 Force needed to draw flue gases is the draught. This force may be due to small pressure difference in the stream or the current of gas or air that causes the flow to take place. In stoves, to support combustion, it is necessary to supply a quantity of air and to remove the products of combustion by means of draught. In stove there is generally natural draught applied. Natural draught is the difference between the pressure of a column of hot gases in the chimney or stack and an equal column of cold outside air. To calculate the draught induced in chimney  following assumptions are made: 1.      Nitrogen is present but does not take part in chemical reaction. Volume of hydrogen burnt is also small and consequently volume of steam formed is also small. 2.      The volume of products of combustion in chimney is equal to the volume of the air supplied for combustion, provided that both volumes are taken at the same temperature. Mass of the air supplied per kg of fuel = m kg Mass of chimney flue gases = (m+1) kg From the above assumption, the volume of 1 kg of flue gases at 273K is equal to that of 1 kg of air at 273 K. Volume of air at 273 K = Or, v = 0.7734 m3/kg Let, Ta = absolute temperature of outside air, K Tf =mean absolute temperature of the chimney gases, K Then, Volume of outside air at Ta, va= Volume of m kg of air at Ta, va = Volume of (m+1) kg of chimney gases at Tf, Vcg = Density of air at Ta, = = ra = 1.293 kg/m3 Density of Chimney gases at temperature, Tf rf = kg/m3 Let,      H = height of the chimney             h = draught measured in water column Pressure exerted by a column of hot flue gas of H meter, Pf = rf*g*H      = *g*H   N/m2 Pressure exerted by a column of cold air of H meter, Pa = ra*g*H      = 1.293* N/m2 Difference in static pressure       = P                                                 = Pa-Pf                                                 = g*H (ra-rf)   P= 353*g*H ( – * )    The draught in terms of h mm of water column h = H (ra-rf) h= 353*H ( – * ) mm of H2O 