Modern Engineering Thermodynamics

9.5 Nozzles, Diffusers, and Throttles 285
Notice that Eq. (9.17) is now written completely in terms of directly measurable physical quantities (m, c, T, v, p,
and V).
Similarly, if the fluid flowing through these devices is an ideal gas with constant specific heats c p and c v ,then
from Chapter 7, we have
s out − s in = c v ln T out
T in
+ R ln v out
v in
(7.36)
and, from Chapter 6, we have
= c p ln T out
T in
− R ln p out
p in
(7.37)
h out − h in = c p ðT out − T in Þ (6.22)
Combining Eqs. (9.12) and (7.37) gives the adiabatic modified energy rate balance equation for an ideal gas
with constant specific heats as
Entropy production rate of an ideal gas in an adiabatic nozzle, diffuser, or throttle
_S P
adiabatic
= _m c p ln T out
− R ln p
out
> 0 (9.18)
T in p in
ideal gas
In the case of diffusers, p out > p in and Eq. (9.18) requires that T out > T in , as in the case of incompressible fluid flow.
However, for nozzles and throttling devices, p out < p in and T out can be either greater or less than T in , depending on
the value of the Joule-Thomson coefficient (see Eq. (6.25)). Combining Eqs. (9.13), (9.16), (7.37), and (6.22)
gives the combined nonadiabatic first and second law relation for an ideal gas with constant specific heats as
Equation ð9:18Þ with the heat transfer rate evaluated using the ERB
ideal
_S P = _m c p ln T out
− R ln p
out
> 0 (9.19)
gas T in p in
− c p
T out − T in
T b
− V2 out − V 2 in
2g c T b
Finally, if the fluid flowing through these devices is neither an incompressible fluid nor an ideal gas, then Eqs.
(9.13) and (9.16) can still be combined and rearranged to give a combined nonadiabatic first and second law
relation of the form
The entropy production rate for a general fluid in a nozzle, diffuser, or throttle
_S P = − _m
ðh out − T b s out Þ− ðh in − T b s in Þ+ V2 out −
V2 in
> 0 (9.20)
T b
2g c
EXAMPLE 9.2
Determine the rate of entropy production as 0.2000 lbm/s of liquid water at 50.00°F, 95.00 psia flows through the nozzle
on the end of a garden hose and exits at 14.70 psia. The inlet and outlet diameters of the nozzle are 1.000 and 0.2500 in.,
respectively. Assume that the flow through the nozzle is too fast to allow a significant heat transfer to occur. 2
Solution
First, draw a sketch of the system (Figure 9.2).
The unknown is the rate of entropy production ( _S P = ?) forthis
system. The material is liquid water and the thermodynamic
station conditions are:
System boundary
S P = ?
1 2
Nozzle
m = 0.2000 lbm/s
of liquid water
Station 1
p 1 = 95:00 psia
T 1 = 50:00°F
Nozzle
ƒƒƒ!
process Station 2
p 2 = 14:70 psia
FIGURE 9.2
Example 9.2.
(Continued )