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Hello.
This is my for post and i have tried to look everywhere for the answer but getting very confused.
What's the equation that will tell me the air coming off the fan coil unit when the flow and return temps is 82/71degrees. Taking a on-coil temperature of 20degress.
Is there an equation to work out the temperature of the air coming off the fan coil unit?
Thanks for your help
Regards
Robert
I am sure others may come up wit an equation, but surely if the heating medium (presuming water?) is dropping by aroung 11 deg, the the majority of that heat should end up in the air being passed over the coil, with some losses of course.
So with an on temp of 20, I would expect off coil of 30ish.
Not the precise calculation you asked for, but it kept me amused for a few minutes!!
:cool:
You need air flow and water flow for calculation!
Formula is Q=m*c*(T2-T1) both for primary and secondary.
m is mass of media
c is specific heat capacity of media
I fear such an equation should be a complicated one (but, sertainly, there should be such. In consideration must be taken: speed of the flow of the medium inside the coil, speed of the air through the coil, temperatures, power of the coil (its size)). Without the equation I would suggest the axle boxes on the fan's motor - which is probably bearingless - may be worn and its speed is slowed.
Hello Nike 123. There were no posts apart from that of the Robert's, when I posted. Fine formula. Do I understand it correctly that Q foir water should equal Q for air? That is , for instance, if Qw = 10*20*(82-71), (numbers for "m" and "c" here are arbitrary), for our unit's Qa = 5*10*(? -20) we find T2 by substitution.
Quote:
Originally Posted by Yuri B.
Formula is same but media at primary side and secondary side is not same.
If primary side loosed energy than secondary side gained that energy.
It remain only to calculate from known data how much energy is transferred.
For example:
Primary side=
100l/h water temperature difference of 5K, water entering temperature 45°C
Take that density of water is rounded to 1000 kg/m3 for this example purpose.
m of that 100l water = 100kg
Water specific heat c = 1,163 Wh/kgK
So we have heat transfer of 1,163 x 100 x 5 = 581Wh
We could say that same energy is given to secondary side.
Than (by substitution)
ΔT air=Q/m(air)*c(air)
Let say that we have at secondary side 200m3/h air flow and air entering temperature is 25°C, altitude is sea level and humidity of air is 65%. That gives us air density of 1,16 kg/m3
m(air) is = 203,2kg
c (air) is = 0,2884 Wh/kgK
ΔT air=581 / 203,2 x 0,2884 = 9,92K
That mean for air entering 25°C air leaving will be 34,92°C
That works only for heating. For cooling we need to calculate sensible and latent heat of air.
I think that is basic.
I use software of fan coil manufactures for such calculations. By comparing above calculation with their software results it appear that there is no need to add efficiency in calculation.
Hello.
Could we do a simple example;
For Water if we take Flow Temp as 82, Return Temp as 71degress, Specific Heat Capacity 4186JKG-1K-1 and Mass 1000KG/m3
For air if we take the supply temp at 20degress and Specific Heat Capacity 1000JKG-1K-1 and Mass 1.2KG/m3
The water flow rate is 0.2l/s and the air flow rate is 100l/s
How do we work out the off-coil temperature?
Please feel free to suggest other figures is these are not accurate.
Thanks
Robert
Oh just saw the above, one step ahead... nice one!
Quote:
Is there an equation to work out the temperature of the air coming off the fan coil unit?
T2=T1+((m*c*(T2-T1))/(m*c))
red is water
black is air
Quote:
That works only for heating. For cooling we need to calculate sensible and latent heat of air.
Nike 123.
Re-check your calcs, specific heat of water is 4.192 Kj/Kg.K The specific heat of air is 1.0 Kj/Kg.K, density of standard air is 1.2 Kg/m3, so effectively use 1.2 Kj/litre. K.
Otherwise Robertheppell will be barking up the wrong tree and be totally confused.
magoo
Magoo, convert kj to Wh!Quote:
Originally Posted by Magoo
Conversion factor is 0.277777
4,192kj=1,164Wh
I don't see error in my calculations.
Hi Nike 123. thanks for tip will be very useful.
magoo