Unformatted text preview: = 0.92 P2 = 80 kPa a 80 kP T2a
2s T2s FIGURE 7–55
Schematic and Ts diagram
for Example 7–16. 2a s s2s = s1 SOLUTION A sketch of the system and the Ts diagram of the process are given
in Fig. 7–55.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 The
inlet kinetic energy is negligible. 4 The nozzle is adiabatic.
Analysis The temperature of air will drop during this acceleration process because some of its internal energy is converted to kinetic energy. This problem can
be solved accurately by using property data from the air table. But we will assume constant specific heats (thus sacrifice some accuracy) to demonstrate their
use. Let us guess that the average temperature of the air will be about 800 K.
Then the average values of Cp and k at this anticipated average temperature are
determined from Table A–2b to be Cp 1.099 kJ/kg · K and k 1.354.
(a) The exit velocity of the air will be a maximum when the process in the nozzle involves no irreversibilities. The exit velocity in this case is determined from
the st...
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 Spring '09
 Thermodynamics, Energy, Entropy, entropy change

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