Fans and blowers Principles

1. Fan is a machine used to add energy to the gaseous fluid to increase its
pressure. Fans are used where low pressures (from a few mm of water to
50 mm Hg) and comparatively large volume are required. They run at relatively low speed, the casing and impeller usually built of sheet iron.
FAN TYPES
1) AXIAL FLOW FANS - the flow of the gases is parallel to the fan shaft.
a. tube axial
b. vane axial
c. Propeller
2) RADIAL OR CENTRIFUGAL FLOW FANS- the flow of gases depends
upon the centrifugal action of the impeller or rotor.
a. Straight blades
b. Forward curved blades
c. Backward curved blades
d. Double curved blades

2. Propeller Fan
Tubeaxial Fan
Centrifugal Fan
Air in
Housing
Rotor
Motor
Air out
Vaneaxial Fan

11. COMMON USES OF FANS
1. Ventilation and air conditioning
2. Forced and induced draft service for boilers
3. Dust collection
4. Drying and cooling of materials
5. Cooling towers
6. Mine and tunnel ventilation
7. Pneumatic conveying and other industrial process work
Head Calculations
2
1
discharge
suction
For a fan ∆Z = 0 ; ∆PE = 0 and Q = 0, because fans are designed to
overcome fluid friction. No cooling system is needed due to small temperature
differential between suction and discharge.

12. From Bernoulli’s energy theorem
1. For fans installed with both suction and discharge duct
2
2
P2 − P1 v 2 − v 1
ht =
+
γ
2g
m of gas
2. For fans installed with only a suction duct; P 2 = 0 gage
2
2
0 − P1 v 2 − v 1
ht =
+
γ
2g
m of gas
3. For fans installed with only discharge duct; P 1 = 0 gage and v1 = 0
2
P2 v 2
ht = +
γ 2g
m of gas

13. let
P2 − P1
hs =
m of gas
γ
2
2
v 2 − v1
hv =
2g
m of gas
ht = hs + hv m of gas
Where:
hs - static head at which a fan operates, m of gas
hv - velocity head at which a fan operates, m of gas
ht - total head added to the fluid, m of gas

14. Head Conversion: From m of gas to m of water
hw =
γ gh g
γw
=
ρ gh g
ρw
m of water
htw = hsw + hvw
Where:
h - stands for ,total head, static head or velocity head
w - refers to water; g - refers to gas

17. FAN POWER
FP = Qγ whtw KW
STATIC POWER
SP = Qγ whsw KW
where Q - capacity in m3/sec
γw - specific weight of water (gage fluid) in KN/m 3
htw - total head in m of WG
hsw - static head in m of WG
FP - total fan power in KW
SP - Static power in KW
Static Power - is that part of the total air power, that is
used to produced the change in static head.

18. FAN EFFICIENCY
FP
η =
x 100 %
F BP
STATIC EFICIENCY
SP
η =
x 100%
S BP
BP - Brake or shaft power in KW

19. FAN LAWS
A. Variation in speed and impeller diameter
Q ∝ ND3
H ∝ N2D2
B. Variation in impeller Speed
Q ∝ N ; H ∝ N2 ; Power ∝ N3
C. Variation in impeller size; Tip speed = C ; ρ = C and
same proportions; H = C
Q ∝ D2 ; Power ∝ N2 ; N ∝ 1/D
D. Variation in impeller size; N = C; ρ = C ; Same proportions
Q ∝ D3 ; Power ∝ D5 ; H ∝ D2 ; Tip Speed ∝ D
E. Variation in density; Q = C; N =C; D = C; system = C
H ∝ ρ ; Power ∝ ρ
F. Variation in Density; D = C; H = C
1
Q∝
; Power ∝
ρ
1
1
; N∝
ρ
ρ

20. G. Variation in density; m = C;D = C; system = C
1
Q∝ ; H ∝
ρ
1
1
1
; N ∝ ; Power ∝ 2
ρ
ρ
ρ

21. A certain fan delivers 340 m3/min of air at a static
pressure of 25.4 mm WG when operating at a
speed of 400 RPM and requires an input of 3 KW.
If in the same installation 425 m3/min of air are
desired, what will be the new Q, hsw and Fan power
required? (40 mm WG;500 RPM;6 KW )

22. Q1 = 340m 3 / min
hs1 = 0.025m of H 2 O
N1 = 400 RPM
BP1 = 3 KW
Q 2 = 425 m 3 /min
N2 =
BP2 =
From Fan Laws
Q ∝ N; h ∝ N 2; P ∝ N 3
Q2 N 2
=
Q1 N1
425 N 2
=
340 400
N 2 = 500 RPM
2
h s2  500 
=

25  400 
hs 2 = 39.1 mm WG
BP2  500 
=

3
400 

BP2 = 6 KW
3

23. BLOWERS
Blower is a machine used to compressed air or gas by centrifugal force to a
final pressure not exceeding 241 KPa gage. Usually blower has no cooling
system or it is not water cooled.
COMPRESSION OF GASES
The design of blower is usually based upon either an adiabatic or isothermal
compression.
A. For Adiabatic or Isentropic Compression:
P
P2
P1
2
PVk = C
1
V

24. k −1
k
T2  P2 
=  
T1  P1 
 
 P k −1k

kP1Q  2
 
W=
− 1
P 
k − 1  1 



W = QγH
where Q = V1
Q - capacity in m3 /sec
- adiabatic head in meters
H
 P  k −1k 
1000kRT1  2
 
H=
− 1 m of gas
g( k − 1)  P1 

 



25. B. For Isothermal Compression:
P
P2
2
P1
PV = C
1
V
P1V1 = P2 V2 = C
P2
P2
W = P1Q ln = mRT1ln KW
P1
P1
W = QγH KW
1000RT1 P2
H=
ln meters
g
P1
where
H - isothermal head in meters
Q - capacity in m3/sec
g - gravitational acceleration in msec2

26. Efficiency:
A. Adiabatic or Isentropic Efficiency
Isentropic Work
ηk =
x 100%
Actual Work
B. Isothermal Efficiency
Isothermal Work
ηI =
x 100%
Actual Work
RATIO OF THE ADIABATIC TEMPERATURE RISE TO THE
ACTUAL TEMPERATURE RISE
 P k −1k 
T1  2 
− 1
P 
 1 


Y=  '
( T2 − T1 )

27. RELATIONSHIP FOR CORRECTING PERFORMANCE CURVES
1. Volume Flow
QB NB
=
Q A NA
2. Weight Flow
mB  NB   P1B
=   
m A  N A   P1A
 
  T1A
 
 T
  1B




3. Pressure Ratio
 P k −1k 
 2 
− 1
P 
2
 1 


 B  NB   T1A
=  
N   T
k −1
 P  k 
 A   1B
2
 
− 1
P 
 1 


A
P2
= rp (pressure ratio)
P1





28. 4. Head
2
HB NB
= 2
HA NA
5. Brake Power
BPB  NB 
= 
BPA  N A 
 
3
 P1B

P
 1A
 T1A

 T
 1B




 P k −1k 
 2 
− 1
P 


BPB  P1B  QB   1 
B

=  
BPA  P1A  Q A   k −1k 
 
 P
 2 
− 1
P 
 1 


A
Where:
1 - suction
2 - discharge
A - 1st condition
B - 2nd condition
R - gas constant, KJ/kg-K
P - absolute pressure in KPa
ρ - density, kg/m3
T - absolute temperature, K
H - head, m γ - specific weight, KN/m3
Q - capacity, m3/sec
BP - brake power, KW
N - speed, RPM
W - work, KW
m - mass flow rate, kg/sec

35. Copyright: YURI G. MELLIZA
324619CYE