Modern Engineering Thermodynamics

16.10 Nozzle and Diffuser Efficiencies 681
by using Eqs. (16.13), (16.36), and (16.37). This results in the equation
p osy
p osx
=
h
k + 1
2 M2 x / 1+ k − 1
2k
k + 1 M2 x − k − 1
k + 1
2 M2 x
1/ k−1
i k/ ðk−1Þ
ð Þ
= 1:00
3:00 = 0:333
However, it is quite tedious to solve this equation for M x without using a computer, and since we have a direct entry
in Table C.19 at this value of p osy /p osx , we use this table in our solution. From Table C.19 at p osy /p osx = 0.333, we read
M x ≈ 2.98 and M y ≈ 0.476. From Table C.18 at M = M x = 2.98, we find that A/A ≈ 4.16. But our nozzle only has an
A e /A* = 2.00, so the shock wave must be in the exit plane; therefore, p exit = p E = 0.281 atm, M exit = M E = 2.20, and
T exit = 0.050813(293.15) = 148.96 K. The pressure readjustment from p exit to p B occurs outside the exit (see region c to d in
Figure 16.21). Finally, the exit velocity is given by
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
V exit = M exit c exit = M exit kg c RT exit
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
= ð2:20Þ
ð1:40Þð1Þ½286 m 2 /s ð
2 . KÞŠð148:96 KÞ
= 537 m/s
Exercises
35. Use Table C.18 to determine the exit Mach number and exit pressure in Example 16.13 for an exit to throat area ratio of
the converging-diverging nozzle of 6.78962. Assume all the other variables remain unchanged. Answer: M exit = 3.50, and
p exit = 39.33 kPa.
36. Suppose the back pressure in Example 16.13 is increased from 1.00 atm to 2.16261 atm. Use Table C.19 to determine
the values of M x and p exit . Answer: M x = 2.00 and p exit = 1.695 atm.
37. Determine the exit temperature and velocity in Example 16.13, if the upstream stagnation temperature is reduced from
20.0°C to 0.00°C and all the other variables remain unchanged. Answer: T exit = 139 K and V exit = 519 m/s.
16.10 NOZZLE AND DIFFUSER EFFICIENCIES
Inefficiencies in nozzles and diffusers result from irreversibilities that occur within their boundaries. Shock waves
and fluid friction (viscosity) in the wall boundary layer are the most common types of irreversibilities that occur.
If a nozzle or diffuser is not designed with exactly the correct wall contour, oblique shocks, boundary layer
separation, and turbulence destroy the nozzle’s performance.
Because nozzle and diffuser performance depend on their internal irreversibilities, we can base their efficiency
equation on the second law of thermodynamics by taking the isentropic nozzle and diffuser to be 100% efficient.
Then, since the function of a nozzle is to convert pressure (or thermal energy in the case of an ideal gas) into
kinetic energy, we can define its efficiency η N to be (see Figure 16.24)
η N =
Actual exit kinetic energy
Isentropic exit kinetic energy at the actual exit pressure
= ðV2 exit /2g cÞ actual
ðV 2 exit /2g cÞ isentropic
= ðV2 exit /2g cÞ actual
ðh inlet − h exit Þ s
and using Eqs. (7.38), (16.10), and (7.40), this can be written as
Nozzle efficiency
= ðV2 exit /2g cÞ actual
c p ðT inlet − T exit Þ s
ðk − 1Þ
ðV exit /c inlet Þ 2
η N =
2
(16.39)
1 − ðp exit /p inlet Þ ðk−1Þ/k
If the inlet velocity is veryp
slow, thentheentrancecanbetakentobethe isentropic stagnation state, or
p ínlet = p os and c inlet = c os =
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
kg c RT os : The p exit and V exit terms in Eq. (16.39) are the actual exit pressure and exit
velocity values that must be determined from measurements on the actual nozzle. Typical efficiencies for welldesigned
nozzles vary from 0.90 to 0.99 at high flow rates.
We can also define a nozzle velocity coefficient C v for the nozzle in a similar way:
Actual exit velocity
C v =
Isentropic exit velocity at the actual exit pressure = p ffiffiffiffiffi η N
(16.40)