Hi Keith, rule of thumb in what sense? Heat loss is roughly proportional to temperature, so increasing the temperature difference with ambient by ~43% means ~43% more heat loss.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

I agree, the losses should increase by at least that much. Increases in convective and radiative heat losses will be greater but conductive heat loss though the walls probably dominates.

Steady state heat transfer is based on the ratio of absolute temperatures.
So 300F vs 400F is about 12% higher, so I would expect about 12% more power required.
At higher temps you would need to also factor in radiation loss, but not at these temps.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

Conductive heat loss is based on differential temperature no?

So assuming the outside is at say 70F, then the difference is 330/230 = 43% more?, so with a bit of extra convection etc, say 50% more heat loss?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

Convective and conductive heat transfers are proportional to temperature differences with the environment, so I picked 70F; that makes the new temperature delta 330F vs. 230F, hence ~43% more

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

Just what I was looking for and I'd say that even feels right against what I was seeing in the heat cycling. Seemed about 50% more cycles per time period.

I was visiting a bakery and was surprised to see 7 ovens idling. Some at 300F some at 400F. I asked around trying to discern when they were actually going to run product thru them. It was 1am. The general consensus was, "The baker will be in at 6AM".

Shortly after the president stumbled by and I mentioned, "you have 7 ovens idling here for 5 hours, maybe 4.5 hours if instead they were turned on 30 minutes before the baker shows up."

I asked, "Do you know how much it costs to idle one of these ovens per hour?"
Response, "I have no idea but that would be really nice to know but I have no idea how to figure that out."

So using a stopwatch I figured out how much gas one that was idling at 300F used. Then used the gas tariff tier tables and came up with the cost of idling at 300F.

1/3 Therm or $0.43/hour. A lot less than I expected cost wise, (if not planet-wise). I want to give them the cost but wanted to provide an approximate difference between 300F and 400F too.

Thanks everyone.

Keith Cress
kcress - http://www.flaminsystems.com

Natural gas, at this instant in history, is extremely cheap, mainly because it's a waste byproduct of various forms of oil production.

Nevertheless, that number seems low to me, but, maybe that's because they're idling. Something like this supposedly runs 110,000 BTU/(hr? the datasheet seems to be in error) The corresponding electric oven claims 18 kW power consumption, but electric would be more efficient in this case. Assuming, say, 50% of max heat for 5 hours and 7 ovens gives me 19 therms

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

Thanks IR.

These ovens are larger and look like this:



You wheel a six foot tall 4 foot square rack in that the oven lifts off the floor and rotates the entire time.

They're 275,000BTU/hr which I'm sure you realize means the burner power not the oven hourly consumption rate.

They run 14.3 cycles per hour and each cycle runs 31.5 seconds

I get 451 seconds of burn per hour. (That would be 275,000btu/hr)
100,000BTUs = 1 Therm

Here 1 Therm, in the quantity they use, costs $1.26

Keith Cress
kcress - http://www.flaminsystems.com

We had what seemed like an acre of fluorescent lighting on a timer, set to turn off at 6pm, but some people worked late in the office. I had a small lamp in my cube, so no problem, but one guy's office was simply sectioned off from the open area and had no other lighting. I calculated he was burning a dollar or two a day by coming out and flicking the timer over-ride switch. So I made a suggestion that some offices (naming no names) might get a floor lamp. Instead, the VP of the division tried to get a new $5000 lighting control solution for a different facility and, to get even more credit, that's what he directed a bunch of summer interns, (who were never allowed to have contact with any engineers) to investigate. Plus he named the intern program after himself.

While a great hoopla was made of the interns working on his self-named tiny company, they eventually discovered the other facility already had a timer lighting control installed and had just never bothered to use it. Big savings there. I wonder how much the interns cost.

So it remained that once a year the lighting controller that plunged me, and that guy who got an office, into darkness an hour early for two weeks because the facilities manager left an hour early and cared not one whit about the start of daylight saving time, and then on every remaining day, that guy would get to walk the 100 feet to the override and 100 feet back to burn a dollar a day (for him, workaholic that he was) so about $300 a year, and never got that $150 floor lamp. I was there about 25 years, so nearly $8,000, not including the summer heat load extraction for the A/C.

Typical, but keep burning those KWs.

Lighting is not the only areas where customers could reduce consumption. Is also possible for compressed air, AC, heat, etc. Problem is management just can't see the problem.

And I see it our offices also.

Heat loss from an oven will also involve convection. Convection requires the determination of the coefficient of convective heat transfer on the inside and outside surfaces of the oven. For more information about convective heat transfer coefficient you should consult engineering handbooks and heat transfer text books.