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In this page you'ill learn about the course on "thermal power plant" for beginning engineer's level to professional level. All contents in this page are for learning purpose only.

 

Thermal Power Plant


 

Thermal Power Plant Performance

The performance of any thermal power plant depends on so many factors like coal quality, heat rate, cycle efficiency, design etc. The efficiency of any plant is not constant always but reduces years of years. Generally a coal fired thermal power plant efficiently performs for 25 years if all the required maintainance is done time to time. 

Heat rate - “Heat rate is the heat required to generate one unit of electrical energy.” Heat rate in a thermal power plant is measure constitute for its thermal efficiency. The plant output should be higher with minimum heat requirement so we can say that the heat rate must be as low as possible.

                        The property of heat generation by fuel/coal depends on the calorific value. Also the CV(Calorific Value) depends on the elemental and non-elemental composition of coal. The analysis of coal composition are divided into two categories:

  1. Proximate analysis – In this analysis non-elemental composition are calculated. The non-elemental composition are fixed carbon, volatile matter, moisture and ash contents.
  2. Ultimate analysis – In this analysis elemental composition are calculated. The elemental composition are carbon(C), hydrogen(H), sulphur(S), nitrogen(N), oxygen(O), phosphorus(P), chlorine(Cl) etc.

Grading of coal is the measurement of heat content of the coal. It is measured in calorific value for each kilogram of coal weight.

Grade

Calorific Value (kCal/kg)

A

> 6200

B

5601 – 6200

C

4941 – 5600

D

4201 – 4940

E

3361 – 4200

F

2401 – 3360

G

1301 – 2400

Below ‘G’ grade coal are generally not used in any power plant to maintain the efficiency. High grade coals are generally mixed and blended with low grade coal. Before using the coal, it is passed from coal washery system where lot of waste substances of coal are removed by washing.

In a thermal power plant, unit's heat rate is required to power the turbine but practically it is not possible because heat loss occured in most of the parts like turbine extraction, vents, drains, tubes and a major part of heat is consumed in condenser side. Hence unit heat rate can never be equalize turbine heat rate.

Let's find out which formula that can be used to calculate the terms like heat rate, calorific values, efficiency etc. Then we'ill do some calculation on all these.

We know that Efficiency  η = output/input

   Efficiency  η = THR/UHR                 where    THR = turbine heat rate and UHR = unit heat rate

UHR (Q x CV)/MW           where    Q = coal flow rate, CV = calorific value and MW = plant output in MW

  GCV is calculated by Dulong’s formula,           GCV = (33820 x C) + 144000( H – O/8) + (9.204 x S)


 


Qu. A steam power plant has 20% overall efficiency. To generate each unit of electricity 900gm of coal is required. Calculate the calorific value of coal.

Sol.                             CV(Calorific Value) = ?

900gm = 0.9 kg                   

Heat content by 0.9kg of coal = 0.9 x CV

                                    Efficiency  η = output/input= 860kCal/Heat input

                                      0.20 = 860/(0.9 x CV)

CV= 860/(0.9 x 0.2)

CV = 4777.78  kCal/kg

 

 


Qu. A 250MWsteam power plant consumes 160T/hr(tonns per hour) with calorific value of coal is 3300kCal/kg. Calculate the overall efficiency.

Sol.                             Efficiency  η = ?

CV(Calorific value) = 3300kCal/kg

Coal flow Q = 160T/hr

Plant capacity MW = 250              

Heat content by 0.9kg of coal = 0.9 x CV

                                    UHR(unit heat rate) = (Q x CV)/MW

                                                 = (160 x 3300)/250

                                                 = 2112 kCal/kWh

Efficiency  η = output/input = 860kCal/unit heat rate

             η = 860/2112 = 0.40719

      η = 40.719%

 

 


Qu. Calculate the amount of coal required in TPH(Tonns per hour) for a 600MW, 37.5% efficient thermal power plant if the calorific value of coal is 3150kCal/kg.

Sol.                             CV(Calorific value) = 3150kCal/kg

Coal flow Q = ?

Efficiency  η = 37.5% = 0.375

Plant capacity = 600 MW  

                                    UHR(unit heat rate)= 860/η

                        UHR = 860/0.375 =2293.33 kCal/kWh

                                    UHR = (Q x CV)/MW

                       Q = (MW x UHR)/CV

                                    Q = (600 x 2293.33)/3150 = 436.8 T/hr

 

 


Qu. Calculate the GCV(gross calorific value) of coal for which the following are the percentage obtained by ultimate analysis.

Moisture       18%

Ash                 8%

Carbon           59%

Oxygen           8.4%

Hydrogen      3.7%

Nitrogen        1.2%

Sulphur         1.7%

Sol.                        GCV is calculated by Dulong’s formula :

                        GCV = (33820 x C) + 144000( H – O/8) + (9.204 x S)

In the above formula, we can see that we need only percentage of C, H, O and S to calculate the GCV.

Now put the values in the Dulong’s formula,

                        GCV = (33820 x 59) + 144000 x (3.7 – (8.4/8)) + (9.204 x 1.7)

                                 = 2377 kJ/kg

Er. Sameer Gupta

Hi,

I'm Sameer Gupta and I am a working engineer. I am from Midmida(Raigarh) at Chhattisgarh, India. I graduated(2008-2011) in Electrical Engineering from college SSCET-I Bhilai and have post graduation in thermal power plant engineering. Playing with electrical,electronics & embedded system design is my hobby and passion. I have always loved playing with all these things since my childhood. I don't believe in marks and degree without knowledge.

 

                                      I started this site to share my ideas & projects with all viewers over this site. Electronics is one most important thing in the modern age of science and everyone should have the knowledge about this. This site provides practicles, project designing & knowledge on embedded world for all beginners & leaders. If you want to share your knowledge with me or you just want to say me hello please contact me without hesitation.

Various units and their relationship used in power plant


Energy Conversion :    1HP(Horse power) = 746 Watts = 0.746 Kilo Watts

1 Joule = 1 watt-second

1 kWh = 1000 watts × 3600 seconds = 36 × 105 joules =3,412 Btu = 860 kcal = 1Unit of electricity

1 Watt = 3.412 Btu/hour

1 Calorie = 4.184 Joules

1 kcal/Kg = 1.8 Btu’s/lb.

1 Btu = 252 Calories = 1,055 Joules

1 Million Btu = 293.1 kWh

1 Btu/lb. = 2.3260000 kJ/kg = 0.5559 Kilocalories/kg

1Chu ( Centigrade Heat Unit)= 1.8Btu = 1899 joules

1 Therm = 10,54,80,400 joules  = 100,000 Btu (British Thermal Units)

 

Mass Conversion :        1 kgf (kilogram-force )= 9.8 N (Newtons)

1 lb = 0.45359237 kg

1 kg = 1000g = 2.20462262 lb

 

Pressure Conversion : 1 bar =  105 Pa(Pascal)  =  1.01 kg/cm2 =  14.50 psi(pounds per square)

1 kg/cm2  =  0.98 bar =  980.665 mbar (milli bar) =  98.06 kPa = 0.098 MPa

     =  10000 mmwc (mm of water column)

     =  735.56 mm-Hg  =  29 in-Hg

                          =  14.22 psi (lb/in2)

 1 atm(atmospheric pressure) =1.033kg/cm2 = 1.012bar

 

Temperature Conversion : Centigrade degrees = (5/9) X (Fahrenheit degrees - 32)

   Fahrenheit degrees = (9/5) X (Centigrade degrees) + 32

   Kelvin degrees = Centigrade degrees + 273

 

Volume Conversion : 1 gallon = 3.78541 litre

       1 m3  = 1000 litre

 

Length Conversion :    1 meter = 100 cm = 3.28084 feet = 1.09361 yard = 39.3701 inch

1 foot = 12 inch

1 yard = 36 inch = 3 feet

               1 inch = 2.54 cm 

1 mile = 1609.34 meter

 

Multipliers :    Mega = 106

Giga = 109

Tera = 1012

Centi = 10-2

Milli = 10-3

Micro =10-6

Nano = 10-9

Pico = 10-12

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