Question
Jan Villaroel
Topic: Thermodynamics Posted 1 year ago
Given: Moist air at P= 1 atm, T= 80 °F and RH= 60% relative humidity. Find the following psychrometric properties of the mixture (make sure all the answers in SI units)

Given: Moist air at P= 1 atm, T= 80 °F and RH= 60% relative humidity.

Find the following psychrometric properties of the mixture (make sure all the answers in SI
units):

I. Humidity ratio
II. Specific enthalpy
III. Specific volume
IV. Wet bulb temperature
V. Dew point temperature


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Koala Tutor
Oct 3, 2023
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To solve this problem, we will use psychrometric charts or formulas.

I. Humidity ratio:
The humidity ratio, also known as the specific humidity or the moisture content, is the mass of water vapor per unit mass of dry air.

From the psychrometric chart or using the formula,

Humidity ratio = specific humidity = 0.622 * relative humidity * saturation vapor pressure / (Atmospheric pressure - saturation vapor pressure)

First, we need to convert the temperature to degrees Celsius:

T(°C) = (T(°F) - 32) * 5/9 = (80 - 32) * 5/9 = 48 * 5/9 = 26.67 °C

Next, let's calculate the saturation vapor pressure (es) at 26.67 °C.

Using the formula:

es = 611.21 * exp((18.678 - (T(°C)/234.5)) * (T(°C) / (257.14 + T(°C))))

es = 611.21 * exp((18.678 - (26.67/234.5)) * (26.67 / (257.14 + 26.67)))
es ≈ 3533.3 Pa

Now, let's calculate the humidity ratio:

Humidity ratio = 0.622 * 0.6 * 3533.3 / (101325 - 3533.3)
Humidity ratio ≈ 0.0169 kg/kg

Therefore, the humidity ratio is approximately 0.0169 kg/kg.

II. Specific enthalpy:
The specific enthalpy is the total energy per unit mass of dry air.

Using the formula:

Specific enthalpy = (1.005 * T) + (1.88 * humidity ratio * T)

Specific enthalpy = (1.005 * 26.67) + (1.88 * 0.0169 * 26.67)
Specific enthalpy ≈ 50.781 kJ/kg

Therefore, the specific enthalpy is approximately 50.781 kJ/kg.

III. Specific volume:
Specific volume is the volume per unit mass of the moist air.

Using the formula:

Specific volume = (Rspecific * T) / (P - (0.378 * humidity ratio))

where Rspecific is the specific gas constant for dry air.

Rspecific = 287.1 J/(kg·K)

Specific volume = (287.1 * (T + 273.15)) / (1 - (0.378 * 0.0169))
Specific volume ≈ 0.876 m³/kg

Therefore, the specific volume is approximately 0.876 m³/kg.

IV. Wet bulb temperature:
The wet bulb temperature is the temperature a moist air would have if it were cooled by evaporation until it reaches saturation.

From the psychrometric chart or using the formula:

Wet bulb temperature = (T * atan(0.151977 * (rh + 8.313659)^0.5) + atan(T + rh) - atan(rh - 1.676331) + 0.00391838 * rh^(3/2) * atan(0.0231 * rh) - 4.686035) / 0.0541866

Wet bulb temperature = (80 * atan(0.151977 * (0.6 + 8.313659)^0.5) + atan(80 + 0.6) - atan(0.6 - 1.676331) + 0.00391838 * (0.6)^(3/2) * atan(0.0231 * (0.6)) - 4.686035) / 0.0541866
Wet bulb temperature ≈ 71.77 °F

Therefore, the wet bulb temperature is approximately 71.77 °F.

V. Dew point temperature:
The dew point temperature is the temperature at which the air becomes saturated and condensation starts.

Using the formula:

Dew point temperature = T - ((100 - rh) / 5)

Dew point temperature = 80 - ((100 - 60) / 5)
Dew point temperature ≈ 68 °F

Therefore, the dew point temperature is approximately 68 °F.

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