Atmospheric air at 35^∘ C with a relative humidity of 10 %, is too warm and also too dry. An air

Atmospheric air at 35^∘ C with a relative humidity of 10 %,...

Atmospheric air at $35^{\circ} \mathrm{C}$ with a relative humidity of $10 \%,$ is too warm and also too dry. An air conditioner should deliver air at $21^{\circ} \mathrm{C}$ and $50 \%$ relative humidity in the amount of $3600 \mathrm{m}^{3} / \mathrm{h}$ Sketch a setup to accomplish this. Find any amount of liquid (at $20^{\circ} \mathrm{C}$ ) that is needed or discarded and any heat transfer.

Key Concepts

Psychrometrics

Psychrometrics is the study of the thermodynamic properties of moist air, including temperature, relative humidity, dew point, and humidity ratio. This concept is fundamental for understanding how air-conditioning systems alter air conditions by simultaneously affecting both temperature and moisture content.

Sensible and Latent Heat Transfer

Sensible heat transfer refers to the energy exchanged due to a temperature change without a phase change, while latent heat transfer involves energy absorbed or released during a phase change (such as condensation or evaporation). In air-conditioning processes, both types of heat transfer are managed to achieve the desired temperature and humidity conditions.

Cooling Coils and Condensation

Cooling coils are used in air-conditioning systems to lower air temperature. When air is cooled below its dew point, water vapor condenses on the coils, effectively removing moisture from the air. This dual effect of cooling and dehumidification is essential in controlling both sensible and latent loads.

Humidification and Dehumidification Processes

Humidification involves adding water vapor to the air, whereas dehumidification removes excess moisture. Air-conditioning designs often incorporate both processes to precisely control indoor air quality by adjusting the moisture content to desired levels.

Air Mixing and Reheating

Air mixing and reheating are techniques used to achieve precise control over final air conditions. After primary cooling and dehumidification, air may be reheated or mixed with another air stream to attain the required temperature and humidity, ensuring occupant comfort and system efficiency.

Transcript

00:01 Hi, everybody.

00:02 So for this one, we need to find any amount of liquid that is needed for needed or discarded for any heat transfer.

00:11 So again, we need to find the pressure of v1, which is going to be 5pg2, partial pressure, by the way, guys.

00:20 Okay, so it's going to be 0 .1 times 5 .628, which is 0 .5628 kilopascals.

00:34 And we can find the absolute humidity as 0 .622.

00:39 And i'm just going to use absolute humidity pressure.

00:42 I'm not going to rewrite this out the equation for everybody.

00:47 So you should be able to know it in your book, or you can just look at the context clues that i am writing right here.

00:56 0 .5628 equals 0 .003474.

01:07 And now we have your phi 2 is at 50%.

01:14 And we can just find your pv2 at 5 times partial pressure of g2 equals 0 .5 times 2 .505, which equals 1 .252525k pascels.

01:38 And we can find the absolute humidity at 0 .622.

01:43 Again, at 1 .2525, divided by 100 minus 1 .2525.

01:53 2525 equals 0 .0778.

02:02 Okay.

02:04 And we can find the mass airflow rate, the mass airflow rate at p .a2 v.

02:15 Divided by r -a -t -2.

02:18 And we get 101 .325 minus 1 .25.

02:26 And that's one here we get 0 .287 times 21 plus 273 .15 equals 1 .185 equals 1 .1853 kilograms per second.

02:51 And so now we have your mass.

02:59 Of the liquid equals mass of a times your w1, 2 equals 1 .185, 3 times 0 .0078, minus 0 .003474.

03:33 Okay.

03:34 Your mass of your liquid is 0 .001503 kilograms per second...

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