Hi. I am going to be installing my gas forge in my shop (walk out,
unfinished basement), and I am planning on having an exhaust fan in a
hood to suck out the exhaust/gases. I was wondering about using a
galvanized hood for such a setup? (Actually I was thinking that a
trash-can cut in 1/2 would make a pretty good and cheap one). My
forge is covered w/ BBQ high-temp paint, and this hasn't burned off
(except a little around the opening), so I'm wondering if the
relatively low exhaust temps will burn off the Zn? Should I paint the
hood with the BBQ paint?

Also, while we are here - I did a quick calculation and figured that I need to exhaust for about 275 CFM, does this seem reasonable for a forge burning about a BBQ tank/6hours? (I.e. I'm groping for a sanity check). My calc was thus, and it's been a while:

My forge burns a 20lb BBQ tank/6 hours (conservative). C3H8 == 0.044kg/mol

20lb * 2.2kg/lb * 1mol/0.044kg = 1000mol C3H8/6 hours

At STP 1mol occupies 22.4L

22.4L/mol

That is the volume of C3H8 at STP in the tank, but we to burn it we need 5 02/1 C3H8, which is

5 * 793ft^3 == 3968ft^3 of O2,

but air is 20% O2, so we need

3968ft^3 / 0.20 = 19841ft^3 Air at STP to burn a 20lb BBQ tank. But the exhaust is about 5x less dense, so that would be 99205ft^3 exhaust gas.

It takes 6 hours to burn the tank, so to remove that much gas we need to exhaust at 16534 ft^3/hour == 275 CFM.

Wow, that seems like a lot (previously I had forgotten to consider the incease in volume of hot air). It's 1/16th the volume of my shop. But I know that bathroom fans can run 150-200CFM, so I guess it's not completely out of order.

Anyone see a problem with this calc/reasoning (I corrected it twice while writing it out - see it helps to talk about such things ;) )? Of course I'll err on the high side and run a bigger fan and have ample fresh air intakes and have the CO/GAS detectors, but I was just trying to get a feel for the volume of air needed, and I (naturally) want to make sure that we are operating in a safe manner.

Thanks for your input,

don

Also, while we are here - I did a quick calculation and figured that I need to exhaust for about 275 CFM, does this seem reasonable for a forge burning about a BBQ tank/6hours? (I.e. I'm groping for a sanity check). My calc was thus, and it's been a while:

My forge burns a 20lb BBQ tank/6 hours (conservative). C3H8 == 0.044kg/mol

20lb * 2.2kg/lb * 1mol/0.044kg = 1000mol C3H8/6 hours

At STP 1mol occupies 22.4L

22.4L/mol

*** 1000mol ***0.265gal/L * 0.1337ft^3/gal = 793ft^3 at STPThat is the volume of C3H8 at STP in the tank, but we to burn it we need 5 02/1 C3H8, which is

5 * 793ft^3 == 3968ft^3 of O2,

but air is 20% O2, so we need

3968ft^3 / 0.20 = 19841ft^3 Air at STP to burn a 20lb BBQ tank. But the exhaust is about 5x less dense, so that would be 99205ft^3 exhaust gas.

It takes 6 hours to burn the tank, so to remove that much gas we need to exhaust at 16534 ft^3/hour == 275 CFM.

Wow, that seems like a lot (previously I had forgotten to consider the incease in volume of hot air). It's 1/16th the volume of my shop. But I know that bathroom fans can run 150-200CFM, so I guess it's not completely out of order.

Anyone see a problem with this calc/reasoning (I corrected it twice while writing it out - see it helps to talk about such things ;) )? Of course I'll err on the high side and run a bigger fan and have ample fresh air intakes and have the CO/GAS detectors, but I was just trying to get a feel for the volume of air needed, and I (naturally) want to make sure that we are operating in a safe manner.

Thanks for your input,

don