damn- I shoot that guy a email. Seems he carries alot of that stuff...
I wired in a 240V 30A circuit, I am using a 5500W ULWD RIPP element on a three wire, three prong cord.
Virtuous:
"Something is wrong with that GFCI then and I would have it checked. What that GFCI should be doing is monitoring current leaving L1 and then returning through L2 and vice versa depending on phase. if there is a difference of approx 4ma between the two it should trip. If you pull one of the lines off the gfci and run it to a 110 device with a neutral you have created a current imbalance between L1 and L2 and it should trip. Thats if your run the 110 device AFTER the gfci of course, if it is on the supply side it doesn't matter."
Perhaps I am confused (very likely). I have 50 amp 120/240 VAC Square D Homeline GFCI breaker in my load center. From this is wired a 4 prong, 50 amp, 240 vac outlet. I have a 50 amp power cord that runs to my enclosure containing my power distribution block. From this distribution block my two hot leads are divided to power two 4500 watt elements using 240 vac and a pump and pid using 120 vac (using one hot limb). My understanding of a 120/240 vac GFCI is that it detects diffferences in current between the hot leads and the neutral lead. If the difference is greater than 5 milliamps, it trips. The way my unit is wired, the algebraic sum of the current from the hot leads should equal that in the neutral unless I have an inappropriate connection to ground. I am not sure why you think my GFI should trip.
I am far from an expert on these matters. However, I had my plans and my system checked by a friend who is both an electrical engineer and a licensed electrician. I have been using it for two years without a problem. All that said, I really don't want to get electricuted.
2 SSRs are not overkill, but not required either. With one SSR, you are only breaking one leg of the heater. You will always have VOLTAGE at the heater, but not current.
Same principal as when some inept electrician mis-wires a light in the house and puts the switch in the neutral leg. Turn the switch off and the light goes off as expected. But change out the bulb and stick your finger in the socket, it will still eat you for lunch! Just remember, NEVER WORK ON THE HEATER WITH POWERCONNECTED BECAUSE ONE LEG WILL ALWAYS BE HOT!
No worries or problems bull8042 I also respect you replies. I'm in the same room as you, a just retired union IBEW electrician mostly industrial and commercial. 30 year pin next week. I must admit us wiremen and EE's have had disagreements on what is on the blueprints vs installing items which is a reality thing on the job sites. At times an EE's idea will not work or fit as designed.
I must correct you on one item above, many years ago up to app the early 1920's era or before they did switch the neutral. Yes the center of the socket was hot all the time.
I ran across this in an old 1898 house that was converted to electric when it became available way back then years ago plus it still had the old gas lines in the attic. I was working during the summer in the hot attic sweating away dripping in my glasses laying across gas pipes cutting wires with my side cutters then wow a damn big surprise. Across my arm and body and out thru my little friend down low. I got a big buzz as that was the best ground you can find a gas pipe.
They also had plugs hot on the right not left slot back then.
Nuff said, lets keep driving Pol crazy, J/K.
KSR3.80
This will cover you with no problem and is rated for switching a real load.
Please stay away from the residential stuff with a project like this....
That's a great find item as well price.
With this only one PID is required with a manual control on it,
switch the 240 volt power from your HLT to the boil kettle after the SSR's or a SSRD.
You also need to add a small DPDT switch to switch the temp probes
from the MLT to boil kettle back to PID for your heat control.
Found this with the P/N in the title:
Temperature Controller (CD101)
Main feature:
1. Input signal: Thermocouple: K, E, S, B, J, T, R, N thermal resistance: Pt100, Cu100, Cu50, BA2, BA1, GPt100 mini measure remote sending manometer: 0~400ohm linear signal: 0~20mV, 0~75mv, 0~200mV, 0~5V, 1~5V, 0~10mA, 4~20mA frequency 0~10KHz
2. Accuracy: +/-0. 5%F. S +/-1 word
3. Control way: Fuzzy PID control, states control, manual control
4. Control output: Relay contact point: AC250V 3A (resistance) logical voltage DC 0. 5V. Start pulse over zero: Optical couple control silicon output 1A 600V start pulse on transport: Optical couple control silicon 1 A 600V, 0~10mA, 0~20mA current output (below load resistance 600 ohms)
5. Alarm type: High limit alarm, low limit alarm, upper high limit alarm, below low limit alarm
6. Transmitter output: 0~10mA, 4~20mA, 0~5V, 1~5V
8. Power output: DC24V(used in double line transmitter)
9. Working power: AC220V+/-10%, 50/60Hz, AC85~264V 50/60Hz
10. Outline dimension: 48 x 48x 100mm(or 48 x 48 x 65)(width x height x depth)
11. Install dimension: 45 x 45mm
bull8042; I did not see any price for that PID unless I missed it?
I see about everything now comes from China especially in the electronics field. UL approved? BTW before a wireman a licensed A&P mechanic, 75 gas war got me in late 76 hence a job change.
I will not have a single thermocouple on my PID probably.
My HLT is already controlled and switched with a JC419 controller. (1500W heater)
I will only be using the PID for switching the SSr for the RIPP 5500W ULWD element in the kettle... phew! I have learned A LOT in a few days here guys, thanks so much for all of the info!!!!
I cannot wait to get this up and running and get some pics for yall to see what you have helped me build! Good thing we are getting Holiday Bonuses at the airline this year, I need to defer a couple hundred $$$$ to my projects.
I would be careful with that switch. The contact rating looks like is 24A, but you are switching a heater which is purely inductive with a big inrush. I would seriously consider the contacts ability to handle the load if the SSR is on when the switch is opened or closed.
The ratings state 2HP-240V-277V for a motor which is also an inductive load.
hp x 746
E x %Eff x pf
equates to 6.78A at 100% efficiency and a power factor of 1. Your heater element is not that efficient, so the current capacity would be less.
bull8042:
Isn't an electric heating element a resistive load? I use this switch on a 4500 watt element which would pull less than 21 amps. The max amperage rating of the switch is 24 amps. I understand why it would not be adequate for a 5500 watt element. Wouldn't it be safe for a 4500 watt element?