Energy, fuels, and chemicals 



3191 



(table 26-5). If we assume a boiler efficiency of 63 percent, then 11 .7 lb per hour 

 of green wood would be required to produce 1 BoHP: 



33,480 Btu/hr 



1 BoHP = 



(boiler efficiency) (available heat of fuel, Btu/lb) 

 33,480 Btu/hr 



11.7 Ib/hr 



(0.63)(4,540 Btu/lb) 



Today, in the forest products industry the most common expression of steam 

 delivery is number of pounds of steam delivered per hour (PPH). The unit is not 

 an absolute measurement as steam at different pressures and temperatures will 

 have different amounts of energy. For example, steam at 165 psi absolute and 

 550°F contains heat energy of 1,118 Btu/lb more than feed water at 212°F. 

 Under such a system, the amount of green pine-site hardwood (average available 

 heat, 4,540 Btu/lb) required to deliver 50,000 lb of steam per hour with a boiler 

 efficiency of 63 percent is equal to (50,000 lb steam/hr) (1,118 Btu/lb steam) ^ 

 (0.63) (4,540 Btu/lb wood), or 19,544 lb of wood per hour. 



Overall boiler efficiency is expressed as energy output divided by energy 

 input: 



output 



Percent efficiency = x 100 (26-8) 



input 



This energy input equals the dry weight of fuel multiplied by the higher heating 

 value of the fuel: 



Input = nif X q^ (26-9) 



nif = dry weight of fuel burned, Ib/hr 

 qh = higher heating value of fuel, Btu/lb 



The output equals the steam generated multiplied by the difference in heat 

 contents of the feed water and steam leaving the boiler: 



Output = mg (h, - h2) (26-10) 



nig = steam generated, Ib/hr 

 hi = heat content of steam leaving boiler, Btu/lb 

 h2 = heat content of feed water, Btu/lb 



The instrumentation and procedures required to measure efficiency can be- 

 come complicated. However, the process of calculating overall efficiency can be 

 greatly simplified by assuming that everything going into the furnace comes out 

 as mass or energy (Hughes 1976). Then, boiler output does not have to be 

 measured and overall boiler efficiency can be expressed as: 



input - losses 



Percent efficiency = x 100 (26-11) 



input 



Corder (1973) fists the sources of boiler heat loss as follows: (a) moisture in 

 fuel; (b) moisture formed from hydrogen in fuel; (c) dry stack gas loss; (d) 

 incomplete combustion, radiation, and unaccounted sources. 



