Issues with fuel consumption in gas boilers
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This post was last edited by HG-LXG on 2016-1-14 at 15:22. A 3-ton gas boiler has a feed water temperature of 80 degrees; how much natural gas is required to raise this temperature to 140 degrees? The calorific value of natural gas is taken as 8500 kcal/m3, and the boiler efficiency is assumed to be 90%. The volume of water at the normal water level in the boiler is 7.5 cubic meters
A 3-ton gas boiler has a power output of around 2100 KW; it consumes approximately 210 cubic meters of gas, and with a coefficient of 1.1, this amount increases to 230 cubic meters.
Is this calculation correct? It seems to be the gas consumption per hour for a 3-ton boiler; I’m referring to the fuel consumption at temperatures between 80 and 140 degrees
It is a gas boiler with a capacity of 3 tons (in terms of rated evaporation rate)
This post was last edited by xyc114 on 2016-1-15 at 15:39. Known saturated temperature T(c): 140. The resulting saturated pressure P(Mpa) is 0.361064895665928. The enthalpy of saturated water h’+(KJ/kg) is 588.863900292985, while the entropy of saturated water s’+(J/kg·K) is 1.73842090353667. The enthalpy of saturated steam h”+(KJ/kg) is 2734.53584430752, and its entropy s”+(J/kg·K) is 6.93035776759428. The specific volume of saturated water v’+(KJ/kg) is 0.00107996800859291, whereas that of saturated steam v”+(KJ/kg) is 0.509865240937056. ********************************************************************* P(Mpa): 0.361; T(c): 80. H(KJ/kg): 335.163615902933; S(J/kg·K): 1.07504563587094; V(m3/kg): 0.00102904050960609. X: 0. H = 2734.53584430752 – 335.163615902933 = 2400 KJ/kg. With an evaporation rate of 3 tons per hour, the total heat output of the boiler is Q = (3000 * 2400 / 3600) / 0.9 = 2221.76 kW. The calorific value of natural gas is 8500 kcal/m3, which is equivalent to 8500 * 4.12 = 35020 KJ/m3. The amount of natural gas consumed per hour is (2221.76 / 35020) * 3600 = 228 m3/h