Say I have a 100W and 300W heating element that are both drawing 5A of current. The 100W heating element has an applied voltage of 20V, and the 300W heating element has an applied voltage of 60V. Would the 300W heating element heat up faster? I originally thought that all that mattered was the current passing through the heating element that dictated the heating time, but I want to fully understand how power plays a role.
-
\$\begingroup\$ No general relationship like that. Wattage vs physical size of the heater vs volume to be heated is what matters. If your 100W element is 1/3 the size of your 300W element as they often are if they are the same series they heat up at the same rate given the 100W element is heating 1/3 the volume of product. If they are heating the same volume of product it heats up slower. No straightforward relationship for heat trahsfer since it is about concentration and equilibrium. \$\endgroup\$– DKNguyenAug 6, 2021 at 13:12
-
\$\begingroup\$ If heats up means rate of increase in temperature, it depends on the material's physical properties as well. \$\endgroup\$– Mitu RajAug 6, 2021 at 13:17
-
1\$\begingroup\$ $$P\cdot t = E=mc\Delta T $$, 5A*20V=100W; 5A*60V=300W, so if they both heat the same capacity then 300W would heat 3x times faster. \$\endgroup\$– Marko BuršičAug 6, 2021 at 13:17
-
3\$\begingroup\$ Given the right context, either can heat up faster. A 60W headlamp bulb (5A,12V) is a heating element that heats up really fast, a 300W immersion heater in a cup of water will take minutes. So you need to know the thermal mass and cooling to answer. \$\endgroup\$– user16324Aug 6, 2021 at 13:23
3 Answers
What's missing from your question is the heat capacity of each heater. What might be missing is that you want to heat something up with the element -- I'm going to assume that's not what you're talking about. Well, and about a thousand other related things that may or may not pertain.
To simplify things, assume that you have heating elements that experience exactly the same temperature rise across the whole of the element, and assume that the only two things that differ is that they consume different amounts of power, and they have different heat capacities.
To reduce the math, let's further assume that you don't care about final temperature (we'd need yet another heat-related constant if so) -- let's say "heating up fast" means how fast the temperature rises when you first turn the thing on.
Say that heater 1 is a 100W heater, and has a heat capacity of 100 Joules/Kelvin. When you turn it on, its temperature will start rising at one degree C per second (because \$100 \mathrm{W} = 100 \mathrm{\frac{J}{s}}\$, and so \$100 \mathrm{W} / \left(100\mathrm{\frac{J}{K}} \right ) = 1 \mathrm{\frac{K}{s}}\$, and if you're talking temperature rise, a change of one C is the same as a change of one K).
Now say that heater 2 is your 300W heater, and has a heat capacity of 300 J/K. The initial rate of temperature rise will be the same.
Now let's change this around: say you beat your brains out making a 100W element that has the absolute least amount of material in it. Just for fun, say you've managed to make a 100W element that has a heat capacity of 1 J/K. If you power up that element, you'll see a rise of 100 degrees C per second. You can get higher rises yet -- when you get a slip of paper printed by a thermal printer, that printing happened because of heating elements that heat and cool very fast indeed, and ink in the paper that's activated by heat).
So -- can a 100W heating element heat up as fast as a 300W heating element -- absolutely.
Now, if you want to heat something up with your element, the story is different -- in that case, you care about the thermal capacity of the element plus the thermal capacity of the thing. In that case, once you get your elements small enough that their temperature rise is governed more by the thing you're heating than by the construction of the element itself -- the 300W element will win the race.
If the heaters are physically essentially identical in size and materials (say one uses a thinner wire but it's mushed into ceramic powder and sheathed in identical Inconel so the mass difference is negligible- then the 300W heater will begin heating at 3x the rate (in degrees per second) as the 100W heater.
They will likely take about the same amount of time to get within a smidgen of their final temperature, but the 300W heater will end up much hotter. In a world where conductive heat transfer is the only effect (say in a vacuum and radiation is negligible) the temperature rise of the 300W heater would be about 3x that of the 100W heater. In other words, if the 100W heater topped out at 50°C with a 25°C ambient, the 300W heater would top out at about 100°C.
All this falls apart when the heaters are physically different, when convection and radiation (nonlinear effects) are taken into account and perhaps when varying thermal conductivity and heat capacity are taken into account. (Not something we normally run into in electronics, but heat capacity drops to almost zero as the temperature approaches abs zero).
In the case of a closed-loop control, the higher power heater could reach the setpoint much faster, but might tend to overshoot more, especially if the sensor placement or controller tuning is not great. Sizing the heater properly will help optimize that.
Given that you are using three times the voltage for three times the power it seems that you are using the same gauge wire for both the 100 W and 300 W heater elements. In this case it should be clear that the power per unit length will be the same in both cases. Why? Because resistance per unit length will be the same and \$ P = I^2 R \$.
You may get some heat being conducted away through the terminals and this would be more significant on a short element more than a long one.
The steady state temperature will be reached when power in = heat lost to surroundings. If both elements have similar heat loss per unit length then the heat-up time (rather than heating time) will be the same.