Modern Engineering Thermodynamics

12.4 Psychrometrics 417
Finally, the specific heat ratio of the mixture is
(Note that k m ≠∑w i k i because of its definition as a ratio.)
k m = c pm
c vm
= 4:61
2:78 = 1:66
Exercises
7. Rework Example 12.4 for a dive to 200. m below the surface of the water, where the pressure is 2.063 MN/m 2 . Answer:
χ O2
= 0:0103, χ He = 0.990, R m = 1.94 kJ/kg·K, c vm = 2.93 kJ/kg·K, c pm = 4.87 kJ/kg·K, and k = 1.66.
8. Rework Example 12.4 for an argon-oxygen mixture subject to the same condition of p O2 = 0:0212 MN/m 2 . Answer:
χ O2
= 0:0196, χ He = 0.980, R m = 0.209 kJ/kg·K, c vm = 3.08 kJ/kg·K, c pm = 5.13 kJ/kg·K, and k = 1.66.
Ideal gas mixture equations are used to produce property values in thermodynamic problems just as though the
mixture is a single unique gas. This is illustrated in the next example.
EXAMPLE 12.5
Determine the power per unit mass flow rate required to isentropically compress the helium-oxygen mixture described in
Example 12.2 from atmospheric pressure (0.101 MN/m 2 ) and 20.0°C to1.08MN/m 2 in a steady flow, steady state process.
Assume the mixture has constant specific heats.
Solution
The unknown here is the power per unit mass flow rate _W/ _m m . Since this is an open system, the energy rate balance for a
steady state, steady flow, single-inlet, single-outlet system is (neglecting kinetic and potential energy effects)
ERB (SS, SF, SI, SO)
and, since an isentropic process is normally also adiabatic,
For an ideal gas in an isentropic process, Eq. (7.38) gives
and, using the results of Example 12.4, this gives
Then,
_Q − _W + _m m ðh m1 − h m2 Þ = 0
_W/ _m m = h m1 − h m2 = c pm ðT m1 − T m2 Þ
T m2 = T m1 ðp m2 /p m1 Þ ðkm−1Þ/km
T m2 = ð20:0 + 273:15 KÞð1:08/0:101Þ 0:66/1:66 = 803 K = 530°C
_W/ _m m = c pm ðT m1 − T m2
Þ = ½4:61 kJ/ðkg .KÞ
Šð293 − 803 KÞ = −2350 kJ/kg
Exercises
9. Determine the isentropic power per unit mass flow rate required in Example 12.5 if the inlet temperature is 10.0°C
rather than 20.0°C. Answer: _W/ _m m = −2030 kJ/kg.
10. If the exit pressure in Example 12.5 is increased to 2.06 MN/m 2 , determine the required isentropic power input per unit
mass flow rate. Answer: _W/ _m m = −3110 kJ/kg.
11. Determine the aergonic (zero work) heat transfer required to cool the compressed mixture in Example 12.5 from 530°C
back to 20.0°C again. Answer: _Q/ _m m = −2350 kJ/kg.
12.4 PSYCHROMETRICS
Psychrometrics is the study of atmospheric air, which is a mixture of pure air and water vapor at atmospheric
pressure. 4 The pure air portion of an air–water vapor mixture is commonly called dry air; consequently,
atmospheric air is said to consist of a mixture of dry air and water vapor. Both the air and the water vapor in
4 The term psychrometer is from the Greek psychros (cold) and meter (measure).