# Thread: Enthalpy conversion btu/lb to kj/kg

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## Enthalpy conversion btu/lb to kj/kg

Can someone explain the conversion of enthalpy from btu/lb to kj/kg?

Every website I search says to multiply by 2.326, but when I do some checks (using web based calculator or psychrometric charts) the results do not work.

For example;
77F, 50%RH = 29.3 btu/lb
25C (i.e. 77F) 50%RH= 50.29 kj/kg
and 29.3 x 2.326 <> 50.29!

I am using an Alerton system which has an Enthalpy block which only supports F and btu/lb and I am working in SI (metric) units.

2. 1 Btu/lb = 2.326 kJ/kg
http://www.engineeringtoolbox.com/un..._per_unit_mass

Not sure where you're getting 50.29 from.

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Information required for proper conversion is 1 calorie is heat reqquired to raise 1 gram of water by 1 Deg C and 1 BTU is amount of heat require to raie 1 LB of water by 1 deg F . 1 calorie = 4.187 Joule . 1 Deg F = (9/5 C ) + 32. All this information gives all information to think upon.

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Information required for proper conversion is 1 calorie is heat reqquired to raise 1 gram of water by 1 Deg C and 1 BTU is amount of heat require to raie 1 LB of water by 1 deg F . 1 calorie = 4.187 Joule . 1 Deg F = (9/5 C ) + 32. All this information gives all information to think upon.

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Originally Posted by integrationx
Can someone explain the conversion of enthalpy from btu/lb to kj/kg?

Every website I search says to multiply by 2.326, but when I do some checks (using web based calculator or psychrometric charts) the results do not work.

For example;
77F, 50%RH = 29.3 btu/lb
25C (i.e. 77F) 50%RH= 50.29 kj/kg
and 29.3 x 2.326 <> 50.29!

I am using an Alerton system which has an Enthalpy block which only supports F and btu/lb and I am working in SI (metric) units.
That's a good question. Luckily for me, when I've done work where they use SI, they're still OK with BTU/LB. This explains it though. "You can not convert from metric to imperial enthalpy or vice versa directly."

6. forget osa/ra enthalpy comparison for economizer control

small difference in rh/temp accuracy equals large energy/capacity loss

better to use dewpoint transmitter from rotronics or airtest

open economizer only if,
osa db temp is < 72 F AND
call for cooling is present AND
osa dewpoint is < than 52.5 F

cheers
lb

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Actually, this question is very similar to converting temperatures between Fahrenheit and Celsius. Since they cross 0 at different temperatures, you cannot simply multiple by 9/5 and 5/9. You have to move the margin by adding/subtracting 32.
If you examine your psych charts, you will notice that enthalpy is 0 at 0 degrees for both SI and Imperial charts. You need to add/subtract the enthalpy at 32 Deg F and 0% rh. From my ASHRAE book, this would be 7.686 BTU/lb.
Therefore 29.3 BTU/lb = (29.3 -7.686) x 2.326 = 50.27 KJ/KG

8. Originally Posted by integrationx
Every website I search says to multiply by 2.326, but when I do some checks (using web based calculator or psychrometric charts) the results do not work.
you are absolutely right, ...... and the conversion factor is correct. How come ??

In essence the enthalpy is a meaningless number. What you are interested in is what is the enthalply difference between 2 conditions

Let's take your example
77F, 50%RH = 29.3 btu/lb
25C (i.e. 77F) 50%RH= 50.29 kj/kg
We will plot this condition on a psychrometric chart and call it point "A"

We will heat this air to 95F (or 35C)
On the psychrometric chart, heating is a horizontal line to the right and we will call this point "B"
On the Imperial psychrometric chart I find that RH = 28% .... Enthalpy = 34 btu/lb
On the Metric psychrometric chart I find that RH = 28% .... Enthalpy = 62 kj/kg

So the amount of heat required for 1 lbs of air to go from point "A" to point "B" is:
34 btu/lb - 29.3 btu/lb = 4.7 btu/lb
or.... 62 kj/kg - 50.29 kj/kg = 11.7 kj/kg

4.7 btu/lb * 2.326 = pretty close to 11.7 kj/kg

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