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 Author: dr funda Time: 01/17/00 20:17 PST This is a reply to message no. 58 by wbornsea Reply | Original Message | New Topic | List Topics | List Messages on This Topic
 Current Topic:Steam Production Hi, To answer your question, first we need to assume a starting state. For simplicity, let's assume that we're starting with water at 100 ? and 200 psi. We'll call this State 1. The boiler needs to raise the temperature of the water to the saturated liquid temperature corresponding to 200 psi.  From the steam tables, we find this to be 381.8 ?. We'll call this point State 2. Next, the boiler will convert all the saturated liquid to saturated vapor (at the same pressure and temperature), a condition that we'll call State 3. The heat input required to produce saturated vapor in State 3 from the initial condition of State 1 is determined by the change in enthalpy: q = h3 - h1??[Btu/lbm] We'll record the enthalpy values at each of three states, pulling the data from Steam Tables: State 1: 200 psi, 100 ? water: ?h1 = 68 Btu/lbm ?s1 = 0.129 Btu/lbm-? State 2: 200 psi, 381.8 ? saturated water: ?h2 = 356 Btu/lbm ?s2 = 0.544 Btu/lbm-? State 3: 200 psi, 381.8 ? saturated vapor: ?h3 = 1199 Btu/lbm ?s3 = 1.546 Btu/lbm-? The total change in enthalpy is 1199 Btu/lbm - 68 Btu/lbm, or 1131 Btu/lbm. And since the efficiency of the boiler is 80%, the input should be 1414 Btu/lbm Note that we could have also determined the heat input requirement from a change in entropy argument, as opposed to a change in enthalpy, and get the same answer: State 1 to State 2 : (s2-s1)?T2+T1)/2 State 2 to State 3 : (s3-s2)?i>T2 Important: The temperatures have to be degrees Rankine in the above calculation. Hope this helps, eFunda Staff
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