# Solution 180 btumin t1 50f 300 lbmmin t2 240f mixing

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Unformatted text preview: 240°F. The mixture leaves the chamber at 20 psia and 130°F, and heat is lost to the surrounding air at 70°F at a rate of 180 Btu/min. Neglecting the changes in kinetic and potential energies, determine the rate of entropy generation during this process. SOLUTION 180 Btu/min T1 = 50°F 300 lbm/min T2 = 240°F Mixing chamber P = 20 psia T3 = 130°F FIGURE 7–68 Schematic for Example 7–20. We take the mixing chamber as the system (Fig. 7–68). This is a control volume since mass crosses the system boundary during the process. We note that there are two inlets and one exit. Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus ∆mCV 0, ∆ECV 0, and ∆SCV 0. 2 There are no work interactions involved. 3 The kinetic and potential energies are negligible, ke pe 0. Analysis Under the stated assumptions and observations, the mass and energy balances for this steady-flow system can be expressed in the rate form as follows: · m in Mass balance: Energy balance: · → 0 (steady) 0 ∆m system · → 0 (steady) ∆E system 1442443 · m out · · E in E out 14243 Rate of net energy transfer by heat, work, and mass · m2 · m3 0 Rate of change in internal, kinetic, potential, etc., energies · E in · m 2h2 · m 1h1 · → m1 · E out · m 3h3 · Q out · (since W 0, ke pe 0) Combining the mass and energy...
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## This document was uploaded on 11/28/2012.

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