BTU calculation

Myoungmike · 01-28-11, 12:56 PM

Anybody have a quick formula to calculate BTUs that I'm getting from the well water in my open loop GSHP? Entering water temp is 52 F, and discharge temp is about 38 F, pumping 12 GPM.

Thanks.

Mike

RobertSmalls · 01-28-11, 05:09 PM

Simple.

Heat flux equals mass flow rate times specific heat times change in temperature.
Q' = m' * K(H2O) * deltaT
Q' = 12gal/min*8(lb/gal)*4.2kJ/(kg*K)*(52°F-38°F)

Drop that whole line minus the Q'= into Wolfram Alpha, and it spits out the answer in any unit you like.
12gal/min*8(lb/gal)*(4.2kJ/(kg*K))*(52°F-38°F - Wolfram|Alpha)

81000BTU/hr, or 24kW.

mrd · 02-26-11, 08:12 AM

While doing some reading, I wondered how effective a solid metal heat pipe would be at transferring heat, instead of pumping fluid around. I threw some numbers together. Just a silly exercise, but I thought I'd throw it on this thread for kicks.

I found this formula for conductive heat transfer:
Q = k A dT t / L

dT temp diff
t transfer duration
A cross sectional area
L length
k thermal conductivity, J / (s m degreesC)

I found the thermal conductivity of aluminum = ~120 btu/(hr ft degreesF)
I converted it to ~208 watt/meter/K, which is equivalent units to k above.

I decided to see numbers for meeting a 20kbtu/h load.
20kbtu/h / 3.414 = ~5860 J/s

Q = k A dT t / L
Q / t / k = A dT / L
5860 / 208 = A dT / L
28.17 = A dT / L

From this I made a spreadsheet to see the required cross-sectional area of aluminum pipe to handle a given dT and length

length in meters, delta T in celcius, and the resultant data is the diameter of solid heat pipe, in inches.

The smallest is a 2 foot diameter pipe, a mere 1 meter long, with a deltaT of 1000 degrees C, to just meet a 20kbtu/h load ..

I wonder if I did something wrong, that seems ludicrous.

RobertSmalls · 02-26-11, 08:31 AM

I think you did something wrong. I did the following calculation:

Let k=200W/m*K, A=1m², l=1m, q'=6kW.
Find dT.
q'=k*A*dT/l
6000W=200W/(m*K)*1m²*dT /1m
30K=dT

So for each meter that you have to move 6kW through a 22.2" diameter Al rod, you need 54°F temperature differential. Mass transfer is a dramatically more effective way to move heat than conduction. Speaking of mass in motion, how much heat would you lose to convection off an uninsulated Al rod that's 54°F above ambient?

mrd · 02-26-11, 10:40 AM

The SQRTPI function did not do what I thought it did.

01-28-11, 12:56 PM	#1
Myoungmike Lurking Renovator Join Date: Jan 2011 Location: Camden, NY Posts: 7 Thanks: 0 Thanked 0 Times in 0 Posts	BTU calculation Anybody have a quick formula to calculate BTUs that I'm getting from the well water in my open loop GSHP? Entering water temp is 52 F, and discharge temp is about 38 F, pumping 12 GPM. Thanks. Mike

02-26-11, 08:12 AM	#3
mrd Apprentice EcoRenovator Join Date: Feb 2010 Location: Milford, DE Posts: 106 Thanks: 5 Thanked 9 Times in 9 Posts	While doing some reading, I wondered how effective a solid metal heat pipe would be at transferring heat, instead of pumping fluid around. I threw some numbers together. Just a silly exercise, but I thought I'd throw it on this thread for kicks. I found this formula for conductive heat transfer: Q = k A dT t / L dT temp diff t transfer duration A cross sectional area L length k thermal conductivity, J / (s m degreesC) I found the thermal conductivity of aluminum = ~120 btu/(hr ft degreesF) I converted it to ~208 watt/meter/K, which is equivalent units to k above. I decided to see numbers for meeting a 20kbtu/h load. 20kbtu/h / 3.414 = ~5860 J/s Q = k A dT t / L Q / t / k = A dT / L 5860 / 208 = A dT / L 28.17 = A dT / L From this I made a spreadsheet to see the required cross-sectional area of aluminum pipe to handle a given dT and length length in meters, delta T in celcius, and the resultant data is the diameter of solid heat pipe, in inches. The smallest is a 2 foot diameter pipe, a mere 1 meter long, with a deltaT of 1000 degrees C, to just meet a 20kbtu/h load .. I wonder if I did something wrong, that seems ludicrous. Attached Thumbnails

02-26-11, 10:40 AM	#5
mrd Apprentice EcoRenovator Join Date: Feb 2010 Location: Milford, DE Posts: 106 Thanks: 5 Thanked 9 Times in 9 Posts	The SQRTPI function did not do what I thought it did. Attached Thumbnails

01-28-11, 05:09 PM	#2
RobertSmalls Journeyman EcoRenovator Join Date: May 2009 Location: Buffalo, NY Posts: 344 Thanks: 3 Thanked 17 Times in 15 Posts	Simple. Heat flux equals mass flow rate times specific heat times change in temperature. Q' = m' * K(H2O) * deltaT Q' = 12gal/min8(lb/gal)4.2kJ/(kgK)(52°F-38°F) Drop that whole line minus the Q'= into Wolfram Alpha, and it spits out the answer in any unit you like. 12gal/min8(lb/gal)(4.2kJ/(kgK))(52°F-38°F - Wolfram\|Alpha) 81000BTU/hr, or 24kW.

02-26-11, 08:31 AM	#4
RobertSmalls Journeyman EcoRenovator Join Date: May 2009 Location: Buffalo, NY Posts: 344 Thanks: 3 Thanked 17 Times in 15 Posts	I think you did something wrong. I did the following calculation: Let k=200W/mK, A=1m², l=1m, q'=6kW. Find dT. q'=kAdT/l 6000W=200W/(mK)1m²dT /1m 30K=dT So for each meter that you have to move 6kW through a 22.2" diameter Al rod, you need 54°F temperature differential. Mass transfer is a dramatically more effective way to move heat than conduction. Speaking of mass in motion, how much heat would you lose to convection off an uninsulated Al rod that's 54°F above ambient?