Air heater sizing. 1

[itsmoked]

I'm faced with heating the inside of a sphere about 4 feet in diameter. I realize that I could put in a heater that would heat it up to the required 300F in 30 seconds but the process takes about an hour so there is no rush to reach setpoint. I want a PID to be happy about controlling this, not stuck with an overly twitchy system prone to horrible overshoot.

There will be air flow thru the sphere so that will be the biggest part of the heat loss picture. This will also allow for a reasonable thermal counter to the heater.

So my question. Is there any rule of thumb for sizing an air heater for robust but manageable control?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

Keith,

Wouldn't it be the other way around, i.e., size the drive according to the performance of the plant, which is the sphere?

You need a minimum of about 375W just for the steady-state convective loss, if it's a high thermal conductivity sphere.

The control system has to be sized around the disturbance response, i.e., how accurately do you control the temperature and what temporal response time you need.

The initial rampup will then be whatever you get from the control system and the plant response.

TTFN

FAQ731-376

 

[itsmoked]

IR; Thanks for that 375W value! I had yet to figure out how to even calculate it. Yes - it's just an uninsulated steel sphere.

I would assume that 375W value does NOT include radiated losses to surrounding room temperature surfaces?

It's a hard one to nail down, as in most obtuse designs..
It's a sphere with a rotating sphere inside it that is in-set an inch or so. Riding inside the inner sphere is 200 kilos of nuts. The idea is to roast the nuts at an accurate temperature. Since they have circulating air on both sides they provide no heat-loss insulation but do certainly contribute to thermal mass in a first order manner. Specific heat is something like 1415.3 J/kg K,
0.338038597 Btu/lb F.

Temp control to +/- 2F.
Response I have little clue. I'd like to see the product hit 300F in about 10 minutes. I could run two heating rates if I had to. One for heat up and one for maintenance.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

Correct, no radiation.

That sounds a bit tricky. Getting the air to temp is not necessarily a problem. Getting the product to temp without exceeding your spec temp would be an issue. Since heat flow is driven by temperature difference, the instant your air reaches 300°F, you've hit your max heat flow, and the flow to the product decreases from then on, unless it's overdriven.

As for the calculation, I used htc*area*230°F(deltaT), where area was 1/2 of the surface area of the ball; it's a swag, and htc was 2.5 W/m^2-K; sorry for the mixed units, I use Mathcad and it fixes the stuff on the fly. This assues that the external surface temp is 300°F, so it might be a bit lower with the thermal conductivity of the steel.

Oh, my bad, the radiation is WAY higher. sigma*emiss*area*(Ts^4-Tamb^4) gives about 2.1kW, with emiss=0.7, full surface area of the sphere, and 749R for surface temp, and 549R for ambient temp.

TTFN

FAQ731-376

 

[itsmoked]

Naw the 300F is where I want the product at not the maximum air temp. So I have room to move the product's temp about.

Ah, This was why I asked, "just convection?", because I was having a serious problem believing a ball 4ft in diameter would only require 375W to be at 300F. A 300W light bulb probably has a surface temp of only about 300F.

So are you saying 2.1kW to keep the surface at 300F with convection and radiation losses? I'd guess more like 7kW.

Thanks.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

Unfortunately, the answer could be pretty much anything in that range, depending on a variety of assumptions. It's one of those things that you keep iterating on the model until you are satisfied or tired ;-)

The more refined the model, the closer you get, but at some point, it'll be too complicated to get the gross feel.

In any case, since you have some leeway, you'd basically pump the hottest allowable air through the system, with as much flow as you can get through the product. I assume you'll need some sort of agitation on the nuts to get better circulation? 10 minutes still sounds challenging.

Is there a specific reason for the mechanical architecture? One would think that a conveyor belt pizza oven configuration would be more viable and repeatable. You could control both the temperature and duration quite well in that situation, in addition to having as much parasitic heat loss.

TTFN

FAQ731-376

 

[itsmoked]

The inner sphere rotates churning them.

The horrible mechanical config is a major headache but it's 'owned' and it's 'crazy ornate' for show reasons. So I'm stuck with it.

A rotating cylinder would be so much easier to deal with.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

OK, so the other, more obstinant, problem would be the actual Joule heat of the nuts.

200kg*1415J/kg-K*230°F/10min = 60kW!!!!

So, that's just the brute force calc of the amount of power required to raise the 200 kilos of nuts 230°F above a 70°F room temperature.

It's a lot of amps, a lot of BTU/hr. It comes out to 205 kBTU/hr, which is about 5 times the capacity of a single, heavy-duty commercial cooktop burner. Or, comparable to this tankless water heater:

http://www.noritz.com/professionals/products/view/0931_series_asme_tankless_water_heater/

TTFN

FAQ731-376

 

[itsmoked]

Groan.
Very illuminating.

Doesn't even account for the metal mass nor the fact that the whole thing is losing heat as it spools up.

Thanks IR.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

Well, bear in mind that the calc was tied to the 10 min warmup time, and the mass of the nuts. Changes to either would change the amount of power required.

TTFN

FAQ731-376