The document discusses various formulas for calculating electrical motor efficiency, speed, torque, horsepower, and other metrics. It provides formulas for calculating motor efficiency based on shaft output power measured in watts or horsepower. Formulas are also given for calculating primary and secondary resistance losses, synchronous speed based on supply frequency and poles, braking torque, work, full-load torque, and horsepower based on voltage, current, and efficiency. Minimum nominal efficiencies are listed for NEMA Design B electrical motors of different horsepower ratings.

1. Electrical Motor Efficiency
Electrical motor efficiency is the ratio between the shaft output power - and the electrical
input power.
Electrical Motor Efficiency when Shaft Output is measured in Watt
If power output is measured in Watt (W), efficiency can be expressed as:
ηm = Pout / Pin (1)
where
ηm = motor efficiency
Pout = shaft power out (Watt, W)
Pin = electric power in to the motor (Watt, W)
Electrical Motor Efficiency when Shaft Output is measured in Horsepower
If power output is measured in horsepower (hp), efficiency can be expressed as:
ηm = Pout 746 / Pin (2)
where
Pout = shaft power out (horsepower, hp)
Pin = electric power in to the motor (Watt, W)
Primary and Secondary Resistance Losses
The electrical power lost in the primary rotor and secondary stator winding resistance
are also called copper losses. The copper loss varies with the load in proportion to the
current squared - and can be expressed as
Pcl = R I2 (3)
where
Pcl = stator winding - copper loss (W)
R = resistance (Ω)
I = current (Amp)

2. Iron Losses
These losses are the result of magnetic energy dissipated when when the motors
magnetic field is applied to the stator core.
Stray Losses
Stray losses are the losses that remains after primary copper and secondary losses, iron
losses and mechanical losses. The largest contribution to the stray losses is harmonic
energies generated when the motor operates under load. These energies are dissipated
as currents in the copper windings, harmonic flux components in the iron parts, leakage
in the laminate core.
Mechanical Losses
Mechanical losses includes friction in the motor bearings and the fan for air cooling.
NEMA Design B Electrical Motors
Electrical motors constructed according NEMA Design B must meet the efficiencies
below:
Power Minimum Nominal
(hp) Efficiency1)
1-4 78.8
5-9 84.0
10 - 19 85.5
20 - 49 88.5
50 - 99 90.2
100 - 124 91.7
> 125 92.4
1)
NEMA Design B, Single Speed 1200, 1800, 3600 RPM. Open Drip Proof (ODP) or
Totally Enclosed Fan Cooled (TEFC) motors 1 hp and larger that operate more than 500
hours per year.

3. Options:
- Useful Formulas Formulas
- Transformer Formulas
Calculating Motor Speed:
A squirrel cage induction motor is a constant speed device. It cannot operate for any
length of time at speeds below those shown on the nameplate without danger of burning
out.
To Calculate the speed of a induction motor, apply this formula:
Srpm = 120 x F
P
Srpm = synchronous revolutions per minute.
120 = constant
F = supply frequency (in cycles/sec)
P = number of motor winding poles
Example: What is the synchronous of a motor having 4 poles connected to a 60 hz power
supply?
Srpm = 120 x F
P
Srpm = 120 x 60
4
Srpm = 7200
4
Srpm = 1800 rpm
Calculating Braking Torque:
Full-load motor torque is calculated to determine the required braking torque of a motor.
To Determine braking torque of a motor, apply this formula:
T = 5252 x HP
rpm

4. T = full-load motor torque (in lb-ft)
5252 = constant (33,000 divided by 3.14 x 2 = 5252)
HP = motor horsepower
rpm = speed of motor shaft
Example: What is the braking torque of a 60 HP, 240V motor rotating at 1725 rpm?
T = 5252 x HP
rpm
T = 5252 x 60
1725
T = 315,120
1725
T = 182.7 lb-ft
Calculating Work:
Work is applying a force over a distance. Force is any cause that changes the position,
motion, direction, or shape of an object. Work is done when a force overcomes a
resistance. Resistance is any force that tends to hinder the movement of an object.If an
applied force does not cause motion the no work is produced.
To calculate the amount of work produced, apply this formula:
W=FxD
W = work (in lb-ft)
F = force (in lb)
D = distance (in ft)
Example: How much work is required to carry a 25 lb bag of groceries vertically from
street level to the 4th floor of a building 30' above street level?
W=FxD
W = 25 x 30
W = 750 -lb
Calculating Torque:
Torque is the force that produces rotation. It causes an object to rotate. Torque consist of
a force acting on distance. Torque, like work, is measured is pound-feet (lb-ft). However,
torque, unlike work, may exist even though no movement occurs.
To calculate torque, apply this formula:

5. T=FxD
T = torque (in lb-ft)
F = force (in lb)
D = distance (in ft)
Example: What is the torque produced by a 60 lb force pushing on a 3' lever arm?
T=FxD
T = 60 x 3
T = 180 lb ft
Calculating Full-load Torque:
Full-load torque is the torque to produce the rated power at full speed of the motor. The
amount of torque a motor produces at rated power and full speed can be found by using a
horsepower-to-torque conversion chart. When using the conversion chart, place a straight
edge along the two known quantities and read the unknown quantity on the third line.
To calculate motor full-load torque, apply this formula:
T = HP x 5252
rpm
T = torque (in lb-ft)
HP = horsepower
5252 = constant
rpm = revolutions per minute
Example: What is the FLT (Full-load torque) of a 30HP motor operating at 1725 rpm?
T = HP x 5252
rpm
T = 30 x 5252
1725
T = 157,560
1725
T = 91.34 lb-ft
Calculating Horsepower:
Electrical power is rated in horsepower or watts. A horsepower is a unit of power equal to
746 watts or 33,0000 lb-ft per minute (550 lb-ft per second). A watt is a unit of measure
equal to the power produced by a current of 1 amp across the potential difference of 1

6. volt. It is 1/746 of 1 horsepower. The watt is the base unit of electrical power. Motor
power is rated in horsepower and watts.
Horsepower is used to measure the energy produced by an electric motor while doing
work.
To calculate the horsepower of a motor when current and efficiency, and voltage are
known, apply this formula:
HP = V x I x Eff
746
HP = horsepower
V = voltage
I = curent (amps)
Eff. = efficiency
Example: What is the horsepower of a 230v motor pulling 4 amps and having 82%
efficiency?
HP = V x I x Eff
746
HP = 230 x 4 x .82
746
HP = 754.4
746
HP = 1 Hp
Eff = efficiency / HP = horsepower / V = volts / A = amps / PF = power factor
Horsepower Formulas
Example
To Find Use Formula
Given Find Solution
HP = 240V x 20A x 85%
HP = I X E X Eff.
HP 240V, 20A, 85% Eff. HP 746
746
HP=5.5
I = 10HP x 746
I = HP x 746 10HP, 240V,
I I 240V x 90% x 88%
E X Eff x PF 90% Eff., 88% PF
I = 39 A
To calculate the horsepower of a motor when the speed and torque are known,
apply this formula:
HP = rpm x T(torque)
5252(constant)

7. Example: What is the horsepower of a 1725 rpm motor with a FLT 3.1 lb-ft?
HP = rpm x T
5252
HP = 1725 x 3.1
5252
HP = 5347.5
5252
HP = 1 hp
Calculating Synchronous Speed:
AC motors are considered constant speed motors. This is because the synchronous speed
of an induction motor is based on the supply frequency and the number of poles in the
motor winding. Motor are designed for 60 hz use have synchronous speeds of 3600,
1800, 1200, 900, 720, 600, 514, and 450 rpm.
To calculate synchronous speed of an induction motor, apply this formula:
rpmsyn = 120 x f
Np
rpmsyn = synchronous speed (in rpm)
f = supply frequency in (cycles/sec)
Np = number of motor poles
Example: What is the synchronous speed of a four pole motor operating at 50 hz.?
rpmsyn = 120 x f
Np
rpmsyn = 120 x 50
4
rpmsyn = 6000
4
rpmsyn = 1500 rpm

8. E = Voltage / I = Amps /W = Watts / PF = Power Factor / Eff = Efficiency / HP =
Horsepower
AC/DC Formulas
To Find Direct Current AC / 1phase AC / 1phase AC 3 phase
115v or 120v 208,230, or 240v All Voltages
Amps when HP x 746 HP x 746 HP x 746 HP x 746
Horsepower is Known E x Eff E x Eff X PF E x Eff x PF 1.73 x E x Eff x PF
Amps when kW x 1000 kW x 1000 kW x 1000 kW x 1000
Kilowatts is known E E x PF E x PF 1.73 x E x PF
Amps when kVA x 1000 kVA x 1000 kVA x 1000
kVA is known E E 1.73 x E
Kilowatts IxE I x E x PF I x E x PF I x E x 1.73 PF
1000 1000 1000 1000
Kilovolt-Amps IxE IxE I x E x 1.73
1000 1000 1000
Horsepower I x E x Eff I x E x Eff x PF I x E x Eff x PF I x E x Eff x 1.73 x PF
(output) 746 746 746 746
Three Phase Values
For 208 volts x 1.732, use 360
For 230 volts x 1.732, use 398
For 240 volts x 1.732, use 416
For 440 volts x 1.732, use 762
For 460 volts x 1.732, use 797
For 480 Volts x 1.732, use 831
E = Voltage / I = Amps /W = Watts / PF = Power Factor / Eff = Efficiency / HP =
Horsepower
AC Efficiency and Power Factor Formulas
To Find Single Phase Three Phase
746 x HP 746 x HP
Efficiency
E x I x PF E x I x PF x 1.732
Input Watts Input Watts
Power Factor
VxA E x I x 1.732
Power - DC Circuits
Watts = E xI

9. Amps = W / E
Ohm's Law / Power Formulas
P = watts
I = amps
R = ohms
E = Volts
Voltage Drop Formulas
2 x K x I x L K = ohms per mil foot
VD =
Single Phase CM
(2 or 3 wire) 2K x L x I (Copper = 12.9 at 75°)
CM= VD
(Alum = 21.2 at 75°)
1.73 x K x I x L Note: K value changes with temperature. See Code chapter 9,
VD= CM Table 8
L = Length of conductor in feet
Three Phase
1.73 x K x L x I
CM= VD I = Current in conductor (amperes)
CM = Circular mil area of conductor
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Eng-Tips Forums wilg (Mechanical) 23 Dec 03
SWAG here!
13:28
The equation for Fan Horsepower is
BHP = (cfm x static friction x specific gravity)/(6356 x
motor efficiency)....How is the 6356 derived ?
KHilgefort (Mechanical) 24 Dec 03 9:33
I usually see this figure in terms of eff = (volume flowrate
in cfm) X (pressure in inches of water) / 6356 / (brake
horsepower).
R.A. Wallis in "Axial Flow Fans" similarly puts the
equation as Shaft h.p. = (5.2) X (in. water) X (cu. ft/min) /
(33,000 X efficiency).
If you convert the horsepower units to ft-lb/min (33,013
per hp) and convert the pressure to lb/ft^2 (5.202 per inch
H2O), you get 33013/5.202 = 6346. This at least gets
you to a dimensionless value for the efficiency.
lilliput1 (Mechanical) 24 Dec 03
13:35
The efficiency to get bhp is fan efficiency, not motor
efficiency. A good apporoximation for fan efficiency is
0.65
For pumps
bhp = (gpm x ft wg TDH)/(3960 x eff)

11. Here alse 0.65 is a good approximation for pump
efficiency.
On projects we HVAC engineers have to give electrical
loads to the Elect Engineer on the project. We use the
above formula + guick estimate from experience of
pressure drops to come up with motor hp. We want to be
safe at the initial stage so we always pick mhp
conservatively. Later on as plans are developed, actual
pressure drops are calculated & fans/pumps are selected
on actual fan/pump curves.
quark (Mechanical) 24 Dec 03
23:56
There are two mistakes in your equation. First one, as
suggested by lilliput, it is fan efficiency you have to use to
get bhp. Secondly you can omit specific gravity as this is
inbuilt in the equation.
1HP = 33000 ft-lbf/min = 33000 (cuft/min)x(lbf/sq.ft)
and 1 inch. wc = 5.192 lbf/sq.ft (this is where sg is
included)
therefore, 1 HP = 33000 (cu.ft/min)x inches wc/5.192
which gives, 1HP = 6356 cfm x inches wc
So fan BHP = cfmx dp across fan/(6356xeff. blower)
PS: I HP is the amount of power required to lift a weight
of 76 kgs to a height of 1 meter in 1 second. I don't know
exact definition in IP units but you can convert to get the
above said value.

12. Basic Motor Formulas And Calculations
The formulas and calculations which appear below should be used for estimating
purposes only. It is the responsibility of the customer to specify the required motor Hp,
Torque, and accelerating time for his application. The salesman may wish to check the
customers specified values with the formulas in this section, however, if there is serious
doubt concerning the customers application or if the customer requires guaranteed
motor/application performance, the Product Department Customer Service group should
be contacted.
Rules Of Thumb (Approximation)
At 1800 rpm, a motor develops a 3 lb.ft. per hp
At 1200 rpm, a motor develops a 4.5 lb.ft. per hp
At 575 volts, a 3-phase motor draws 1 amp per hp
At 460 volts, a 3-phase motor draws 1.25 amp per hp
At 230 volts a 3-phase motor draws 2.5 amp per hp
At 230 volts, a single-phase motor draws 5 amp per hp
At 115 volts, a single-phase motor draws 10 amp per hp
Mechanical Formulas
Torque x
Torque in lb.ft. HP x 5250 rpm
120 x Frequency
HP = rpm =
= rpm No. of Poles
5250
Temperature Conversion
Deg C = (Deg F - 32) x 5/9
Deg F = (Deg C x 9/5) + 32
High Inertia Loads
WK2 x rpm
t= WK2 = inertia in lb.ft.2
308 x T av. t = accelerating time in sec.
WK2 x rpm T = Av. accelerating torque
T= lb.ft..
308 x t
inertia reflected to motor = Load Inertia Load rpm

13. Motor rpm 2
Synchronous Speed, Frequency And Number Of Poles Of AC Motors
120 x f P x ns 120 x f
ns = f= P=
P 120 ns
Relation Between Horsepower, Torque, And Speed
Txn 5250 HP 5250 HP
HP = T= n=
5250 n T
Motor Slip
ns - n x
% Slip =
100
ns
Code KVA/HP Code KVA/HP Code KVA/HP Code KVA/HP
A 0-3.14 F 5.0 -5.59 L 9.0-9.99 S 16.0-17.99
B 3.15-3.54 G 5.6 -6.29 M 10.0-11.19 T 18.0-19.99
C 3.55-3.99 H 6.3 -7.09 N 11.2-12.49 U 20.0-22.39
D 4.0 -4.49 I 7.1 -7.99 P 12.5-13.99 V 22.4 & Up
E 4.5 -4.99 K 8.0 -8.99 R 14.0-15.99
Symbols
I = current in amperes
E = voltage in volts
KW = power in kilowatts
KVA = apparent power in kilo-volt-amperes
HP = output power in horsepower
n = motor speed in revolutions per minute (RPM)
synchronous speed in revolutions per minute
ns =
(RPM)
P = number of poles
f = frequency in cycles per second (CPS)
T = torque in pound-feet
EFF = efficiency as a decimal
PF = power factor as a decimal
Equivalent Inertia

14. In mechanical systems, all rotating parts do not usually operate at the same speed. Thus,
we need to determine the "equivalent inertia" of each moving part at a particular speed of
the prime mover.
The total equivalent WK2 for a system is the sum of the WK2 of each part, referenced to
prime mover speed.
The equation says:
WK2EQ = WK2part Npart
Nprime mover 2
This equation becomes a common denominator on which other calculations can be based.
For variable-speed devices, inertia should be calculated first at low speed.
Let's look at a simple system which has a prime mover (PM), a reducer and a load.
WK2 = 900 lb.ft.2
WK2 = 100 lb.ft.2 WK2 = 27,000 lb.ft.2
(as seen at output shaft)
PRIME MOVER 3:1 GEAR REDUCER LOAD
The formula states that the system WK2 equivalent is equal to the sum of WK2parts at the
prime mover's RPM, or in this case:
WK2EQ = WK2pm + WK2Red. Red. RPM + WK2Load Load RPM
PM RPM 2 PM RPM 2
Note: reducer RPM = Load RPM
WK2EQ = WK2pm + WK2Red. 1 + WK2Load 1
3 2 3 2
The WK2 equivalent is equal to the WK2 of the prime mover, plus the WK2 of the load.
This is equal to the WK2 of the prime mover, plus the WK2 of the reducer times (1/3)2,
plus the WK2 of the load times (1/3)2.
This relationship of the reducer to the driven load is expressed by the formula given
earlier:
WK2EQ = WK2part Npart
Nprime mover 2

15. In other words, when a part is rotating at a speed (N) different from the prime mover, the
WK2EQ is equal to the WK2 of the part's speed ratio squared.
In the example, the result can be obtained as follows:
The WK2 equivalent is equal to:
WK2EQ = 100 lb.ft.2 + 900 lb.ft.2 1 + 27,000 lb.ft.2 1
3 2 3 2
Finally:
WK2EQ = lb.ft.2pm + 100 lb.ft.2Red + 3,000 lb.ft2Load
WK2EQ = 3200 lb.ft.2
The total WK2 equivalent is that WK2 seen by the prime mover at its speed.
Electrical Formulas
Alternating Current
To Find
Single-Phase Three-Phase
HP x 746 HP x 746
Amperes when horsepower is known
E x Eff x pf 1.73 x E x Eff x pf
Kw x 1000 Kw x 1000
Amperes when kilowatts are known
E x pf 1.73 x E x pf
Kva x 1000 Kva x 1000
Amperes when kva are known
E 1.73 x E
I x E x pf 1.73 x I x E x pf
Kilowatts
1000 1000
IxE 1.73 x I x E
Kva
1000 1000
I x E x Eff x pf 1.73 x I x E x Eff x pff
Horsepower = (Output)
746 746

16. I = Amperes; E = Volts; Eff = Efficiency; pf = Power Factor; Kva = Kilovolt-amperes;
Kw = Kilowatts
Locked Rotor Current (IL) From Nameplate Data
577 x HP x KVA/HP
Three Phase: IL =
E
See: KVA/HP Chart
1000 x HP x KVA/HP
Single Phase: IL =
E
EXAMPLE:Motor nameplate indicates 10 HP, 3 Phase, 460 Volts, Code F.
577 x 10 x (5.6 or
IL = 6.29)
460
70.25 or 78.9 Amperes (possible
IL =
range)
Effect Of Line Voltage On Locked Rotor Current (I L) (Approx.)
ELINE
IL @ ELINE = IL @ EN/P x
EN/P
EXAMPLE:Motor has a locked rotor current (inrush of 100 Amperes (IL) at the rated
nameplate voltage (EN/P) of 230 volts.
What is IL with 245 volts (ELINE) applied to this motor?
IL @ 245 V. = 100 x 254V/230V
IL @ 245V. = 107 Amperes
Basic Horsepower Calculations
Horsepower is work done per unit of time. One HP equals 33,000 ft-lb of work per
minute. When work is done by a source of torque (T) to produce (M) rotations about an
axis, the work done is:
radius x 2 x rpm x lb. or 2
TM
When rotation is at the rate N rpm, the HP delivered is:

17. radius x 2 x rpm x
TN
HP = lb. =
5,250
33,000
For vertical or hoisting motion:
HP WxS
= 33,000 x E
Where:
W = total weight in lbs. to be raised by motor
S = hoisting speed in feet per minute
overall mechanical efficiency of hoist and gearing. For purposes of
E =
estimating
E = .65 for eff. of hoist and connected gear.
For fans and blowers:
Volume (cfm) x Head (inches of
HP = water)
6356 x Mechanical Efficiency of Fan
Or
Volume (cfm) x Pressure (lb. Per sq. ft.)
HP =
3300 x Mechanical Efficiency of Fan
Or
Volume (cfm) x Pressure (lb. Per sq.
HP = in.)
229 x Mechanical Efficiency of Fan
For purpose of estimating, the eff. of a fan or blower may be assumed to be 0.65.
Note:Air Capacity (cfm) varies directly with fan speed. Developed Pressure varies with square of fan
speed. Hp varies with cube of fan speed.
For pumps:

18. GPM x Pressure in lb. Per sq. in. x Specific
HP = Grav.
1713 x Mechanical Efficiency of Pump
Or
GPM x Total Dynamic Head in Feet x
HP = S.G.
3960 x Mechanical Efficiency of Pump
where Total Dynamic Head = Static Head + Friction
Head
For estimating, pump efficiency may be assumed at 0.70.
Accelerating Torque
The equivalent inertia of an adjustable speed drive indicates the energy required to keep
the system running. However, starting or accelerating the system requires extra energy.
The torque required to accelerate a body is equal to the WK2 of the body, times the
change in RPM, divided by 308 times the interval (in seconds) in which this acceleration
takes place:
WK2N (in lb.ft.)
ACCELERATING TORQUE =
308t
Where:
N = Change in RPM
W = Weight in Lbs.
K = Radius of gyration
Time of acceleration
t=
(secs.)
WK2 = Equivalent Inertia
308 = Constant of proportionality
Or
TAcc = WK2N

19. 308t
The constant (308) is derived by transferring linear motion to angular motion, and
considering acceleration due to gravity. If, for example, we have simply a prime mover
and a load with no speed adjustment:
Example 1
PRIME LOADER
LOAD
WK2 = 200 lb.ft.2 WK2 = 800 lb.ft.2
The WK2EQ is determined as before:
WK2EQ = WK2pm + WK2Load
WK2EQ = 200 + 800
WK2EQ = 1000 ft.lb.2
If we want to accelerate this load to 1800 RPM in 1 minute, enough information is
available to find the amount of torque necessary to accelerate the load.
The formula states:
1000 x
WK2EQN 1800000
TAcc = or 1800 or
308t 18480
308 x 60
TAcc = 97.4 lb.ft.
In other words, 97.4 lb.ft. of torque must be applied to get this load turning at 1800 RPM,
in 60 seconds.
Note that TAcc is an average value of accelerating torque during the speed change under
consideration. If a more accurate calculation is desired, the following example may be
helpful.
Example 2
The time that it takes to accelerate an induction motor from one speed to another may be
found from the following equation:

20. WR2 x change in rpm
t=
308 x T
Where:
T = Average value of accelerating torque during the speed change under consideration.
t = Time the motor takes to accelerate from the initial speed to the final speed.
WR2 = Flywheel effect, or moment of inertia, for the driven machinery plus the motor rotor in lb.ft.2
(WR2 of driven machinery must be referred to the motor shaft).
The Application of the above formula will now be considered by means of an example.
Figure A shows the speed-torque curves of a squirrel-cage induction motor and a blower
which it drives. At any speed of the blower, the difference between the torque which the
motor can deliver at its shaft and the torque required by the blower is the torque available
for acceleration. Reference to Figure A shows that the accelerating torque may vary
greatly with speed. When the speed-torque curves for the motor and blower intersect
there is no torque available for acceleration. The motor then drives the blower at constant
speed and just delivers the torque required by the load.
In order to find the total time required to accelerate the motor and blower, the area
between the motor speed-torque curve and the blower speed-torque curve is divided into
strips, the ends of which approximate straight lines. Each strip corresponds to a speed
increment which takes place within a definite time interval. The solid horizontal lines in
Figure A represent the boundaries of strips; the lengths of the broken lines the average
accelerating torques for the selected speed intervals. In order to calculate the total
acceleration time for the motor and the direct-coupled blower it is necessary to find the
time required to accelerate the motor from the beginning of one speed interval to the
beginning of the next interval and add up the incremental times for all intervals to arrive
at the total acceleration time. If the WR2 of the motor whose speed-torque curve is given
in Figure A is 3.26 ft.lb.2 and the WR2 of the blower referred to the motor shaft is 15
ft.lb.2, the total WR2 is:
15 + 3.26 = 18.26 ft.lb.2,
And the total time of acceleration is:
WR2 rpm1 rpm2 rpm3 +--------- rpm9
+ +
308 T1 T2 T3 + T9
Or
t = 18.2 150 + 150 + 300 + 300 + 200 + 200 + 300 + 100 + 40
6

21. 46 48 47 43.8 39.8 36.4 32.8 29.6 11
308
t = 2.75
sec.
Figure A
Curves used to determine time required to accelerate induction motor and blower
Accelerating Torques
T1 = 46 lb.ft. T4 = 43.8 lb.ft. T7 = 32.8 lb.ft.
T2 = 48 lb.ft. T5 = 39.8 lb.ft. T8 = 29.6 lb.ft.
T3 = 47 lb.ft. T6 = 36.4 lb.ft. T9 = 11 lb.ft.
Duty Cycles
Sales Orders are often entered with a note under special features such as:
"Suitable for 10 starts per hour"
Or
"Suitable for 3 reverses per minute"
Or
"Motor to be capable of accelerating 350 lb.ft.2"
Or
"Suitable for 5 starts and stops per hour"
Orders with notes such as these can not be processed for two reasons.

22. 1. The appropriate product group must first be consulted to see if a design is available that will
perform the required duty cycle and, if not, to determine if the type of design required falls within
our present product line.
2. None of the above notes contain enough information to make the necessary duty cycle calculation.
In order for a duty cycle to be checked out, the duty cycle information must include the following:
a. Inertia reflected to the motor shaft.
b. Torque load on the motor during all portions of the duty cycle including starts, running
time, stops or reversals.
c. Accurate timing of each portion of the cycle.
d. Information on how each step of the cycle is accomplished. For example, a stop can be
by coasting, mechanical braking, DC dynamic braking or plugging. A reversal can be
accomplished by plugging, or the motor may be stopped by some means then re-started in
the opposite direction.
e. When the motor is multi-speed, the cycle for each speed must be completely defined,
including the method of changing from one speed to another.
f. Any special mechanical problems, features or limitations.
Obtaining this information and checking with the product group before the order is
entered can save much time, expense and correspondence.
Duty cycle refers to the detailed description of a work cycle that repeats in a specific time
period. This cycle may include frequent starts, plugging stops, reversals or stalls. These
characteristics are usually involved in batch-type processes and may include tumbling
barrels, certain cranes, shovels and draglines, dampers, gate- or plow-positioning drives,
drawbridges, freight and personnel elevators, press-type extractors, some feeders,presses
of certain types, hoists, indexers, boring machines,cinder block machines, keyseating,
kneading, car-pulling, shakers (foundry or car), swaging and washing machines, and
certain freight and passenger vehicles. The list is not all-inclusive. The drives for these
loads must be capable of absorbing the heat generated during the duty cycles. Adequate
thermal capacity would be required in slip couplings, clutches or motors to accelerate or
plug-stop these drives or to withstand stalls. It is the product of the slip speed and the
torque absorbed by the load per unit of time which generates heat in these drive
components. All the events which occur during the duty cycle generate heat which the
drive components must dissipate.
Because of the complexity of the Duty Cycle Calculations and the extensive engineering
data per specific motor design and rating required for the calculations, it is necessary for
the sales engineer to refer to the Product Department for motor sizing with a duty cycle
application.
Last Updated September
Chapter 10: Fans and Drives
10.1. Theory and Applications

23. 10.2. Commissioning Fans and Drives
10.2.1. Functional Testing Field Tips
Key Commissioning Test Requirements
Key Preparations and Cautions
Time Required to Test
10.2.2. Design Issues Overview
10.3. Typical Problems
10.3.1. Static pressure requirements in excess of design
10.3.2. Improper belt drive system adjustment
10.4. Testing Guidance and Sample Test Forms
10.5. Supplemental Information
10.1. Theory and Applications
The fan is the heart of the air handling system since it is one of the most significant energy users
in a building. Commissioning and re-commissioning fans and drives is a key factor for ensuring
that a building�s efficiency goals are met over the life of the building.
There are both indirect and direct components to a fan�s energy consumption. The indirect
component relates to the system the fan serves. The fan must impart enough energy to the air
stream to overcome the system�s resistance to flow. This energy consumption can be
significantly altered by:
� Fan installation considerations like system effect
� Duct and fitting design and their related pressure drops
� Component pressure drops
� Duct system leakage
� Duct system thermal loss
These topics are discussed in Chapter 11: Distribution, and Chapter 13: Return, Relief and
Exhaust System.
The direct fan energy component relates to how efficiently the fan can covert the energy going
into its prime mover (usually electricity into a motor) into air flow and pressure in the fan system.
This energy consumption is a function of the following items:
� Fan efficiency
� Motor efficiency
� Drive system efficiency and adjustment
The fan horsepower equation (Equation 10.1) is a function of several fundamental components:
flow rate, static pressure, fan efficiency, and motor efficiency. Application of this equation to fan
system analysis is discussed in detail in Appendix C: Calculations. Commissioning efforts should
be targeted at these factors to ensure system efficiency, performance, and reliability.
Equation 10.1 Fan Horsepower

24. The fundamental technology associated with the fans, coils, casings and other major system
components is well established. Most of the advances in technology that improve the
performance of these components are related to the drive systems and control systems, not with
the components themselves. Drive and control systems can be readily upgraded as technological
improvements warrant. This easy upgrade is in direct contrast with a more complex machine like
a chiller where technology changes are continually improving performance, and where this
technology is generally an integral part of the machine�s package.
If one examines a reasonably well-maintained 50-year-old airfoil centrifugal fan and a similar
unit right off of the production line, one will probably see only modest performance differences.
Over time, the shaft and/or bearings in the older fan may have required replacement, and the
wheel periodic cleaning. However, it is likely that the fan is capable of moving air as efficiently
as the newer fan. Through attention to proper maintenance and equipment location (indoors rather
than outdoors), fans can be a lasting component of the air handling system.
10.2. Commissioning Fans and Drives
The following sections present benefits, practical tips, and design issues associated with
commissioning an air handler�s fans and drives.
10.2.1. Functional Testing Field Tips
Key Commissioning Test Requirements lists practical considerations for functional testing. Key
Preparations and Cautions address potential problems that may occur during functional testing
and ways to prevent them.

25. Key Commissioning Test Requirements
Fan energy constitutes a significant portion of a building's overall energy consumption. Even
minor improvements in efficiency can have a major impact on the energy consumption pattern
associated with a building. A well executed commissioning plan for the fans and their
associated drive systems ensures that the systems are set up for peak efficiency and that this
efficiency will persist. The fan and drive control should reliably integrate with the overall
system control strategy in a manner that provides the intended function and level of
performance.
1 Verify the fan size and capacity. Capacity tests results should be evaluated in the light of the
accuracy of in instrumentation and the actual conditions at the time of the test.
2 Backdraft dampers need to be tested for proper operation. Non-motorized dampers must open
and close freely without binding. Motorized dampers must be connected to the DDC control
system and verified that they are commanded open prior to fan operation.
3 Verify that network failures do not result in unsafe operating modes. The recovery from the
failure should safely return the drive to the network.
4 Verify that drive settings and adjustments provide for safe and reliable system operation at
peak efficiency levels in all operating modes.
Key Preparations and Cautions
Cautions
1 Applicable cautions as outlined in Functional Testing Basics should be observed.
2 Safety and interlock testing, verification of some of the drive settings, and loop tuning efforts
will place the system at risk. Appropriate precautions and procedures should be in place to
protect the personnel and machinery involved in the test process, including plans for
quickly aborting the test.
3 Any belt drives have been adjusted and aligned.
5 All safeties, interlocks, and alarms are programmed (or hard-wired, if applicable) and
function correctly, regardless of VFD operating position (i.e. hand, auto, by-pass).
6 If necessary, the motor shaft is grounded.
7 Distribution system pressure drops do not exceed design expectations. This is typically
performed while conducting construction observation. If changes increase distribution
system pressure drop, ensure all equipment still receives design flow rate.
8 Verify all VFD operating parameters are correct for the application, including acceleration
and deceleration times and minimum speed setting.
Test Conditions
1 Tests that are targeted at verifying design parameters and settings for the fan and its
enclosure can generally be performed after the assembly of the air handling unit but prior to
its start-up.
2 Other tests targeted at the interlocks and fundamental control functions, loop testing and
tuning, and capacity testing will require that the air handling system be operational and
moving the design volume of air, but not necessarily fully under control. Safety systems
should be operational to protect the machinery and occupants in the event of a problem

26. 10.2.2. Design Issues Overview
The Design Issues Overview presents issues that can be addressed during the design phase to
improve system performance, safety, and energy efficiency. These design issues are essential for
commissioning providers to understand, even if design phase commissioning is not a part of their
scope, since these issues are often the root cause of problems identified during testing.
Does the unit have good access for control installation, maintenance, and
component replacement?
Access to the fan and its related components is critical for ensuring the persistence of energy
efficiency and other commissioning related benefits.
1 Piping should be arranged to ensure that access panels are not blocked, service routes remain
open, and components such as coils and fan shafts can be removed and replaced without
shutting down adjacent systems or central plant equipment.
2 Fan scrolls should be provided with access doors to allow the wheel to be inspected and
cleaned.
3 Coils should be provided with space between then and access to that space to facility
inspection and cleaning and allow for the installation of control elements in their proper
location. For example, space is required between a preheat coil and the next coil downstream
to allow the freezestat to be installed downstream of the preheat coil (which by design will
see subfreezing entering air temperatures and should be capable of handling them safely).
Have variable speed drive installation and operation requirements been taken into
account?
1 Most VFD manufactures have some specific requirements regarding the length, routing, and
general configuration of the power circuit from the drive to the motor. Failure to pay attention
to the requirements can cause operational problems in the electrical system and in severe
cases, cause failures in switchgear, drives, and transfer switches.
2 Many VFDs can be damaged if they start against a reverse spinning motor. This condition is
likely to occur in parallel fan systems, even if they are equipped with backdraft dampers. No
damper is 100% leak proof, and it does not take much reverse flow to set a fan wheel in
motion. Most drives also have a feature to handle reverse flow, usually called DC injection
braking. The process pulses the motor with a DC signal before starting and accelerating it.
The DC signal brakes the rotating armature. Usually there are adjustments that need to be
made to tailor this feature to the load served in addition to activating it. Verifying this feature
is properly set and functioning should be part of the commissioning process both during the
pre-start checks as well as the functional tests.
3 Many drives are supplied with bypass contactors that allow the motor to run at full speed if
the drive fails. In some cases, the system could be damaged by full speed fan operation when
the loads were configured for minimum flow conditions.
4 The drive should be configured and wired to ensure that all safety interlocks are effective in
all possible selector switch configurations (local, auto, hand, inverter, bypass, etc.). Some
drives are arranged to allow the safety interlocks to be effective when the drive is operating
but not effective if the drive is bypass. Some drives can also be configured so that if they are
placed in the local mode, any external interlock (external to the drive circuit board) will be
ignored. This feature may be desirable in process applications, but it is highly undesirable in
most HVAC applications. Verifying that the drive is properly set and functioning should be

27. part of the commissioning process both during the pre-start checks as well as the functional
tests.
Are the VFDs and the motors compatible?
Motors that are not rated for VFD applications may have a reduced life if used with a VFD. In
retrofits, it is desirable to evaluate the motor�s capabilities relative to the drive. Even it budget
constraints prevent a motor replacement when the drive is installed, the potential for a future
problem and early failure can be anticipated. In new installations, the drives and motors should be
coordinated to be compatible with each other.
Does the VFD shaft need to be grounded?
The variable voltages, magnetic fields and harmonics associated with VFD operation can induce
currents in the motor shaft that have no path to ground other than through the bearings for most
conventional motors. Evidence suggests that these eddy currents can lead to premature bearing
failures, perhaps in a matter of years on some motors. Shaft grounding kits installed on the motor
provide a direct path from the shaft to ground via a brush system.
Is the drive arrangement suitable for the application?
Given the wide array of drive options available, it is important to tailor the selection to the
application.
1 If direct drives are applied, then fan speed adjustment for balancing purposes will have to rely
on less efficient approaches like discharge dampers, or will require that a variable speed drive
be included as a part of the package.
2 A variable speed drive on a constant volume fan may represent false economy. While it does
minimize balancing efforts and eliminate the need for a final sheave change or adjustment to
set the fan speed, the drive results in a loss in fan system efficiency that increases in
magnitude as the speed is reduced (see Chapter 10: Fans and Drives Supplemental
Information). The drive also introduces operating complexity, first cost, potential electrical
system harmonic problems, and multiple failure mode possibilities into the system. These
issues coupled with the efficiency reduction will probably outweigh any modest savings in
balancing costs achieved.
3 Variable pitch sheaves provide flexibility and a good intermediate stepping stone between
start-up and final balanced speed as a system is brought on line. But some of their
disadvantages may make the installation of a fixed pitch sheave as the final step in the
balancing effort a desirable feature to include in the project.
4 During design review, verify that the fan and drive capacity is properly sized so that the VFD
will operate near 100% speed at full load (do not use the VFD as a throttling device).
Could the fan motor run in the wrong direction?
For most axial fans, if the impeller were to run in the opposite direction, it would move air in the
opposite direction. With centrifugal fans, running the impeller backwards will still provide flow
in the correct direction, but the performance will be degraded significantly.
Reverse flow or back-draft through most fan wheels will cause them to spin in the reverse
direction. Forward curved fan wheels will spin in the wrong direction if air is blown through them
in the right direction but they are not energized. For most single phase motors, if the motor is
spinning in the wrong direction when power is applied, the fan will simply run in the wrong
direction. The rotational direction of most three phase motors used for HVAC applications is tied

28. to the phase rotation established by the way the windings are connected to the distribution
system. Thus, if the motor is spinning backwards when voltage is applied, it will reverse and run
in the proper direction. Problems can occur with variable speed drives when they attempt to start
against a reverse rotating motor.
Systems with operating conditions that could cause backflow should be designed and installed to
safely and reliably deal with any problems. Both normal and failure modes need to be considered.
Common examples of situations where backflow potential exists include:
1 Systems with parallel fans or air handling units. Don�t forget that parallel fan terminal units
have fans that are essentially in parallel with the supply fan.
2 Systems with series fans: the supply and exhaust fans associated with a 100% outdoor air
systems and the fans in series powered fan terminal boxes relative to the supply fan.
Does the air handler specification include desirable options?
Most fans and air handling units are available with an array of options, some of which are
desirable in most installations and others of which are only required for special installations.
Examples include:
1 Access doors in casings and fan scrolls.
2 Lubrication lines extended to be accessible from the exterior of the unit.
3 Baseline vibration characteristics measured at the factory.
4 Premium efficiency motors.
5 Special vibration isolation provisions.
6 Scroll drains (essential for exhaust fans located outdoors and discharging in the up-blast
configuration)
7 Factory installed back draft dampers.
8 Non- sparking or explosion proof construction for hazardous locations.
9 Special coatings for handling abrasive or corrosive fluid streams.
10.3. Typical Problems
The following problems are frequently encountered with fans and drives.
10.3.1. Static pressure requirements in excess of design
A typical problem found during commissioning or retro-commissioning is high static pressure in
the fan system. In creating excess static pressure that is not required to operate the system, a fan
wastes significant amounts of energy. This problem arises because fan selections often fall into a
range where there is a difference between the design brake-horsepower (bhp) requirement and the
actual motor horsepower installed due to the standard horsepower ratings available in motor
product lines. The difference between available sizes can become quite significant for larger fans.
For example, a fan with an 82 bhp motor requirement would probably come with a 100 hp motor.
If the fan was unable to deliver design flow against the installed system static requirement, then
there would be a lot of margin for speeding the fan up to achieve the design requirement without
overloading the motor (assuming the operating point did not end up in a different fan class
requirement). This safety net may be desirable, as the excess motor capacity allows problems to
be solved in the field. But, the added energy consumed by the fan beyond that intended by the
design will become an energy burden that will persist for the life of the system.

29. Extra diligence during design and construction can prevent conditions that add unanticipated
static pressure to the system, thus averting the need to run the fan at an operating point in excess
of design. If the balancing team discovers that they have excess system static pressure, there are
ways to lower static pressure that will allow the system to function at or near its intended design
point rather than adding on ongoing energy burden to the project by simply throwing energy at
the problem. An example of such a situation is contained in Example 2 in Appendix C :
Calculations.
10.3.2. Improper belt drive system adjustment
While simple in concept, there are some critical parameters associated with the installation and
adjustment of this belt drive systems that are often ignored, resulting in belt failures, poor
performance, noise, reduced equipment life and energy waste.
Alignment of the drive and motor sheaves is a critical step in the belt installation process.
Without proper alignment, belts will run less efficiently, wear out more quickly, and, in extreme
cases, be thrown off the drive sheaves.
Over-tensioning the belts can cause problems with bearings and shafts due to the excessive loads
imposed. In addition, new belts will stretch during the first 8 to 24 hours of operation; belts that
have been properly set initially will require re-tensioning after they have run. This contingency is
often overlooked to the detriment of the drive system efficiency.
Multiple belt drives will function best if factory matched belt sets are installed. This ensures that
the drive loads are equally distributed between all of the belts, equalizing wear and life.
The T.B. Woods Company offers a very good guide to proper belt drive selection and adjustment
on their web site.
xx
10.4. Testing Guidance and Sample Test Forms
Click the button below to access all publicly-available prefunctional checklists, functional test
procedures, and test guidance documents referenced in the Testing Guidance and Sample Test
Forms table of the Air Handler system module.
AHU Testing Guidance and Sample Test Forms
xx
10.5. Supplemental Information
Supplemental information for fans and drives has been developed to provide necessary
background information for functional testing.

30. Chapter 10: Fans and Drives
Supplemental Information
10.1. Fans
10.2. Capacity control strategies
10.3. Drive systems and arrangements
Figures
Figure 10.1: Typical VFD Efficiency vs. Speed
Figure 10.2: Motor and Drive Arrangement Block Access
10.1. Fans
Although fans come in a wide variety of designs, shapes, sizes, and configurations. They
generally, they fall into two categories:
� Centrifugal fans This type of fan imparts kinetic energy to the air primarily by
centrifugal force. In essence, the air is drawn into the center of the fan wheel where it is
captured and contained by blades. These parcels of air are then �flung� to the periphery of
the wheel.[1] The wheel itself can have an inlet on one side (Single Width, Single Inlet or
SWSI) or an inlet on both sides (Double Width, Double Inlet or DWDI). The design of the
blades on the wheel can have a significant impact on efficiency, performance and cost.
Common designs are forward curved, backward included, airfoil, and radial. [2]
� Axial fans This type of fan uses aerodynamic effects to impart velocity to the air as it
passes through the impeller. Generally, the air travels along the axis of the fan and impeller as
compared the centrifugal design where the air enters the impeller by flowing parallel to the
shaft, but exits the impeller radially relative to the shaft. Generally, the impeller for this type
of fan will be resemblance to an airplane propeller, but with many more blades.
10.2. Capacity control strategies
Regardless of the design, the rotating nature of the fan wheel can create significant structural
loads on the shaft, wheel, bearings, and housing. Issues related to these factors are accounted for
in the fan class rating. A fan with a wheel that is rated Class II has a higher speed and pressure
capability than the same fan with a wheel that is rated Class I. Therefore, some caution must be
used when changing fan speeds in the field to be sure that the new operating point is still within
the fan�s class rating.
There are a variety of techniques used to control fan capacity. The most common include:
� Discharge dampers Dampers located on the outlet of the fan can simply throttle the fan.
Basically, the discharge damper increases/makes worse, the system effect associated with the
fan outlet, thereby degrading its performance. Generally, this is probably the least expensive
but also the least desirable approach due to the efficiency implications of a damper on the fan
discharge. It can also be quite noisy.

31. � Inlet vanes Inlet vanes modify the performance of the fan by �pre-swirling� the air as
it enters the eye of the fan. This has the effect of changing the shape of the fan performance
curve as can be seen in Figure 16 of Chapter 13 of the 2000 ASHRAE Systems and
Equipment Handbook. This approach is much more desirable than a discharge damper, but
not as desirable as a variable speed approach. The emergence of affordable and reliable
variable speed drive technology has displaced this approach, but when VAV systems first
emerged, it was a common means of achieving the required capacity control and is still found
on many existing systems or on systems where the variable speed drives have been
eliminated by a value engineering effort.
� Blade pitch Varying blade pitch is a efficient but mechanically complex approach to
controlling capacity on axial fans. The effect is very similar to a speed change as can be seen
from Figure 17 in Chapter 13 of the 2000 ASHRAE Systems and Equipment Handbook. Most
fans that use this approach require additional maintenance in the form of periodic lubrication,
inspection, and over-haul of the mechanical vane pitch control system.
� Variable speed Currently, this is probably the most common approach to controlling fan
capacity due to its efficiency, mechanical simplicity, and steadily improving first cost.
Commonly referred to as VSD technology (for Variable Speed Drive), it is not necessarily
mean VFD technology, which is a subset. Before modern electronic technology made semi-
conductor based Variable Frequency Drives a practical and affordable reality, there were a
variety of more exotic approaches used including:
� Variable speed DC motors These were complex and costly and were usually found
only on industrial or very large commercial applications.
� Hydro-mechanical clutches This technology employed hydraulics and a clutch
system to vary the speed of the output shaft relative to the input shaft. They too were not
common on commercial HVAC systems and tended to have relatively high mechanical
losses.
� Variable pulley systems Often termed �pulley pincher� drives[3], these systems
did find somewhat wide application on commercial HVAC systems. The devices
functioned by moving the sides of an adjustable drive pulley towards or away from each
other. This changed the effective pitch diameter of the pulley, and thus, the output speed.
While capable of modulating speeds, the devices tended to be hard on belts and had
relatively high mechanical losses.
� Solid-state variable frequency drives Typically called VFDs or invertors[4],
current technology drives of this type provide a nearly ideal solution to the fan capacity
control problem. In and of themselves, they tend to be more efficiency than some of the
other approaches (see Figure 10.1 for a typical efficiency plat), but they also tend to
maintain the fan efficiency at or near the selected efficiency as they vary its capacity by
changing speed. However, this is not without its complications, but paying careful
attention to design and commissioning issues can readily overcome any problems and the
advantages typically outweigh the disadvantages.
Figure 10.1: Typical VFD Efficiency vs. Speed

32. While efficiency does decay with load, these drives will generally deliver better efficiency and
less decay than some of the other alternatives like variable pulley systems.
Regardless of the technique used, capacity control systems will subject the fan and its
components to a wide array of continuously varying performance conditions. The interaction of
the multiple operating points with the fan components, system components, and building can lead
to a number of surprising and unanticipated problems, especially for larger fans with a lot of
power. Examples include:
� On one late 1990�s project the resonance between the large air handling unit fans and the
building resulted in vibration in the building�s structural system under certain operating
conditions.
� In a semiconductor facility, resonance between process exhaust fans and sensitive
machinery in the process clean room caused quality control problems.
� Over the years, there have been multiple occurrences of fan failure related to resonate
frequency problems including axial fans shedding blades and centrifugal fan wheels
disintegrating.
These problems can be difficult to predict and often show up as commissioning issues. Often, the
most viable approach to solving them is to make sure the design incorporates features that will
allow you to solve the problem if it occurs. For example, avoiding operation at the triggering
condition can solve most of these types of problems. And, most current technology variable speed
drives will allow you to program in multiple frequency ranges that the drive will �jump over�

33. as it is commanded through its speed range. Thus, ensuring that the drives that will be supplied
for your project include this feature can give the start-up and commissioning team the tools they
need to solve this type of problem when if it crops up. Another desirable feature to include in the
project is vibration analysis and documentation under a variety of operating modes for large fans,
especially if they will be operating at variable capacities and speeds. It is also possible to do tests
on the building structure to determine it�s resonate frequency and then use that information for
setting up the drive systems. The project structural engineer may also be able to predict the range
of resonate frequencies anticipated for the structure and this information can be reviewed by the
rest of the team in light of the anticipated operating parameters for the system to allow potential
problems to be identified and addressed during design.
10.3. Drive systems and arrangements
In all but direct drive applications, some sort of sheave or pulley and belt system will typically be
associated with a fan and its motor. It is not uncommon for one of these pulleys to be supplied in
an adjustable configuration to allow the speed of the fan to be easily adjusted by the balancing
contractor in the field. While desirable from this standpoint, there are several draw backs to
adjustable pulleys or sheaves:
� Belt service life Most V-belts will provide the best service life if they run with their
outside perimeter (the flat part at the open end of the V cross section) slightly above the edge
of the sheaves they are installed on. If an adjustable pitch sheave requires significant
adjustment, it is not uncommon for the outside perimeter of the belt to run below the top of
the sheave sides. This results in extra wear on the belt and can reduce service live
significantly.
� Loss of setting Despite its advantages, adjustability can also be the downfall of
adjustable pitch sheaves. It is not uncommon for the balanced setting of the sheave to be lost
inadvertently when the belts are replaced, especially if the mechanic performing the work has
not been trained regarding adjustable sheaves and mistakenly thinks that the adjustability
feature is a convenient way to tension the belt(s) or make the set of belts that they happen to
have with them fit. As a result, the once balanced system ends up out of balance and
performance suffers. If the new setting delivers less air than was intended, then capacity
problems may show up at a later date when design loads show up on the system. If the
settings deliver more air than was intended, energy can be wasted, especially if the system is
one of the constant volume reheat systems frequently found in hospital or process
environments. Both problems can lead to pressure relationship problems if the misadjusted
fan happens to be an exhaust fan. If the exhaust is hazardous, a loss of airflow can create a
dangerous condition in the area served by the fan that may not be immediately detected.
Fans and there prime movers come in a variety of mounting arrangements. AMCA Standard 99-
86 illustrates these along with other standards related to fans and air handling units including
dimensioning, motor positions, etc. This information can also be found in most manufacturers fan
catalogs. Usually, one or more of the following considerations will dictate the specific
arrangement:
� The needs of the HVAC process, prime mover and drive system Some HVAC
applications may be sensitive to potential by-products from the drive system and there-for,
may which to place the entire drive assembly outside of the conditioned air steam. Similarly,
certain HVAC process may be a hostile environment for belts or motors and installing them
outside of the air stream will improve their serviceability and service life. This can be
particularly important for exhaust systems handling hazardous and/or explosive or flammable
materials where a motor in the air stream could be a source of ignition.

34. � The arrangement of the fan By their nature, the arrangement of some fans precludes
some of the drive arrangements. In addition, physical constraints of the fan installed location
may place limitations on the type of drive arrangement that might be used.
Figure 10.3: Motor and Drive Arrangement Block Access
On this new construction project, access to the inlet side of this SWSI fan, which was difficult to begin with,
will be totally blocked by the belts and belt guard between the motor and shaft (red circles). It was too late
to solve the problem on this project but a different arrangement may have prevented it. On this project, the
maintenance staff will need to remove the belt and drives to inspect the fan wheel.
� Service requirements Some arrangements may make service of the motor or fan wheel
impossible in the installed location or may block access to some other component in the fan
room (see Figure 10.3)
� Balancing A belt and pulley system provides a convenient way to adjust fan speed for
balancing purposes. Direct drive fans do not have this option and require other methods to
adjust for final balance such as adjustable blade pitch or a variable speed drive. Adjustable
blades to not have to be automated but are labor intensive to set as compared to a sheave
change. Variable speed drives are attractive from an ease of use standpoint, but add
unnecessary cost, complexity, and failure modes to a constant volume system.
� Heat gains Because they are doing work on the air stream and air is compressible, all fans
will show a temperature rise across them, even if the motor is not in the air stream. This
temperature rise is called fan heat and can be calculated by converting the fan brake
horsepower into btu�s per hour and then solving the following equation for the fan
temperature rise:
If the motor is located in the air stream, then the motor efficiency losses will also show up as
a part of the temperature rise.[5] For large fans with large motors, this can be a significant load
on the system that could be avoided by locating the motor outside of the air stream. These

35. advantages have to be weighed against the complications this can introduce for some
arrangements in terms of sealing the drive shaft where it penetrates the casing and vibration
isolation.
� Vibration and sound isolation The method by which vibration and sound isolation will
be accomplished can also affect arrangement selection. Mounting the entire fan and drive on
an isolation mount will allow the assembly to be further soundproofed by locating it inside an
acoustically treated fan casing at the cost of placing the motor in the air stream. By their
nature, direct drive fan usually have their vibration isolation problems addressed by the motor
mounting arrangement. A hidden but sometimes significant aspect of the vibration isolation
technique relates to how the equipment will be seismically restrained (see Functional Testing
Basics: Supplemental Information for details).
It is becoming increasingly common for manufactures to provide two parallel fans in packaged
equipment. Usually space constraints, redundancy requirements, or both drive this design. When
employed, there are several issues that need to be considered.
� Backdraft Even if the intent of the design is to always run two fans, it is quite likely that
at some point in time a failure in the power system, drive system or fan itself will result in
one fan needing to operate while the other sits idle. Backdraft dampers are commonly
employed to prevent air from the active fan from re-circulating into the inactive fan.
However, if not carefully applied, there can be some operational difficulties that will show up
during the commissioning process.
� Surge When two identical fans are operated in parallel, there is a potential for surge to
occur between the two fans.[6] This is because it is very difficult to create two fans that are
exactly identical and then get them operating at exactly the same point on their performance
curve. Since the fans are coupled to the same system and but that system places them at
slightly different points on their operating curves, pressure fluctuations can occur as the fans
shift around and interact, trying to find a mutually agreeable operating point. The effects from
this can range from unnoticeable to noise to (in rare cases) fan damage.
Chapter 18 of the 2000 ASHRAE Systems and Equipment Handbook, AMCA publications 99-86,
200, 201, 202-88, and 203, and the Trane Fan Engineering Handbook are all excellent resources
for additional detailed information regarding the topics outlined above.
[1]
It�s the same effect you experienced as a child on the merry-go-round at the playground.
[2]
While less common than the other designs, radial blade fans are sometimes found in exhaust systems, especially
exhaust systems that handle materials like dust or other particulate matter or where high pressures are required.
[3]
For those who are wood workers, this is basically the same approach as is used for varying speed on a Shopsmith�
multipurpose tool.
[4]
This is a reference to the electronic process going on in most drives; basically the drives take alternating current,
rectify it to direct current perform their �magic�, and then invert the direct current to create an alternating current
out put with the desired frequency and other electrical characteristics necessary to control the motor.
[5]
This temperature rise can be calculated in the same manner as the fan heat but the motor horsepower (vs. fan brake
horsepower) at the current operating condition is used.
[6]
This should not be confused with the surge that can occur in a single fan if it is operated at a point on its curve
where the pressure difference across it fights with the fans ability to generate that pressure difference causing
sporadic flow reversals through the impeller.