# Blower power terms: What are the parameters to quantify ac blower input requirements and output capabilities?

Blower performance, efficiency, and output are partially quantified by power terms that account for ac electric input, resistance, reactance, impedance, airpower, and output in airwatts. Power equations for ac setups relate three types of electrical power — namely, true power, reactive power, and apparent power — to the scalar quantities of resistance, impedance, and reactance.

Recall from circuit fundamentals that Ohm’s law, electrical power dissipated as Joule heating, and true power are:

**V=IR and P=I ^{2}R and P=E^{2}/R**

Where V= Voltage; I= Current; P = True power in watts; R = Resistance; and E = Potential difference in Volts. True power is the portion of power into the motor design (and its inherent resistance) available to output mechanical work. Compare that with reactive power. Reactive power expresses how much a blower motor appears to store and discharge power. It is expressed in

Volts-Amperes-Reactive:

**Q=I**

^{2}X=E^{2}/XWhere Q = Reactive power and X = Motor-circuit reactance.

The last power term to quantify blower performance is apparent power. Expressed in Volt-Amperes, apparent power is the vector sum of reactive and true power ... in other words, voltage into the ac motor multiplied by all input current:

**S=I**

^{2}Z and S=E^{2}/Z and S=IE

Where S = Apparent power and Z = Total circuit impedance.

Where S = Apparent power and Z = Total circuit impedance.

To understand the relationships between true power, reactive power, and apparent power, consider an example application with a realistic model of a load that includes reactive and resistive elements. Assume we have a circuit with an ac input of 120 V at 60 Hz; I = 1.410 A; the load’s L = 160 MH with XL = 60.319 Ω; and the load’s R = 60 Ω.

In this case,

**true power P = i**

^{2}R = 119.365 W; reactive power Q = I^{2}X = 119.998 VAR; and apparent power S = I^{2}Z = 169.256 VA.These values then express a blower’s power factor — the ratio of the blower’s true power drawn P to the apparent power S that the blower draws. That’s why a larger VAR makes for lower power factor — mathematically expressed as a percent (x 100) of the cosine ø of the angle between vectors S and P.

Only a system’s overall true power in watts can output useful work. Even so, the utility supplies into which a blower is integrated must generate and distribute the actual draw — that of apparent power. All utilities meter watts but some also measure the overall power factor of the facilities they service — and impose surcharges for power factors of 75% or lower.

One caveat here: A blower with a low power factor can indicate low efficiency, but remember that these designs always integrate into systems that include the treated environment as well as other electrical equipment ... and the system’s overall power factor is what’s more relevant. Sometimes drive-based approaches and improving other design elements (such as resistance-based heating; power transformers; stray inductance associated with system wiring; and electric-motor subtypes) can more significantly boost an installation’s power factor. Some setups benefit from input power-factor correction modules (PFC modules) that combine IGBTs, diodes, and inducting magnetics to boost the setup’s power factor as well.

We’ve outlined how industry quantifies blower-motor energy input and use. Industry quantifies blower output in terms of air power in units of airwatts. This expresses the effectiveness of a blower’s airflow. The standard airwatt formula is defined by ASTM International in the organization’s document ASTM F558 – 13:

**P=0.117354∙F∙S**

Where P = Power in airwatts; F = Rate of airflow in cubic feet per minute (denoted ft3/min. or CFM) and S = Suction capacity expressed as a pressure in units of inches of water. Note that one airwatt equals 0.9983 W.

Scandal and resolution in the air-mover industry

Scandal and resolution in the air-mover industry

The origins of the airwatt calculation in an interesting one. The need for airwatts arose from confusion and misleading claims in the vacuum-cleaner industry. Some decades ago, vacuum manufacturers quantified products’ performance in terms of power and horsepower. Some stretched the meaning of power terms and used the values of their vacuum-motors’ peak-horsepower ratings in literature targeted to nontechnical consumers.

Of course, peak-horsepower ratings aren’t sustainable for more than a second or two at most before motor damage occurs. In the context of application design, the value is relatively useless. Not surprisingly, the situation spurred lawsuits over motor-performance shortfalls. In response to this trend — and in a move to prevent general misconceptions — the American Society for Testing and Materials (ASTM) established the standard airwatts expression to quantify vacuum-cleaner performance. Today, the entire air-moving industry uses the measure of air power — as well as flow versus vacuum CFM in the unit of inches of water.

So what is the origin of the 0.117354 factor in the airwatts equation? In fact, the value is based on standardized parameters from ASTM. Power is the rate of work in a time: P = F·v, where P = power; F = force; and v = velocity, ft/min. Air power is the net time rate of work (in watts) by an airstream expending energy on airflow by a unit under set air resistance: AP = 745.7/33000 F·v, where AP = Air power; and F = Force generated by the airstream passing through the orifice, lb.

The constant 745.7/33000 maintains units:

1 watt= 33,000 ft lb745.7 min.

For an airstream through a set orifice:

F=112phsA

The constant 745.7/33000 maintains units:

1 watt= 33,000 ft lb745.7 min.

For an airstream through a set orifice:

F=112phsA

Where F = Force generated by airstream passing through the orifice, lb; p = Density of water at 68°F and 62.3205 lb/ft3; hs = Differential pressure at standard conditions, in. of water; and A = Orifice area cross-section, ft2. The 1/12 constant maintains units:

F (lb)=1 ft/12 in.(p)lbft/3hs(in.) A (ft

F (lb)=1 ft/12 in.(p)lbft/3hs(in.) A (ft

^{2})**Air velocity is V = Q/A**, where velocity of air stream passing through the orifice, ft/min.; and Q = Flow rate at standard air density and temperature, CFM (and not reactive power as in blower electrical-power equations). Substituting equations into the air-velocity expression, we get p = 62.3205 lb/ft3 and AP = 0.117354 hs·Q — hence the 0.117354 factor in the equation to calculate airwatts.

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