FANS & BLOWERS

Introduction
Fans and blowers provide air for ventilation and industrial process requirements. Fans
generate a pressure to move air (or gases) against a resistance caused by ducts, dampers, or other
components in a fan system. The fan rotor receives energy from a rotating shaft and transmits it to
the air.

The Difference between Fans, Blowers, and Compressors

Fans, blowers and compressors are differentiated by the method used to move the air, and
by the system pressure they must operate against. As per American Society of MechanicalEngineers
(ASME) the specific ratio - the ratio of the discharge pressure over the suction pressure– is used for
defining the fans, blowers and compressors (see Table 5.1)

Fan Types
Fan and blower selections depend on the volume flow rate, pressure, type of material
handled, space limitations, and efficiency. Fan efficiencies differ from design to design and also by
types. Typical ranges of fan efficiencies are given in Table 5.2.
Fans fall into two general categories: centrifugal flow and axial flow. In centrifugal flow, airflow
changes direction twice - once when entering and second when leaving (forward curved, backward
curved or inclined, radial) (see Figure 5.1).
In axial flow, air enters and leaves the fan with no change in direction (propeller, tubeaxial,
vaneaxial) (see Figure 5.2).

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Centrifugal Fan: Types
The major types of centrifugal fan are: radial, forward curved and backward curved (see
Figure 5.3).
Radial fans are industrial workhorses because of their high static pressures (upto 1400 mm
WC) and ability to handle heavily contaminated airstreams. Because of their simple design, radial
fans are well suited for high temperatures and medium blade tip speeds.
Forward-curved fans are used in clean environments and operate at lower temperatures.
They are well suited for low tip speed and high-airflow work - they are best suited for moving large
volumes of air against relatively low pressures.
Backward-inclined fans are more efficient than forward-curved fans. Backward-inclined
fansreach their peak power consumption and then power demand drops off well within their useable
airflow range. Backward-inclined fans are known as "non-overloading" because changes in static
pressure do not overload the motor.

Axial Flow Fan: Types

The major types of axial flow fans are: tube axial, vane axial and propeller (see Figure 5.4.)
TUBEAXIAL fans have a wheel inside a cylindrical housing, with close clearance between
blade and housing to improve airflow efficiency. The wheel turn faster than propeller fans, enabling
operation under high-pressures 250 – 400 mm WC. The efficiency is up to 65%.
VANEAXIAL fans are similar to TUBEAXIALS, but with addition of guide vanes that improve
efficiency by directing and straightening the flow. As a result, they have a higher static pressure
with less dependence on the duct static pressure. Such fans are used generally for pressures up to
500 mmWC. Vaneaxials are typically the most energy-efficient fans available and should be used
whenever possible.

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 Propeller fans usually run at low speeds and moderate temperatures. They experience a
large change in airflow with small changes in static pressure. They handle large volumes of air at low
pressure or free delivery. Propeller fans are often used indoors as exhaust fans. Outdoor applications
include air-cooled condensers and cooling towers. Efficiency is low – approximately 50% or less.

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Common Blower Types
Blowers can achieve much higher pressures than fans, as high as 1.20 kg/cm 2. They are also
used to produce negative pressures for industrial vacuum systems. Major types are: centrifugal
blower and positive-displacement blower.
Centrifugal blowers look more like centrifugal pumps than fans. The impeller is typically gear-
driven and rotates as fast as 15,000 rpm. In multi-stage blowers, air is accelerated as it passes
through each impeller. In single-stage blower, air does not take many turns, and hence it is more
efficient.
Centrifugal blowers typically operate against pressures of 0.35 to 0.70 kg/cm2 , but can
achieve higher pressures. One characteristic is that airflow tends to drop drastically as system
pressure increases, which can be a disadvantage in material conveying systems that depend on a
steady air volume. Because of this, they are most often used in applications that are not prone to
clogging. Positive-displacement blowers have rotors, which "trap" air and push it through housing.

Figure 5.5 Centrifugal Blower

Positive-displacement blowers provide a constant volume of air even if the system

pressure varies. They are especially suitable for applications prone to clogging, since they can
produce enough pressure - typically up to 1.25 kg/cm2 - to blow clogged materials free. They turn
much slower than centrifugal blowers (e.g. 3,600 rpm), and are often belt driven to facilitate speed
changes.

Figure 5.5 Positive-displacement blowers

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FANS CALCULATIONS & PERFORMANCE
1. Static Pressure & Static Head
Static Pressure is the pressure caused by its degree of compression. It is equal to the
difference between the static outlet/discharge pressure and the static inlet/suction pressure.

Static Head (hs):

𝑃𝑑−𝑃𝑠
hs = , usually expressed as head of water
ɣ
2. Dynamic Pressure & Velocity Head
Dynamic Pressure is the pressure caused by movement

Velocity Head (hv):

hv = [Vd2– Vs2] / 2g, usually expressed as head of air

Where: Vd = discharge velocity and Vs = suction velocity

To convert head of air to head of water:

hAρA = hW ρW ρA = PA/RATA
For Standard air:
ρA = density of air P = 101.325 kPa
ρW = density of water t = 21.1oC

3. Total Pressure and Total Head

Total Pressure is the pressure owing to compression and movement.
Total Head (ht) is the sum of static head and velocity head

ht = hs + hv

4. Fan Capacity (Q)

Q = AV

5. Air Power
Air Power (AP) is the power output of the fan or power delivered by the fan

AP = Q ɣℎ𝑡

Where: Units Units

AP = Air Power kW kW
Q = fan capacity m3/s m3/s
ht = total head meter of water meter of air
ɣ = 𝑠𝑝𝑒𝑐 𝑤𝑡 9.807 kN/m3 (std) 0.01177 kN/m3 (std)

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 6. Brake Power
Brake Power (BP) is the power input to the fan or the power delivered to the fan.

𝐴𝑃
BP = ƞ𝑚 = 𝑓𝑎𝑛 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
ƞ𝑚

7. Static Efficiency (ƞ𝑠) is the mechanical efficiency multiplied by the ratio of the static head
tothe total head

Ƞ𝑠 = ƞ𝑚 [hs/ht]

FAN LAWS:
The three basic fan laws which corresponds all fan functional principles:

1. Fan Speed Variations (Constant fan size, Constant density)

Q1/Q2 = N1/N2 H1/H2 = [N1]2/[N2]2 P1/P2 = [N1]3/[N2]3

2. Fan Size Variations (Constant speed, Constant density)

Q1/Q2 = [D1]3/[D2]3 H1/H2 = [D1]2/[D2]2 P1/P2 = [D1]5/[D2]5

3. Air Density Variations (Constant fan size, Constant speed)

Q1 = Q2 H1/H2 = ρ1 / ρ 2 P1/P2 = ρ1 /ρ 2

Where:
N = fan rotational speed in rpm
H = head in meters or ft
D = fan blade diameter
ρ = Air density
P = Power (usually Brake Power)

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 PROBLEMS AND SOLUTIONS: FANS & BLOWERS
Problem 1
A ventilating air enters a fan through a duct 1.6m 2. in area. The inlet pressure is 2.54cm of
water less than atmospheric pressure (vacuum). The air leaves the fan thru a duct 0.87m 2 in area
and the discharge static pressure is 7.68cm of water above atmospheric pressure. If the air density
is 1.2 kg/m3 and the fan delivers 9.5m3/s of air, find:
a. The total head in meters of water and in meters of air
b. The Air Power
c. The fan efficiency if the power input to the fan is 12.5kW

Given: Q = 9.5m3/s ρ = 1.2 kg/m3

As = 1.6m2 Ps/ ɣ = -2.54cm water
Ad = 0.87m2 Pd/ ɣ = 7.68cm water

Required: Find
a. ht (total head) in meters of water & in meters of air
b. AP (Air Power) in kW
c. ƞ𝑚 (𝐹𝑎𝑛 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦) if BP = 12.5kW

Solution:
a. ht = hs + hv

𝑃𝑑−𝑃𝑠
hs = = 7.68cm – (-2.54cm)
ɣ

hs = 10.22cm of water = 0.1022m of water

hv = [Vd2 – Vs2] / 2g

Vd = Q/Ad = 9.5m3/s / 0.87m2 = 10.92m/s

Vs = Q/As = 9.5m3/s / 1.6m2 = 5.937m/s

hv = [(10.92m/s)2 – (5.937m/s)2] / 2(9.807m/s2)

hv = 4.28m of air

Convert hv in meters of air to hv in meters of water:

hvAirρAir = hvWaterρWater

hvWater = 4.28m (1.2 kg/m3 / 1000 kg/m3) = 0.00513m of water

ht = 0.1022 + 0.00513

ht = 0.1073m of water

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 Convert ht in meters of water to ht in meters of air:

htAirρAir = htWaterρWater

htAir = 0.1073m (1000 kg/m3 / 1.2 kg/m3)

htAir = 89.44m of air

b. AP = Q ɣℎ𝑡

Using ht in meter of water Using ht in meter of air

AP = Qɣℎ𝑡 AP = Qɣℎ𝑡
AP = 9.5m3/s(9.807kN/m3)(0.1073m) AP = 9.5m3/s(0.01177kN/m3)(89.44 m)
AP = 10kW AP = 10kW

c. ƞ𝑚 = 𝑓𝑎𝑛 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

𝐴𝑃
BP =
ƞ𝑚

𝐴𝑃 10
ƞ𝑚 = x 100 = x 100
𝐵𝑃 12.5

ƞ𝒎 = 80%
Problem 2

An industrial fan delivers 9.45 m 3/s of air through a 0.9m x 1.22m outlet duct. Static
pressure at discharge is 127mm of water. Air temperature is 20oC and barometric pressure is
101.325 kPa. Determine:
a. The Air Power in kW
b. The power input to the fan if the mechanical efficiency is 62%

Given: Q = 9.45m3/s

Ad = 0.9m x 1.22m = 1.098m2 Ps/ ɣ = 0 (not given assume zero)

Pd/ ɣ = 127mm of water
Patm = 101.325 kPa tair = 20oC

Required: Find
a. AP (Air Power) in kW
b. BP (Air Power) in kW if ƞ𝑚 = 62%

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Solution:
a. AP = Q ɣℎ𝑡

ht = hs + hv

𝑃𝑑−𝑃𝑠
hs = = 127mm of water
ɣ

hs = = 0.127m of water

hv = [Vd2 – Vs2] /2g

Vd = Q/Ad = 9.45m3/s / 1.098m2 = 8.606m/s

Vs = Q/As = 0 (no area given, assume zero)

hv = [(8.606m/s)2 – (0)2] / 2(9.807m/s2)

hv = 3.775m of air

Convert hv in meters of air to hv in meters of water:

hvAirρAir = hvWaterρWater

Solving for air density:

ρA = PA/RATA = 101.325 / (0.28708)(293)

ρA = 1.2 kg/m3

hvWater = 3.775m (1.2 kg/m3 / 1000 kg/m3) = 0.00454m of water

ht = 0.127m + 0.00454m

ht = 0.1315m of water

AP = Qɣℎ𝑡 = 9.45m3/s(9.807kN/m3)(0.1315m)
AP = 12.19kW

b. BP (Air Power) in kW if ƞ𝑚 = 62%

BP = 19.66kW

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Problem 3
A blower draws 3000 ft3/min of air through a duct 12 inches in diameter with suction pressure
of 3 inches water below atmospheric. The air is discharged through a duct 10 inches in diameter
against a pressure of 2 inches water above atmospheric. The air is measured at 70 oF and30.2”Hg.
Determine:
a. The total head in ft of water
b. The air horsepower. Use specific weight of water equal to 62.34 lb/ft3

Given: Q = 3000 ft3/min

Ds = 12 in. = 1ft Ps/ ɣ = -3 inches water
Dd = 10 in. = 0.833ft Pd/ ɣ = 2 inches water
Patm = 30.2 in.Hg tair = 70oF

Required: Find
a. ht (total head) in ft water
b. AP (Air Power) in horsepower

Solution
a. ht = hs + hv

𝑃𝑑−𝑃𝑠
hs = = [(2 inches water)– (-3 inches water)]
ɣ

hs = 5 inches water = 0.4166 ft. water

hv = [Vd2– Vs2 ] / 2g
𝝅
Vd = Q /Ad = 3000 ft3/min / (0.833ft)2 x 1min / 60s = 91.673 ft/s
4
𝝅
Vs = Q /As = 3000 ft3/min / (1 ft)2 x 1min / 60s = 63.66 ft/s
4

hv = [(91.673ft/s)2 – (63.66ft/s)2] / 2(32.2ft/s2)

hv = 67.56 ft. air

Convert hv in ft air to hv in ft water:

hvAirρAir = hvWaterρWater

Solving for air density:

ρA = PA/RATA

ρA = 30.2 [14.7psi / 29.92] x [144in2 / 1ft2] / [(53.34 ft-lb/lb-R)][(70+460)R]

ρA = 0.07557 lb/ft3

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 hvWater = 67.56 ft (0.07557 lb/ft3 / 62.34 lb/ft3) = 0.0819 ft. water

ht = 0.4166 ft + 0.0819 ft

ht = 0.4985 ft. water

b. AP = Q ɣℎ𝑡
1 ℎ𝑝
AP = 3000 ft3/min (62.34 lb/ft3)(0.4985ft) 𝑥
33,000 𝑓𝑡−𝑙𝑏/𝑚𝑖𝑛
AP = 2.825 hp

Problem 4
A fan delivers 4.7m 3/s of static pressure of 5cm of water when operating at a speed of 400
rpm. The power input required is 2.96kW. If 7.05m3/s are desired in the same fan installation, find:
a. The new static head in cm of water
b. The new power input to the fan

Given: hs1 = 5cm of water N1 = 400rpm BP1 = 2.96kW

Q1 = 4.7 m3/s Q2 = 7.05 m3/s same fan

Required: Find
a. hs2
b. BP2
Solution:

Fan Law: Fan Speed Variations (Constant fan size, Constant density)
Q1/Q2 = N1/N2 H1/H2 = [N1]2/[N2]2 P1/P2 = [N1]3/[N2]3
Q1/Q2 = N1/N2
(4.7 / 7.05) = 400rpm / N2

N2 = 600rpm

hs1/hs2 = [N1]2/[N2]2

(5cm of water / hs2) = [400]2/[600]2

hs2 = 11.43 cm of water

BP1/BP2 = [N1]3/[N2]3

2.96kW / BP2 = [400]3/[600]3

BP2 = 10kW

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