It's all about HP heaters (Feed water heaters) in Power plants

 1-Why do you use HP heaters in power plants?

HP heaters are used for heating the feed water, which will contribute in increasing cycle efficiency as well as reduction in fuel consumption

2-What is the design code of HP heaters?

HEI, ASME SEC VIII Div-I IBR

3-What are the advantages of using feed water heaters in power plants?

• Fuel consumption reduces
•  Reduce heat losses in the condenser
•  Lower emissions as fuel use is reduced due to improved heat rate
• Decreases the plant heat rate & hence increases the plant efficiency

4-What are the different types of feed water heaters?

Open type feed water heaters & closed type feed water heaters

5-What do you mean by open type of feed water heaters?

In open type feed water heaters steam directly mixes with feed water. Steam pressure used is lower < 5 kg/cm2

6-What do you mean by closed type of feed water heaters?

In closed type feed water heaters steam directly mixes with feed water. Steam pressure used is high > 5 kg/cm2, these are shell & tube type heat exchangers

7-What are the pipe lines connected to HP heaters?

• Bleed steam inlet line
• Feed water inlet line
• Condensate outlet line
• Feed water outlet line
• Feed water box drain & vent lines
• Shell zone drain & vent

Procedure for Boiler gauge glass line up

8-What are the MOCs of Tubes & shells used in closed type of HP heaters?

Tubes: SA 213 TP 304 (Feed water pressure up to 170 kg/cm2)

Shell: SA 516 Gr.70

9-What are the different zones of HP heaters?

• Desuper heating zone
• Condensing zone
• Sub cooling zone

10-Where does the maximum heat transfer occur out of all zones?

Maximum heat transfer occurs at condensing zones

11-Why the name Sub-cooling has come?

Here condensed steam from condensing zone is cooled by feed water entering by convective heat transfer method.

12-What is the function of drain coolers in HP heaters?

Drain Coolers are employed because of heat consumption improvement in case of drain introduction into the lower heater through the control valve.

13-Why the feed water inlet line connection is at the bottom & outlet line is at the top?

          

Feed water inlet line & outlet line are connected in such a way that to separate desuperheating zone, condensing zones & sub cooling (drain cooler) zones

14-Where do the fixed & sliding supports of HP heaters are located?

Fixed support: Towards feed water inlet & outlet line

Sliding support: Opposite side of feed water line connections

15-What would be the velocity of feed water in HP heaters tubes?

It’s around 0.6 to 0.8 m/sec

16-What is the effect of high/low condensate level in heaters shell?

Higher the condensate level lower is the performance of heater and vice versa. Heater level is always maintained in between 30–50%.

17-Briefly explain the condensate level control in HP heaters?

One of the most common causes of tube failures in a feed water heater (FWH) is the improper control of the internal liquid level, which also can cause operational and maintenance costs that might lead to premature replacement. These problems are not new, they have been experienced by many utility plants throughout the industry during the past 50 years. However in many cases, the resulting damaging phenomenon has seldom been totally understood and the loss of corporate knowledge and failure of some utilities to identify and rectify level control problems continues to bring this issue to the forefront of root causes of FWH operational failures.

In general, the performance of the Drain Cooler (DC) Zone is tied to the operational parameter of Drain Cooler Approach (DCA). DCA is a good indication of whether the DC Zone is operating properly or not, it is not the only parameter that should be considered. DCA is a measurement of temperatures

The pressure of the drains also must be known in order to determine the degree of sub cooling and whether there is a potential for flashing, either within the DC itself or the downstream piping before the level control valve. Flashing and two-phase flow in either of these areas can cause significant damage to the heater.

It is important to remember that the drain cooler is designed to be a water-to-water exchanger. It must remain that way to function properly. Any admission of vapour into the zone typically results in problems. This might be a result of a low liquid level in which steam is admitted directly from the condensing zone into the DC zone, the result of flashing within the DC zone itself, or can be the result of leakage into the zone via the endplate or shroud cracks.

18-What do you mean by drain cooler approach (DCA) in HP heaters?

DCA is the temperature difference between the drains (steam condensate) leaving the heater and the temperature of feed water entering the heater. For more cycle efficiency DCA value should be small.

19-A HP heater is used to heat the feed water from 160 °C to 180 °C by using turbine bleed steam at 15 kg/cm2 and 320 °C. The condensate returning from heater is at 170 °C, calculate the DCA of heater.

We have,

DCA = Temperature of condensate leaving the heater – Temperature of feed water entering the heater

DCA = 170 - 160 = 10 °C

Note: For best performance, heaters are designed to get DCA 3 to 5 °C at full operation capacity.

20-What do you understand by Terminal Temperature Difference (TTD)?

It is the difference between the saturation temperature at the operating pressure of the inlet steam to the heater and the temperature of the feed water leaving the heater. For more cycle efficiency TTD value should be small.

21-A HP heater is used to heat the feed water from 110 °C to 160 °C by using MP steam at pressure 13 kg/cm2 at temperature 280 °C, calculate the TTD.

We have,

TTD = Saturation temperature of inlet steam - Feed water outlet temperature

Saturation temperature of inlet steam at 13 kg/cm2g pressure = 195.6 °C

TTD = 195.6 - 160 = 35.6 °C

Note: For best performance, heaters are designed to get TTD 3 to 5 °C at full operation capacity.

22-How the DCA does affects condensate level of HP heaters?

An increase in DCA,HP heater level decreases & vice versa

23-What is flashing in heaters? How does it occur?

Flashing, by definition, is the change in state of liquid to vapour. While in most cases this change of state results from the addition of heat (as in the boiler) in a FWH the most common cause of flashing is a result of a reduction in pressure (or pressure drop). Pressure drop might be a result of the geometry of the Drain Cooler Entrance window, the fact that the drains must travel around the tubes and change direction many times due to the baffling arrangement and also due to changes in elevation and elbows in the downstream piping. If the liquid drains are not sub cooled enough, any one of these pressure drops could result in flashing and two-phase flow. Two-phase flow is known to cause problems to piping, tubing, the cage and the shell, especially in the case of carbon steel components.

24-What do you mean by fouling in heat exchangers?

Deposition of any undesired material on heat transfer surfaces is called fouling. Fouling may significantly impact the thermal and mechanical performance of heat exchangers. Fouling is a dynamic phenomenon which changes with time. Fouling increases the overall thermal resistance and lowers the overall heat transfer coefficient of heat exchangers. Fouling also impedes fluid flow, accelerates corrosion and increases pressure drop across heat exchangers.

Different types of fouling mechanisms have been identified. They can occur individually but often occur simultaneously.

Scaling:

• Particulate/Sedimentation Fouling
• Corrosion Fouling
• Chemical Fouling
• Freezing Fouling

25-What are the problems associated with HP heaters?

•  Initial two phase mixture & hammering
• Tubes failure due to wrong operation
• Level fluctuation & leakages
• Overfeeding of steam & feed water
• Operating the heaters above the operating & design parameters

Read Power plant standard operating procedure

26-Write down the initial charging process of feed water heater (HP heater)

Steps:

• Ensure all the maintenance activities on HP heaters are completed
• Ensure all inlet & outlet valves of heaters are healthy
• Ensure all field instruments are healthy
• Keep open all water box & shell side vents & drains are  open
• Ensure steam condensate outlet valve is open

Water side

• Crack Open the feed water outlet valve
• Then crack open the feed water inlet valve
• Allow to vent out the air
• Then gradually open the outlet & inlet feed water valves
• Then close the water box vents & drain valves

Steam side

• Ensure steam line drains are in opened condition
• Ensure steam parameters are as per desired values
• Ensure no water in steam line drains
• Crack open the steam inlet valve to HP heater
• Ensure there are no water particles in drain & vent line of shell. If found clear, then close the valves
• Then gradually open steam inlet valves & allow for stabilization

27-A HP heater is used to heat the 200 TPH feed water from 160 °C to 180 °C by using turbine bleed steam at 15 kg/cm2 and 320 °C. The condensate returning from heater is at 170 °C, calculate the quantity of steam used.

 Given that,

Qf = 200 TPH

Tf1 = 160 °C

Tf2 = 180 °C

Hg  at pressure 15 kg/cm2 & temperature 320 °C =735.3 kcal/kg

Enthalpy of condensate water Hf = 171 kcal/kg

Heat lost by the steam = Heat gained by feed water

Ms X (Hg-Hf) = Mw X (Tf2-Tf1)

Ms X (735.3-171) = 200 X (180-160)

Ms = 7.08 TPH

28-What are the precautions to be taken for safe operation of HP heaters?

Precautions:

• Operate the HP heaters as per SOP
• Take utmost care during initial charging
• Do not operate the heaters beyond the operating pressure & temperature
• Bypass the HP heaters during Boilers Hydraulic tests
• Conduct routine preventive maintenance

Boiler Feed Pumps Design factors & Pump Capacity calculation

Zebronics Zeb-County Bluetooth Speaker with Built-in FM Radio, Aux Input and Call Function (Black)

                                          

Turbine oil and transformer oil standard testing parameters

Boiler calculations for Boiler operation engineer (BOE) exam

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