7

Assuming you mean evaporator as part of a vapour compression cycle, then because the fluid enters as a liquid and leaves as a gas, which means that the temperature can be the same but the internal energy changes due to the latent heat of vapourisation.


3

The evaporator is usually in the area to be cooled ie where the people work, read etc As the compressor takes space and tends to be noisy it is located away from that area. That makes the system in the « active » area smaller and quieter.


2

Mixing two media (air and water) just averages the temperature between them (proportionally to their respective heat capacities and total volumes) - through classic conduction. Additionally, evaporation absorbs a large amount of heat, reducing the resulting temperature (of both media) on exit. Rapid flow of large amounts of air speeds up evaporation ...


2

Please note the following fact about dehumidifiers: A dehumidifier is basically a room air conditioner which works by cooling the inlet air down just to the point where the water dissolved in it begins to condense out of solution. This temperature is called the dew point temperature. It does this by blowing the air over a set of cold tubes with refrigerant ...


2

What you describe is quite common where there isn't any problem with power and noise. It requires a bit more sophistication to manage the superheat. And yes, you need to quiet the thing some. I replaced a '60s era 20 ton compressor in a restaurant's dining room AC closet while they were serving dinner. This is also a good idea near the coast to keep the salt ...


1

I would start from the amount of liquid being sprayed, because your goal is to evaporate the liquid and leave the solute(powder). Once you know that you can calculate the heat required to add to the air, and calculate the approximate temperature of the hot air (so that you can monitor with a control system). Assuming you have a liquid of 1 kg that you need ...


1

Start with specifications. What temperature, what flow rate, from those two and the specific heat capacity of air you can work out the power required. $$ P = \Delta T \times m \times SHC $$ where $P$ is the power in watts, $m$ is the mass per second and $SHC$ is the specific heat capacity of air. You will also need the density of air if you want to convert ...


1

If your primary objective is to increase evaporation, you are more likely to decrease the temperature of the water rather than increase it. If you can accept the reduction of temperature, consider that many otherwise-stagnant ponds use fountains to provide for evaporation and agitation to prevent algae growth. If your tank is large enough, a small aquarium ...


1

I'm not an expert on swamp coolers, but I would try to keep all the evaporation outside of your dorm room. Here's an example: You build a heat exchanger between inside and outside, then you mist the exterior with water. You'd put fans on both sides preferably. If you mist enough, the exterior will reach the dew point temperature of the exterior environment. ...


1

Foundations (Energy Balance) For the Room Consider the system shown in the picture below. Air flows through a room volume (area and height) with a residence time. Heat is provided by people inside the room. The steady state energy balance equation becomes $$ \dot{m}_a\left( \tilde{H}_{a,out} - \tilde{H}_{a,in} \right) = \dot{q}_p \\ \left(\frac{A h}{t_{...


1

I'm going to take the formal question statement as "Is evaporative cooling for a pool effective?" There are a handful of factors that need to be considered. First and foremost though, is wet bulb temperature. The wet bulb temperature is theoretically the coldest temperature that you will ever be able to achieve utilizing a wet cooling system (a ...


1

The temperature sounds correct, the energy balance does not. I get 405°C for the compressor discharge but the enthalpy increases from 2547 kJ/kg to 3289 kJ/kg, for a work input to the gas of 743 kJ/kg, not 560 kJ/kg. I don't remember why you need to use enthalpy instead of internal energy in a compressor. But it is always correct and just as easy to look up,...


1

No, no, no, guys. Vapor compression recovers the latent heat of condensation that is normally lost to cooling water in a condenser. It does this by taking vapors that would normally go to a condenser and compressing them, raising their temp and pressure, then reinjecting them into the process. Since the energy in latent heat for water/steam is 5 or 6 times ...


1

What is the general range of relative humidity? You won't get anywhere unless the air is reasonably dry. Next, you want to maximize the evaporative area or most of the water will just roll off the roof. Essentially, you are looking at turning your roof into a radiator aka heat exchanger. This will involve bonding a lot of finstock or evaporative towers ...


1

If you are only interested in temperature, flow rate, and drying rate, then you could consider the following quantities for a given air temperature $T$: How much heat you need for evaporation (based on how much water is on the glassware)? How much heat you need to bring the entire mass of glassware to that temperature? What volume of air do you need to ...


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