For those who are interested in brewery embedded control, I am hoping to go beyond showing people how to assemble an embedded control system using PIDs and off-the-shelf parts. I am talking about understanding what is going on under the hood from an electrical and an embedded control software point of view, which will allow a brewer to build a brewery control system that can be monitored remotely via any computer with a web browser.

As an introduction, let's look at two basic electrical principles that we will encounter in electric brewing; namely, Watt's and Ohm's laws.

Ohm's Law

E = I x R, where E = electromotive force (measured in volts), I = current (measured in amps), and R = resistance (measured in ohms)

Watt's Law

P = E x I, , where P = power (measured in watts), E = electromotive force (measured in volts), and I = current (measured in amps)

By manipulating these two formulas algebraically, we can solve most of the electricity-related problems encountered in electrical brewing when using single-phase electric power.

For example, how many amps of current do we need on a 240 VAC line to run a 5500W heater element?

Re-writing Watt's Law to solve for I yields:

I = P / E

P = 5500W

E = 240VAC

I = 5500 / 240 = 22.917 amps (which would require the installation of at least a 30 amp electrical circuit)

One has probably read a posting on the Electric Brewing forum where Kal has stated that one has to divide the power output on a 240VAC heater element by 4 when using it on a 120VAC line, why?

First, lets determine the resistance of a 5500W, 240VAC heater element using Ohm's Law

Re-writing Ohm's Law to solve for R yields:

R = E / I

We calculated the amount of current drawn by this element above; therefore, all we need to do is plug in the values for E and I

R = 240 / 22.916 = 10.473 Ohms

Here's a trick that allows us to determine the power drawn by a device using only E and R.

Let's re-write Ohm's Law to solve for I.

I = E / R

If P = E x I and I = E / R, then P = E x E / R, or simply P = E^{2} / R

Setting E to 120 (the element used on a 120VAC line) and R to 10.473 (the resistance of the element) yields:

P = 120^{2} / 10.48 = 14400 / 10.473 = 1375W, which is 1/4th of 5500W

Why doesn't a 5500W element draw 2250W on a 120V circuit? Well, we cannot drop more than 120VAC between the hot (black) and neutral (white) wires and our resistance is 10.473 Ohms.

I = 120 / 10.473 = 11.458 amps, which is half as much current as is drawn by the element on a 240VAC circuit

If we are using half as much voltage and drawing half as much current, then we are drawing 1/4th the power drawn by the element on a 240VAC circuit.