Answer :
The amount of moisture in the air by weight, the amount of water air can hold at various temperatures is as follows (you'll see why later): 30° C: 30 grams per cubic metre of air. 20° C: 17 grams per cubic metre of air. 10° C: 9 grams per cubic metre of air.
a temperature of 30 and a dew point of 30 will give you a relative humidity of 100%, but a temperature of 80 and a dew point of 60 produces a relative humidity of 50%.
Every home is different, but a level between 30 and 40 per cent humidity is typically ideal for keeping your home warm and comfortable in the winter, without leaving condensation on the windows. In the summer, that level can be higher, between 50 and 60 per cent.
While 30-40% is a good baseline for humidity levels in winter, you'll have to go through a period of adjusting temperatures and your humidifier to find your perfect setting. But once you do, there's no better feeling!
Concept:
For saturated air relative humidity [tex]$(\phi)$ is $100 \%$ i.e. $P_v=P_{v s}$[/tex]
i.e. partial pressure of water vapour in the moist air is equal to the saturation pressure of the water vapour
Specific Humidity: It is the ratio of the mass of water vapour to the mass of air at a given temperature and volume.
Specific humidity [tex]$\omega=\frac{m_v}{m_a} \times \frac{P_v}{P_t-P_v}$[/tex]
where [tex]$P_v$[/tex]is the partial pressure of water vapour and [tex]$P_t$[/tex] is the total pressure of moist air.
Calculation:
If we find the specific humidity before the cooling coil and after the cooling coil then the difference between two will give the amount of water condensed in the cooling coil.
[tex]$\phi=80 \%=0.8, m_v=18 \mathrm{~g} / \mathrm{mol}, \mathrm{m}_{\mathrm{a}}=28.94 \mathrm{~g} / \mathrm{mol}, \mathrm{P}_{\mathrm{t}}=105 \mathrm{kPa}$[/tex]
At[tex]$30^{\circ} \mathrm{C} P_{v s}=4.24 \mathrm{kPa}$[/tex]
Relative humidity [tex]$(\phi)=\frac{P_v}{P_{v s}}$[/tex]
[tex]$\begin{aligned}& \therefore P_v=\phi \times P_{v s}=0.8 \times 4.24 \\& \Rightarrow P_v=3.392 \mathrm{kPa}\end{aligned}$[/tex]
Specific humidity[tex]$(\omega)=\frac{m_v}{m_a} \times \frac{P_v}{P_t-P_v}$[/tex]
[tex]$(\omega)_{\text {moistair }}=\frac{18}{28.94} \times \frac{P_v}{105-P_v}$\therefore \omega_{\text {moistair }}=\frac{18}{28.94} \times \frac{3.392}{105-3.392}=0.0207 \mathrm{~kg} / \mathrm{kg}$[/tex]
of dry air
Now after the exit of the cooling coil
At[tex]$15^{\circ} \mathrm{C} P_{\mathrm{vs}}=1.7 \mathrm{kPa}$[/tex]
For saturated air[tex]$\phi=1$[/tex]
[tex]$\Rightarrow P_{\mathrm{V}}=P_{\mathrm{vs}}=1.7 \mathrm{kPa}$\omega_{\text {saturated air }}=\frac{18}{28.94} \times \frac{1.7}{100-1.7}=0.0107 \mathrm{~kg} / \mathrm{kg}$[/tex]of dry air
[tex]$\therefore$ Mass of water condensing $=\omega_{\text {moist air }}-\omega_{\text {saturated air }}$\Rightarrow$ Mass of water condensing $=0.0207-0.0107=0.010 \mathrm{~kg} / \mathrm{kg}$[/tex]of dry air
[tex]$\therefore$[/tex] Mass of water condensing [tex]$=\omega_{\text {moist air }}-\omega_{\text {saturated air }}$[/tex]
[tex]$\Rightarrow$ Mass of water condensing $=0.0207-0.0107=0.010 \mathrm{~kg} / \mathrm{kg}$ of dry air[/tex]
Mass of water condensing out from duct[tex]$=10 \mathrm{gm} / \mathrm{kg}$ of dry air[/tex]
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