Modern Engineering Thermodynamics

16.10 Nozzle and Diffuser Efficiencies 683
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The nozzle’s discharge coefficient C d is determined from Eq. (16.41) as
C d =
ðρAVÞ actual
=
ðρAVÞ isentropic
ðρ e V e Þ actual
ðρ e V e Þ isentropic
Now,
ðρ exit Þ actual = p exit /RT exit =
370:4 kN/m 2
= 0:686 kg/m3
½2:077 kN.m/ ðkg.KÞŠð260:1KÞ Since the flow is choked, M exit = 1.0, and the isentropic exit temperature and density can be determined from
Eqs. (16.18) and (16.20) as
and
ðT exit Þ s = T = T os ½2/ðk + 1ÞŠ = ð283:7Þ½2/2:67Š = 212:5K
ρ exit
ð Þ s = ρ os ½2/ ðk + 1ÞŠ 1/ðk–1Þ = ðp os /RT os Þ½2/ ðk + 1Þ
Š 1/ðk–1Þ
ð456:2 kN/m 2 Þð2/2:67Þ 1/0:67
=
= 0:503 kg/m3
½2:077 kN.m/ ðkg.KÞŠð283:7 KÞ and
p
ðV exits Þ s = c exit j s = ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
kg c RT ð c Þj s
p
= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1:67Þð1Þ½2077 m 2 /s ð
2 . KÞŠð212:5 KÞ
= 859 m/s
then,
ð0:686 kg/m3Þð474:8 m/sÞ
C d =
ð0:503 kg/m 3
= 0:754
Þð859 m/sÞ
Exercises
38. Your technician reports that there was an error in the sensor used to measure the nozzle exit velocity in Example 16.14.
The correct exit velocity is 426.3 m/s, not 474.8 m/s. Determine the new values for the nozzle’s efficiency, velocity
coefficient, and discharge coefficient, assuming all the other variables remain unchanged. Answer: η N = 0.772,
C v = 0.878, and C d = 0.677.
39. Oops, the technician in charge of the nozzle test in Example 16.14 now tells you that the velocity sensor reading is
correct, it is the exit temperature sensor reading that was in error. The correct nozzle exit temperature is 271.5 K, not
260.1 K. Determine the new values for the nozzle’s efficiency, velocity coefficient, and discharge coefficient, assuming all
the other variables remain unchanged. Answer: η N = 0.957, C v = 0.978, and C d = 0.722.
40. The nozzle in Example 16.14 is retested using air instead of helium. The same inlet conditions are used, but the
following new exit conditions are measured: V exit = 190.1 m/s, p exit = 430.3 kPa, and T exit = 270.5 K. Determine the new
values for the nozzle’s efficiency, velocity coefficient, and discharge coefficient under the air test. Answer: η N = 0.444,
C v = 0.666, and C d = 0.964.
The function of a diffuser, on the other hand, is to convert kinetic energy into pressure. Therefore, we define its
efficiency η D as
η D =
Isentropic enthalpy increase at the actual exit stagnation pressure
Inlet kinetic energy
= h es − h inlet
Vinlet 2 /2g ð cÞ = h es − h inlet
h oi − h inlet
where h es is the enthalpy at the actual exit stagnation pressure but at the same entropy as the inlet state (see
Figure 16.25).
For an ideal gas, the diffuser efficiency becomes
η D = c pðT es − T inlet Þ
V 2 inlet /ð2g cÞ
= T inletðT es /T inlet − 1Þ
V 2 inlet /ð2g cc p Þ