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TRASH101
20-02-2009, 03:07 PM
Does anyone have an equation to kg/hr from an electrode boiler to kj/kg or kw.

I look forward to any ideas :)

Thanks

narkom
20-02-2009, 07:49 PM
For 1kg/hr of vapor you need 0.725kW of energy.

TRASH101
23-02-2009, 08:37 AM
thanks for your reply Narkom

I believe that the 0.725kw is the electrical load to produce 1kg of water vapor in one hour but what I need is to calculate the change in enthalpy in the air due to the super heated vapor being introduced.

If you could help me further I would be thankful.

nike123
23-02-2009, 01:05 PM
thanks for your reply Narkom

I believe that the 0.725kw is the electrical load to produce 1kg of water vapor in one hour but what I need is to calculate the change in enthalpy in the air due to the super heated vapor being introduced.

If you could help me further I would be thankful.
http://www.engineeringtoolbox.com/enthalpy-moist-air-d_683.html

TRASH101
23-02-2009, 04:11 PM
Thanks Nike

Thats the sort of thing I was looking for

I thought I'd already checked the toolbox. I must be getting old.