Fundamental Parameters
(1) Pressure
The air layer above the earth forms atmospheric pressure. Atmospheric pressure:
• sea level 14.692 psi
• elevation of 6600 ft, 11.513 psi
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(note: pressure inch mercury)
A, b —Table 3-2 in M&P Partial pressure (Dalton’s Law)
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Component/partial pressure: p = px + p2 + p3+…
Gibbs Dalton’s Law for Moist air:
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Pa—dry air component (constant)
Pv—vapor component (change with moisture content)
Note: When applying ideal gas law to each component of a mixture (e. g., moist air), should use partial pressure for the component.
For component i: PjVj=RjTj
Where Pi is partial pressure for component i.
One Ibm H20 vapor in 100 lbm dry air at standard pressure.
(a) What is pv vapor pressure? (b) What is saturation T at this pv?
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(b) Saturation: State of maximum concentration for mixture components.
(2) Temperature
Temperature is the macro results of molecular kinetics.
0th law of thermodynamics
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Example
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(3) Humidity Ratio. W
Definition: W/ = — (Kg/ Kgdm. ajr) ma
I. e. 1 kg dry air + w kg water vapor = (1+W) kg moist air O
Where P is the atmospheric pressure. Because pv « P, thus W °c pv
(4) Relative Humidity d)
Thermodynamic fluid states
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P R |
Vat |
)or |
T — Tsat saturated vapor |
T»Tsat Gas |
T>Tsat superheated vapor
X100% = — X100% Ps |
Pv = partial pressure of the water vapor in the air
Ps = partial pressure of the water vapor in a saturated mixture under the same temperature
Dry air: 0=0% Saturated air: 0=100%
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Moist air: W = 0.622——
P~P,
Saturated air: Ws = 0.622——
P-Ps
. w = Pv P~Ps =a P~P Ws ps P-pv P-p,
W P-p
=———— — x 100%
WsP-Ps
Since P» pv and P >>ps
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Further
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Determine the humidity ratio of moist air at a temperature of 24°C and a relative humidity of 50% at a standard pressure latm
Given: T, (])
Find: W
Solution:
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(5) Dewpoint Temperature, Ta
Td — the saturated temperature of a given mixture at the same pressure and humidity ratio.
Find Td of the air in the above example.
Solution:
At the dew point temperature: air mixture 0=100%
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(6) Enthalpy
Enthalpy of the moist air = enthalpy of the dry air + enthalpy of the water vapor Enthalpy is energy per unit mass.
I = ia + W iv
Ia = Cp. a T iv = ig + Cp;v T
Where
Cp, a = specific heat of dry air kJ/(kg°C), Btu/(lb°F)
Cp, v = specific heat of water vapor kJ/(kg°C), Btu/(lb°F)
Ig = the enthalpy of saturated water vapor at 0°C or 0°F.
Ig = 2501.3 kJ/kg at 0°C; ig = 1061.2 Btu/lb at 0°F.
Therefore, we have
I = Cp, aT + W(ig + Cp, vT)
Find the enthalpy of the air mixture in the above example.
Solution:
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Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning