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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 steadyflow 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 steadyflow 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.
 Spring '09

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