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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
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
m in Mass balance:
Energy balance: · → 0 (steady) 0
· → 0 (steady)
m out ·
E in E out
14243 Rate of net energy transfer
by heat, work, and mass ·
m3 0 Rate of change in internal, kinetic,
potential, etc., energies ·
m 2h2 ·
m 1h1 ·
→ m1 ·
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.
- Spring '09