Modern Engineering Thermodynamics

12.8 Psychrometric Enthalpies 427
Exercises
24. Use the defining equation for h # to calculate the psychrometric enthalpy of air with a dry bulb temperature of 100.°C
and a relative humidity of 81.0% at a total pressure of 0.101 MPa. Answer: h # = 227 kJ/(kg dry air).
25. Use the defining equation for h # to calculate its value for air at 100.°C and a relative humidity of 69.1% at a total
pressure of 0.200 MPa. Answer: h # = 998 kJ/(kg dry air).
26. Use the defining equation for h # to calculate its value for air at 100.°F and a relative humidity of 52.6% at atmospheric
pressure. Compare your value with that found in Chart D.5. Answer: h # = 48.2 Btu/(lbm dry air).
Using the psychrometric enthalpy values in the energy
rate balance on the adiabatic saturator (see Figure 12.2)
gives
or
_m a h # 1 + _m w2h w2 − _m a h # 3 = 0
h # 1 + ω 2h w2 = h # 3
Typically, ω 2 h w2 ≪ h 1 , so that we obtain h # 1 ≈ h# 3 ,and
since both h a and h w depend only on temperature at low
pressures, lines of constant h # are parallel to lines of constant
T WB , as shown in Figure 12.8. The following example
illustrates the processes of dehumidification in an air
conditioning application.
p w
h # Scales
T WB Scales
T DB
φ = 100%
T WB = constant
h # = constant
FIGURE 12.8
Reading the psychrometric enthalpy from the psychrometric chart.
ω
EXAMPLE 12.11
A new paint-drying system requires air with a dry bulb temperature of 35.0°C and a relative humidity of 80.0% to be cooled
and dehumidified to a dry bulb temperature of 20.0°C and a relative humidity of 40.0%. Determine the heat transfer rate
per unit mass flow rate of dry air required to carry out this process.
Solution
The cooling and dehumidification process is illustrated in Figure 12.9.
For an aergonic (zero work) system, an energy rate balance on the air in the paint
booth gives
_Q = _m a h # 3 − _m ah # 1 − _m w2h w2
Neglecting the magnitude of the thermal energy associated with the condensate
water (i.e., setting _m w2 h w2 = 0), we get
_Q / _m a = h # 3 − h# 1
Values for psychrometric enthalpy are easily found on the psychrometric chart, Chart D.6.
Looking at this chart, we find the intersection of the lines for T DB1 = 35.0°C and for
ϕ 1 = 80.0%. Then, we follow the diagonal line upward and to the left until we intersect
the psychrometric enthalpy axis and read h # 1
≈ 110 kJ/kg dry air. Using a similar technique
at T DB3 = 20.0°C and ϕ 3 = 40.0%, we find that h 3 # ≈ 35 kJ/kg dry air. The heat
transfer rate per unit mass flow rate of dry air is
_Q / _m a = h # 3 − h# 1 − 35 − 110 = −75 kJ/kg dry air
Cooling and
condensation 1−2
φ 1
1
Reheating
φ
2−3
3 < φ 1
ω
p w
T DB T DB3 < T DB1
FIGURE 12.9
Example 12.11.
φ = 100%
2 3
This means that 75 kJ of thermal energy must be removed from every kilogram of (dry) air that passes through the system.
Exercises
27. Use Chart D.6 to determine the initial state psychrometric enthalpy h # 1
of the air in Example 12.11 if it is at 40.0°C
instead of 35.0°C at the same relative humidity. Answer: h # 1 ≈ 140 kJ/kg dry air.
28. If the final state in Example 12.11 is at a dry bulb temperature of 22.0°C and a relative humidity of 20.0%, use Chart
D.6 to find the psychrometric enthalpy of this state. Answer: h # 2
≈ 30 kJ/kg dry air.
29. Rework Example 12.11 for an inlet dry bulb temperature of 100.°F at a relative humidity of 70.0% and an exit dry bulb
temperature of 70.0°F at a relative humidity of 10.0%. Answer: _Q / _m a ≈−37:5 Btu/lbm dry air.