Fan Static vs Total Efficiency: Put Outlet Velocity Pressure on the Same Boundary
Two selection sheets report different efficiency percentages for apparently the same duty. Before choosing the higher number, check which pressure appears in the numerator and which power appears in the denominator. Fan total efficiency and fan static efficiency can both be correct for one fan at one operating point. Mixing their pressure boundaries produces an incorrect power estimate.
1. Define the fan boundary before the efficiency
Use stations 1 and 2 at the defined fan inlet and outlet. At either station, total pressure pₜ equals static pressure pₛ plus velocity pressure pᵥ. For the low-speed approximation used here, pᵥ = ρV²/2, based on the relevant mean velocity and air density. Fan total pressure FTP = pₜ₂ − pₜ₁.
In the AMCA convention, fan velocity pressure is the velocity pressure at the specified fan outlet. Fan static pressure FSP = FTP − pᵥ₂. Substituting the station pressures gives FSP = pₛ₂ − pₜ₁. It is therefore generally different from the simple difference between outlet and inlet static pressures.
AMCA Publication 803-26 defines these fan-pressure quantities and the specified outlet area. The boundary matters: the outlet of a rated fan assembly may include a particular diffuser or accessory. A downstream duct area chosen from the building drawing cannot automatically replace the outlet area used for the fan rating.
2. Use matching pressure and power definitions
For an illustrative incompressible calculation, fan total efficiency ηₜ = Q × FTP/Pshaft, and fan static efficiency ηₛ = Q × FSP/Pshaft. Q is in m³/s, pressure in Pa and mechanical input power Pshaft in watts. The product of m³/s and Pa is watts. A value of 72% enters a calculation as 0.72.
The two efficiencies are related by ηₛ = ηₜ × FSP/FTP when both use the same operating point and mechanical input-power basis. AMCA’s laboratory fan-testing definitions distinguish mechanical fan input from fan-with-drive input and include a compressibility coefficient in the complete total-efficiency definition. The equations here set that coefficient to one; do not extend that simplification to high-pressure applications without checking it.
Read the selection-sheet notes. A percentage based on mechanical shaft input cannot be compared directly with an overall percentage based on input to a motor-and-drive assembly. Similarly, an AHU’s external static pressure is a unit connection boundary; it is not automatically the pressure developed by its internal fan. Internal components and casing losses must be reconciled separately.
3. Worked example: one fan, two valid efficiencies
Assume an original hypothetical selection with Q = 5.00 m³/s, density ρ = 1.20 kg/m³, defined outlet area A₂ = 0.400 m², fan total pressure FTP = 900 Pa and mechanical shaft input of 6.250 kW. Assume steady, low-speed operation, negligible compressibility effects and consistent rated pressure stations. This is an accounting example, not a fan product rating.
Outlet mean velocity V₂ = 5.00/0.400 = 12.50 m/s.
Outlet velocity pressure pᵥ₂ = 1.20 × 12.50²/2 = 93.75 Pa.
Fan static pressure FSP = 900 − 93.75 = 806.25 Pa.
Total efficiency ηₜ = 5.00 × 900/6250 = 0.720, or 72.0%.
Static efficiency ηₛ = 5.00 × 806.25/6250 = 0.645, or 64.5%.
Original constant-density example: Q = 5.00 m³/s, ρ = 1.20 kg/m³, defined outlet area 0.400 m² and shaft input 6.250 kW. Both matched pressure-efficiency pairs give the same shaft power.
Both percentages describe the same shaft input. The 7.5-percentage-point difference does not identify a deterioration or an improvement. It reflects the exclusion of outlet velocity pressure from the static-efficiency numerator. AMCA’s fan-curve explanation shows the same distinction between static-pressure, total-pressure and efficiency curves.
Reconstruct shaft input using either matched pair: 5.00 × 900/0.720 = 6250 W, or 5.00 × 806.25/0.645 = 6250 W. If someone divides total-pressure air power by static efficiency instead, the result becomes 5.00 × 900/0.645 = 6977 W, about 11.6% high. Conversely, static-pressure air power divided by total efficiency gives 5599 W, about 10.4% low.
4. Why two static taps give a different answer
Extend the example with inlet area A₁ = 0.625 m² and inlet total pressure pₜ₁ = −120 Pa relative to a common reference. Inlet mean velocity is 8.00 m/s, giving pᵥ₁ = 38.40 Pa. Therefore inlet static pressure pₛ₁ = −158.40 Pa. Because FTP is 900 Pa, outlet total pressure pₜ₂ = 780 Pa and outlet static pressure pₛ₂ = 686.25 Pa.
The static-to-static rise is 686.25 − (−158.40) = 844.65 Pa. Fan static pressure remains 806.25 Pa. Their difference is the 38.40 Pa inlet velocity pressure. Substituting 844.65 Pa as FSP would silently change the rating definition. Algebraically, pₛ₂ − pₛ₁ = FSP + pᵥ₁ under the assumptions above.
Actual site measurements introduce additional issues: nonuniform profiles, station locations, intervening duct losses and instrument uncertainty. A nearby pair of pressure readings is not automatically a measurement at the rating boundary. Specify the measurement method and corrections before treating a reconstructed efficiency as a verified fan test.
5. Do not count outlet velocity pressure twice
Outlet kinetic energy may be partly converted to static pressure in a suitable downstream expansion, or dissipated farther along the system. Neither outcome creates additional total pressure. Keep the entire system calculation consistent: if the stated requirement is already total pressure on the correct boundary, adding the fan outlet velocity pressure again overstates the requirement.
If starting from a stated static requirement, first establish its definition and locations before converting it. A complete conversion is not simply “add a velocity pressure” whenever the word static appears. The unit connection area, fan rating outlet and final discharge can be different sections with different velocities and intervening losses.
AMCA’s discussion of system effect explains how nearby inlet and outlet arrangements can impair installed performance. A neat pressure conversion cannot repair an installation mismatch; review the selected arrangement and its applicable performance data separately.
6. Make a like-for-like selection review
Record airflow, density, speed and the exact duty point rather than comparing peak efficiencies.
Identify the fan inlet/outlet boundary, test arrangement and accessories included in the rating.
Obtain the rated outlet area or outlet velocity and reconcile FTP, FSP and outlet velocity pressure.
Identify the power denominator and use the corresponding efficiency definition.
Recalculate input power with a matched pressure-efficiency pair and reconcile the result with the supplier’s value.
Check the installed inlet/outlet arrangement and the broader system pressure path before making the selection.
For an AHU review, the paid Air Handling Unit (AHU) Selection & Fault Diagnosis Handbook is a related reference. The decisive record is still the selected fan’s duty sheet with its pressure stations, outlet area and power basis. Once those match, the efficiency comparison becomes meaningful.
Comments