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I have a task to create a software that calculates how much evaporative coolers are needed to maintain a certain temperature in a given room. I need to create a simple form, where variables are next :

1 - Location (i can fetch dry and wet bulb temperatures) 2 - Desired Temperature 3 - Air changes per hour 4 - Area of the room (m2) 5 - Number of people in the room

I am working with AOLAN evaporative coolers, and i have all the specification from them for the devices used. I just can't wrap my head around the calculations. So my question(s) are rather simple : How do i calculate the amount of coolers needed to cool down an area with all the given parameters up there.

I would love if someone could give me guidelines into how to achieve this.

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  • $\begingroup$ Are you familiar with energy balances on control volumes? Is the heat load to the room only due to the people in the room, or does it include heat from the outside? $\endgroup$
    – Jeffrey J Weimer
    Oct 5, 2018 at 14:09
  • $\begingroup$ All i know if the airflow of the device, which is 13.000. I have to input the room size, and let's say amount of people and that they generate x heat $\endgroup$
    – Vulegend
    Oct 5, 2018 at 14:10
  • $\begingroup$ I need to create something similar to this, but nevermine the code and the software, i am interested in the formulas. biocool.info/dimensionado-climatizacion-evaporativa $\endgroup$
    – Vulegend
    Oct 5, 2018 at 14:14
  • $\begingroup$ Depends on the humidity, they are about worthless above 80%. $\endgroup$
    – blacksmith37
    Oct 5, 2018 at 14:37
  • $\begingroup$ Ye ye, i know. But what i need to know is how many i need to cool down a room, some formulas for that. Like - i can calculate refrigerating power of one of my coolers, which at 13.000 airflow is about 42Kw`. And i know that from dry bulb and wet bulb temperature my dry bulb cool air leaving the cooler is 26 degrees for example. Now, what i don;t know if how to apply all of this to let's say a 1000 square meter room, and how much of these i would need to cool it. $\endgroup$
    – Vulegend
    Oct 5, 2018 at 14:49

1 Answer 1

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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.

picture of room

The steady state energy balance equation becomes

$$ \dot{m}_a\left( \tilde{H}_{a,out} - \tilde{H}_{a,in} \right) = \dot{q}_p \\ \left(\frac{A h}{t_{TO}}\right) \left(\frac{M_a p}{RT_{in}}\right) \tilde{C}_{p,a}\left(T_{out} - T_{in} \right) = N_p\hat{\dot{q}}_p $$

This says, the heat generated causes an enthalpy change in the air flow (out - in). The enthalpy change of an ideal gas causes it to undergo a temperature change.

You have the room area. You also need its height. This gives volume. Divided by residence time gives volumetric flow. With ideal gas density you get mass flow. The specific heat and molar mass of air are known. The pressure of the inlet air is known. The output temperature of your AC unit is the inlet temperature to the room. You might also adjust the flow rate of air in to the room. This is the residence time. You know the number of people and the heat per person. The remaining factor is the outlet temperature. This is the desired room temperature.

Your goal is to have a set room temperature. As more people enter the room, you will need either to increase the flow rate (decrease residence time) at the same inlet air temperature or you will need to decrease the inlet air temperature to the room (the outlet temperature of your AC unit) at the same air flow rate.

Sizing the AC Unit(s)

The minimum cooling load on the AC units is $\dot{q}_p$. You must remove at least the heat generated by the people in the room. The CoP of an AC is $CoP = \dot{q}/\dot{w}$, where $\dot{w}$ is the work required. This is to first order the electrical power input multiplied by an efficiency factor for the heat pump. Each AC takes $W$ watts power at an efficiency of $\epsilon$. The net result to establish the minimum number of units is

$$ \dot{q}_p = N_u\ CoP\ \epsilon\ W = N_p \hat{\dot{q}}_p $$

Summary

The approach might be as follows:

  • Determine the minimum number of AC units to meet the cooling load based on the efficiency, CoP, and power load of a unit as well as the number of people in the room and their heat output.

  • Determine whether the AC units will meet the demand to maintain a desired room temperature based on the rated output temperature and air flow rate of the units.

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  • $\begingroup$ Hello sir, it has been a long time. I've done quiet a bit of work with your help over here, and it works quiet well with the AC units. I have to adapt it to work with evaporative coolers - something like this : biocool.info/dimensionado-climatizacion-evaporativa Any chance you could give me some leads in how they've done it and the direction i need to go to. Thanks! $\endgroup$
    – Vulegend
    Nov 28, 2018 at 12:53
  • $\begingroup$ I am not familiar with evaporative coolers. I would not know where to start. $\endgroup$
    – Jeffrey J Weimer
    Nov 28, 2018 at 14:07
  • $\begingroup$ I am having this as the data, if you take a look at the screenshot prntscr.com/lo4b0f I can calculate the "refrigerating capacity" of my evaporative cooler. What would that exactly be, the CoP or something else? $\endgroup$
    – Vulegend
    Nov 28, 2018 at 15:30
  • $\begingroup$ And i can calculate the output temperature of an evaporative cooler as well $\endgroup$
    – Vulegend
    Nov 28, 2018 at 15:33

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