E3.6 Steam Superheater

Example 3.6: Oil (C₁₂H₂₆) is burned with 20% excess air. Completely combusted gases enter a steam superheater at 1550 °F and exit at 1150 °F. Steam enters at 400 psi and is 92% saturated. It is required to generate superheated steam at a temperature of 650 °F. Assume that there is no pressure drop in the superheater and specific heat of the product gases is 0.25 Btu/(lb·°F). Determine the amount of steam that could be produced per lb of oil burned. Molecular weight of oil = 170. Combustion of oil proceeds as follows

Solution Basis: 1 lbmole of oil:

CO₂ produced = 12 lbmol = 528 lb

H₂O produced = 13 lbmol = 234 lb

O₂ required (stoichiometric) = 18.5 lbmol

O₂ in the product = 0.2(18.5)(32) = 118.4 lb

N₂ in the product = 3.764(1.2)(18.5) = 83.56 lbmol = 2340 lb

Mass of the gas per lb of oil, W:

W = (528 + 234 + 118.4 + 2340)/170 = 18.942 lb

Sensible heat given by product gases to wet steam, Q_h:

Q_h = WC_p(T₁ - T₂) = 18.942(0.25)(2010 - 1610) = 1894 Btu

Heat absorbed by steam, q_c:

q_c = 1335.9 - (424.2 + 0.92(780.4)) = 193.732 Btu/lb

Amount of steam produced per lb of oil burned, w_steam:

w_steam = Q_h/q_c = 1894/193.732 = 9.777 lb