Modern Engineering Thermodynamics

9.5 Nozzles, Diffusers, and Throttles 287
EXAMPLE 9.3
Suppose 0.800 kg/s of argon flows at 93.0 m/s through an insulated diffuser from 97.0 kPa, 80.0°C to 101.3 kPa. Assuming
the argon to be an ideal gas with constant specific heats, determine the rate of entropy production within the diffuser.
Solution
First, draw a sketch of the system (Figure 9.3).
The unknown is _S P = ?, and the material is argon gas. The
thermodynamic station conditions are
Station 1
p 1 = 97:0kPa
T 1 = 80:0°C
Diffuser
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process Station 2
p 2 = 101:3kPa
The modified energy rate balance equation for this system is
_Q − _W = 0 = _m h 2 − h 1 + V2 2 −
V2 1
2g c
Assuming V 2 ≈ 0, and using the ideal gas auxiliary formula (Eq. (6.22)) with data for argon from Table C.13b,we
find that
Then, Eq. (9.18) gives
_S P = _m c p ln T 2
− R ln p
2
T 1 p 1
= 2:47 J/ðs.KÞ = 2:47 W/K
T 2 = T 1 + V2 1
ð93:0 m/sÞ 2
= ð80:0 + 273:15 KÞ+
2g c c p 21 ð Þ½523 J/ ðkg.KÞŠ
= 353:15 + 8:27 = 361:42 K
= ð0:800 kg/s
h
Þ ½523 J/ ðkg.KÞŠln 361:42
353:15 − ½ 208 J/ ð kg .KÞŠln 101:3 i
97:0
In this example, both the pressure and the temperature of the gas increase as it passes through the diffuser.
Exercises
7. Determine the entropy production rate in Example 9.3 when the mass flow rate of the argon is increased from 0.800 kg/s
to 1.30 kg/s. Keep the values of all the other variables the same as they are in Example 9.3. Answer: _S P = 4:01 W/K.
8. Determine the entropy production rate in Example 9.3 as the inlet velocity is increased from 93.0 m/s to 155 m/s. Keep
the values of all the other variables the same as they are in Example 9.3. Answer: _S P = 19:1 W/K.
9. Could the gas in Example 9.3 be changed from argon to air, keeping the values of all the other variables (except c p and
R) the same as they are in Example 9.3? Hint: Check the sign of _S P . Answer: No. When the values of c p and R for air are
used along with the values of the other variables given in Example 9.3, Eq. (9.18) gives _S P = −0:187. Since _S P < 0 in this
case, the process as described in Example 9.3 with air replacing argon cannot occur.
1
V 1 = 93.0 m/s
FIGURE 9.3
Example 9.3.
S P = ?
Diffuser
2
System boundary
m = 0.800 kg/s
of argon
EXAMPLE 9.4
Suppose 0.100 lbm/s of Refrigerant-134a is throttled across the expansion valve in a refrigeration unit. The R-134a enters the valve
as a saturated liquid at 100.°F and exits at 20.0°F with a quality of 53.0%. If the inlet and exit velocities are equal, determine
a. The entropy production rate inside the valve if the valve is not insulated and has an isothermal external surface
temperature of 60.0°F.
b. The entropy production rate inside the valve if it is insulated and assuming it has the same inlet conditions and exit
temperature as just stated.
c. The percent decrease in the entropy production rate of part a brought about by adding the insulation in part b.
Solution
First, draw a sketch of the system (Figure 9.4).
The unknowns are _S P uninsulated , _S P
insulated , and the percent decrease in _S P due to the insulation. The material is R-134a.
The thermodynamic station conditions are
Station 1
x 1 = 0:00 kPa
T 1 = 100:°F
Throttling
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process
Station 2
x 2 = 0:530 ðonly in part aÞ
T 2 = 20:0°F
(Continued )