**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.

**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

**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%

**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

**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**= 0.86 X 100/0.93 = 92.5%

_{vacuum}**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.

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**

Very informative thank you

ReplyDelete