I'm designing a heater to run off of a 12V battery at a field weather station. As I have no formal electrical training, I'm having a challenge coming up with an effective power budget. Note: This is similar to other questions posed on this Stack Exchange, but I haven't found anything that's quite answered my question. I also know that I could measure these once my circuit's assembled, but I'm hoping to avoid the cost of buying the components if I can get help deducing the answer beforehand.
Heat will be supplied from a resistive element designed to keep household pipes thawed. This element is designed to run off mains power, i.e. 120 VAC, and consumes 7 W/ft under this configuration. By my calculations, this means (7 W / 120 V =) 58.3 mA/ft of current. There should be no problem switching to DC input according to the best answer on this thread, but when I switch to battery power, I don't know how power/current draw will change. I envision 2 cases:
Case 1: Element draws same power at lower voltage, requiring 10X higher current.
7 W / 12 V = 583 mA draw
Case 2: Element draws same current at lower voltage, giving 1/10 power output.
12 V * 58.3 mA = 0.7 W
I need to know how this will behave so that I can budget battery power effectively. If the heater draws the same power at a higher current cost, my batteries will need more powerful/frequent recharging. If the heater draws lower power, my batteries will survive longer, but I'm assuming that this will come at the expense of poorer heating ability.
UPDATE SEPT 27: In searching for a 12V DC heat tape, I found the following link: http://www.oemheaters.com/t-dc-powered.aspx, which definitely steps through this discussion well, and provides a formula for the new power consumption of a 120V device running on 12V. The formula is as follows:
actual P = rated P * (applied V^2) / (rated V^2)
When you plug in my numbers:
actual P = 7W * (12V^2) / (120V^2) = 0.07W or 1/100 the power, as answered below…
Best Answer
Actually, it will draw 1/10 of the current at 1/10 voltage, producing 1/100 (1%) of the original power and heat!
This is because of Ohm's law: voltage = current * resistance (E=IR). Using algebra, I=E/R. Because the voltage, E is 1/10 while resistance, R is constant, I is 1/10.
Power = voltage * current. Because both voltage and current are 1/10 of the original value, power is 1/100 of the original value.
Your heater will produce 1% of the heat at 12V. Sorry.