15-Equipments efficiency calculation in power plant

Efficiency is the ratio of useful energy output to the total energy input  to the system or equipment

So, efficiency = Useful energy output X 100 / Total energy input.

This, energy can be Electrical, heat, pneumatic, hydraulic etc

1-Furnace efficiency:

The efficiency of a furnace is the ratio of useful output to heat input. The furnace efficiency can be determined by both direct and indirect method. The efficiency of the furnace can be computed by measuring the amount of fuel consumed per unit weight of material produced

Furnace efficiency = Heat released in furnace X 100 / Fuel energy supplied

Solved example:

Calculate the furnace efficiency of a Boiler which releases 18  Mcal/hr heat & consumes coal fuel 10 TPH, take fuel GCV as 2250 kcal/kg

ηfurnace = 18 X 1000000 X 100 / (10 X 1000 X 2250) = 80%

2-Boiler efficiency:

Boiler efficiency is calculated by two methods

1-Direct method:

A-Boiler feed water & attemperator water is at same temperature

Boiler efficiency in %=Steam flow X (Steam enthalpy –Feed water Enthalpy) X 100 / (Fuel GCV X Fuel consumption)

B- Boiler feed water & attemperator water is at different temperature

Boiler efficiency in %=( Steam flow X Steam enthalpy –Feed water flow X Feed water Enthalpy) X 100 / (Fuel GCV X Fuel consumption)

Note:

• Blow down water loss is not considered

• Steam used for soot blowers is not considered

• L1-Heat loss due to dry flue gas.

• L2-Heat loss due to moisture content in burning fuel.

• L3-Heat loss due to moisture content in combustion and spreading air.

• L4-Heat loss due to formation of water from hydrogen present in fuel.

• L5-Heat loss due to conversion of carbon into carbon monoxide.

• L6-Heat loss due to unburnt in bottom and fly ash.

• L7-Heat loss due to radiation

• L8-Heat loss due to convection & other un measurable

• L9-Heat loss due to soot blowing

• L10-Heat loss due to blow down.

• Losses in bearings

• Losses in oil seals

• Losses in Gears

• Losses in lubrication due to churning effect.

Power generation cost calculation...

Examples:

Calculate the efficiency of 100 TPH Boiler operating at 88 kg/cm2G pressure & temperature 515 deg C, consumes 17 TPH coal whose GCV is 5000 kcal /kg & is supplied with feed water at temperature 165 deg C. Assume no blow down loss.

Solution:

Steam Flow Qs = 100 TPH

Steam enthalpy at above operating parameters (refer steam table) Hs = 817.7 kcal/kg

Feed water enthalpy at temperature 165 deg C, Hf = 166.5 dkcal/kg

Coal consumption Mc = 17 TPH

Coal GCV = 5000 kcal/kg

Now ηboiler = Qs X (Hs-Hf) X 100 / (5400 X Mc)

                   = 100 X  (817.7-166.5) X 100 / (5400 X 17) = 70.93%

In this method water required for attemperating is more

Calculate the efficiency of 100 TPH Boiler operating at 88 kg/cm2G pressure & temperature 515 deg C, consumes 17 TPH coal whose GCV is 5000 kcal /kg & is supplied with feed water 92 TPH at temperature 165 deg C & 10 TPH water for attemperating at temperature 105 Deg C.

Solution:

Steam Flow Qs = 92 TPH

Steam enthalpy at above operating parameters (refer steam table) Hs = 817.7 kcal/kg

Feed water flow Qf = 92 TPH

Feed water enthalpy at temperature 165 deg C, Hf = 166.5 dkcal/kg

Attemperator water flow Qa = 10 TPH

Attemperator water enthalpy at temperature 105 deg C Ha = 106 kcal/kg

Coal consumption Mc = 17.5 TPH

Coal GCV = 5000 kcal/kg

Now Boiler efficiency = (Qs X Hs-(Qf X Hf + Qa X Ha)) X 100 / (5400 X Mc)

   ηboiler= 100 X  817.7-(166.5 X 92 + 10 X 106) X 100 / (5400 X 17.5) = 69.19%

In this method water required for attemperating is less.

2-Indirect method

This method is also called as heat loss method. In this total heat losses in the Boilers is subtracted from a number 100.

This method gives exact efficiency of Boiler. Small errors in readings will not lead to much difference. However it needs more data to calculate the efficiency

Boiler efficiency = 100-Losses

Losses in Boilers are

Note: Boiler efficiency calculation does not include losses L9 & L10

In Bagasse based power plants, heat loss due to moisture is more whereas in Coal based power plants heat loss due to dry flue gas is more

It's all about HP heaters

Example 

A boiler generates steam 80 TPH at 66 kg/cm2 and 485 °C. Mesured O2, CO and CO2 in flue gas are 8%, 850 ppm and 12% respectively. Ash analysis shows unburnt in fly ash and bottom ash are 10.5% and 3% respectively, GCV of fly ash and bottom ash are 695 kcal/kg and 1010 kcal/kg respectively. Coal analysis shows carbon 50%, Hydrogen 3.2%, Oxygen 8.2%, Sulphur 0.4%, Nitrogen 1.1%, Ash 19% and moisture 18.1 and its GCV is 4100 kcal/kg. Then calculate the Boiler efficiency. Consider ambient air, flue gas out let temperature are 30 and 150 °C respectively, humidity in ambient air is 0.02 kg/kg of dry air.

From the given data, boiler efficiency can be calculated from indirect method.

We have, Theoretical air requirement = (11.6 X %C + 34.8 3 (%H2 - %O2/8) + 4.35 X %S)/100…Kg/kg of fuel

Therefore, Th air requirement will be = (11.6 X 50 + 34.8 X (3.2 - 8.2/8) + 4.35 X 0.4)/100

                                                 = 6.57 kg of air/kg of coal.

Given that,O2 in flue gas is 5%

We have, Excess air = (O2%/(21 - O2%)) X 100

                                 = (5/(21 - 8)) X3 100 = 38.5%

Total air supplied = (1 + EA/100) X Th air

                              = (1 + 38.5/100) X 6.57

                              = 5 9.1 Kg/Kg of Coal

Actual mass of dry flue gas generated during combustion is,

Mass of CO2 in flue gas + Mass of N2 in flue gas + Mass of N2 in combustion air + Mass of O2 in flue gas + Mass of SO2 in flue gas.

= ((Carbon in fuel X MW of CO2)/MW of carbon) + Mass of N2 in fuel + (Total air X N2 in air/100) + ((Total air - Th air) X 23/100) +((SO2 in fuel X MW)/MW of Sulphur)

= (0.5 X 44/12) + 0.011 + ((9.1 X 77)/100) + ((9.1 - 6.57) X 23/100) + (0.004 3X 64)/32

  Mass of dry flue gas Mg is 9.44 kg/kg of coal.

Where Molecular weight of CO2, Carbon, SO2 and Sulphur are 44, 12, 64 and 32 respectively.

To find out the boiler efficiency need to calculate all the different losses

L1 = % of heat loss due to dry flue gas

= Mg X Cp X (Tf - Ta)/GCV of fuel

= 9.44 X 0.24 X (150 - 30) X 100/4100

L1 = 6.63%

L2= Heat loss due to moisture in fuel

        L2 = M X (584 + Cp X (Tf - Ta) X 100)/GCV of fuel

             = (0.181 X (584 + 0.45 X (150 - 30)) X 100)/4100

         L2= 2.81%

L3=Heat loss due to formation of water from Hydrogen present in fuel

L3 = (9 X H2 X (584 + Cp X (Tf - Ta))) X 100/GCV of fuel.

L3 = (9 X 0.032 X (584 + 0.45 X (150 - 30) X 100)/4100)

L3 = 4.48%

L4= Heat loss due to moisture in air

L4 = (Total air X humidity X Cp X (Tf - Ta) X 100)/GCV of fuel

L4 = (9.1 X 0.02 X 0.45 X (150 - 30)) X 100/4100

L4 = 0.24%

L5= Heat loss due to partial conversion of Carbon to Carbon Monoxide

L5 = (((%CO X C)/(%CO + %CO2)) X (5654/GCV of fuel)) X 100

L5 = (((0.0850 X 0.5)/(0.085 + 12)) X (5654/4100)) X 100… Converted CO ppm to % (% 5 ppm/10000)

L5 = 0.48%

L6= Heat loss due to radiation and convection are considered 1–2%, it depends on age and insulation of the boilers.

L7= Heat loss due to unburnt in fly ash

% of Ash in coal = 19%

Unburnt in fly ash = 10.5%

GCV of fly ash = 695 kcal/kg

Amount of fly ash in 1 kg of coal = 0.105 X 0.19

                                                      = 0.012 kg/kg of coal

Heat loss due to unburnt = 0.012 X 695

                                          =5 8.34 kcal/kg

% of Heat loss due to unburnt in fly ash L7 = 8.34 X 100/4100

L7 = 0.2%

L8= Heat loss due to unburnt in bottom Ash

% of Ash in coal = 19%

Unburnt in bottom ash = 3%

GCV of bottom ash = 1010 kcal/kg

Amount of bottom Ash in 1 kg of coal = 0.03 X 0.19

 0.0057 kg/kg of coal

Heat loss due to unburnt = 0.0057 X 1010

= 5.75 kcal/kg of coal

% of Heat loss due to unburnt in fly ash = 5.75 X 100/4100

L8 = 0.14%

So Boiler efficiency is 100 - (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8)

= 100 - (6.63 + 2.81 + 4.48 + 0.24 + 0.48 + 1 + 0.2 + 0.14)

  ηboiler = 84.02%

Factors considered for Boiler Engineering/Design

Specific terms used in power plant

3-Economiser Effectiveness/Efficiency Economiser effectiveness is calculated as

ηEco. = (Economiser outlet feed water temperature Two-Economiser inlet feed water temperature Twi)  X 100 / (Economiser inlet flue gas temperature Tfi- Economiser inlet feed water temperature Twi)

Example:

 Calculate the economiser effectiveness, whose feed water inlet & outlet temperatures are 200 Deg C & 290 Deg C respectively & flue gas inlet & outlet temperatures 400 deg C & 230 deg c respectively.

Solution:

Twi = 200 deg C

Two = 290 deg C

Tfi = 400 deg C

Tfo = 230 deg C

ηEco = (Two-Twi) X 100 / (Tfi-Twi)

ηEco = (290-200 ) X 100 / (400-200)

ηEco= 45%

4-Air Preheater (APH) Effectiveness/Efficiency

APH effectiveness is calculated on gas side & air side.

APH gas side efficiency

ηAPHg = (Flue gas inlet temp.Tfi-Flue gas outlet temp.Tfo) X 100 / (Flue gas inlet temperature tfi-Air inlet temperature Tai)

APH air side efficiency

ηAPHa = (Air outlet temp.Tao-Air inlet temp.Tao)) X 100 / (Flue gas inlet temperature tfi-Air inlet temperature Tai)

Example:

A tubular APH has air inlet & outlet temperatures are 25 deg c & 185 deg C & flue gas inlet & outlet temperatures are 230 deg C & 145 deg C. Calculate the APH effectiveness on Gas side & Air side

Solution

Given that

Tai = 25 deg C

Tao = 185 deg C

Tfi = 230 deg c

Tfo = 145 deg C

APH gas side efficiency

ηAPHg =(Tfi-Tfo) X 100 / (Tfi-Tai)

ηAPHg = (230-145) X 100 / (145-25) = 70.83%

APH air side efficiency

ηAPHa = (Tao-Tai) X 100 / (Tfi-Tai)

ηAPHa = (185-25) X 100 / (230-25) = 78.04%

5-ESP efficiency:

ESP efficiency is calculated as

Efficiency of ESP ηESP = 1-eˆ(-AV/Q) X 100

Where,

A = Surfacing area of the collecting plate in M2

V = Migration velocity of the particle in m/sec.

Q = Volume flow rate of flue gas in m3/sec.

Example:

An ESP handles total flue gas at the rate of 80 m3 /sec., it has total collecting surface area 5890 m2, calculate the efficiency of ESP if ash particles migration velocity is 0.077 m/sec.

Solution:

Given that,

A = 5890 M2

V = 0.075 m/sec.

Q = 80 m3/sec.

Efficiency of ESP = 1–eˆ (-AV/Q) X 100

                            = 1– eˆ (-5890 X 0.077/80) X 100

                   η_ESP = 99.98%

6-HP heater effectiveness

It is calculated as temperature range of steam / Temperature range of feed water

Example:

A HP heater is been used to raise the feed water temperature from 125 deg C to 145 deg C by using Turbine bleed steam at inlet temperature 325 deg C, calculate the HP heater effectiveness. Consider the HP heater condensate out let temperature is 165 deg C

Solution

Twi = 125 deg C

Two = 145 deg C

Tsi = 325 deg C

Tco =165 deg C

 HP heater effectiveness =( Tsi-Tco) / (Two-Twi) = (325-165) / (145-125)  =8

7-Deaerator (D/A) efficiency:

The main purpose of the Deaerator is to remove the dissolved gases in boiler feed water mainly oxygen. So its efficiency is calculated based on its capacity to remove O2 from the feed water.

       η _D/A = (Concentration of Oxygen in inlet water(Ci)-Concentration of oxygen in outlet water (Co)) X 100 /(Concentration of Oxygen in inlet water(Ci))

Example:

Calculate the efficiency of Deaerator if inlet & outlet oxygen concentrations of D/A are 20 ppm & 0.005 ppm respectively.

   η D/A = (Ci-C0) X 100 / Ci

   η D/A = (20-0.005) X 100 / 20 = 99.97%

8-Turbine efficiency:

Overall efficiency

Turbine overall efficiency is calculated as the ratio of power out put from the turbine to the heat input to the Turbine.

   η_Turbine = Power generation in kcal X 100/  Heat input in Kcal

Example: Calculate the overall efficiency of a 5 MW back pressure turbines, operating at 67 kg/cm2 pressure & 495 deg C temperature. Consider specific steam consumption (SSC) of the Turbine is 7.5 & steam is exhausted at pressure 1.8 kg/cm2 & temperature 180 deg C

Solution:

Given that

Power generation = 5 MW

Convert it into kcal, we have 1 KW = 860 kcal

Therefore 5 X 1000 X 860 = 4300000 kcal

Steam inlet enthalpy at operating pressure & temperatures is Hi =813 kcal/kg

Exhaust Steam enthalpy Ho =675 kcal/kg

Steam flow at Turbine inlet Qs = Power generation X SSC = 5 X 7.5 = 37.5 TPH

ηTurbine = Power generation in kcal X 100/  Heat input in Kcal

ηTurbine = Qs X (Hi-Ho) = 37.5 X 1000 X (813-675) = 5175000

ηTurbine = 4300000 X 100 / 5175000 = 83.1%

Turbine cycle efficiency can be calculated as

ηTurbine = 860 X 100 / Turbine heat rate

Reasons for increase in Turbine specific steam consumption (SSC)

Example-2: Calculate the cycle efficiency of 55 MW Turbine operating at 110 Kg/cm2 pressure & 540 degree C temperature. Consider feed water temperature 210 deg C & SSC 3.8

Solution:

Enthalpy of inlet steam at operating parameters (Refer steam table) Hs = 827 kcal/kg

Enthalpy of feed water = 215 kcal/kg

Turbine heat rate = Steam flow Qs X (Steam enthalpy Hs-Feed water enthalpy Hw) / Power generation

 Steam flow Qs = Power generation X SSC = 55 X 3.8 =209 TPH

THR = 209 X (827-215) / 55 =2325.6 kcal/kwh

Turbine efficiency = 860 X 100 / THR

ηTurbine = 860 X 100 / 2325.6 =36.97%

9-Power plant efficiency:

Again power plant efficiency is calculated based on heat output & heat input

Power plant efficiency = 860 X 100 / Heat rate

Example:

Calculate the efficiency of 100 MW power plant which consumes 65 TPH of coal having GCV 5200 kcal/kg.

Solution:

First calculate the plant gross heat rate (PGHR),

PGHR = Fuel consumption X GCV / Power generation

PGHR = 65 X 5200 / 100 = 3380 kcal/kwh

Ηplant = 860 X 100 /3380 = 25.44%

Note: Heat rate of cogeneration power plants is calculated as

Cogen heat rate (CHR)=((Fuel consumption X GCV + Heat content in return condensate + Heat content in makeup water - Sum of heat content in process steam))/Power generation.

10-Condenser efficiency:

It is given as

η_condenser = Actual cooling water temperature rise X 100 / Maximum possible cooling water temperature rise

ηcondenser = (To-Ti) X 100 /(Ts-Ti)

To = Cooling water outlet temperature in deg C

Ti = Cooling water inlet temperature in deg C

Ts = Saturation temperature at exhaust in deg C

Example

A down flow type surface condenser has vacuum -0.85 kg/cm2 condenses 85 TPH steam at cooling water inlet and outlet temperatures 25 °C and 36 °C respectively, calculate the condenser efficiency.

Solution:

Ti = 25 deg C

To = 36 deg C

Ts at pressure -0.85 kg/cm2 = 58 deg C

ηcondenser = (To-Ti) X 100 /(Ts-Ti)

ηcondenser = (36-25) X 100 / (58-25) = 33.33%

11-Vacuum efficiency:

It is the ratio of actual vacuum in the condenser to the maximum possible vacuum that can be achieved.

Actual it is not possible to create 100% vacuum in any system

     η_vacuum= Actual vacuum in condenser X 100/Maximum Obtainable vacuum in the condenser

Example:

Exhaust steam from condenser enters at 47 °C, if the vacuum gauge of condenser reads -0.86 kg/cm2, find the vacuum efficiency.

Solution

Given that,

Condenser pressure = -0.86 kg/ cm2

So exhaust steam temperature = 47 °C

From steam tables, partial pressure of steam at exhaust temperature Ps =0.105 kg/cm2

Maximum obtainable vacuum by considering atmospheric pressure as 1.033 kg/cm2

= 1.033 - 0.105= 0.93 kg/cm2

Vacuum efficiency = (Actual vacuum in condenser X 100)/Max. obtainable vacuum.

         η_vacuum= 0.86 X 100/0.93 = 92.5%

12-Gear box efficiency:

Gear box efficiency is the ratio of out power to the input power

                   η_Gearbox= Output power X 100 / Input power

Gear box efficiency cannot be 100%, there is always losses in terms of friction.

Some potential losses in gear box are

Example:

A helical gear box is used to drive a fuel feeding system, the input power of the gear box is 9.5 KW & output power is 8.7 KW, calculate GB efficiency

η_Gearbox = Output power X 100 / Input power =8.7 X 100 / 9.5 =91.5%

13-Pump efficiency:

Pump efficiency is the ratio of to the pump hydraulic power to the Pump shaft power.

Example:

A pump is consuming 20 KW to deliver 72 M3/hr of water at height 55 meter, calculate its efficiency.

Pump hydraulic power Ph = Flow in m3/sec X Total head X 9.81 X water density / 1000 

Ph = (72/3600) X 55 X 9.81 X 1000 /1000

Ph =10.79 KW

ηpump = Hydraulic power X 100 / Shaft power

ηpump = 10.79 X 100 / 200 = 53.95%

14-Cooling tower (CT) efficiency:

η_{CT = (CT inlet water temperature Ti-CT outet water temperature To) X 100 /(CT outet water temperature To-WBT)}

η_{CT =(Ti-T0) X 100 / (To-WBT)}

CT efficiency can also be written as

η_{CT = Range X 100 /(Range +Approach)}

_{Where Range is temperature difference between CT inlet & outlet water}

_{Approach is the temperature difference between CT outlet water & wet bulb temperature (WBT)}

15-Fans efficiency:
Fans efficiency can be Mechanical efficiency or Static efficeincy.These are calculated based on static pressure & total pressure.

Static efficiency of the fan ηsfan= (Air flow in M3/sec X Static pressure in mmwc X 100) / (102 X Input power to fan shaft in KW)

Similarly mechanical efficiency can be calculated as

Mechanical efficiency of the fan ηfan= (Air flow in M3/sec X Total pressure in mmwc X 100) / (102 X Input power to fan shaft in KW)

Example:

A boiler ID fan consumes 220 KW power to sucky 60 m3/sec flue gas at static pressure 280 mmwc, calculate its static efficiency.

Solution:

Ps = 220 KW

Q = 60 m3/sec

Static pressure Hs = 280 mmwc

Static efficiency of the fan ηsfan= Q X Hs X 100 / 102 X Ps

                                         ηsfan = 60 X 280 X 100 / (102 X 220) =74.8%

Also read efficiency & Heat rate calculation of power plants

Heat rate & Efficiency of power plants

Also read 16-Perfect reasons for increasing the fuel consumption of Boilers

Questions & Answers on AFBC Boilers

Power plant SOPs

Best practices to reduce the Auxiliary power consumption (APC) in Sugar based Cogeneration plants.

Auxiliary power consumption is directly related with power plant profit. Lesser the APC more will be the power export & hence more profit. So it becomes duty of every power plant professions to strive to reduce APC wherever possible.

Below table gives the area wise APC in Bagasse/coal based power plants

Sl No.	Area	APC in %
1	Boiler fans	35-38%
2	Boiler feed water pumps	35-36%
3	Turbo generator & its auxiliaries	10-12%
4	Bagasse feeding system	2-2.5%
5	Coal feeding system	1%
6	Bagasse handling system	4-4.5%
7	Coal handling system	1.5- 2%
8	Ventilation, AC & Air Compressors	4-5%
9	Water treatment plant	2-3%

Note: Bagasse based Cogeneration plant have incorporated coal handling system as supporting system & also run the power plant in non-season days. So APC for bagasse handling/feeding & coal handling/feeding has been considered separately.

Now a day there is huge scope for reduction of plant APC. Here scope of APC reduction at various areas of plant has been discussed.

A-Boiler:

1-Improve the combustion efficiency:

Combustion efficiency improvement will directly relate to the air consumption. Improved combustion will reduce load on FD, SA & ID fans, hence power consumption will reduce. In cogeneration plants or ion any power plants Boiler fans consume around 35-38% of total auxiliary power consumption.

In order to reduce the air requirement & to improve the combustion efficiency need to concentrate on bellow areas

• Select good quality of fuel, if using bagasse the moisture should be 49 to 50%, as the moisture increases excess air required will also increases. This loads the FD, SA & ID fans more than requirement.
• Arrest all leakages in air & flue gas paths
• Modify combustion air ducts to reduce resistance to air flow

16-Perfect reasons for increasing the fuel consumption of Boilers

Opportunities for energy conservation in power plant...

2-Boiler fans:

As discussed above Boiler fans consume 35-38% of total APC, so need to concentrate on Boiler fans efficient operation.

• Operate boiler fans in VFD mode at optimum speed
• Incorporate inlet guide vanes to the system
• For high speed fans, ensure the silence fitted at suction size has sufficient opening & at lesser height
• Ensure all inspection & man way doors are sealed properly
• Ensure clearance between inlet cone & impeller suction neck is minimum
• Monitor draught losses in flue gas duct, air ducting, APH, Economiser & ESP regularly
•  Follow lubrication schedule, replace damaged bearings to reduce vibrations & bearing temperature. High vibration bearings consume more power

3-Boiler feed water pumps (BFP)

Boiler feed water consume around 35% of total APC of the plant. Hence it is utmost important to reduce the BFP power consumption. Attention must be given on following points to reduce BFP power consumption.

• Select optimum capacity BFPs, Under load running pumps have low efficiency
• Better to install HT drives for BFP
• Incorporate VFD to BFPs
• Operate BFP in Auto mode based on discharge header pressure & drum pressure
• Do not operate the pump with discharge valve throttled
• Ensure BFP ARC valve has no leakage
• Ensure BFP has sufficient suction pressure
• Maintain Deaerator level on higher set point side.
• Replace worn out balance & counter balance discs
• Schedule pumps servicing as per OEM guidelines. And replace impeller wear rings or impellers if clearance found more. More clearance between wear ring & impeller will lead to higher power consumption
• Follow lubrication schedule, replace damaged bearings to reduce vibrations & bearing temperature.
• Do not uses BFP discharge water for de-super heating of LP process steam, instead use CEP discharge water
• Clean suction strainers regularly

4-ESP

• Implement hopper heater automation by thermostat
• Optimize the charge ratio of Transformer

Questions & Answers on ESP troubleshooting

B-Fuel feeding system:

Fuel feeding system consumes around 2-4% of total APC. Hence it need to give attention to reduce APC in this area. Following are the some listed action points to reduce APC.

• Incorporate VFDs to all fuel feeding systems, as fuel feeding system never run on 100% load.
• Select planetary type gear boxes to fuel feeding system, as planetary gear boxes found of higher efficiency
• Follow regular preventive maintenance to reduce wear, tear & friction of rotating parts, which ultimately lead to more power consumption
• Replace all loose V belts. Loose V belts lead to increased power consumption

C-Turbo generator & Auxiliary:

Turbine auxiliary consume around 10-12% of total APC of the plant. Following are the areas where we can monitor the APC

• Maintain lube & control oil filters clean
• Maintain optimum lube oil temperature
• Maintain cooling tower fills & drift eliminators clean to get required cooling water temperature
• Incorporate VFDs to all cooling water pumps
• Set CT blade angle at 11-14 degree.
• Replace CT fan aluminium blades by FRP
• Ensure cooling tower surrounding area is free of trees & structures for free flow of air
• Incorporate VFDs to Condensate extraction pumps (CEP)
• Ensure hot well make up is going on from gravity water from feed tank or surge tank
• Schedule regular cleaning of heat exchangers like steam condenser, oil cooler, ejectors & Generator air cooler & gland steam condensers
• Replace TG building exhaust fans by turbo ventilators

D-Fuel handling (Coal & Bagasse handling):

• Ensure motors selected for all Bagasse belt conveyors are of optimum rating
• Incorporate planetary gear boxes to belt & chain conveyors
• Replace all damaged idlers regularly. Damaged idlers lead to more friction in the system
• Ensure Vertical gravity take up height is optimum (1 to 1.5% of conveyor length)
• Ensure belt scrapers (cleaners) & skirt boards are not over rubbing the belt to avoid friction & wear/tear
• Follow regular preventive maintenance & lubrication schedule for conveyor pulleys bearings & idlers
• Remove saturated bagasse/coal from deck plates of conveyors regularly to avoid friction

It's all about HP heaters

E-Compressors/blowers & Ventilation/air conditioning system

For air compressors optimise discharge air pressure & set loading & unloading times as per requirement

• Clean air filters regularly
• Replace all globe valves by gate valves or ball valves
• Ensure compressor discharge air line is of correct size to avoid pressure drop in the line
• For pneumatic ash handling system operate ash handling plant in probe mode or else optimize cycle time & conveying time
• Incorporate VFDs to ventilation system
• Optimize the usage of Air conditioning system in offices, meeting halls etc
• Avoid compressed air for cleaning applications

F-Water treatment plant

Thumb rules water treatment plant

• Ensure enough capacity of tanks for DM water storage, so that the operation time of the plant can be reduced
• Plan to get more quantity of return condensate from process
• Optimize boiler & CT blow down to reduce pumping power
• Incorporate drain condensate recovery system to reduce pumping power as well as to conserve thermal energy
• Operate DM plant at its full capacity & carryout regeneration as per out put OBR given by OEM to reduce pumping power as well as water consumption
• Use treated N-pit water as  service water

Other:

1-Motor

• Provide proper ventilation to the motors. For every 10 °C increase in motor operating temperatures over recommended peak, the motor life is estimated to be halved.
•  
• Synchronous motors are more suitable to improve power factor.
•  
• Balance the three phase power supply, an unbalanced voltage can increase motor input power by 3–5%.
•  
• Ensure the motor proper rewinding, an improper rewinding could lead to efficiency reduction.
• Ensure proper alignment between motor and load ends (fans, pump, gear box, blower etc.) to avoid more power consumption and failures.

2-Lighting

Incorporate timers for plant lighting systems like conveyors, street, bagasse & coal yards etc

Replace all CFL & incandescent bulbs by LED bulbs

Read POWER PLANT CALCULATIOS

50-Most frequently asked Questions Answers on Bearings

1-Define the term Bearing

A bearing is a machine element which reduces friction between two members, having relative motion with respect to each other. It’s a device used to support and guide rotating, oscillating or sliding shaft.

2-What are the functions of bearings?

• Supports & guides the rotating parts
• Reduces the friction & hence no or less wear & tear in moving parts
• Reduces the noise
• Reduces the lubricants consumption
• Reduces the power consumption of an equipment where it is being used

Its all about HP heaters

3-Where do you find the application of bearings?

Bearings found applications in almost all type of industries from a small trimmer to big & big Machines.

4-What type of loads act on bearings?

Bearings can take in both axial, radial directions with small misalignment

What are the main types of bearings used in industries?

Main types of bearings are Journal bearing & Rolling contact bearings

5-Briefly explain the Journal bearings used in Industries

Journal or plain bearings consist of a shaft or journal which rotates freely in a supporting metal sleeve or shell. There are no rolling elements in these bearings.

It is a cylindrical bush, add up of suitable material and containing properly machined ID and ODs.It is a part of shaft or pin that rotates inside the bearing. They handle high load and velocities because metal to metal contact is minimal due to oil film. Operation is smoother. They require large supply of lubricating oil. For high speed need forced cooling/lubrication there may be possibilities of failures of bearings in start up and shutdown.

6-Where do you find the applications of Journal bearings?

Journal bearings used High speed & high load machines.

In power plant journal bearings found application in;

• Steam Turbine
• Boiler feed pumps
• Slat chain conveyors

7-What are the two types of Journal bearings?

• Oil lubricated: Used for high speed & high load carrying machines
• Grease lubricated journal bearings: Used for High load & low speed machines

8-What are the different materials used for the manufacturing of Journal bearings?

Materials used are

Copper & its alloys, generally Gun metal, bronze

White metal & Babbitt metal: A tin base alloy containing 88% of tin, 8% of antimony and 4% of copper & bismuth

9-Why do you select the copper based alloys for journal bearings?

Because they have low coefficient of friction

10-What is the maximum operating temperature of Journal bearings?

Journal bearings can be operated up to the temperature 100-105 Deg C

11-Classify the rolling contact bearings.

A. Ball Bearings:

• Deep groove ball bearing
• Angular contact ball bearings
• Self-aligning ball bearing
• Thrust ball bearing

B. Roller Bearings:

• Spherical roller bearing
• Cylindrical roller bearing
• Taper roller bearing

C. Thrust Roller and Needle Bearings

12-What are the parts of rolling contact bearings?

Parts of Bearings:

• Outer race
• Inner race
• Cage
• Rolling elements

13-What is the material of composition of rolling contact bearings?

It is hardened Chromium steel

14-What is the hardness of bearing materials/parts?

Hardness is up to 55-60 HRC

15-How do you specify the Rolling contact bearings?

Rolling contact bearings are specified as;

• Bearing bore size
• Bearing outer diameter
• Bearing width
• Bearing cage type & material
• Bearing clearance

16-What do you mean by Bearings Prefixes & Suffixes? & Explain some of bearing suffixes

• Prefixes are mainly used to identify bearing rolling component. 
• Suffixes identify special designs, variants & characteristics, which differ in some way from the original design or from the current basic design

Bearing suffix:

• Z: Steel shield on one side of bearing
• 2Z: Steel shields on both the sides of bearing
• ZZ: Rubber shields on both the sides of bearing
• K: Tapered bearing bore in the ratio of 1:12
• W33: Lubricating groove and 3 holes on outer race
• W33X: Lubricating groove and 6 holes on outer race
• C3: Normal bearing clearance (Clearance more than C2)
• C4: More bearing clearance (Clearance more than C3)
• J: Pressed steel cage
• F: Machined steel cage
• M/Y: Machined brass cage/Pressed brass cage
• NR: Bearing with snap ring

Power plant Maintenance Calculations

17-What is the maximum operating temperature of Rolling contact bearings

It is 85 to 90 degree C for grease lubricated bearings

18-What are the various loads considered in bearings designing?

Basic static loads, involves mainly Dynamic (C) & Static loads (Co)

19-When the static load occurs on bearings?

Static load occurs on following conditions

• When bearings are under load & stationary for long time
• When bearings rotate < 10 RPM
• When bearings are performing slow oscillating movements

20-How do you identify the ball bearings, taper roller bearings, Spherical roller bearings & Angular contact bearings?

Bearings can be identified based on their starting nos

• Ball bearings Nos start with 1 & 6
• Spherical roller bearings 2
• Angular contact bearings 7
• Taper roller bearings 3
• Cylindrical roller bearings start with NU

21-How do you nomenclature the bearing 6205 2Z

Nomenclature:

• 6 indicates the type of bearing
• 2 Indicates OD & Width sizes of the bearings
• 05 indicates the bore size of the bearing
• 2Z is a Suffix, bearing having metal shields on its both sides

22-What is the difference between the bearing No.6205 & 6305

6205 has lesser OD & width than 6305 bearing, however both bearings have same bore

23-Write an example of bearing number for self aligning ball bearings

1219, 2207 etc

24-Write an example of Spherical roller bearing?

22220 EK/C3, 23215 EK/W33

25-Write an example of Angular contact ball bearing

7205, 7305

26-Write an example for taper roller bearing

30305, 31205

Note: Suffixes can be added as per specific requirement

27-Write an example for cylindrical roller bearing

NU 203 ECP, NU 2204

28-Calculate the bore diameter of Deep groove ball bearing 6208 C3

Bearing bore size is calculated as 08 X 5 = 40 mm

Similarly for bearings 6315 2Z, Bore size = 15 X 5 = 75 mm

29-Calculate the shaft size for a bearing 22222 K/C3 having tapered bore & sleeve thickness 5mm

In tapered bore spherical roller bearings 22222 K/C3,

Shaft size = (Last two digits) 22 X 5 -2 X Sleeve thickness =110 -2 X 5 = 100 mm

30-Why it is necessary to maintain minimum load on Bearings?

It has been learned from experience that bearings require a minimum applied load to insure traction for the rolling elements so they roll as the shaft starts to rotate. If the balls or rollers do not roll, they will skid on the moving raceway, wiping away the lubricating oil, and causing damage to the rolling element O.D.s and raceway surfaces. This is called skidding and the resultant damage is referred to as smearing, which will shorten bearing life.

A good approximation of the minimum load for each is:

Pmin = 0.02 x C

where:

Pmin = required minimum equivalent load on the bearing, radial load for radial bearings and thrust load for thrust bearings.

C = Bearing Dynamic Capacity

31-Why the bearings operating at higher temperature are having lower life?

Bearing Dynamic and Static Capacities will reduce at high operating temperatures. The main reason is the reduction of raceway and rolling element hardness at high temperatures.

32-What are the potential reasons for bearings failure?

Following are the main reasons for bearing failure

Why do vibrations occur in machines???

• Bearings overloading
• Wrong or miss application of the bearings
• Misalignment
• Lack of lubrication or improper lubricant
• Contaminants in lubrication

Read Why do bearings fail???

• Over lubrication
• Operating bearing at higher temperature & vibration for long time
• More or less bearing clearance
• Jerk or fatigue loads
• Improper methods of bearing installation
• Bearing manufacturing defect

Know about valves.

33-List down the reasons for bearing seize

• Lack of lubrication
• Less clearance
• Over speed
• Contaminant in lubricant

34-List down the steps for installation/mounting of new bearing

• Ensure all the tools are in hand before installing the bearing.
• Clean the shaft and bearing.
• Measure and record the values for correct size, roundness and surface roughness.
• Unpack the bearing.
• For special purpose lubrication, clean the bearing with low pressure jet of kerosene.
• After cleaning apply the correct method of fitting Press fit, adapter sleeve, withdrawal sleeve and thermal expansion type

Note:

• For ball bearings, bearings need to apply the heat up to 90-100 Deg on induction bearing heater for easy installation.
• Do not use oil bath for bearing heater
• Do not cross the bearing or oil temperature above 100 deg C
• For fitting the bearing use gun metal or nylon rod & hammering should be done on inner race only.

35-List down the procedure for bearings dismounting.

• Select the proper tool and person
• Decide the method of removal
• Use special wrenches, if required.
• Use puller, Induction heater and Oil injection method.
• Apply force to inner ring: If removing from shaft and to outer race if removing from housing.

Note:

• Apply load on outer race for bearing removing, do not apply load on inner race
• If required heating, plan to apply the heat on bearing only not on shaft. May wrap wetted cloth on shaft to avoid shaft expansion

IBR forms & Acts

36-List down the Do’s and Don’ts for bearings handling and storing.

Following are the Do’s and Don’ts for bearing:

DO’s

• Keep the bearings in protective packing till they are mounted.
• Cover the assembled bearings appropriately to avoid dust ingress.
• Store the bearings in dry area.
• Use tubes, Puller, Hydraulic nuts, and other tools for removal/fixing of the bearings.
• Use correct amount of specified lubrication at right time with correct procedure.

Don’ts

• Over lubrication is a silent killer for bearings. Do not over lubricate.
• Never allow welding at bearings that will damage the contact surfaces.
• Don’t let the bearing stand upright, store them flat on their sides.
• Do not use water to clean the bearings.

37-What do you mean by the term bearing clearance?

Bearing clearance is the clearance or gap between outer race or inner race and rolling elements. And in journal bearings it is the clearance between shaft and bearing liners.

38-How do you calculate the bearing clearance of bearing having bore diameter D?

As a thumb rule Bearing clearance minimum = 0.00185 X D

Bearing clearance maximum = 0.00254 X D

39- What do you mean by basic rating life of a Bearing?

 Basic rating life of a bearing is the life that corresponds to 90% reliability using standard materials & manufacturing quality at normal operating conditions.

It is given by

L10 = (C/P)^e

Where C is bearing dynamic load in KN and P is dynamic equivalent load (KN) in axial or radial directions.

e = Exponent (e = 3 for ball and 10/3 for roller bearings).

40-What is meant by bearing life and how do you calculate it?

Bearing life is the number of revolutions that 90% group of bearings can run without causing flaking due to rolling fatigue.

Bearing life L10 = (C/P)^e X 10⁶/60 N

41-What factors do affect the bearings life?

Following factors affect on bearings life

• Operating speed
• Operating environment
• Load on bearing
• Type of lubricant

42-What is the misalignment tolerance for ball bearings?

Deep groove ball bearing : 2 Minutes of arc (1 Degree=60 minutes of arc)

Self aligning ball bearing

• 2.5 Deg-SR1200
• 3 Deg-SR1300
• 2.5 Deg-SR2200
• 1.5  Deg-SR 2200 2RS1
• 3 deg-SR 2300
• 1.5 Deg-SR 2300 2RS1

Angular contact ball bearing: No much misalignment tolerance is allowed

43-What is the misalignment tolerance for spherical roller bearings?

• Spherical roller bearing
• series 21300 - 1°
• series 22300 - 2°
• series 23100 - 1.5°
• series 24100 - 2.5°

44-What is the misalignment tolerance for taper roller bearings?

Taper roller bearing: 3 Minutes of arc

45-What is the misalignment tolerance for thrust bearings?

Thrust bearings: Thrust Roller bearing-2 to 2.5 degree & Cylindrical thrust bearing-No misalignment

46-What are the operating, alarm & trip level vibrations for bearings?

• Operating level vibrations:1 to 4 mm/sec
• Alarm level vibrations: 4 to 6 mm /sec
• Trip level Vibrations: > 6 mm/sec

47-How do you calculate the grease required for re-lubrication of a Bearing?

Grease required for re-lubrication in grams is = Bearing OD X Bearing width X 0.005

48-On what factors bearing re-lubrication & frequency depends?

• Bearing operating speed
• Bearing operating environment
• Operating temperature
• Bearing operating vibrations
• Bearing operating hours

49-What causes bearing currents?

When voltage is present on the motor shaft it can overcome the insulating effect of the bearing lubrication film. This cause a current flow that results effectively in electric discharge machining (EDM) of the bearing, causing premature wear and ultimately, early failure. 

50-where & why do you use the insulated bearings?

Insulated bearings are used in Alternators & VFD motors. Insulated bearings are fitted at Non drive end side of the machine.

The purpose of insulated bearing for Motors with VFD drives or the Alternators is to prevent the flow of circulating current in a closed circuit through bearing and ground and consequently prevent the bearing from damage.

An important new solution for motors in IEC frame sizes 132 to 250 is to install a motor shaft grounding brush that directs the current to the ground via the brush, rather than through the bearing. This protects  the motor itself and the complete installation. The brush can be pre-installed on new motors by specifying a variant code. Or it can be retro-fitted on site.

Turbine oil and transformer oil standard testing parameters