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FAN CONCEPT & ANALYSIS IN CEMENT
INDUSTRY
BY –NITIN ASNANI (Certified Energy Auditor by BEE)
(CEA-28964)
Static Pressure
❑Static Pressure
❑Pressure exerts due to compression of air
❑Independent of gas velocity:- P
❑Changes as per system resistance
System Resistance?
Sum of static pressure is known as System resistance
Independent of Gas velocity?
System Resistance
Numerical no 1
Find the system resistance across mill?
a) Find the system resistance across mill?
b) System resistance across mill inlet to fan inlet?
Solution:-
Static pressure at mill inlet +Static pressure above nozzle +static pressure on sep
body= Mill outlet pressure
Independent of Gas Velocity?
Circular duct
Gas
Static Pressure is same at both
points.
Velocity Pressure is minimum Velocity Pressure is greater than
point1
Point1
Point2
Dynamic Pressure
❑Pressure exerts due to gas motion
❑Dependent upon Velocity:-1/2 ρ v2
❑Changes as per velocity profile inside the duct
❑Maximum at Centre:-
THUMB RULES
▪ Average dynamic pressure equals to 90 % of center value
▪ Due to eddy's formation the value may increase at side wall
▪ If flow is laminar then Dynamic pressure value increase from side wall to
centre.
Gas properties
❑Density of gases decreases as temperature increases
❑Total volumetric flow rate increases by increasing temperature.
❑Mass flow rate is independent of temperature
❑Actual Density/Normal Density *m3/hr=0.60/1.42*50,0,000
Nm3/hr.-Volume of gas standard temperature & pressure
m3/hr.-Volume of gas at temperature& pressure
Standard Gas Density
• Density of Air-1.29 kg/Nm3
• Density of Flue gases:-1.42 kg /Nm3
• Density of Flue gases at high oxygen Concentration:-1.35 kg/Nm3
• Average Density calculation by formulae
Barometric Pressure
• Atmospheric pressure is also known as Barometric pressure.
• Pressure caused by the weight of air above us
• Barometric pressure decreases with increasing Altitude(Reference
point-sea level)
• Effect on Fan power & compressor
P(h)=barometric pressure in kPa
h = Height above sea level in m
P(h)=barometric pressure in kPa
P(h)=101.325exp(-0.00012h)
Density Correction factor
• Density at temperature & pressure is calculated by below formula:-
• Þt=Density at Standard temperature& pressure
• Þn=Density at temperature & pressure
• T= Temperature in degree Celsius
• Pressure in mmwg
Þn= {(Barometric pressure +static pressure)X273}X Þt
{ (273.15+T)X Atmospheric pressure}
Numerical Problem-1
• Find the density of air?
Given:-
❑Temp of cooler vent air:-250 degree C
❑Static pressure inside the duct: -30 mmwg
❑Density of air:-1.29 kg/Nm3
❑Atmospheric pressure:-10336 mmwg
❑Height above sea level:-300 m
Solutions
Step1:-Find Barometric pressure
Barometric Pressure = 101.325 x exp(-00012 x 300)
= 97.74 kPa
= 97.74x101.97
= 9966mmwg
Step2:-Apply density correction formula
= {(9966+(-30))x(273)/(273+250)x(10336)X1.29
= 0.65 kg/m3
Concept of pitot Tube
Static Pressure measurement
For Static pressure measurement;-
➢Insert the pitot tube inside the duct in any position and keep one end
of manometer is open to the atmosphere because SP is independent
of velocity.
➢Try to take readings as per methodology which will be discussed in
further slides
Dynamic pressure measurements:-
• Gas Flow
➢Put the pitot tube in along the gas direction.
➢Connect both end (+ or -) of manometer and then take readings from side wall of duct to
center.
➢Change the position(yellow end upwards now) of pitot tube for calibration point of view
in such a way the both readings should be come same .
➢Measurements should be carried by S type pitot tube for dust laden gas.
➢L type for clean air(Generally for bag filters)
Methodology to take correct readings
Numerical No-2
Find the average dynamic pressure of gas flow inside the horizontal duct?
Given readings:-
11,12,10,9
Solution
Average={ (sqrt(11)+sqrt(12)+sqrt(10)+sqrt(9)) }2
{ 4}
= 10.47 mmwg
Gas Flow Measurements
V = Pitot constant xSqrt{(2x9.8xdynamic pressure)}
{ (density at particular temp)}
g = 9.8 m/s2
Dynamic pressure = mmwg
Density = kg/m3
Q = AXV
• A = Area(m2)
• V = Velocity(m/s)
• Q = Total volumetric flow rate-(m3/s)
Numerical -
Find preheater gas flow in m3/hr,Nm3/hr, kg/hr,Nm3/kg clinker?
Given
Temp = 280 degree C
SP = -550 mmwg
dynamic Pressure = 12 mmwg
Duct diameter = 3.2 m
Height above sea level = 330m
density of flue gases = 1.42kg/Nm3
Clinker production = 4000TPD
Pitot constant = 0.85
Solution
Step 1 :-
Find barometric Pressure-
= Barometric Pressure
= 101.325 x exp(-00012 x 330)x101.97
= 9931 mmwg
Step 2:-
Find Density correction factor:-
Step2:-Apply density correction formula
= {(9931+(-550))x(273)/(273+270)x(10336)X1.42
= 0.64 kg/m3
Solution
Step 3:-
Find the velocity?
V = Pitot constant X Sqrt{(2x9.8xdynamic pressure)}
{ (density at particular temp)}
V = 0.85Xsqrt(2x9.8x12/0.64)
= 16 .34 m/s
Q = A XV
A = 3.14X3.5X3.5/4
Q = 16.34X9.62
= 16.34X9.62X3600
Solution
= 565764 m3/hr
Flow (kg/hr) = 565764x0.64
= 359961 kg/hr
Nm3/hr = m3/h x(corrected density/density at STP)
Nm3/hr = 565764 x(0.64/1.42)
= 253493 m3/hr
Nm3/kg Clink = 253493/4000x1000/24
= 1.52 Nm3/kg clinker
Concept of fan
• A centrifugal fan is a mechanical device for moving air or other
Gases, with rotating Impellers.
• They use the kinetic energy of the Impellers / the rotating blade to
increase the pressure of the air/gas stream which in turn moves
them against the resistance caused by ducts, dampers and other
components
• A rotary machine which maintains a continuous flow of air at a
pressure ratio not normally exceeding 1.3
• A device for moving air which utilizes a power driven rotating
impeller. A fan shall have at least one inlet opening and at least one
outlet opening. The openings may or may not have elements for
connection /duct
Fans Classification
• Simple Definition
• Machines to move gases (or) gases mixed with small solid
particles
• Operation of fans & pumps similar
• Broad Classification
• Centrifugal
• Axial
Difference between Fans & Blowers
• The property that distinguishes a centrifugal fan from a blower
is the pressure ratio it can achieve.
• A blower in general can produce higher pressure ratio.
• As per American Society of Mechanical Engineers (ASME) the
specific ratio – the ratio of the discharge pressure over the
suction pressure – is used for defining the fans and blowers
Difference Between Fans, Blowers
and Compressors:-AS per ASME
ILLUSTRATION
MAJOR PROCESS FANS
• 30% to 40% of the plants total electrical energy consumption are from
Major Process fans
• Major Process Fans:
• Pre Heater Fans
• Raw Mill Fan
• Coal Mill
• Raw Mill BH / ESP Fan
• Cooler Vent / ESP Fans
• Cement Mill Fans (CVRM)
• Raw Mill / Coal Mill / CVRM Booster Fans
• Clinker Cooler Fan
Thumb Rules
• FAN is a Rotary Machine
• Fan has a Pre- designed Capacity (Flow & Pressure)
• Fan works against a SYSTEM
• Fan utilizes a driving equipment & consumes Power,
against the work done
• Fan being a “ROTARY MACHINE”, helps in the movement
of Gas in a system from one location to another location.
Concept of System Resistance
• System resistance is the sum of Pressure losses in the system
• System resistance is a function of configuration of ducts, bends &
pressure drop across the equipment’s.
• The system resistance varies with the square of the volume of air
flowing through the system.
• The system resistance increases substantially as the volume of air
flowing through the system increases; square of air flow.
Resistance in the system
• Process Equipment
• Dampers
• Duct Losses ( due to duct layout / sizing)
• Cyclones
• ESP /BH
• Other sources of Resistance:
• Loss of energy due to sudden change in Gas velocity
• Improper change in Cross sectional area
• Change in direction of flow
Pressure HEAD VS FLOW
Head ∝ Volume 2
FAN LAW
Thumb Rules
Numericals
Given:-
Q1=300000 m3/hr;N1=800 RPM if rpm is reduced by 10% then find Q2?
Sol :-
Apply Fan law
N1/N2 = Q1/Q2
Q2 = N2XQ1/N1
Q2 = 300000 X 720/800
Q2 = 270000 m3/hr.
% = (300000-270000)/(300000)
= 10%
Flow is reduced by 10%
Numericals
Given:-
SP1=400 mmwg;N1=800 RPM if rpm is reduced by 10% then find SP2?
Solution :-
Apply Fan law
(N1/N2)2 = SP1/SP2
SP1/SP2 = (N1/N2)2
SP1/SP2 = (800/720)2
SP2 = 400/1.23
SP2 = 324 mmwg
% = (400-324)/(400)
Head reduced by19%
Fan Characteristic Curve
System curve and Fan curve
• System curve:
• Represents performance of duct system
• Shows the relationship between the flow rate and the
pressure required to create that flow through the duct
system
• Fan curve:
• Represents performance of fan
• Shows the relationship between the flow rate and pressure
created by the rotation of a fan impeller
System Curve & Fan Curve
Operating Point
The intersection of system curve and fan curve determine the operating point for the system
Factors Affecting Fan Performance
❖RPM
❖Diameter
❖Density
❖Temperature
Fan Efficiency
Formulae
Calculation of Fan Head & Power Saving
Fan head = Fan O/L-Fan I/L
Power Saving = Pressure Drop X Electrical Power
Fan Head
Numericals
N7:-If an existing raw mill old cyclones replace with new one and get the benefit
of 30 mmg then how much electrical saving will be achieved in terms of
SEC(kWh/ton of material)?
Given
Raw mill TPH = 300 TPH
Electrical Power of Fan = 2000 kW
Fan O/L pressure = -10 mmwg
Fan inlet Pressure = -1000 mmwg
Solution:-
Step 1
Calculate Fan Head
= ((-10 )-(-1000))
= 990 mmwg
Numericals
Step 2:-
How to Calculate pressure drop saving in terms of Power ?
= Pressure Drop X Electrical Power
Fan Head
= (30/990) X Electrical power
= 60 kW
SEC = 60/300
Saving = 0.20 kWh/ton of material
Fan Curve Analysis
Option-1
If actual Pressure is higher than the design pressure the operating point
shifts towards left side- Volume reduces
• Maximum Operating Pressure should be less than “Peak pressure” to avoid
getting into surging Zone.
Option-2
• Operating Pressure – Too Low
• Operating point shifts to right- efficiency level goes down.
• Design Efficiency is good – Operating Efficiency can be good by initiating
actions.
• Design Efficiency is Low- Operating Efficiency Low
OPTION 3
• A Fan with no margin on Parameters having margin on Motor
• Look for possibility of up gradation
• B. Fan with no margin on Parameters & no margin on Motors Look
for creating margin within the system
▪ Gas flow being function of process – look for reduction in system
Pressure.
▪ Once margin is created on system pressure, we get margin on Motor
GAS VELOCITY PROFILE
• Fan Inlet & Outlet velocities should be compatible to System duct
velocity(difference should not be greater than 8 m/s)
• Higher is the difference between Fan Inlet/Outlet velocity…& Duct
velocity Higher will the turbulence
• Higher the turbulence…more will be the pressure drop..
Fan efficiency Calculations….
Solutions
Given Q = 28651m3 /hr.
Static pressure at fan inlet = -185 mmwg
Static pressure at fan outlet = -8 mmwg
Electric power = 30.60 kW
Motor efficiency = 90%
Step 1:-
First calculate Fan head
Fan head = Fan O/L-Fan I/L = ((-8 )-(-185)) = 177 mmwg
Solutions
Step 2:-
Then, calculate Mechanical Power
Q=m3 /hr; g=9.8 m/sec2
Mechanical power = QXFan headX9.8 /3600X1000
= 28651x177x9.8 /3600x1000
= 13.80 kW
= 13.80/(30.60X0.90) = 50.10%
How to Calculate Damper losses?
Fan Controlled Techniques
• Is it off the correct capacity/head?
• How is the fan controlled
• Damper throttling
• Guide vane control
• Speed control – GRR / SPRS / VFD/ON/OFF
Various Control Methods
GRR Losses Calculations
• Saving Calculations
• Fan1 rated RPM = 980
• Fan1 operating RPM = 845
• Slip = 1-(operating RPM/rated RPM) Slip = 0.138
• GRR Losses = Slip*(1-(Slip*Slip))*Present power
= 0.138*(1-(0.138*0.138))*1012
= 136.76 kW
Fan Retrofication Concept
• Retrofitting refers to the addition of new technology or features to
older systems.
• This is the up gradation of the Fan performance for improved capacity
or reduction in specific power consumption.
• During Retrofitting, look at the past performance of the fan and then
retrofit with the high efficiency impeller into the existing fan casings to
achieve highest performance.
• Practically it is replacement of Impeller within existing casing, to suit
the upgraded parameters with higher operating efficiency level
Fan Retrofication Concept
• Can the capacity/head be further fine tuned?
• What is the efficiency of fan?
• Can it be replaced by high efficiency fan
• Is it possible to reduce system resistance – Suction /
discharge side
• Can fan operation be interlocked with process operation?
ESP Exhaust fan-Case Study
• Fan Designed for 300 TPD Kiln output
• Fan Rating :-1,10,000 m3 /h, 200 mm WC, 110 kW
• Fan Operating With DC drive
• Present Operation – 450 TPD kiln output
• Actual fan parameters -1,48,000 m3 /h, 110 mmwg
ESP VENT
FAN CURVE
Fan retrofication opportunities
• Observations:-
• Fan Rated for 78.5% Efficiency
• Present Efficiency - Only 45%
• Fan efficiency is deteriorated because mismatch between design head
with operating head.
• Design Head:- 200 mmwg
• Operating Head :-110 mmwg
Fan retrofication opportunities
• Recommendations:-
• Replace Fan with new Correct head fan & flow(Efficiency -80%)
Specifications
• Q = 150000 m3 /hr
• Head = 120 mmwg
• Efficiency = 80%
Fan efficiency Concept –How to increase?
• Optimum margin in head & flow
• Higher efficiency
• Operation close to best efficiency point
• Suitable fan control system
Conclusion –Fan retrofication
• Assess the Process requirement
• Analyze the Fan design data
• Compare the actual performance data with design
• Fan Power is a function of Volume & Pressure
• Volume & Pressure- function of Process & system
• So we must investigate the System & try to optimize to reduce Fan
power.
• Look even beyond Fan & identify potential.
Considerations
❑Low Design & operating efficiency of the existing Fan
❑Up gradation of process system needs higher parameters
High / Low Margin
The scope should be limited to:
a) Impeller b) Shaft c) Inlet cone
2. Existing Static parts, e.g. Casing /pedestal etc. to be utilized
3. No change in Foundation
4. No change in Bearings / Housing (inventory aspect)
5. Shaft may remain identical, as far as possible
6. Retrofit Impeller to be designed to utilize Existing Motor.
7. Cost effective
8. Minimum time to install the Retrofit Impeller shaft
FAN MODIFICATIONS
• Increase in cooler ESP Stack height
• Reduction of suction pressure at cooler fan inlets through bell mouth
• Suction box modifications of process fans by reducing velocity at fan
inlet.
• Rectify the cone gap of cooler fans
• Reduce the pressure drop across the system through CFD Study.
Cooler exhaust fan in a Cement plant
Reducing SEC OF Cooler exhaust fan in a Cement plant
by Increasing Stack height
• When cooler upgradation
• Exhaust height to be increased
• From 23 m to 60 m
• Additional draft - 65 mm
• Net benefit - 50 mmwg
• Power consumption reduced by nearly 50%
• Saving potential 60 kW
Reducing SEC OF Cooler exhaust fan in a Cement plant by
Increasing Stack height
Reducing SEC OF Cooler exhaust fan in a Cement plant by
Increasing Stack height
Calculations
Suction side pressure drop in cooler fans by doing bell mouth
• Extensive discussions held
• Fan suppliers
• Fan experts
• Suction pressure drop
• Not more than 15-20mmwg
• Cooler fans invariably very high - 70-90 mmwg
• Intentional
Suction side pressure drop in cooler fans by doing bell mouth
Suction side pressure drop in cooler fans by doing bell mouth
• Typically flow measurement
• using Piezometer ring
• Additional pressure difference for accuracy
• Measures to be taken
• Reduce inlet pressure drop
• Measure air flow
• For reducing the inlet pressure drop
• Inlet Air velocity - 10-12 m/sec
• Reduction in entry loss – using bell mouth
Suction side pressure drop in cooler fans by doing bell mouth
Suction side pressure drop in cooler fans by doing bell mouth
❑Aerofoil in cooler fans
Initial apprehension – dust
Prone for wear due to dust
❑ Extensively practiced in power plants FD fans
Inlet Pressure – not more than 20 mm WC
Saving potential - 6-7%
Suction side pressure drop in cooler fans by doing bell mouth
❑Aerofoil in cooler fans
Initial apprehension – dust
Prone for wear due to dust
❑ Extensively practiced in power plants FD fans
Inlet Pressure – not more than 20 mm WC
Saving potential - 6-7%
CASE STUDY
Reduce the pressure drop in cement mill1 sep vent fan duct through
CFD Study
Path Lines Colored by Velocity Magnitude AS IS before duct modification
(High Velocity Profile-Turbulence)
Path Lines Colored by Velocity Magnitude after duct modification
FAN INLET SUCTION BOX MODIFICATION
FAN INLET SUCTION BOX MODIFICATION
❑ Difference between duct & suction box velocity is on higher side that creates
excessive pressure drop.
❑ Increase the width of suction box in such a way that the difference between duct
and fan inlet velocity could be minimized.
❑ Difference between the duct & Fan inlet velocity should be less than 7-8 m/s
❑ New Suction boxes should be designed by less than 30m/s
❑ Should not be greater than 35% of existing width(Limitation)
Saving potential:-42 kW & 35 mmwg
FAN INLET SUCTION BOX MODIFICATION
IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION
Different factors impacts the efficiency of a fan:
❑Build-up inside the impeller or between the impeller and the inlet
cone
❑Wears between the impeller and the inlet cone
❑ Inappropriate gap between impeller and inlet cone
IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION
-STEP1
IMPROVE FAN EFFICIENCY-STEP2
▪ In any centrifugal fan the gaps
between the rotating part and inlet;
the overlap of the inlet to the wheel is
crucial.
▪ Otherwise the internal recirculation of
air/gas increases (from positive side
outside the wheel to negative inside
the wheel) and thus increases the
wasted energy as well the fan’s
capacity to move gas.
IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION
-STEP3-BUILDUPS
Check all build-ups Between impeller and inlet cone
❑ This build up has a strong influence on fan efficiency.
❑It is common to be able to increase fan efficiency from 3 to 5 points with a
thorough cleaning.
❑Clean manually with a small scraper around the inlet cone until you can see again
the iron casing. Clean also the anti-rotating plate fixed on the inlet cone. They
avoid turbulences and are very important.
SOLUTION
Inside the impeller
A high pressure water jet can be used to
remove build-up
IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION
-STEP4-CHECK WEAR
Cone Gap of fan
FAN CONE GAP NORMS
CONE GAP NORMS
TYPES OF FAN ENTRY
AXIAL FAN
• Axial-flow fans produce lower pressure than centrifugal fans, and exhibit a
dip in pressure before reaching the peak pressure point
• Applications-Cooling tower, Ventilation fans
THANK YOU

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FAN CONCEPT & ANALYSIS IN CEMENT INDUSTRY

  • 1. FAN CONCEPT & ANALYSIS IN CEMENT INDUSTRY BY –NITIN ASNANI (Certified Energy Auditor by BEE) (CEA-28964)
  • 2. Static Pressure ❑Static Pressure ❑Pressure exerts due to compression of air ❑Independent of gas velocity:- P ❑Changes as per system resistance System Resistance? Sum of static pressure is known as System resistance Independent of Gas velocity?
  • 4. Numerical no 1 Find the system resistance across mill? a) Find the system resistance across mill? b) System resistance across mill inlet to fan inlet? Solution:- Static pressure at mill inlet +Static pressure above nozzle +static pressure on sep body= Mill outlet pressure
  • 5. Independent of Gas Velocity? Circular duct Gas Static Pressure is same at both points. Velocity Pressure is minimum Velocity Pressure is greater than point1 Point1 Point2
  • 6. Dynamic Pressure ❑Pressure exerts due to gas motion ❑Dependent upon Velocity:-1/2 ρ v2 ❑Changes as per velocity profile inside the duct ❑Maximum at Centre:- THUMB RULES ▪ Average dynamic pressure equals to 90 % of center value ▪ Due to eddy's formation the value may increase at side wall ▪ If flow is laminar then Dynamic pressure value increase from side wall to centre.
  • 7. Gas properties ❑Density of gases decreases as temperature increases ❑Total volumetric flow rate increases by increasing temperature. ❑Mass flow rate is independent of temperature ❑Actual Density/Normal Density *m3/hr=0.60/1.42*50,0,000 Nm3/hr.-Volume of gas standard temperature & pressure m3/hr.-Volume of gas at temperature& pressure
  • 8. Standard Gas Density • Density of Air-1.29 kg/Nm3 • Density of Flue gases:-1.42 kg /Nm3 • Density of Flue gases at high oxygen Concentration:-1.35 kg/Nm3 • Average Density calculation by formulae
  • 9. Barometric Pressure • Atmospheric pressure is also known as Barometric pressure. • Pressure caused by the weight of air above us • Barometric pressure decreases with increasing Altitude(Reference point-sea level) • Effect on Fan power & compressor P(h)=barometric pressure in kPa h = Height above sea level in m P(h)=barometric pressure in kPa P(h)=101.325exp(-0.00012h)
  • 10. Density Correction factor • Density at temperature & pressure is calculated by below formula:- • Þt=Density at Standard temperature& pressure • Þn=Density at temperature & pressure • T= Temperature in degree Celsius • Pressure in mmwg Þn= {(Barometric pressure +static pressure)X273}X Þt { (273.15+T)X Atmospheric pressure}
  • 11. Numerical Problem-1 • Find the density of air? Given:- ❑Temp of cooler vent air:-250 degree C ❑Static pressure inside the duct: -30 mmwg ❑Density of air:-1.29 kg/Nm3 ❑Atmospheric pressure:-10336 mmwg ❑Height above sea level:-300 m
  • 12. Solutions Step1:-Find Barometric pressure Barometric Pressure = 101.325 x exp(-00012 x 300) = 97.74 kPa = 97.74x101.97 = 9966mmwg Step2:-Apply density correction formula = {(9966+(-30))x(273)/(273+250)x(10336)X1.29 = 0.65 kg/m3
  • 14. Static Pressure measurement For Static pressure measurement;- ➢Insert the pitot tube inside the duct in any position and keep one end of manometer is open to the atmosphere because SP is independent of velocity. ➢Try to take readings as per methodology which will be discussed in further slides
  • 15. Dynamic pressure measurements:- • Gas Flow ➢Put the pitot tube in along the gas direction. ➢Connect both end (+ or -) of manometer and then take readings from side wall of duct to center. ➢Change the position(yellow end upwards now) of pitot tube for calibration point of view in such a way the both readings should be come same . ➢Measurements should be carried by S type pitot tube for dust laden gas. ➢L type for clean air(Generally for bag filters)
  • 16. Methodology to take correct readings
  • 17. Numerical No-2 Find the average dynamic pressure of gas flow inside the horizontal duct? Given readings:- 11,12,10,9 Solution Average={ (sqrt(11)+sqrt(12)+sqrt(10)+sqrt(9)) }2 { 4} = 10.47 mmwg
  • 18. Gas Flow Measurements V = Pitot constant xSqrt{(2x9.8xdynamic pressure)} { (density at particular temp)} g = 9.8 m/s2 Dynamic pressure = mmwg Density = kg/m3 Q = AXV • A = Area(m2) • V = Velocity(m/s) • Q = Total volumetric flow rate-(m3/s)
  • 19. Numerical - Find preheater gas flow in m3/hr,Nm3/hr, kg/hr,Nm3/kg clinker? Given Temp = 280 degree C SP = -550 mmwg dynamic Pressure = 12 mmwg Duct diameter = 3.2 m Height above sea level = 330m density of flue gases = 1.42kg/Nm3 Clinker production = 4000TPD Pitot constant = 0.85
  • 20. Solution Step 1 :- Find barometric Pressure- = Barometric Pressure = 101.325 x exp(-00012 x 330)x101.97 = 9931 mmwg Step 2:- Find Density correction factor:- Step2:-Apply density correction formula = {(9931+(-550))x(273)/(273+270)x(10336)X1.42 = 0.64 kg/m3
  • 21. Solution Step 3:- Find the velocity? V = Pitot constant X Sqrt{(2x9.8xdynamic pressure)} { (density at particular temp)} V = 0.85Xsqrt(2x9.8x12/0.64) = 16 .34 m/s Q = A XV A = 3.14X3.5X3.5/4 Q = 16.34X9.62 = 16.34X9.62X3600
  • 22. Solution = 565764 m3/hr Flow (kg/hr) = 565764x0.64 = 359961 kg/hr Nm3/hr = m3/h x(corrected density/density at STP) Nm3/hr = 565764 x(0.64/1.42) = 253493 m3/hr Nm3/kg Clink = 253493/4000x1000/24 = 1.52 Nm3/kg clinker
  • 23. Concept of fan • A centrifugal fan is a mechanical device for moving air or other Gases, with rotating Impellers. • They use the kinetic energy of the Impellers / the rotating blade to increase the pressure of the air/gas stream which in turn moves them against the resistance caused by ducts, dampers and other components • A rotary machine which maintains a continuous flow of air at a pressure ratio not normally exceeding 1.3 • A device for moving air which utilizes a power driven rotating impeller. A fan shall have at least one inlet opening and at least one outlet opening. The openings may or may not have elements for connection /duct
  • 24. Fans Classification • Simple Definition • Machines to move gases (or) gases mixed with small solid particles • Operation of fans & pumps similar • Broad Classification • Centrifugal • Axial
  • 25. Difference between Fans & Blowers • The property that distinguishes a centrifugal fan from a blower is the pressure ratio it can achieve. • A blower in general can produce higher pressure ratio. • As per American Society of Mechanical Engineers (ASME) the specific ratio – the ratio of the discharge pressure over the suction pressure – is used for defining the fans and blowers
  • 26. Difference Between Fans, Blowers and Compressors:-AS per ASME
  • 28. MAJOR PROCESS FANS • 30% to 40% of the plants total electrical energy consumption are from Major Process fans • Major Process Fans: • Pre Heater Fans • Raw Mill Fan • Coal Mill • Raw Mill BH / ESP Fan • Cooler Vent / ESP Fans • Cement Mill Fans (CVRM) • Raw Mill / Coal Mill / CVRM Booster Fans • Clinker Cooler Fan
  • 29. Thumb Rules • FAN is a Rotary Machine • Fan has a Pre- designed Capacity (Flow & Pressure) • Fan works against a SYSTEM • Fan utilizes a driving equipment & consumes Power, against the work done • Fan being a “ROTARY MACHINE”, helps in the movement of Gas in a system from one location to another location.
  • 30. Concept of System Resistance • System resistance is the sum of Pressure losses in the system • System resistance is a function of configuration of ducts, bends & pressure drop across the equipment’s. • The system resistance varies with the square of the volume of air flowing through the system. • The system resistance increases substantially as the volume of air flowing through the system increases; square of air flow.
  • 31. Resistance in the system • Process Equipment • Dampers • Duct Losses ( due to duct layout / sizing) • Cyclones • ESP /BH • Other sources of Resistance: • Loss of energy due to sudden change in Gas velocity • Improper change in Cross sectional area • Change in direction of flow
  • 32. Pressure HEAD VS FLOW Head ∝ Volume 2
  • 35. Numericals Given:- Q1=300000 m3/hr;N1=800 RPM if rpm is reduced by 10% then find Q2? Sol :- Apply Fan law N1/N2 = Q1/Q2 Q2 = N2XQ1/N1 Q2 = 300000 X 720/800 Q2 = 270000 m3/hr. % = (300000-270000)/(300000) = 10% Flow is reduced by 10%
  • 36. Numericals Given:- SP1=400 mmwg;N1=800 RPM if rpm is reduced by 10% then find SP2? Solution :- Apply Fan law (N1/N2)2 = SP1/SP2 SP1/SP2 = (N1/N2)2 SP1/SP2 = (800/720)2 SP2 = 400/1.23 SP2 = 324 mmwg % = (400-324)/(400) Head reduced by19%
  • 38. System curve and Fan curve • System curve: • Represents performance of duct system • Shows the relationship between the flow rate and the pressure required to create that flow through the duct system • Fan curve: • Represents performance of fan • Shows the relationship between the flow rate and pressure created by the rotation of a fan impeller
  • 39. System Curve & Fan Curve
  • 40. Operating Point The intersection of system curve and fan curve determine the operating point for the system
  • 41. Factors Affecting Fan Performance ❖RPM ❖Diameter ❖Density ❖Temperature
  • 43. Calculation of Fan Head & Power Saving Fan head = Fan O/L-Fan I/L Power Saving = Pressure Drop X Electrical Power Fan Head
  • 44. Numericals N7:-If an existing raw mill old cyclones replace with new one and get the benefit of 30 mmg then how much electrical saving will be achieved in terms of SEC(kWh/ton of material)? Given Raw mill TPH = 300 TPH Electrical Power of Fan = 2000 kW Fan O/L pressure = -10 mmwg Fan inlet Pressure = -1000 mmwg Solution:- Step 1 Calculate Fan Head = ((-10 )-(-1000)) = 990 mmwg
  • 45. Numericals Step 2:- How to Calculate pressure drop saving in terms of Power ? = Pressure Drop X Electrical Power Fan Head = (30/990) X Electrical power = 60 kW SEC = 60/300 Saving = 0.20 kWh/ton of material
  • 46. Fan Curve Analysis Option-1 If actual Pressure is higher than the design pressure the operating point shifts towards left side- Volume reduces • Maximum Operating Pressure should be less than “Peak pressure” to avoid getting into surging Zone. Option-2 • Operating Pressure – Too Low • Operating point shifts to right- efficiency level goes down. • Design Efficiency is good – Operating Efficiency can be good by initiating actions. • Design Efficiency is Low- Operating Efficiency Low
  • 47. OPTION 3 • A Fan with no margin on Parameters having margin on Motor • Look for possibility of up gradation • B. Fan with no margin on Parameters & no margin on Motors Look for creating margin within the system ▪ Gas flow being function of process – look for reduction in system Pressure. ▪ Once margin is created on system pressure, we get margin on Motor
  • 48. GAS VELOCITY PROFILE • Fan Inlet & Outlet velocities should be compatible to System duct velocity(difference should not be greater than 8 m/s) • Higher is the difference between Fan Inlet/Outlet velocity…& Duct velocity Higher will the turbulence • Higher the turbulence…more will be the pressure drop..
  • 50. Solutions Given Q = 28651m3 /hr. Static pressure at fan inlet = -185 mmwg Static pressure at fan outlet = -8 mmwg Electric power = 30.60 kW Motor efficiency = 90% Step 1:- First calculate Fan head Fan head = Fan O/L-Fan I/L = ((-8 )-(-185)) = 177 mmwg
  • 51. Solutions Step 2:- Then, calculate Mechanical Power Q=m3 /hr; g=9.8 m/sec2 Mechanical power = QXFan headX9.8 /3600X1000 = 28651x177x9.8 /3600x1000 = 13.80 kW = 13.80/(30.60X0.90) = 50.10%
  • 52. How to Calculate Damper losses?
  • 53. Fan Controlled Techniques • Is it off the correct capacity/head? • How is the fan controlled • Damper throttling • Guide vane control • Speed control – GRR / SPRS / VFD/ON/OFF
  • 55. GRR Losses Calculations • Saving Calculations • Fan1 rated RPM = 980 • Fan1 operating RPM = 845 • Slip = 1-(operating RPM/rated RPM) Slip = 0.138 • GRR Losses = Slip*(1-(Slip*Slip))*Present power = 0.138*(1-(0.138*0.138))*1012 = 136.76 kW
  • 56. Fan Retrofication Concept • Retrofitting refers to the addition of new technology or features to older systems. • This is the up gradation of the Fan performance for improved capacity or reduction in specific power consumption. • During Retrofitting, look at the past performance of the fan and then retrofit with the high efficiency impeller into the existing fan casings to achieve highest performance. • Practically it is replacement of Impeller within existing casing, to suit the upgraded parameters with higher operating efficiency level
  • 57. Fan Retrofication Concept • Can the capacity/head be further fine tuned? • What is the efficiency of fan? • Can it be replaced by high efficiency fan • Is it possible to reduce system resistance – Suction / discharge side • Can fan operation be interlocked with process operation?
  • 58. ESP Exhaust fan-Case Study • Fan Designed for 300 TPD Kiln output • Fan Rating :-1,10,000 m3 /h, 200 mm WC, 110 kW • Fan Operating With DC drive • Present Operation – 450 TPD kiln output • Actual fan parameters -1,48,000 m3 /h, 110 mmwg
  • 59.
  • 61.
  • 62.
  • 63. Fan retrofication opportunities • Observations:- • Fan Rated for 78.5% Efficiency • Present Efficiency - Only 45% • Fan efficiency is deteriorated because mismatch between design head with operating head. • Design Head:- 200 mmwg • Operating Head :-110 mmwg
  • 64. Fan retrofication opportunities • Recommendations:- • Replace Fan with new Correct head fan & flow(Efficiency -80%) Specifications • Q = 150000 m3 /hr • Head = 120 mmwg • Efficiency = 80%
  • 65.
  • 66. Fan efficiency Concept –How to increase? • Optimum margin in head & flow • Higher efficiency • Operation close to best efficiency point • Suitable fan control system
  • 67. Conclusion –Fan retrofication • Assess the Process requirement • Analyze the Fan design data • Compare the actual performance data with design • Fan Power is a function of Volume & Pressure • Volume & Pressure- function of Process & system • So we must investigate the System & try to optimize to reduce Fan power. • Look even beyond Fan & identify potential.
  • 68. Considerations ❑Low Design & operating efficiency of the existing Fan ❑Up gradation of process system needs higher parameters High / Low Margin The scope should be limited to: a) Impeller b) Shaft c) Inlet cone 2. Existing Static parts, e.g. Casing /pedestal etc. to be utilized 3. No change in Foundation 4. No change in Bearings / Housing (inventory aspect) 5. Shaft may remain identical, as far as possible 6. Retrofit Impeller to be designed to utilize Existing Motor. 7. Cost effective 8. Minimum time to install the Retrofit Impeller shaft
  • 69. FAN MODIFICATIONS • Increase in cooler ESP Stack height • Reduction of suction pressure at cooler fan inlets through bell mouth • Suction box modifications of process fans by reducing velocity at fan inlet. • Rectify the cone gap of cooler fans • Reduce the pressure drop across the system through CFD Study.
  • 70. Cooler exhaust fan in a Cement plant
  • 71. Reducing SEC OF Cooler exhaust fan in a Cement plant by Increasing Stack height • When cooler upgradation • Exhaust height to be increased • From 23 m to 60 m • Additional draft - 65 mm • Net benefit - 50 mmwg • Power consumption reduced by nearly 50% • Saving potential 60 kW
  • 72. Reducing SEC OF Cooler exhaust fan in a Cement plant by Increasing Stack height
  • 73. Reducing SEC OF Cooler exhaust fan in a Cement plant by Increasing Stack height Calculations
  • 74. Suction side pressure drop in cooler fans by doing bell mouth • Extensive discussions held • Fan suppliers • Fan experts • Suction pressure drop • Not more than 15-20mmwg • Cooler fans invariably very high - 70-90 mmwg • Intentional
  • 75. Suction side pressure drop in cooler fans by doing bell mouth
  • 76. Suction side pressure drop in cooler fans by doing bell mouth • Typically flow measurement • using Piezometer ring • Additional pressure difference for accuracy • Measures to be taken • Reduce inlet pressure drop • Measure air flow • For reducing the inlet pressure drop • Inlet Air velocity - 10-12 m/sec • Reduction in entry loss – using bell mouth
  • 77. Suction side pressure drop in cooler fans by doing bell mouth
  • 78. Suction side pressure drop in cooler fans by doing bell mouth ❑Aerofoil in cooler fans Initial apprehension – dust Prone for wear due to dust ❑ Extensively practiced in power plants FD fans Inlet Pressure – not more than 20 mm WC Saving potential - 6-7%
  • 79. Suction side pressure drop in cooler fans by doing bell mouth ❑Aerofoil in cooler fans Initial apprehension – dust Prone for wear due to dust ❑ Extensively practiced in power plants FD fans Inlet Pressure – not more than 20 mm WC Saving potential - 6-7%
  • 81. Reduce the pressure drop in cement mill1 sep vent fan duct through CFD Study
  • 82. Path Lines Colored by Velocity Magnitude AS IS before duct modification (High Velocity Profile-Turbulence)
  • 83. Path Lines Colored by Velocity Magnitude after duct modification
  • 84. FAN INLET SUCTION BOX MODIFICATION
  • 85. FAN INLET SUCTION BOX MODIFICATION ❑ Difference between duct & suction box velocity is on higher side that creates excessive pressure drop. ❑ Increase the width of suction box in such a way that the difference between duct and fan inlet velocity could be minimized. ❑ Difference between the duct & Fan inlet velocity should be less than 7-8 m/s ❑ New Suction boxes should be designed by less than 30m/s ❑ Should not be greater than 35% of existing width(Limitation) Saving potential:-42 kW & 35 mmwg
  • 86. FAN INLET SUCTION BOX MODIFICATION
  • 87. IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION Different factors impacts the efficiency of a fan: ❑Build-up inside the impeller or between the impeller and the inlet cone ❑Wears between the impeller and the inlet cone ❑ Inappropriate gap between impeller and inlet cone
  • 88. IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION -STEP1
  • 89. IMPROVE FAN EFFICIENCY-STEP2 ▪ In any centrifugal fan the gaps between the rotating part and inlet; the overlap of the inlet to the wheel is crucial. ▪ Otherwise the internal recirculation of air/gas increases (from positive side outside the wheel to negative inside the wheel) and thus increases the wasted energy as well the fan’s capacity to move gas.
  • 90. IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION -STEP3-BUILDUPS Check all build-ups Between impeller and inlet cone ❑ This build up has a strong influence on fan efficiency. ❑It is common to be able to increase fan efficiency from 3 to 5 points with a thorough cleaning. ❑Clean manually with a small scraper around the inlet cone until you can see again the iron casing. Clean also the anti-rotating plate fixed on the inlet cone. They avoid turbulences and are very important.
  • 91. SOLUTION Inside the impeller A high pressure water jet can be used to remove build-up
  • 92. IMPROVE FAN EFFICIENCY BY INTERNAL INSPECTION -STEP4-CHECK WEAR
  • 93. Cone Gap of fan
  • 94. FAN CONE GAP NORMS
  • 96. TYPES OF FAN ENTRY
  • 97. AXIAL FAN • Axial-flow fans produce lower pressure than centrifugal fans, and exhibit a dip in pressure before reaching the peak pressure point • Applications-Cooling tower, Ventilation fans