Required kW for cooling and dehumidification + reheating

In summary: AHU performance,- outside air temperature and humidity.For your specific case, the calculation would be as follows:1. Determine the dimensions of the cooling coil.2. Calculate the wattage needed to run the cooling coil at its rated speed.3. Compare the wattage needed to run the cooling coil to the wattage stated on the product packaging. If the wattage needed is greater, then the coil needs to be replaced.In summary, the calculations for a cooling and reheating process in a room with several ventilations per hour require the use of
  • #1
BigNick
5
2
Dear all,

First of all I would like to thank everyone for this great forum. I've been reading the forum for years, but I hadn't registered until today.
My registration on this forum obviously has a reason. I have an issue that I can't quite figure out.

I am trying to calculate how much power I need to cool and dehumidify a room in combination with reheating the air for comfort.
The thermodynamic calculations seem to be correct. However, I'm not sure about the time factor.
My doubts are caused by the fact that there are several ventilations per hour and therefore also varying needed power.

The goal of the calculation is to pass on the required kW for the configuration of the cooling coil and heat coil to the supplier, who makes the coils.

The input data and the progress of the process in the attached images. For simplicity I have omitted all external factors (such as heat loss and, for example, moisture / heat production by people). In addition, it is important to mention that it is a recirculation process. No air is supplied/mixed from outside.

My question is, what number of kW is correct and should I provide to the coil supplier?

Cooling coil:
A: (Q1coolcoil / 2) + (Q2coolcoil / 2) - the division by 2 because each ventilation takes 1800 seconds.
A: (51,135 kW / 2) + (44,975 kW / 2) = 48,055 kW
B: 51,135 kW
C: 44,975 kW

Heating coil:
A: (Q1heatcoil / 2) + (Q2heatcoil / 2) - the division by 2 because each ventilation takes 1800 seconds.
A: (8,493 kW / 2) + (10,769 kW / 2) = 9,631 kW
B: 8,493 kW
C: 10,769 kW

My gut tells me that Im converting answer A to kWh. But I'm not quite sure. Or is this right and is kW expressed in a time frame of 1 hour?

Who can explain to me what mistake I'm making?

Many thanks in advance,

01.PNG
02.PNG
03.PNG
 
Science news on Phys.org
  • #2
Welcome!

Could you please clarify the apparent contradiction between the input 2 changes/hour and “it is a recirculation process. No air is supplied/mixed from outside.”?
Aren’t any sources of latent and sensible heat within the regular load?
 
  • Like
Likes BigNick
  • #3
Thanks Lnewqban,

I will try to clarify the situation. I also made 2 ''sketches'' to express the process, check the attachment.
Situation:
* A room with a volume of 3600 m³.
* The air in this room is treated by an air handling unit that is located in the room.
* For simplicity, we assume no external sensible/latent loads influence the process.
* We assume a perfect airflow. The same cubic meter of air only returns to the AHU after 1800 seconds.
* In 1 hour, 2 recirculations are made.
* At the start condtions in the room are:
- Tdry = 25 °C
- RH = 85 %
* After 1 circulation (1800 secs) the condtions in the room are:
- Tdry = 20 °C
- RH = 80 %
* After 2 circulations (3600 secs) the condtions in the room are:
- Tdry = 15 °C
- RH = 75 %

My qeustions: How do I calculate the required power (kW) for the cooling & heating coil?

I hope my explanation has been clarified.

Thank you in advance,

Vent 1.PNG
Vent 2.PNG
 
  • #4
  • #5
BigNick said:
I am trying to calculate how much power I need to cool and dehumidify a room in combination with reheating the air for comfort.
The thermodynamic calculations seem to be correct. However, I'm not sure about the time factor.
My doubts are caused by the fact that there are several ventilations per hour and therefore also varying needed power.

The goal of the calculation is to pass on the required kW for the configuration of the cooling coil and heat coil to the supplier, who makes the coils.

The input data and the progress of the process in the attached images. For simplicity I have omitted all external factors (such as heat loss and, for example, moisture / heat production by people). In addition, it is important to mention that it is a recirculation process. No air is supplied/mixed from outside.

My question is, what number of kW is correct and should I provide to the coil supplier?
I checked your "Ventilation 1" calculations and the math and process works out. It's a straightforward cooling and reheat process to achieve a desired output temperature and RH from a given input temperature and RH (and airflow). The answers you got in kW are the correct answers.

But, I agree with @Lnewqban that the overall description isn't a standard HVAC processes/calculation. Such calculations are generally done in a steady-state, and this looks like a time-to-cool process. Maybe that is really what you need, but it is very unusual.

For the coil performance to provide the vendor; most I deal with do their own calculations of load based on entering and leaving conditions anyway. So I'd give them both stages and let them figure it out from there. But be aware that the true performance may differ substantially from what you calculated. If, for example, the cooling coil is a chilled water coil, you are likely to get much more cooling for the first stage than from the second, due to the larger approach temperature (difference between air and water temperatures).

BigNick said:
My gut tells me that Im converting answer A to kWh. But I'm not quite sure. Or is this right and is kW expressed in a time frame of 1 hour?
KW is just kW. It's thousands of joules per second. The time parameter is already correctly in seconds from your flow rate. No conversions needed.
 
  • Like
Likes BigNick and Lnewqban
  • #6
Lnewqban said:

Dear Lnewqban,

Thank you very much for the helpful documentation, I really appreciate this!

russ_watters said:
I checked your "Ventilation 1" calculations and the math and process works out. It's a straightforward cooling and reheat process to achieve a desired output temperature and RH from a given input temperature and RH (and airflow). The answers you got in kW are the correct answers.

But, I agree with @Lnewqban that the overall description isn't a standard HVAC processes/calculation. Such calculations are generally done in a steady-state, and this looks like a time-to-cool process. Maybe that is really what you need, but it is very unusual.

For the coil performance to provide the vendor; most I deal with do their own calculations of load based on entering and leaving conditions anyway. So I'd give them both stages and let them figure it out from there. But be aware that the true performance may differ substantially from what you calculated. If, for example, the cooling coil is a chilled water coil, you are likely to get much more cooling for the first stage than from the second, due to the larger approach temperature (difference between air and water temperatures). KW is just kW. It's thousands of joules per second. The time parameter is already correctly in seconds from your flow rate. No conversions needed.

Dear russ_watters,

Thanks for checking the calculation. Anyway, it's good to hear that I'm on the right track.

I've been looking into the time-to-cool process, and it looks like I was looking for this process as well. In combination with the results of the required power, I can form a good idea of the required installation.

Thanks also for pointing out the actual performance of the cooling coil. In this case it is also a chilled water coil. I will discuss with the supplier.
 
  • Like
Likes russ_watters and Lnewqban
  • #7
BigNick said:
Dear Lnewqban,

Thank you very much for the helpful documentation, I really appreciate this!

... In this case it is also a chilled water coil. I will discuss with the supplier.
The beauty of chilled water coils, especially if having six rows, is that the cooling capacity can be modulated via water flow, which keeps a constant temperature, which is good for dehumidification.
 
  • Like
  • Informative
Likes hutchphd, BigNick and russ_watters
  • #8
While we're at it and with this topic, I have one more question.

When I want to reverse the calculation process, I can't quite figure it out.

The data of the incoming air is known.
The air mass flow is known.
The relative humidity after the cooling coil is known.
The cooling power of the cooling coil is known.
The barometric pressure is known.

With this data I can calculate the enthalpy2 by means of:
Enthalpy 2 = Enthalpy1 - (cooling power / air mass flow)
Enthalpy 2 = 44,23 KJ/kg - (30 kW / 4,146 kg/sec)
Enthalpy 2 = 36,99 kJ/kg

Is it possible to calculate the absolute humidity with these 2 known parameters (100% RH and 36.99 kJ/kg)? I've tried various ways but it doesn't work.
When I have solved the absolute humidity I can also calculate the rest.

Thank you in advance!

Question 1.PNG
 
  • #9
BigNick said:
Is it possible to calculate the absolute humidity with these 2 known parameters (100% RH and 36.99 kJ/kg)? I've tried various ways but it doesn't work.View attachment 322416
Where are you getting the properties of air from to begin with? Are you looking them up on a psychrometric chart? If you have any two properties you can find your state point and get the rest from the chart.
 
  • Like
Likes BigNick
  • #10
russ_watters said:
Where are you getting the properties of air from to begin with? Are you looking them up on a psychrometric chart? If you have any two properties you can find your state point and get the rest from the chart.

The properties stated at ''Air in'' are just random (matching) properties as an example. My goal is not to have to use the psychrometric graph to read values. I would like to be able to calculate this after, for example, cooling & dehumidificate moist air.

Ashrae literature has so far helped me a lot to solve formulas/equations. But I can't solve the equation to calculate absolute humidity using enthalpy and relative humidity (and barometric pressure).

Hope you can point me in the right direction :)
 
  • #11
BigNick said:
The properties stated at ''Air in'' are just random (matching) properties as an example. My goal is not to have to use the psychrometric graph to read values. I would like to be able to calculate this after, for example, cooling & dehumidificate moist air.

Ashrae literature has so far helped me a lot to solve formulas/equations. But I can't solve the equation to calculate absolute humidity using enthalpy and relative humidity (and barometric pressure).
There is no simple equation for it. At best it's a curve fit. That's why the chart is used.
 
  • Like
Likes Lnewqban

Related to Required kW for cooling and dehumidification + reheating

What is the formula to calculate the required kW for cooling?

The formula to calculate the required kW for cooling is: Q = m × Cp × ΔT, where Q is the heat load in kW, m is the mass flow rate of the air in kg/s, Cp is the specific heat capacity of air (approximately 1.005 kJ/kg·K), and ΔT is the temperature difference in degrees Celsius.

How do you determine the required kW for dehumidification?

The required kW for dehumidification can be determined by calculating the latent heat load, which is the amount of energy needed to remove moisture from the air. This can be found using the formula: Q = m × Lv × (W1 - W2), where Q is the latent heat load in kW, m is the mass flow rate of the air in kg/s, Lv is the latent heat of vaporization (approximately 2,500 kJ/kg), and W1 and W2 are the humidity ratios before and after dehumidification, respectively.

What factors influence the kW required for reheating after cooling and dehumidification?

The kW required for reheating depends on the temperature to which the air needs to be reheated, the mass flow rate of the air, and the specific heat capacity of the air. The formula is: Q = m × Cp × ΔT, where Q is the heat load in kW, m is the mass flow rate in kg/s, Cp is the specific heat capacity of air, and ΔT is the temperature increase required.

Is it more energy-efficient to use a heat recovery system for reheating?

Yes, using a heat recovery system for reheating can be more energy-efficient. Heat recovery systems capture waste heat from the cooling process and use it to reheat the air, reducing the need for additional energy input and improving overall system efficiency.

How can you optimize the energy usage for cooling, dehumidification, and reheating?

To optimize energy usage, you can employ strategies such as using high-efficiency equipment, implementing variable speed drives for fans and pumps, utilizing heat recovery systems, maintaining proper insulation, and performing regular maintenance to ensure the system operates at peak efficiency. Additionally, integrating advanced control systems can help fine-tune the process and reduce unnecessary energy consumption.

Similar threads

Replies
10
Views
2K
Replies
32
Views
2K
  • General Engineering
Replies
10
Views
3K
Replies
2
Views
2K
Replies
35
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • Materials and Chemical Engineering
Replies
4
Views
2K
Replies
3
Views
2K
  • Mechanical Engineering
Replies
2
Views
2K
Back
Top