30-Things you must know on steam turbine & auxiliaries

1-During cold start up turbine inlet steam  minimum temperature should be saturation temperature at the particular pressure + 50 Deg C.

Example: Turbine operating at 67 kg/cm2 pressure, its inlet steam temperature should be285+50 =335 deg C

2-Distinguishing the Turbine starts up types.

• Cold start up- when HP and IP inner casing temperature is lower or equal 170^oC
• Warm start up- when HP and IP inner casing temperature is lower or equal 430^oC
• Hot start up – when HP and IP inner casing temperature is greater than 430^oC.

Read Power generation phenomenon in STG

3-To open ESV the vacuum should be at least 0.3 Kg/cm2A (-0.73 kg/cm2)

4-In turbine rotors, over speed trip bolt is always fitted at the DE side only. This is because, not weaken the Turbine NDE side shaft, as NDE side shaft size is already made small & drilled for key ways

5-Expansion bellows for lube oil lines are fitted at the Turbine front bearings, as Turbine expansion occurs towards front end

6-In oil cooler heat exchangers,oil pressure is always kept at higher side than water pressure. This is to avoid entry oil water in lube oil system.

7-In many Turbine, control oil (oil used for HP, LP valves actuators & ESV) is used at higher temperature 55-60 deg C.This is because “Actuators & ESV components are operating at very less clearance need low oil viscosity.

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8-Control oil filters are of lesser filter size as compared to lube oil filters. 

Generally filters are designed based on the minimum clearance through which the oil flow, hence lube oil filters are of higher openings (25 to 40 microns) as bearings clearance will be in the range of 200 microns to 500 microns. Control oil filters are of lesser size openings (10 to 25 microns) , as discussed earlier Actuators & ESV components are operating at very less clearance up to 50 microns.

9-Positive displacement Lube oil pumps have in built as well as external PRVs (Fitted at the discharge line).Lube oil pumps (Positive displacement pumps) are always started with discharge valve open unlike centrifugal pumps

10-Emergency oil pump do not have PRVs

11-Emergency oil pumps flow capacity = Main/Auxiliary oil pump X 25%

12-Emergency oil pump pressure = Main oil pump pressure X 30%

13-Control oil pumps flow capacity = Main/Auxiliary oil pump X 10%

14-For lube oil coolers: Cooling water flow = Oil flow X 2

15-For lube oil coolers , Heat load in KW = Cooler surface area X 5.3

16-Turbine lube oil consumes 30 to 35% of total cooling water required for plant auxiliary

17-Generator air cooler consumes 20 to 25% of total cooling water required for plant auxiliary

18-Bearing inlet  oil pressure during high rotor speed (Normal operation) is lesser than that of low speed (During barring gear operation)

19-There is always off set alignment between Turbine rotor & Gear box pinion shaft. This is for accommodating the misalignment during operation, as Gear box is operating at higher oil temperature than Turbine.

        Generally Gear box pinion shaft is kept at lower level (0.15 to 0.3 mm) & offset side depends on the direction of rotation of Turbine shaft viewed from turbine front end. If Turbine rotor is rotating clockwise then offset is towards RHS.

20-Low oil temperature can damage the Turbine bearings: Because;

When temperature decreases too much, oil in the bearing becomes so viscous that it clings to the shaft surface which drags it around the bearing. This makes the oil wedge in the bearing lose. its stability. The pulsating wedge excites high rotor vibration referred to as oil whip or oil whirl.

Too low temperature - and hence, too large viscosity - of the bearing inlet oil causes the bearing oil flow to decrease due to increased friction in the oil supply piping. The reduced oil flow may be too small for adequate cooling, causing bearing overheating and possible damage.

21-Slight sub atmospheric pressure is maintained inside the bearing housing and its drain line by the vapour extraction fans installed on the lube oil tank cover. Why is this pressure maintained? 

First.

To prevent oil mist from escaping past the bearing oil seals into the turbine hall.

Second. 

To prevent accumulation of hydrogen and oil vapour in the lube oil tank atmosphere, which could create an explosion hazard

·   

22-For lube oil:

During normal operation, water is removed from the oil by the oil purifier and the vapour extraction fans. During a long outage, water can also be drained from the bottom of the lube oil tank.

23-Main oil tank level:

The major adverse consequence/operating concern caused by too low tank level is impaired pump performance due to cavitation and possibly vapour locking or gas locking. The lower the oil level, the smaller the suction head of the pumps in the tank. Pump cavitation and eventually vapour locking can result. The lowered level can also lead to ingress of gases from the tank atmosphere into the pump suction piping, and then the pump itself. An excessive accumulation of gases in the pump can decrease its capacity, and finally result in pump gas locking. Too high tank level increases the risk of tank overfill. The resultant oil spill has its own adverse consequences such as an environmental hazard.

24-Rotor lift due to jacking oil pressure

Drive end :0.1 mm & NDE :0.05 mm

The jacking oil pressure at the bearing inlet is not controlled. As the oil is supplied by a positive displacement pump, its pressure rises until the bearing resistance to the oil flow is overcome. This happens when the turbine generator rotor is lifted off the bearings.

25-Wheel chamber pressure = (Turbine inlet pressure X Turbine load in MW X 0.6)/Turbine Capacity in MW

26-Steam condensers has fixed support at cooling water inlet side & sliding support at the opposite side

27-Surface condensers installed at higher elevation are always producing lower vacuum.Power plants installed at  or near the sea produce high vacuum

28-At steam condensers Vacuum breaker valves are provided to bring down the rotor speed to zero as early as possible  
29-Closing time of ESV & HP control valves are 0.3 to 0.4 seconds and 0.4 to 0.6 seconds respectively
30-Pressure of N2 gas in control oil accumulator is 2 to 3 kg/cm2 lesser than control oil line pressure

31-For STG: Bearing temperature trend goes on decreasing from Turbine front to Generator rear end

32-For STG: Bearing Vibrations trend goes on increasing from Turbine front to Generator rear end
33-Velocity of condensate water at ejector & gland steam condenser is 0.5 to 0.7 m/sec & 1 to 1.5 m/sec respectively

Protections & Interlocks in power plants

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Boiler Feed Pumps Design factors & Pump Capacity calculation

Read steam Turbine commissioning

Turbine oil flushing procedure

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