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Solving, Tg,3 = 295 oF. (d) Because a constant mean Cp was assumed for the exhaust gas over the temperature range of interest, simply draw a straight line from 700 oF to 295 oF, with the 295 oF corresponding to a transferred quantity of heat of 1.01 MMBtu/hr. On the steam side, separately determine the heat absorbed by the superheater, the evaporator, and the economizer. These heats are determined by the following equations: Qsteam = (m )(h SH steam - h ) g ) hg (180 psia, saturated steam) = 1196.9 Btu/lb Qsteam = (m Evap steam superheated )(h - h ) Qwater=(m Econ )(h -h water f g f feedwater Substitute the known flow and enthalpy data and solve for these three quantities: Qsteam = (850 lb )(1219.1-1196.9 Btu ) = (850 lb )(22.2 Btu ) = 18,900 Btu SH hr lb hr lb hr Qsteam = (850 lb )(1196.9 - 346.2 Btu ) = (850 lb )(850.7 Btu ) = 723,100 Btu Evap hr lb hr lb hr Q water = (850 lb )(346.2 - 28.6 Btu ) = (850 lb )(317.6 Btu ) = 270,000 Btu Econ hr lb hr lb hr Use these values to calculate cumulative heat duties: Q =Qsteam =18,900Btu=0.019MMBtu at373.1oF 1SH hrhr Q = Q + Qsteam =18,900 + 723,100 Btu = 742,000 Btu = 0.742 MMBtu at 373.1 oF 2 1 Evap hr hr hr Q = Q + Qwater = 742,000 + 270,000 Btu =1,012,000 Btu =1.012 MMBtu at 60 oF 3 2 Econ hr hr hr Plotting these points on the chart below yields the following T-Q diagram. 9-21PDF Image | Fuel Cell Handbook (Seventh Edition)
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