Efficiency of a Boiler

[padovano]

Can somebody please help me with the calculation of the efficiency of a water tube boiler of 17 bars. Its production capacity is 35t/h. The guy before me use to calculate the efficiency using the effluent gas temperature and the excess air. I am thinking of using the heat absorbed/heat furnished. I will be very happy with a detail explanation of which method is best.

 

[pmover]

padovano,

perhaps this website will be of assistance to you.

http://www.heaterdesign.com/design0.htm

see documentation made available within the publications link on main site.

hope this helps.
-pmover

 

[athomas236]

Padovano,

The answer depends on what is important to you.

If its accuracy then the heat loss method is the way to go (gas temperature and excess air); if it is not then the input/output method could be used.

ASME PTC 4 gives the following information regarding the uncertainty (error) for both methods for small gas fired industrial boilers.

Heat loss 0.2 to 0.5 percentage points
Input/output 1.2 percentage points

Best regards,

athomas236

 

[padovano]

thanks Athomas236, accuracy is what I want. I would like to know if this formula is applicable to a boiler with these characteristics i.e a water tube boiler of 17Bars with a production capacity of about 30 t/h with a superheater.
The formula: 100 - (temp.of effluent gas/20 + excess air/2 + 2).

 

[athomas236]

padovano,

I do not know the formula but if you tell me the following, I will check the efficiencies at 30t/h on GCV basis for a range of flue gas temperatures and excess air values:

1. Fuel ultimate analysis and GCV
2. Superheater outlet steam temperature at 30t/h
3. Economiser inlet water temperature at 30t/h
4. Boiler exit flue gas temperature at 30t/h
5. Boiler excess air at 30t/h

Regards,

athomas236

 

[padovano]

Athomas 236,

1-Fuel Analysis: C1=57.71, C2 = 24.36, C3=14.53, C4=3.31, C5=0.07, C6=0.01

2 - Super heater outlet Temperature= 380°C

3 - Economiser inlet water Temperature = 105°C

4 - Boiler exit effluent gas temeprature = 308°C

5 - Excess air = 15%

 

[LSThill]

Please search Google

[PDF] CIBO Energy Efficiency Handbook File Format: PDF/Adobe Acrobat - View as HTML
the members of the CIBO Energy Efficiency Subcommittee for their direction and insight into the process of preparing this Energy Efficiency Handbook; ...

http://www.cibo.org/pubs/steamhandbook.pdf

[PDF] ENERGY EFFICIENCY & INDUSTRIAL BOILER EFFICIENCY An Industry ... File Format: PDF/Adobe Acrobat -
cibo.org/pubs/whitepaper1.pdf

been their did that


L S THILL

 

[athomas236]

padovano,

Based on the data supplied and estimated GCV of 52.673MJ/kg I calculate an efficiency of 78.179% compared with that of 75.1% using your formula.

With gas temp of 350C and XSA of 15%, efficiency is 76.318%.

With gas temp of 308C and XSA of 25%, efficiency is 77.259%.

Combining this data gives the following formula:

Efficiecy = 93.203-0.886*Tg/20-0.928*XSA/10

Regards,

athomas236

 

[padovano]

Thanks Mr Athomas236, I would like to know how the contants, 0.886, 0.928 and 93.203 were obtained. I will be very grateful for this information.

 

[21121956]

Hello everybody:

According with "Boiler Efficiency Calculations Simplified" by V. Ganapathy, for oil fuels #2 and #6, the efficiency is calculated:

Based on the Higher Heating Value

% = 92,9 - (0,001298+0,0195*EA)(Tg-Ta)

Based on Lower Heating Value

% = 99,0 - (0,001383+0,0203*EA)(Tg-Ta)

where: Ta = ambient or Reference temperature.

Tg = exit gas temperature.

EA = excess air factor. For an excess air of 15%, the EA = 1,15.

 

[athomas236]

padovano,

I have a spreadsheet that calculates boiler efficiencies in accordance with ASME PTC4.1.

So I calculated the boiler efficiency based on GCV using the data you provided and obtained a value of 78.179%. Then I increased the excess air from 15% to 25% and obtained 77.259%.

So for change in excess air the change in efficiency is (77.259-78.179)/(25-15)= -0.92/10.

In a similar way I changed the gas temperature from 308C to 350C and obtained an efficiency of 76.318%. So for a change in gas temperature the change in efficiency is (76.318-78.179/(350-308)*20= -0.886. The remaining constant was obtained by algebra.

Regards,

athomas236