# 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.

Example

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?

(a)

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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

 II II U III Table 2.1 Relationship between Different Temperature Scales Relationship Kelvin (K) Celsius (°C) Fahrenheit Rankine (R) Between (°F) Kelvin (K) — K=C +21315 K = —R 9 Celsius (°C) °C = K — 273.15 — "C = —CF — 32) 9 °C = — R -273.15 9 Fahrenheit (°F) 9 T =—°C + 32 5 °F = R-459.67 Rankine (R) R = 9-k 5 ~

Example

 Temperature in Celsius Kelvin Fahrenheit Rankine Water Boiling 100°C 373.15K 212°F 617.67°R Ice Point 0°C 273.15K 32°F 491.67°R Absolute Zero -273.15°C OK -459.67°F 0°R

(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

 P 1 R Liquid T = Tsat

Saturated

 Vapor Liquid

 T — Tsat quality vapor

 T

 1 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%

O Difference between W and 0:

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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Example

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.

Example

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)

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Example

Find the enthalpy of the air mixture in the above example.

Solution:

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