[S. cerevisiae]

I have attempted to keep computers out of my brewing hobby because I brew to get away from computers.  However, I have noticed that a lot of amateur and professional brewers would like to learn more about how embedded systems work.  I have been thinking about writing a book or at least starting a blog or a subforum  that is dedicated to teaching the basics of brewing embedded system design, starting with basic electricity and electronic principles and working methodically through complete computer controlled brewing and/or refrigeration systems.  Is anyone interested in learning about embedded systems design with respect to brewery control?

For those who do not know what the term embedded system means, an embedded system is a computer system that is embedded into a device, usually for the purpose of controlling the device.  A frequently encountered embedded system is the engine control unit (ECU) in one's car.  The ECU is also known as the powertrain control module (PCM).  For those who are not old enough to remember the joys of having to replace points, rotors, distributor caps, and adjust carburetors, the ECU replaced many mechanical engine control systems with electronic actuators and sensors. Engine control is handled via feedback/feed-forward computer algorithms that take inputs from sensors and send control signals to actuators. For example, readings taken from the manifold absolute pressure (MAP) sensor are used to control actuators that feed fuel to the fuel injectors and control ignition timing.   Before the advent of the ECU, cars had to be tuned on a regular basis.


By the way, the BrewPI is an embedded system.  It uses a Dallas Semiconductor DS18B20 one-wire temperature sensor for feedback and SSRs for feed-forward actuators.

[Slowbrew]

I've thought about what it would take to build these types of things but have never had the time to figure it out and actually build them.  I don't think I could add much to a conversation right now but would be interested in reading the information and learning a bit. 

Someday (about 8 years until my kids are all through college and the house is paid for) I may have the time and resources to start hacking.

Paul

[rabeb25]

I have a few of them embedded thingy's 

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-xfp1/v/t1.0-9/10530822_478490855626853_3202753433676354073_n.jpg?oh=c23dac658af0d8da4ecd9de7d5abc5b0&oe=5518F84C&__gda__=1427589829_82c47ca040ebcbf55d0726747992acde
 

and here

http://taplist.noip.me

[S. cerevisiae]

Does that system contain custom embedded software? 

[S. cerevisiae]

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 = E² / R

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


P = 120² / 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.

[rabeb25]

Does that system contain custom embedded software?

Yessir.

[S. cerevisiae]

Yessir.

Very cool!  I hoping to show that embedded control for brewing is not rocket science.  While those who have an electrical/electronic or software background have an leg up, nothing that is encountered in modern embedded control is ridiculously difficult to learn.