BTU calculation

[KEN_KACEL]

I am looking for a formula to give me BTUs based on flow and temperature difference through a heat exchanger. The application is a variable speed pump which will control flow through a heat exchanger liquid-to-air with circulating fan and make sure the correct number of BTUs are removed from the liquid.

The flow will be controled through a PID and based on the variable in BTUs.

I am looking for some control on an outside wood burner (hydronic)

 

[Tom Jenkins]

I had to go back to my old college heat transfer text for this one! The rate of energy transfer through a heat exchanger is:

q = m x Cp x (Tin - Tout)

q = heat flow, BTU/hour

m = mass flow rate, lb per hour

Cp = heat capacity (specific heat) BTU/lb/°F

Tin, Tout = temperature into and out of the heat exchanger °F

 

[Steve Bettles]

Toms answer would give you the total heat lost from the water but this is a long way from being able to calculate the temperature of the air leaving the exchanger. As you increase the air flow rate you get different types of flow( laminar and turbulent etc) which affect the heat transfer from the exchanger wall to air. The easist way may be to produce a graph of heat lost from water to temperature of air, or air flow to air temp . i.e suck it and see.

 

[Mike]

Ken,

go to http://www.thermalinc.com/ this is a great web site for anything that has to do with heating. they have calculators and formulas for just about everything. You don't even have to remember any of the formulas later.

Mike

 

[Mike]

Ken,

Here is the actual address for the formulas page http://www.thermalinc.com/math/index.htm sorry for not giving it to you earlier.

Mike

 

[Allen Nelson]

Conservation of Energy

Toms answer would give you the total heat lost from the water but this is a long way from being able to calculate the temperature of the air leaving the exchanger.

Not so. When Tom first posted the equation, I thought "Which C_P should be used, air or water?

Then I realised that what the equation represents is the loss of energy by whichever medium you are interested in. Applying the Law of Conservation of Energy, the rate of energy loss of the liquid must equal the rate of energy gain in the air (assuming no radiated heat from the heat exchanger itself).

So if you measure the flow rate of the air through the heat exchanger, the infeed and exit temperatures, and look up the Cp_AIR (which might be tricky, as I think that it's a function of pressure, which may in turn be a function of the flow rate), and you'll have the other side of the energy balance.

 

[KEN_KACEL]

THANKS FOR THE "Q" :

MY EQUATION WILL BE-


(TCin-TCout) * (FM gal/min) * 500lbs/hr = Q BTU/hr

 

[shayne]

It's been a few years since my Chem. Eng. classes, however you can try this formula as vague as it may appear:

q=kA(delta T)/ x

where:
q = heat transfer (J/s or W)
A = area of material (m2)
delta T = temperature difference (C or K)(through x)
x = thickness of material (m)
k = thermal conductivity constant (W/mK or W/mC)

the answer for variable q has units of J/s; this can be converted to BTU/s simply using this conversion factor:

1 BTU = 1055 J or 1/1055 BTU = 1 J.

Enjoy!

 

[Tom Jenkins]

Watch your units

"MY EQUATION WILL BE-
(TCin-TCout) * (FM gal/min) * 500lbs/hr = Q BTU/hr"

This isn't going to work. First, you end up with totally ****eyed units. A good rule for any calculation is to write out the equation including the units, and do a simple algebraic cancellation of the units. If what you have left isn't hat yopu want either the formula is wrong or you need a unit conversion factor.

Second, you don't want both gal/min (a volumetric flow rate) and lbs/hr (a "mass" flow rate) in the equation - just lbs/hr.

You can use the equation I provided, just find the heat capacity in Joules/°C/kg or apply the required conversion factors, convert lbs/hr to kg/hr and you are good to go.

1 BTU = 1055 Joules
1 kg = 2.205 lbs

 

[KEN_KACEL]

The units are :

(T2-T1)degrees F * (FM)gal/min* 60 min/hr* 8lbs/gal = btu/hr

I realize dimensionally all units need to be correct

 

[Allen Nelson]

A perfect time to use the [strike ] tag

Ken:

Your units don't jibe:

(T2-T1)degrees F * (FM)gal/min* 60 min/hr* 8lbs/gal

or

       ºF           gal*      min       lbs
(T2-T1)--   *  (FM) ---  * 60 ---  *  8 ---
        1           min       hr        gal
 
[B]=[/B] 
 
                         ºF * [strike]gal[/strike] * [strike]min[/strike] * lbs
(T2-T1) * (FM) * 8 * 60 ---------------------
                         [strike]min[/strike] *  hr * [strike]gal[/strike]
 
[B]=[/B]  
 
(T2-T1) * (FM) * 8 * 60 (ºF lbs / hr)

which is NOT BTU/hr.

1) Are you only going to use 1 significant figure for your density (8 lb/gal) ? That will reduce the accuracy of your result.

2) You need to include the C_P of you liquid. That's where the BTU will come in, and the ºF will drop out.

 

[KEN_KACEL]

ok let's try this again. I will get it yet.

(T2-T1)F * (Cp)BTU/LB/F*(FM)gal/min*(8.34)lbs/gal*(60)min/hr= BTU

where Cp=1

 

[Tom Jenkins]

You get an A-, Ken. The formula is correct, but the final units are BTU/hr. This may seem picky, but over the years I have had so many problems related to unit conversion errors that I AM picky about it!

And, although the heat capacity of water is generally taken as 1, it actually is a function of temperature and pressure as well. The difference is small, but Cp for water varies from 1.001 BTU/lb/°F at 14.7 psia & 32°F to 1.021 BTU/lb/°F at 14.7 psia & 212 °F. It can get as high as 1.45 BTU/lb/°F at 680 °F and 6,000 psi.

And, to close the loop, the Cp of air is generally taken as 0.240 BTU/lb/°F, but as Allen points out it varies with pressure and temperature. Be careful with the Cp of air, as it is sometimes given as MCp, which is the molecular weight of air times the heat capacity.

 

[KEN_KACEL]

Thanks for the help....

I don't mind picky ... fudge factors in calculations do leave alot of room for error , there is no understanding of units and proportions.
PLCs do a good job on monitor and control of physical conditions(temp.,pressure,Volt.,Amp,Power,ect.) with floating point PLCs and the many math functions the physical enviorment can be controled effectively and now with the ever booming MMI the user can be notified with status in floating point with units in all languages.

It is a "strange and wonderful world we live in" ..... more strange then wonderful.

 

[mdonner]

Controlling BTU's

Hi

As you learn from the answers, BTU value is directly proportional to the flow when water (Tin-Tout) is constant.
So monitoring the (Tin-Tout) used for PID input, the PID output will control the pump's rotational speed hence flow.
(Tin-Tout) set point and PID parameters has to be cutomized to the system's characteristics and environmental conditions in mind:
- max BTU achived at max GPM at lowest inlet air temp.
- min BTU at max GPM at higest inlet air temp.

Another solution may use a flow meter and two temp sensors, wired to PLC's analog inputs, using the formula Q=G*Cp*(Tin-Tout) with correct units. If high accuracy is required, you may store in PLC's memory sevral figures of Cp correction factors as a function of averege water temp.