu/lb on one datum; mass rate is lb/h; isentropic efficiency is dimensionless; this is refrigerant work.

NCEES PE Mechanical Reference Handbook v2.1; see the module reference path.

Open targeted practice in the study studio →

Transfer the method

HVAC Equipment and Components · HV-LC-E5

Try a new decision.

Select one

A fixed-diameter fan operates at 1,000 rpm with 5,000 cfm, 2.0 in. water pressure rise and 4.0 hp shaft power. At a comparable point, speed increases to 1,200 rpm. Assume unchanged density and comparable efficiency. The new airflow, pressure rise and shaft power, respectively are most nearly:

A7,200 cfm; 2.88 in. water; 5.76 hpB6,000 cfm; 2.40 in. water; 4.80 hpC4,167 cfm; 1.39 in. water; 2.315 hpD6,000 cfm; 2.88 in. water; 6.912 hp

A lesson check supports practice, not a readiness score. Your response stays in this page session and clears on reload.

Catch the common errors

Applying the speed ratio once to power.
Using electrical input power interchangeably with shaft power when efficiency changes.
Using affinity scaling as a substitute for a static-head system-curve or NPSH check.

Handbook navigation drill

Find it, then explain it.

Search: Pump Affinity Laws

Locate the pump speed-law relationships and the fan section. Find the NPSH discussion in pump characteristics.

Check your navigation

You can identify the exponent and assumptions for each quantity before using the law.

Reference path: §3.7.4–3.7.6, pp. 255–259; §4.5.3, p. 292; §9.3.8, pp. 516–523. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Theory sources

NCEES handbook v2.1, §§3.7.4–3.7.6, 4.5.3 and 9.3.8.
Grondzik and Kwok, Mechanical and Electrical Equipment for Buildings, 12th ed., ch. 12.

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