[cdmaverick]

When 11.3KG formaldehyde(37%) reacts with 8.7KG monoethanolamine(99.7%),what's the heat exchange area?Since the reaction is exothermic,the area seems to be very large,right?

[eugenedakin]

Hi cdmaverick,

I need a little more clarification. You have two liquids (total of 20 Kg), and the reaction is exothermic. Could you help me understand your definition of the heat exchange area? My guess is that you want to remove the heat in some manner. Is there a rate at which you would like the heat dissipated, or a temperature that you do not want the liquid to be above?

Thanks for helping me understand the context of your question.

Sincerely,

Eugene

[cdmaverick]

hi,eugene,thanks for your reply.
yes,as u says,it's an exothermic reaction,we want the reaction temperature not rise above 60℃(140℉).
we design a shell and tube exchanger with area nearly to 0.8M2,but we are not sure it's correction.so we wanna some persons help us testify this data.
thanks.

[eugenedakin]

Hi cdmaverick,

I think a good 'first step' is to calculate out the thermal energy created.

Could you copy and paste your calculations here for me to double-check?

Thank you,

Eugene