calculating kilowatts.

[theapprentice]

I have just installed a new package unit and the tradesman I work with has given me homework of calculating the kilowatts of refrigeration that is taking place. Im after a formula as I have lost one of my main tafe books. We had a 19.2 degree celcius air on the coil and a 27.7 degree celcius air off with 2350l/s of air over the coil. please help me lol
cheers chris.

[coolments]

Ok its been a while, I am sure some one will correct me if i am wrong but here goes.

I assume you got the air on / off wrong way round as the above indicates heating is taking place.

the formula you are looking for is.

Mass x Specific heat capacity x delta T = kw

your mass is: 2.35 cubic meters per second
Specific heat capacity of air is 1.01 per kg
Delta T is (temp difference) 8.5

Therefore 2.35 x 1.01 x 8.5 = 20.175 kw of duty.

[lana]

Hi there,

The theory coolments described is correct if there is only sensible cooling. If there is cooling with DE-humidification then the formula becomes :

Q = m x Dh

m is air mass flow (kg/s)
Dh is air enthalpy difference (kJ/kg)

In order to find the air enthalpy you have to know the air relative humidity or wet bulb. From psychrometric chart you can find the air enthalpy.

Hope this helps.
Cheers

[coolments]

Hi Psychrometric Chart below, hope it helps with your task.

[expat]

Coolments, I'm not sure he made a mistake there as he is way, way down south and only just coming into early spring.

[expat]

Coolments, Lana this chap is only twenty years old and calls himself the apprentice. You two are Jedis masters already, try to remember what it was like when you didn't yet know how much you didn't know.

This in mind theapprentice don't be daunted. Come back and tell us if you find the above graph and equations difficult to dicipher. Lana and Coolments know them backwards and will be able to help you understand; quicker than you think!

[frank]

Ok its been a while, I am sure some one will correct me if i am wrong but here goes.

I assume you got the air on / off wrong way round as the above indicates heating is taking place.

the formula you are looking for is.

Mass x Specific heat capacity x delta T = kw

your mass is: 2.35 cubic meters per second
Specific heat capacity of air is 1.01 per kg
Delta T is (temp difference) 8.5

Therefore 2.35 x 1.01 x 8.5 = 20.175 kw of duty.

You are forgetting to account for density.

Mass x Specific heat capacity x density x delta T = kw

[theapprentice]

na heating is right thats what mode we had it in. I also have one for cooling so Im about to head off to work so we will see how I go.
Thanks everyone

[ozairman]

na heating is right thats what mode we had it in. I also have one for cooling so Im about to head off to work so we will see how I go.
Thanks everyone

If its cooling one you have to do then you will need wet bulb and dry bulb temps so you can calculate the enthalpy change as well

[BigJon3475]

From plotting on the psyc. chart you can then gather latent and sensible loads. If you have the air flow and the enthalpy change it's fairly simple. If you take pressure readings like ESP or TSP you can use the performance chart provided with the equipment to figure out cfm. Now all this changes with the loads inside and outside so you have to take all the readings at one time. Also use the volume of air for each reading before and after the coil. If this is for heating there are charts usually available for both.

[BigJon3475]

Here is a useful tool also from


http://www.handsdownsoftware.com/Downloads.htm


I personally like the Carrier one the best.

[yorkie]

Hi Guys,
Q = m x CP x DT

where m= Air volume/ specific volume (at the air on condition)
CP= Specific heat of air 1.01kg/kg °K
DT= the temperature difference between inlet and outlet


That would calculate a Sensible capacity (ideal for your heating capacity)

If it was in cooling the total capacity would have to include the latent as well

This can be calculated by the amount of moisture removed x the latent heat of evaporation 2460kg.kg @16°C
0r multiply the amount of moisture removed per hour x 0.68333333

ie 2460/3600

i hope i am right as it is late at night