SADB Temperature and Coil Sizing 2

[samoja]

Hello,

Please can somebody explain me how to do following thing:

I am trying to size heating coil on fan power terminal unit and to determine SADB temperature.

If flow thru coil is 2000cfm, 400cfm is OA, envelope losses are 10,000. How can I determine coil size and SADB temperature.
My ideas was to determine coil size based on

Q=1.08*CFM*dT+envelope losses
I also need to determine SADB temperature, if I know that RADB temperature is 70F.

Here is where everything gets complicated, if I dont know SADB temp how can I determine dT

 

[quark]

If I am not missing something, this is a fundamental question and no complexity involved.

Some fundamentals

1. Heat loss from the control space(including ducting) should equal heat gain from the heater.

2. The return air temperature is lower end of the temperature control band in the control space, in case of heating and upper end of the temperature control band, in case of cooling. (For ex. if your space temp. limit is 21 to 25 C, during heating, at no point you want the temperature to be below 21C. The lowest temperature that exists in a control space is near the return)

3. Coil entry temperature should be the weighted average of OA and RA temperatures.

4. Use the sensible heat formula to find out the only one unknown i.e supply air temperature (or coil leaving temperature)

Hope this helps.

 

[samoja]

I still donet get it.
Lets use some numbers so it makes easy to understand.

Coil Flow=1500cfm
OA=400cfm at 55F
Envelope Load=10,000Btu/hr

Q=1.08*CFM*dT+Envelope Load= 1.08*1500*(SADB Temperature-Mixed Air Temp)+10,000Btu/hr
Mixed Air Temp can be calculated, and lets say that is 66F
than our formula is:

Q=1.08*1500*(SADB Temperature-66F)+10,000


It still looks like that I have one equation with two unknowns. What I am missing here?

Thanks

 

[HVACHawaii]

10,000 is the Q. Then solve for SADB

 

[MintJulep]

I agree with HVACHawaii.

 

[mikanical]

You need to determine the ventilation load in order to determine Q. Then you can determine the s/a temperature.

1. Q = 10,000 btuh (skin losses) + Qvent (TBD)

2. determine mixed air temperature. Say the outside winter design is 16 deg F (Vancouver, BC design). Then the mixed air temp would be,

(0.2*16)+(0.8*70)=59.2 deg F (weighted average)

3. the temperature difference for the ventilation load will be the difference between the R/A temp and the Mixed air temp.: 70degF-59.2degF. Therefore, Qvent=1.08*2000*(70-59.2)=23,328 btuh

4. Total heating load is 33,328 Btuh

5. Then determine S/A temp,

S/A={33328/(1.08*2000)}+59.2=74.6 deg F

 

[imok2]

mikanical

Qestion: Where did you get your mixed air temp I believe he said 66 *F, also how do you know that 10,000BTU is skin loses
also where is ventilation mentoned? I believe he said 1500 CFM tru coil It ould be nice if you used his numbers because I don,t understand any of yours (with respect)

MintJulep, how do you determine that 10,000 BTU is Q what am I not seeing This is what I worked up

’ Looks like you have to convert to 10,000 btu/hr to btu/min =167btu/min then you need to convert the 1500cfm to lbs/min = 112 lbs if you start at a mixed air temperature of 66*F and lets say 60%RH then the total heat = 24.74 btu/lb and at the total heat at 69*F and lets say 58,5% RH = 26.184 btu/lb. now we have 112 bs of air and total heat difference of 1.444 btu/lb air or 112 lbs x 1.444 = 162 btu/min close enough to 167btu/lb so I end end up with a 3*F dry bulb. Now lets run the formula and see what it comes out to
Q =162 btu/min x 1.08 x (66*F -69*F) = 525 btu/min = 31,500 btu hr
(Q=1.08*1500*(SADB Temperature-66F)+10,000) I'm assuming that the 10.000 btu represents T2 if it isn't then it's wrong but I always enjoy the exercise

"It still looks like that I have one equation with two unknowns. What I am missing here"?
IF you are to solve the equation you need to know either the total heat of the temp difference

 

[quark]

Presuming 10000 is in btu/hr and that is the heat loss from the control space, to maintain return air temperature of 70F, the room entering temperature will be [10000/(1.08*2000)]+70 = 74.63F

So, when you supply 2000cfm air inside the control space, the temperature drops down to 70F at return as the heat is lost from the space.

Now, the mixed air temperature is 0.2*55+0.8*70 = 67F.

Now, you have to heat 2000cfm air from 67F to 74.63F.

So, heating coil load is 1.08*2000*(74.63-67) = 16480.8 btu/hr

What mikanical did is also the same procedure but in a slightly different way and with a different OA temperature.

 

[imok2]

Quark, thanks for the response I didn't read the beginning good enough A star for your good work

 

[samoja]

I have better undestanding now.

Thank you all for your time and responses.