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Homeroasters - Home Roasting Coffee Community » BUILDING A ROASTER » Building a Coffee Roaster

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electric heat properties: additive?
jedovaty	#1 Posted on 03/03/2012 11:35 PM
1/2 Pounder Posts: 225 Joined: October 31, 2011	I'm trying to wrap my head around how electric heat works.. Let's say there are two identical containers, empty, each has an identical electrical heat source, and a thermometer magically suspended right smack in the middle. We set the heat source so that both containers maintain a temperature of 400F. What happens if we take the heat source from one container and put it into the other? With the temperature in the container stay the same or increase? Would the above matter on how the heating source is set to maintain the temp? Whether voltage or amperage is reduced vs. one of those "on-off-on-off-on-off" infinite switches? Thank you for the physics lesson :) jano


lmclaren	#2 Posted on 03/04/2012 3:11 AM
1/4 Pounder Posts: 159 Joined: March 20, 2011	It depends on how you are controlling the heat source. If you have a fixed wattage that is calculated to offset the losses so that you reach the magic 440f, then doubling the wattage by adding the other heat source will increase the temperature until the losses verses the heat input from the element balance again. The resulting new temperature will be higher but not twice as high as the losses will increase with the increase delta t. If you have feedback and a control circuit, say in the case of a PID controller and a temperature sensor, then the control circuit will adjust the output of the heater until the set point is reached. If you double the wattage (or the elements) then you just reach the set point quicker. I remember reading somewhere that for PID you should have twice as much heat (wattage) as you need to allow the controller to quickly recover from increased load. Does this help? Lee


Dan	#3 Posted on 03/04/2012 8:08 AM
1 1/2 Pounder Posts: 1662 Joined: October 24, 2005	I'm trying to wrap my head around what it is you are confused about! In your comparison, everything is equal except the type of thermostat. Right? If so, then swapping the heater will have no effect. Both containers will still maintain 400?. All thermostats work on the same theory. As the temperature drops more heat is applied to replace what was lost. Some thermostats wait until a lot of heat was lost and then dump a lot of heat back in, usually overshooting your desired temperature, but in the end it averages out; sorta like a home electric oven with 25? or more swings. Other thermostats do the same thing but with smaller swings and are called laboratory or industrial thermostats. Then there are the thermostats that use software to anticipate and predict the needs of the system because they've been programmed or have "learned." These are PIDs and they also turn the heat on-off, but do it very quickly and can easily maintain ? 1?. There are thermostats that vary the current being supplied to the heater. Think of them turning the heater "a little more ON, and a little more OFF," but are probably going to have the same ? 1ۦ resolution as the PID. The real question is how much control do you want versus what you really need.
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JETROASTER	#4 Posted on 03/04/2012 8:45 AM
Administrator Posts: 1780 Joined: March 06, 2010	Moving one source into a shared container means they each work half as much to maintain 400. The control mechanism won't change the physics. Energy in, energy out. However, if you're looking to bend the rules of physics ....I'm in! -Scott


allenb	#5 Posted on 03/04/2012 9:10 AM
Administrator Posts: 3859 Joined: February 23, 2010	Quote jedovaty wrote: I'm trying to wrap my head around how electric heat works.. Let's say there are two identical containers, empty, each has an identical electrical heat source, and a thermometer magically suspended right smack in the middle. We set the heat source so that both containers maintain a temperature of 400F. Heat source = temperature controller or find a constant power level that maintains a temperature balance? What happens if we take the heat source from one container and put it into the other? With the temperature in the container stay the same or increase? If thermostatically controlled and if we're talking about the container now having two heat sources then as Dan stated, temp stays steady. If constant power level then temp goes up. If we're talking about the container with no heat source I'm afraid the temperature will drop quickly. Would the above matter on how the heating source is set to maintain the temp? Whether voltage or amperage is reduced vs. one of those "on-off-on-off-on-off" infinite switches? By voltage or amperage is reduced I'm assuming you're implying either a phase angle or PWM type of temperature controller versus a bimetal type infinite control. If an infinite switch is used and you leave the dial setting the same for both scenarios, the container with no heating element will still drop, the temperature in the one receiving the additional element will go up since an infinite switch can only maintain the duty cycle you set and does not maintain a temperature setpoint. Thank you for the physics lesson :) jano 1/2 lb and 1 lb drum, Siemens Sirocco fluidbed, presspot, chemex, cajun biggin brewer from the backwoods of Louisiana


ginny	#6 Posted on 03/04/2012 5:05 PM
Founder Posts: 3476 Joined: October 24, 2005	Quote I'm trying to wrap my head around how electric heat works if I were you I would not touch a thing!!! -g Edited by ginny on 03/04/2012 5:06 PM


jedovaty	#7 Posted on 03/05/2012 11:02 AM
1/2 Pounder Posts: 225 Joined: October 31, 2011	My apologies for not being more clear or explicit! But I think my question has been answered. To clarify, I am interested in what happens to the container which receives the second heating element. To recap what I have learned what happens to that second heating element: 1. It depends on how the heater is controlled: -- if it's by a temp sensor, then the temperature will stay the same. The amount of "power" each heating element draws will reduce. -- if it's by other means such as an infinite switch or a rheostat/ssr, then the heat will increase if you keep the knob in the same position and add the two heating elements together Given the above, how would one increase the amount of heat available without raising temperature? Or is that not possible?


lmclaren	#8 Posted on 03/05/2012 8:48 PM
1/4 Pounder Posts: 159 Joined: March 20, 2011	You would use a thermostat or (PID) controller to turn of the excess heat when the set point is reached.


jedovaty	#9 Posted on 03/06/2012 4:23 PM
1/2 Pounder Posts: 225 Joined: October 31, 2011	I don't think that would accomplish what I want.. more heat, without raising temp. One way is to reduce the container's size, I think... Is what I want possible? This has brought up another question. Is there any difference, from the bean in the roasting chamber's perspective, between the following methods of environment temp: -- off/on/off/on to keep env temps at desired level -- reducing one of the V-IR components to the electrical element, which would establish a constant heat output Hmmm. I'm making this way more complicated than it should be. Edited by jedovaty on 03/06/2012 4:23 PM

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