0
\$\begingroup\$

I want to design a riddle:
1. 12 metal plates on the wall
2. 8 plates at room temperature.
3. 4 plates at 40-45 Celcius(104-113 Fahrenheit), it is important the plate doesn't exceed the temperature limit for safety purpose.
4. players have to identify the 4 warmed plate by touching.
5. the riddle should be run 8-10 hours continuously a day.

The problem I am trying to solve is how to warm up the 4 plates safely for such duration.

Possible options:
1. Simply have a heater blowing hot air behind the 4 plates.
2. Let the plate act like a resistor and apply power to it.

Any recommendation is appreciated.

\$\endgroup\$
9
  • 2
    \$\begingroup\$ I'd just attach a heater to each plate - which can be as simple as an appropriately rated power resistor. Don't use the plate itself as a resistor, it won't have enough resistance. You'll also want some type of temperature sensor on each plate, so you can make the heater turn off when the plate is warm enough. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 8, 2018 at 23:10
  • 1
    \$\begingroup\$ A riddle or a griddle? ;) \$\endgroup\$
    – brhans
    Commented Mar 8, 2018 at 23:12
  • \$\begingroup\$ Haha, a riddle :) \$\endgroup\$
    – RoastDuck
    Commented Mar 8, 2018 at 23:15
  • \$\begingroup\$ -immibis, thanks for the idea. However, in the case that the temperature sensor fails(I can't assume it works 100% of the time), it would expose danger to the player as the heater won't turn off properly. \$\endgroup\$
    – RoastDuck
    Commented Mar 8, 2018 at 23:19
  • \$\begingroup\$ @RoastDuck in that case you could spec it so that the heater just barely has enough power to get the plate to 40 degrees. That way it's fail safe and you can also do away with the sensor, but it'll take longer to heat up. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 8, 2018 at 23:33

1 Answer 1

Reset to default
3
\$\begingroup\$

enter image description here

Figure 1. A 20 W wire-wound resistor.

The simplest thing would be to just heat the plates using a resistor mounted on the back. You could get a hunch of how much power you will need for this by carefully placing your hand on a 20 W lamp to feel the amount of heat it gives out. That should give you a rough guide to how much power you need per plate. Remember that your plate will be dissipating the power over a wider area so the temperature will be lower.

  • For safety you will need a low voltage supply. Let's say you go for 12 V which are common.
  • Next from \$ P = \frac {V^2}{R} \$ we can calculate the resistance. If we choose a 20 W resistor it will get very hot at the rated power so we should run it at 10 W to make it safe to touch. So rearranging the formula you can calculate \$ R = \frac {V^2}{P} = \frac {12^2}{10} = 14.4 \ \Omega \$. 15 Ω is the nearest standard value.

One or two of those wired in parallel on each plate should do the job.

Connect all your heaters in parallel.

Total current will be \$ I = \frac {V}{R} \cdot n \$ where n is the number of resistors.

\$\endgroup\$
1
  • \$\begingroup\$ A 20 watt resistor connected to a big metal plate won't necessarily get hot. The temperature depends on the power and the thermal conductance of the stuff taking heat away, not so much on the resistor's rating (except that higher rated resistors are bigger), so it should be fine to use a 20-watt resistor at 20 watts... \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Mar 9, 2018 at 2:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.