Moist air at 105 kPa, 30°C and 80% relative humidity flows over a cooling coil in an insulated air-conditioning duct. Saturated air exits the duct at 100 kPa and 15°C. The saturation pressure of water at 30°C and 15°C are 4.24 kPa and 1.7 kPa respectively. Molecular weight of water is 18 g/mol and that of air is 28.94 g/mol. The mass of water condensing out from the duct is ______ g/kg of dry air (round off to 2 decimal places).

The correct answer is 10.01.

Visual Analysis of the Image:
From the schematic diagram in the provided image, we see moist air entering a duct from the left, passing over a cooling coil, and leaving the duct on the right. The values and parameters labeled in the image are:

Inlet conditions (State 1):

Total pressure:
$P_{t1}=105kPa$
Dry bulb temperature:
${\mathrm{DBT}}_1=30°C$
Relative humidity:
${\phi }_1=0.8$
Saturation vapor pressure:
$P_{vs1}=4.24kPa$

Exit conditions (State 2):

Total pressure:
$P_{t2}=100kPa$
Dry bulb temperature:
${\mathrm{DBT}}_2=15°C$
Relative humidity:
${\phi }_2=1$
Saturation vapor pressure:
$P_{vs2}=1.7kPa$

Molar masses:

Molecular weight of water:
$M_{\text{water}}=18g/mol$
Molecular weight of dry air:
$M_{\text{air}}=28.94g/mol$

Step-by-Step Derivation and Calculation:

The humidity ratio (
$\omega$
), representing the mass of water vapor per unit mass of dry air, is given by the formula:

$\omega =\frac{M_{\text{water}}}{M_{\text{air}}}\times \frac{P_v}{P_t-P_v}$

where
$P_v$
is the partial pressure of water vapor and
$P_t$
is the total pressure of moist air.

1. Calculation of Humidity Ratio at Inlet (State 1):
The relative humidity is defined as:

${\phi }_1=\frac{P_{v1}}{P_{vs1}}$

Substituting the values:

$0.8=\frac{P_{v1}}{4.24}\Rightarrow P_{v1}=3.392kPa$

Now, compute the inlet humidity ratio (
${\omega }_1$
):

${\omega }_1=\frac{18}{28.94}\times \frac{3.392}{105-3.392}$

${\omega }_1=0.621976\times \frac{3.392}{101.608}\approx 0.020762kg\; water\; /\; kg\; dry\; air$

Converting to grams per kilogram of dry air:

${\omega }_1\approx 20.76g/kg\; of\; dry\; air$

2. Calculation of Humidity Ratio at Exit (State 2):
Since the air exiting the duct is saturated (
${\phi }_2=1$
), the partial pressure of water vapor is equal to the saturation vapor pressure:

$P_{v2}=P_{vs2}=1.7kPa$

Now, compute the exit humidity ratio (
${\omega }_2$
):

${\omega }_2=\frac{18}{28.94}\times \frac{1.7}{100-1.7}$

${\omega }_2=0.621976\times \frac{1.7}{98.3}\approx 0.010756kg\; water\; /\; kg\; dry\; air$

Converting to grams per kilogram of dry air:

${\omega }_2\approx 10.75g/kg\; of\; dry\; air$

3. Mass of Water Condensing Out:
The amount of water condensed per kilogram of dry air is the difference in humidity ratios between the inlet and the exit:

$\text{Mass of condensed water}={\omega }_1-{\omega }_2$

Substituting the calculated values:

$\text{Mass of condensed water}=20.76-10.75=10.01g/kg\; of\; dry\; air$