Modern Engineering Thermodynamics

682 CHAPTER 16: Compressible Fluid Flow
p osi
p ose
h i = h oi
h
Actual
nozzle
process
Exit actual
specific kinetic
energy,
V 2 e /2g c
h e
h es
p e
s i
s
s e
FIGURE 16.24
The thermodynamic process path of a nozzle plotted on h–s coordinates.
Also, it is common to define a nozzle discharge coefficient C d as
C d =
Actual mass flow rate
Isentropic mass flow rate =
_m actual
ðρAVÞ =
actual
(16.41)
_m isentropic ðρAVÞ isentropic
Typical nozzle discharge coefficients run from 0.60 for sharp-edged nozzles (i.e., orifices) at low flow rates to
0.99 for properly designed nozzles at high flow rates.
EXAMPLE 16.14
Helium enters a newly designed test nozzle at 456.2 kN/m 2 and 283.7 K with a negligible velocity. The exit velocity,
temperature, and pressure are measured at the instant when the nozzle first becomes choked and are found to be 474.8 m/s,
370.4 kN/m 2 , and 260.1 K, respectively. For these conditions, determine the nozzle’s
a. Efficiency.
b. Velocity coefficient.
c. Discharge coefficient.
Solution
■ Equation (16.39) gives the nozzle’s efficiency η N as
k−1
ðV exit /c inlet Þ 2
η N =
2
1 − ðp exit /p inlet
Þ ðk−1Þ/k
For helium, k = 1.67 and R = 2.007 kJ/kg·K, and since the flow enters the nozzle with a negligible inlet velocity, we can
take T inlet ≈ T osi and p inlet ≈ p osi . Then,
p
c inlet ≈ c osi =
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1:67Þð1Þ½2077 m 2 /ðs 2 . KÞŠð283:7KÞ
= 992 m/s
and
0:67
ð474:8/992Þ 2
η N =
2
= 0:957
0:67/1:67
1 − ð370:4/456:2Þ ■
Equation (16.40) quickly gives the nozzle’s velocity coefficient C v as
p
C v =
ffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi
η N = 0:957 = 0:978