Terminal Ends

A few points to consider:

1: read and understand what Redfred is trying to tell you!

2: The 1200 degree rating of your heating element is not the temperature that it will reach an any given application. This is just the maximum temperature that it can reach without failing. What is important to know is how many Watts the element is rated for. This will tell you how much of the "liquid" it can heat to a certain temperature in a given time. It will also tell you how much current the element will draw at a given supply voltage.

3: If you are heating this " liquid" (water? or are you making moonshine?) in a closed system; the first and most important piece of hardware you will need is a temperature/pressure relief valve. The valve should be sized to the volume, and expected temperature and pressure you are planning for and the pressure rating of your plumbing and containers.

4: A closed loop temperature control is a must. Depending on your application this could be as simple as an on/off thermostat such as you have in your home water heater. Better is a temperature sensor (thermocouple or RTD) and a PID controller if precise temperature control is necessary. If this is to be a high temperature and pressure system, you should also use a secondary alarm loop to shut down the power to the system in the event that the primary control loop fails. By" loop" I mean a sensor, a controller, and a power relay.

5: As far as termination is concerned, the most important thing to consider is how much current your heater will draw. If corosion is a consideration, avoid dissimilar metals. Copper to Copper is best. Consider plated copper. Depending on the current, Tin, Silver, or even Gold plating could be used.

Do not take this as instructions to build your system. This is just a few important things that you must consider. Any system design that involves boiling liquids, especially water, that are under pressure requires a thorough understanding of the engineering principlals envolved. A mistake could result in a hole in the ground where your house used to be.

The OP stated only three pieces of information about this heater.

1. That heater is a cord.
2. That the cord is four feet long.
3. That the cord is capable of producing 1200 degrees.

I agree that likely the temperature is a typographic error. Possibly it can only produce 200°C in free air, when no thermal mass is attached. Maybe it is actually a 1200 Watt heater. Then again maybe the fiberglass(?) insulation around the heating element is rated only to a maximum temperature of 1200°C. But my point here is that one of the three pieces of information that the OP provided about this heater is clearly wrong!

Add to this the small piece of information that he wishes to boil water in the confined space of a pipe and I say do not help this attempt. Again the OP maybe not be grasping the hazard of the stated goal because the intent is actually different from the statement. But I will not assist any bomb making ideas here!

Oh, what does any of this have to do with terminal ends, anyway.

As I stated, I agree wholeheartedly with your reservations. As I stated, I am not trying to tell the OP how to build anything. Rather, I am trying to point out the complexity and dangers involved in whatever he is trying to do. If he can't understand our replies, he has no business trying to do what he is apparently trying to do.

Gentleman

This is not a closed system. So no need for a relief valve. Second, the heated liquid is moving water. I understand the volume and the water flow rate will determine the heat temperature necessary to achieve a boiling point. The heating element is supposed to able to reach 1200 degree. The element is 30" long and rated at 150 watts. Attached to copper pipe with cold running water it will not be able to reach the maximum. I wish to regulate the element as necessary to achieve a boiling point of the running water as it flows throough the copper pipe.

What I am looking for is the direction for resources. The terminal connectors, temperature sensor, loop alarm shut off and appropriate water temperature guage for in line flowing water.

Gringogreg, Thank you for your insight. This is not my field of expertise. Thats why its good to ask questions. You have been helpfull.

PS I am not doing this in my house

I often find arrogance a sign of insecurity and a lack of maturity. Sometimes people injoy finding wrong with others. It make them feel better about them selfs.

Now is the time to learn some engineering. I am not going to do the calculations for you. First, as long as the system is open, that is, at atmospheric pressure, temperature is not an issue. As long as there is liquid water present the temperature can never exceed the boiling point 212 F or 100 C no matter how much heat you put into the system. To demonstrate this put a pot of water on your kitchen stove and place a thermometer in it. Heat the water on a low setting until it boils and measure the temperature, 212 degrees F (if you are at sea level) Now turn the stove to High. The water will boil more vigorously, but the temperature will remain the same.

The question is what volume of water do you wish to convert to steam (boil). This alone determines the energy, watts, that you will need. Look these values up, they are readily available, and do the calculations. Just be careful to to account for the units that you are working with. In the end you will arrive at the volume of water that your 150 watts can boil in a given time.

Thank you.

Giving it a moments thought, I doubt that your 150 watts will be enough to make a good cup of tea.

150 Watts is 150 Joules per second, or 35.8 calories per second (assuming zero losses).

80 calories are required to heat one cubic cm of water from room temperature (20°C, 68°F) to sea level boiling (100°C, 212°F). Another 540 calories are required to convert that cubic cm of boiling temperature water to steam (vaporized water). That is a total of 620 calories required to convert one cubic cm of water from room temperature to steam. 620 calories/35.8 calories per second = 17.3 seconds to convert that one cm^3 of room temperature water to steam under ideal conditions. Under real conditions, you will likely loose nearly half of that energy to the surroundings, so it will probably take closer to 30 seconds to boil 1cm^3 of water.

Now a cubic cm of water is roughly 1/5 of a teaspoon, so it will take roughly 150 seconds to boil a teaspoon of water. A cup is roughly 50 teaspoons, so it will take 50 * 150 = 7500 seconds or 2.1 hours to boil a cup of water.

You have no need to control the power to your heater! you need it to run at full power! What you need is to control the trickle of water flowing into the heater. I would suggest you cut the copper tube down to the shortest length that the heater cord can wrap around, and insulate, insulate, insulate!

By now it should be obvious that you need a much higher power than 150 Watts, unless this is for some tiny scientific experiment.

Back to your original question: The most logical connection between the power cord and the heater element is probably a crimp connection, or a pair of stainless steel screws and nuts. Note that the heat is going to degrade any connection over time, even without the salt you mention....

I was just about to post these same calculations but you beat me to it. GA. I am afraid that these figures will dump cold water on his plans. But, at least what his is trying to do is not likely to get him into any hot water.

OK, since I was very snarky earlier, and you do now have a few units that can allow for some calculations let me begin some of your calculations. One watt of power is the release of one Joule of energy per second. It takes about 4.2 joules of energy to raise the temperature of one gram of water one degree Celsius. (This is the another unit of energy size known as a gram calorie.) One gram of water is one milliliter. So to make a 200 ml cup of water originally at room temperature of 25°C up to 100°C (not boiling yet) will take 200 gm * 75°C * 4.2 joules/(gm*°C) = 63,000 joules. Your heating element will produce 150 joules/sec. So 63,000 joules /(150 joules/sec)= 420 sec= 7 minutes. Now that's if every single joule from your heater goes only into the 200 ml of water and nothing else. Now the latent heat to convert 200 gm of 100°C water to 100°C steam (to boil) is 452,000 joules of energy. (This last conversion is not really necessary to calculate because during this time one can call this as boiling water.)

I hope that this helps. As I said before this is some very rudimentary energy transfer calculations. Expect things to go a little slower than this calculation method because some of the heater energy will certainly go to also heating up all of the masses that are touching the water and heater element. Speaking of your heater element, you never did identify the 1200 degree number applies to which temperature scale?

I would consider a different method. Apply a suitable voltage (it won't be very high) to the ends of the copper pipe, and let the pipe be the heater. The principle is the same as using an arc welder to thaw frozen water pipes. Not enough information has been furnished yet, even to attempt any calculations.

Another thought: Do not confuse temperature with heat. They are two different things. The outer atsmophere of the earth has a temperature on the order of a million degrees, but you would be frozen in an instant if you should be exposed to it. On the other hand, imersion in water just a little above your body temperature will cook you.

Thank you all. You have been more than insightful.

Jim

I'm not even an engineer but I noticed that the temperature of 1200 doesn't qualify F or C. 1200 C is above the melting point of copper I believe?

Everyone has assumed you are quoting F degrees. Isn't that how we lost a Mars lander?

Maybe the 1200 degrees means that the heater can only be wrapped around the pipe 3&1/3 times.

I suspect that it's not a coincidence that the melting point of nichrome is near 1400°C. The heating element is probably a nichrome wire or coil wrapped in fiberglass.