[ HPAC Engineering Report ] Fan Selections and Their Energy Impacts

[ HPAC Engineering Report ] Fan Selections and Their Energy Impacts

According to the U.S. Energy Information Administration, commercial buildings account for approximately 20% of annual energy consumption in the United States. Within commercial buildings, a significant amount of energy is used by HVAC systems, including fans. By some estimates, fans account for roughly 15% of the energy used in commercial and industrial buildings and processes. Aerosealing leaky air ducts helps to reduce excess fan work by lowering air duct transmission losses. The accompanying report provides a good overview of how fan selection and operation impact electric energy usage and subsequent fan operating costs.

Report authored by Khalil Kairouz and Dale Price.

Fans and fan systems can be significant consumers of energy in commercial applications and the greatest consumer in industrial ones. Concerned about underperformance, designers tend to compensate for uncertainties by adding capacity to fans. Oversizing fan systems creates problems that can increase costs and decrease reliability. This article explains how to evaluate fans and fan systems and their impact on operating costs. It does so by examining the criteria most commonly used in fan selection:

  • Air-stream characteristics.
  • Fan and system curves.
  • Capacity.
  • Pressure.
  • Power.
  • Efficiency.
  • Fan and system curves.

Air-Stream Characteristics

Fan manufacturers’ published ratings for airflow, pressure, and power are based on laboratory conditions using standard cubic feet per minute (scfm) for airflow and 0.075 lb per cubic foot for fan-inlet-air-stream density (corresponding to dry air at 70°F and atmospheric pressure of 29.92 in. Hg). Normally, they can be used for fan selection and evaluation when the difference between standard density and actual site density at a fan inlet is less than 5 percent. This occurs when:

  • Fan-inlet temperature is within 30°F of the standard +70°F dry air.
  • Fan-inlet pressure is within 12 in. wg of the standard atmospheric pressure of 29.92 in. Hg.
  • Fan-inlet-air-stream moisture content is less than 0.02 lb of water per pound of dry air or the dew point is less than 80°F.
  • Installation is between 1,000 ft below and 1,000 ft above sea level (asl).

In cases in which at least one of the above parameters is not met, inlet-air-stream density should be corrected to actual conditions (actual cubic feet per minute [acfm]). This is done by multiplying the standard density, airflow, pressure, and power by each parameter’s density factor. The density factors will be equal to the actual inlet density divided by the standard inlet density of 0.075 lb per cubic foot. The density factor (Fρ) for each parameter is estimated as follows:

Density-factor temperature, degrees Fahrenheit:

(t) = 530 ÷ (460 + Tinlet)

Density-factor pressure, inches water gauge:

(p) = 407 ÷ (407 + Pinlet)

Density-factor moisture, pounds of water per pound of dry air:

(m) = (1 + w) ÷ (1 + 1.607w)

where:

w = humidity ratio, pounds of water per pound of dry air

Density-factor elevation, feet above sea level:

(e) = [1 − (6.73 × 10−6) × z]5.258

where:

z = elevation, feet

Overall density-correction factor, then, is given by:

Fρ = Fρ(t) × Fρ(p) × Fρ(m) × Fρ(e)

Example 1: Density-factor calculation. For a fan capable of handling 176°F air at an inlet static pressure of –15 in. wg, a humidity ratio of 0.04 lb of water per pound of dry air, and a site elevation of 3,000 ft asl, we will calculate the required density factor and actual air density:

Inlet temperature, 176°F:

(t) = 530 ÷ (460 + 176) = 0.83

Inlet static pressure, −15 in. wg:

(p) = (407 + −15) ÷ 407 = 0.96

Inlet moisture, 0.04 lb water per pound dry air:

(m) = (1 + 0.04) ÷ [1 + (1.607 × 0.04)] = 0.98

Elevation, 3,000 ft asl:

(e) = [1 – (6.73 × 10–6) × 3,000]5.258 = 0.90

Density factor:

Fρ = 0.83 × 0.96 × 0.98 × 0.90 = 0.70

The actual air-stream density at the fan inlet is the product of the density factor and the standard air density (0.075 lb per cubic foot) and is calculated as:

ρ(actual) = 0.70 × 0.075 lb per cu ft = 0.0525 lb per cu ft

Example 2: Fan performance. If a fan were selected from a manufacturer’s standard catalogued ratings for a volumetric-flow rate of 30,000 acfm, a static pressure of 15 in. wg, 1,910 rpm, and 86 hp, but installed at the above conditions, the operating performance would be:

  • Actual volumetric capacity: 30,000 acfm.
  • Actual mass-flow capacity: 1,575 lb per minute (30,000 × 0.0525).
  • Actual static-pressure delivery: 10.5 in. wg (15 in. wg × 0.70).
  • Actual fan-shaft power consumption: 60.2 hp (86 hp × 0.70).
  • Actual inlet-air-stream density: 0.0525 lb per cubic foot (0.70 × 0.075).

Key point: With proper definition of inlet-air-stream conditions for density, temperature, pressure, moisture, and elevation, corrections between a fan manufacturer’s published ratings and actual operating conditions can be made to ensure proper fan selection and performance.

Capacity

Fan capacity can be stated in terms of either volumetric-flow rate or mass-flow rate.

Volumetric-flow rate:

  • Is the normal basis for fan ratings.
  • Normally is stated in cubic feet per minute.
  • Represents the actual volumetric-flow capacity of a fan at a given speed and static pressure.
  • Is constant for any air-stream density.

Mass-flow rate:

  • Normally is stated in pounds per minute.
  • Represents the mass-flow capacity of a fan at a given speed, static pressure, and inlet density.
  • Is calculated by multiplying volumetric-flow rate by actual air-stream density at a fan’s inlet.
  • Varies with changes in air-stream density.

Figure 1 is a simplified illustration of a centrifugal-fan wheel and a substance conveyed by the fan. If each scoop of the fan wheel has a volumetric capacity of 5 cu ft and the speed is a constant 1,000 rpm, the volumetric capacity is 30,000 cfm (5 cu ft per scoop × 6 scoops × 1,000 rpm). The mass-flow rate is:

Figure 1 - Fan Wheel
FIGURE 1. Fan wheel and mound representing a substance conveyed by the fan.

5 cu ft per scoop × 6 scoops × 1,000 rpm × density (pounds per cubic foot) of substance conveyed

Assuming the substance is air at standard conditions and standard density of 0.075 lb per cubic foot, the mass delivery is 2,250 lb per minute (30,000 cfm × 0.075 lb per cu ft).

Assuming the substance is air at the non-standard conditions in examples 1 and 2 with a density of 0.0525 lb per cubic foot, the mass delivery is 1,575 lb per minute (30,000 cfm × 0.0525 lb per cu ft).

If the fan system must maintain a minimum mass-flow rate of 2,250 lb per minute, then the required actual volumetric-flow rate is 42,857 cfm (2,250 lb per min ÷ 0.0525 lb per cu ft) and the fan should be selected on the volumetric basis of 42,857 acfm.

Key point: At constant speed, a fan will deliver a constant volumetric-flow rate and a variable mass-flow rate, with the mass-flow rate varying directly with the density factor, or the ratio of actual density to standard density.

Pressure

In air-moving applications, the most common pressure is static pressure. Still, for the proper selection and operation of fans, one must have a basic understanding of all pressures:

  • Total pressure (TP).
  • Velocity pressure (VP).
  • Static pressure (SP).
  • Fan static pressure (FSP).
  • Fan velocity pressure (FVP).
  • Fan total pressure (FTP).

Figure 2 shows total-, static-, and velocity-pressure components for a short section of ductwork. Total pressure is parallel to and measured in the direction of flow. Velocity pressure is parallel to the direction of flow and can be measured only indirectly as the difference between total pressure and static pressure. Static pressure is exerted equally in all directions and measured perpendicularly to the direction of flow.

Figure 2 - Article AA - Photo C
FIGURE 2. Duct-system pressure components.

Total pressure. Total pressure is the algebraic sum of kinetic energy (velocity pressure) and potential energy (static pressure) at any point in a system. Thus, total pressure represents the total energy at the point of measurement of fluid flow:

TP = VP + SP

Total pressure can be positive or negative, depending on the measurement location.

Velocity pressure. Velocity pressure is kinetic energy in the direction of flow that causes a fluid at rest to flow:

VP = TP − SP

VP = (V ÷ 1,097)2 × ρ(actual)

therefore:

VP = (V ÷ 4,005)2 × Fρ

Velocity pressure is mutually convertible with static pressure. It always is positive, regardless of its measurement location.

Static pressure. Static pressure is potential energy exerted in all directions by a fluid:

SP = TP − VP

Static pressure is mutually convertible with velocity pressure. It can be positive or negative, depending on the measurement location.

Fan total pressure vs. total pressure. While total pressure is the sum of velocity pressure and static pressure, fan total pressure is the increase in total pressure through or across a fan and represents the total work delivered by the fan. As defined in Air Movement and Control Association (AMCA) International standards, fan total pressure normally is not used for fan-rating purposes (this is by custom, not because of any deficiency in the rating technique). However, an understanding of fan total pressure is important for proper selection and field measurement of fans and fan systems. Fan total pressure is expressed as:

FTP = TPfan outlet − TPfan inlet

FTP = (SPoutlet + VPoutlet) − (SPinlet + VPinlet)

When fan-inlet velocity pressure and fan-outlet velocity pressure are the same, the pressures cancel each other, and the fan-total-pressure calculation is simplified to:

FTP = SPoutlet − SPinlet

In this case, fan total pressure simply is the difference between fan-inlet static pressure and fan-outlet static pressure. When a fan is specified on the basis of summing inlet and outlet static pressures (static-pressure rise across the fan), the fan rating is on the fan-total-pressure basis.

Fan velocity pressure. AMCA International defines fan velocity pressure as velocity pressure based on the average velocity at a fan’s discharge. The average velocity at a fan’s discharge typically is calculated as fan outlet velocity divided by fan outlet area (square feet). For standard air, this is:

FVP = (fan outlet velocity ÷ 4,005)2

Fan static pressure vs. static pressure. Fan static pressure as defined by AMCA International is the basis for most published fan ratings today. While static pressure is the difference between total pressure and velocity pressure, fan static pressure is defined as fan total pressure minus fan velocity pressure:

FSP = FTP − FVP

FSP = (TPfan outlet − TPfan inlet) − VPfan outlet

FSP = (SPoutlet + VPoutlet) − (SPinlet + VPinlet) − VPoutlet

Because the outlet velocity pressures are identical and cancel each other:

FSP = SPoutlet − SPinlet − VPinlet

or, more commonly:

FSP = SPoutlet − TPinlet

However, it is not uncommon to find the static pressure used by system designers for fan selection calculated simply as:

SPoutlet − SPinlet

Often in these cases, the fan is oversized by the inlet-velocity-pressure energy component, resulting in excessive energy consumption.

When static pressure used for fan sizing is calculated in this manner, a fan cannot perform exactly on the fan curve because of the velocity pressure at the fan inlet normally not accounted for by the fan manufacturer. Figures 3 and 4 present examples of how to calculate fan pressure for various fan configurations.

Figure 3 - Article AA - Photo D
FIGURE 3. Fan with ducted inlet and outlet.

Key point: For optimum energy utilization and fan performance, follow the fan-static-pressure calculations issued by AMCA International.

Figure 4 - Article AA - Photo E

Example 3: Fan pressure—actual and equivalent. Actual fan pressure normally is defined as the pressure a fan delivers at actual operating conditions, while equivalent fan pressure normally is defined as the pressure a fan delivers at standard conditions (0.075-lb-per-cubic-foot air-stream density).

Because fan manufacturers’ catalogue ratings are based on standard conditions, it is important to make sure a fan is selected to achieve its design pressure at actual operating conditions.

Referring to examples 1 and 2, a fan with a catalogued fan static pressure of 15 in. wg will produce a fan static pressure of only 10.5 in. wg at actual operating conditions. If 15-in.-wg fan static pressure is required at actual operating conditions, a fan must be selected for a catalogued fan static pressure of 21.4 in. wg (15 ÷ 0.7). Normally, the designer would note a requirement of 15 in. wg at standard conditions and 21.4 in. wg at actual operating conditions.

Power, Efficiency And Fan And System Curves

Power

There are three categories of fan horsepower:

  • Chart horsepower.
  • Operating horsepower.
  • Air horsepower.

Chart horsepower. Often referred to as brake horsepower, chart horsepower (CHP) is the power required at a fan shaft to drive a fan. The only loss it may include is bearing drag. It always is stated at standard inlet-density conditions of 0.075 lb per cubic foot and, thus, should be corrected for non-standard inlet-air-stream densities. Corrections are calculated by multiplying the standard density, airflow, pressure, and power by the density factor.

CHP is the horsepower stated by fan manufacturers in published literature and selection programs. In most cases, because it represents only the power required to drive a fan shaft at standard air-stream-density conditions of 0.075 lb per cubic foot, it either understates or overstates actual operating horsepower, depending on the effects of drive-system losses and any change in inlet-air-stream density.

Operating horsepower. Operating horsepower (OHP) is the horsepower consumed by a fan and its driver at actual operating conditions. It includes the energy required by the fan shaft (CHP), the energy required by the driver, and the effects of any change in inlet-air-stream density from standard conditions.

OHP = (CHP × density factor) + drive loss

Example 4: Fan power. If a fan were selected for 30,000 cfm, 15-in.-wg fan static pressure (FSP), 1,910 rpm, and 86 CHP based on a manufacturer’s catalog rating at standard conditions, but installed at the actual conditions in examples 1 and 2 in Part 1 of this article, expected fan performance could be calculated at these actual operating conditions:

  • Density factor: Fρ = 0.83 × 0.96 × 0.98 × 0.90 = 0.70
  • Calculated air-stream density: 0.0525 lb per cubic foot
  • Actual volumetric capacity: 30,000 cfm
  • Actual mass-flow capacity: 1,575 lb per minute
  • Actual FSP: 15 in. wg × 0.70 = 10.5 in. wg
  • Actual fan-shaft power: 86 CHP × 0.70 = 60.2 hp
  • V-belt drive loss from Air Movement and Control Association (AMCA) International: 4 percent of actual motor output power
  • OHP = (CHP × density factor) + drive loss = (86 CHP × 0.70) + (0.04 × 60.2) = 60.2 hp + 2.4 hp = 62.6

If the fan were operated during startup, purge, or shutdown without heat or moisture, fan operating performance would be:

  • Density factor: Fρ = 1temp × 0.96pressure × 1moisture × 0.90elevation = 0.864
  • Calculated air-stream density = 0.0648 lb per cubic foot
  • Actual volumetric capacity: 30,000 cfm
  • Actual mass-flow capacity: 1,944 lb per minute
  • Actual FSP: 15 in. wg × 0.864 = 13.0 in. wg
  • Actual fan-shaft power: 86 CHP × 0.864 = 74.3 hp
  • V-belt drive loss from AMCA International: 4 percent of motor output power
  • OHP = (CHP × density factor) + drive loss = (86 CHP × 0.864) + (0.04 × 74.3) = 74.3 hp + 3 hp = 77.3

Based on a calculated OHP of 62.6, one might decide to select a 75-hp motor. But note that with a drive loss of 4 percent, the primary difference between the CHP (86) and OHP (62.6 or 77.3) is caused by the change in air-stream density. And because air-stream density may vary, particularly during startup, system purge, or shutdown, either control via a fan damper or a variable-frequency-drive (VFD) controller is recommended, or the next-size-larger motor should be selected.

In summary, OHP allows for changes in air-stream composition and the energy losses required to drive a fan. Both OHP and CHP should be considered in motor selection.

In some cases, final motor selection is based not on the above criteria, but on the minimum motor horsepower required to start a fan and bring it up to full speed. This is the effect of the combined wheel, shaft, and driver inertia required of a motor to put a fan in rotation from a resting position.

Air horsepower. Air horsepower (AHP) is the energy added to an air stream per unit time and is considered to be the actual horsepower output of a fan.

AHP = (flow rate × total pressure) ÷ 6,362

Using 30,000 actual cfm and 16-in.-wg actual fan total pressure (FTP):

AHP = (30,000 × 16) ÷ 6,362 = 75.5 hp added to the air stream

Often, it is helpful to calculate the AHP output of a fan to better understand the construction requirements of the fan system and the energy available within the system.

Key points. CHP:

  • Is the energy required to drive a fan shaft as stated in the manufacturer’s published charts and data.
  • Is based on standard air-stream density (0.075 lb per cubic foot).
  • Does not include the energy consumed by a fan drive system (V-belt, motor, or VFD losses).

OHP is the actual horsepower consumed in operation and includes both the energy consumed by a fan and the effects of changes in air-stream density (including drive losses, but not motor and VFD losses).

AHP is the useful energy added to an air stream by a fan.

Proper understanding and calculation of CHP, OHP, and AHP are necessary to fully understand the energy requirements, consumption, and delivery of a fan system.

Efficiency

Fan efficiency is the ratio of horsepower output (AHP) to horsepower input (CHP). It can be evaluated on the basis of FTP or FSP.

Total efficiency is based on FTP, and static efficiency is based on FSP, both as defined by AMCA International. Total efficiency equals AHP divided by CHP, with AHP calculated using FTP. AHP is the same in both cases. Static efficiency is the efficiency that produces the AHP, but is calculated as:

cfm × FSP ÷ 6,352

Figures 5 and 6 represent a typical selection for a backward-inclined wheel at 30,000 cfm and 10-in.-wg FSP, with the selection in Figure 5 based on peak total efficiency, and the selection in Figure 6 based on peak static efficiency.

Figure 5 - Article AA - Photo F
FIGURE 5. Backward-inclined wheel—peak total efficiency.
Figure 6 - Article AA - Photo G
FIGURE 6. Backward-inclined wheel—peak static efficiency.

It can be seen that when a fan is selected based on peak total efficiency—which takes into consideration the efficiency of both static pressure and velocity pressure—sometimes the fan both is smaller in size and operates farther to the right on the fan curve than when the fan selection is based on static efficiency (static pressure only). However, this applies to only housed fans, as unhoused fans typically do not have a means for measuring fan inlet and outlet velocity and, thus, normally are selected based on static efficiency.

Figure 5 shows that, at 30,000 cfm, FSP is 10.0 in. wg and fan velocity pressure (FVP) is 1.0 in. wg ([3,916 ÷ 4,005]2); therefore, FTP is 11.0 in. wg. CHP of 58.3 gives a total efficiency of 0.89. Static efficiency would be 0.81 ([30,000 × 10.0] ÷ [6,352 × 58.3]).

Figure 6 shows that, at 30,000 cfm, FSP is 10.0 in. wg and FVP is 0.65 in. wg ([3,222 ÷ 4,005]2); therefore, FTP is 10.65 in. wg. CHP of 57.0 gives a total efficiency of 0.88. Static efficiency would be 0.83 ([30,000 × 10.0] ÷ [6,352 × 57.0]).

Key point: For the lowest energy consumption, determine the fan requirement in terms of FTP, and maximize the total efficiency. Maximizing fan static efficiency does not necessarily minimize energy consumption.

Fan and System Curves

Fan performance can be predicted using what commonly are referred to as the fan or affinity laws:

  • Fan capacity varies directly with fan speed.
  • Fan pressure varies with the square of the change in fan speed.
  • Fan power varies with the cube of the change in fan speed.

The fan laws are applicable on the basis airflow through a fan is fully turbulent. Thus, in the absence of a connected system, the fan laws always are applicable. And while it is not uncommon to apply the fan laws to fan systems to predict performance, one must be careful to ensure the system to which a fan is connected also is in turbulent flow. If this condition is not met, the fan laws should be applied with care, as they cannot be applied directly to system components, such as deep high-efficiency filters, which may be operating in or near laminar flow. In these cases, a system should be separated into turbulent- and laminar-flow components for separate calculations, which later are joined for a predictable system performance curve.

Figure 7 - Article AA - Photo H
FIGURE 7. System curve representing turbulent-flow conditions in which system pressure varies with the square of system flow.

The system curve in Figure 7 represents turbulent-flow conditions in which system pressure varies with the square of system flow.

Figure 8 - Article AA - Photo I
FIGURE 8. System curve representing laminar-flow conditions in which the pressure component varies directly with capacity.

Example 5: Increase capacity by 10 percent. A fan in a system in which all components are operating in fully turbulent flow is operating at the following conditions: 2,321-rpm fan speed, 15,000-cfm fan capacity, 12-in.-wg FSP, and 35.1 CHP. What change would be required to increase the volumetric capacity by 10 percent? Assume the fan system is operating at standard air conditions.

Fan-law calculations to obtain new operating conditions:

Speed: 2,321 rpm × 1.1 = 2,553 rpm (10-percent increase)

New capacity: 15,000 cfm × 1.1 = 16,500 cfm (10-percent increase)

FSP: 12 in. wg × (1.10)2 = 14.52 in. wg (21-percent increase)

Fan horsepower: 35.1 × (1.10)3 = 46.7 hp (33-percent increase)

Fan curves for the original and revised selections are shown in figures 9 and 10, respectively.

Figure 9 - Article AA - Photo J
FIGURE 9. Fan curve for original selection.
Figure 10 -Article AA - Photo K
FIGURE 10. Fan curve for revised selection.

Example 6: Increase capacity by 10 percent. A fan is operating at the conditions in Example 5; however, the FSP is assumed to be composed of 8 in. wg of components that vary based on fully turbulent flow and 4 in. wg of components that are in non-turbulent flow. Assume laminar flow. The fan speed is 2,321 rpm, the fan capacity is 15,000 cfm, the FSP is 12 in. wg, and the CHP is 35.1. What is the change required to increase the volumetric capacity by 10 percent? Assume the fan system is operating at standard air conditions.

System calculations:

Capacity: 15,000 cfm × 1.1 = 16,500 cfm (10-percent increase)

FSP: Turbulent-flow components follow the fan laws and increase by the square of the change in capacity. Laminar-flow components for this example increase directly proportional to the increase in capacity.

8 in. wg × (1.10)2 = 9.68 in. wg (21-percent increase) + 4 in. wg × 1.1 = 4.4 in. wg (10-percent increase) = 14.08 in. wg (17.3-percent increase)

New fan speed of 2,526 rpm (8.8-percent increase) and new horsepower of 45.4 (29-percent increase) have to be determined by obtaining a new fan curve at the revised flow and pressure condition.

Fan curves for the original and revised selections are shown in figures 11 and 12, respectively.

Figure 11 - Article AA - Photo L
FIGURE 11. Fan curve for original selection.
Figure 12 - Article AA - Photo M
FIGURE 12. Fan curve for revised selection.

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Summary

An understanding of fan total and static efficiencies aids identification and selection of the best fan for an application, while thorough consideration of all fan and system parameters aids energy-efficient design.

Correctly understanding and calculating FSP prevents unintended overdesign and system horsepower consumption.

A fan motor always should be selected based on OHP.

When changes to an existing system need to be made, proper calculation of the new system requirements based on the characteristics of each type of system component (e.g., ductwork, air-pollution-control equipment, after-filter) results in optimal design and prevents unintended overdesign.

# # #

This article originally appeared in HPAC Engineering as a two-part essay. The first part was published on August 23, 2016 and can be viewed on the HPAC Engineering website.

Khalil Kairouz, PhD, PE, is an associate vice president for Carollo Engineers Inc., a consulting firm specializing in water and wastewater treatment. He has bachelor’s and master’s degrees in mechanical engineering from the University of Louisiana at Lafayette and a doctorate from Claremont Graduate University. He has taught in the mechanical-engineering departments of Loyola Marymount University and California State University. Dale Price is president of M&P Air Components Inc., a company specializing in ventilation and air-distribution equipment. He has a bachelor’s degree is mechanical engineering and is an active member of an American Industrial Hygiene Association task force on fan efficiency. He has taught numerous training sessions on fan systems and their effects to employees of state and federal agencies and private industries.

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