Can constant electrical current heat water?

callie123

Gold Member
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Hi, I'm trying to make a realistic scene, in which water is kept under constant electrical charge. If the power is around 75 milliamps, and it's in a stream about twenty feet across by ten feet deep, would this affect water temperature? Thanks!
 

berkeman

Mentor
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Nah. You mention a current (not a power) of 75mA; that is pretty low. And water wouldn't be kept under constant electrical "charge". You could maybe keep a constant voltage acros the stream of 50V or so, and the current that would result would depend on the conductivity of the stream. V=IR, where V is voltage across the resistance R and the current resulting is I. The power is V^2/R or I^2*R.

Hope that helps.
 

callie123

Gold Member
19
8
Thanks, Berkeman. Is there an amount that would affect temperature?
 
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What is an electrical currant? Is it dangerous to eat them? Are they safe to eat if you're wearing rubber boots? :oldbiggrin:

If you ground the bush first, can you pick them without getting electrocuted?
 

callie123

Gold Member
19
8
What is an electrical currant? Is it dangerous to eat them? Are they safe to eat if you're wearing rubber boots? :oldbiggrin:

If you ground the bush first, can you pick them without getting electrocuted?
You sir, have not only caused me to wince at my misspelling, but have also made me laugh. To prevent our readers from going in search of these electrifying bushes, I've corrected the error. Thanks.
 
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You sir, have not only caused me to wince at my misspelling, but have also made me laugh.
I'm happy to hear this, and that you took it in the spirit with which it was intended.
Your electrical "current" reminded me of a time years ago when a friend and I stopped off to get some dinner after a three-day backpack adventure. The waitress asked if we wanted coffee, and my friend asked if it was fresh. The waitress replied, "Yes, it's just been grounded."
 

callie123

Gold Member
19
8
I'm happy to hear this, and that you took it in the spirit with which it was intended.
Your electrical "current" reminded me of a time years ago when a friend and I stopped off to get some dinner after a three-day backpack adventure. The waitress asked if we wanted coffee, and my friend asked if it was fresh. The waitress replied, "Yes, it's just been grounded."
Well, we know where she picked those beans, don't we? :wink:
 

Drakkith

Staff Emeritus
Science Advisor
2018 Award
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Thanks, Berkeman. Is there an amount that would affect temperature?
It depends. A 20 foot stream that's 10 feet deep and moving is a very difficult thing to heat up. Assuming the stream's water speed is roughly 7 ft/s (5 mph), that's over 1400 cubic feet of water moving through the stream every second, or almost 10,500 gallons. To raise the temperature of the stream by 5 degrees F takes 437,500 BTU's, or 128 kW's of power.

This is a lot of power, but not that much. It's equivalent to about 171 horsepower.

To continue, we now need the electrical resistance of the water. But here there's a problem. In a standard circuit, all components are of a small size with clear-cut boundaries. A resistor, for example, has clearly defined boundaries that make it relatively easy to calculate the resistance. A moving body of water, on the other hand, has no clear-cut boundaries except at where you place your electrodes. Basically, the current can run not only straight through the volume of water we just talked about to get to the electrode on the other side, but also up and then back down the stream, or down and then back up the stream.

The net result is that the resistance of the water is very low. Even our volume of water by itself has a resistance of only 0.286 ohms or so (using a resistivity of 1 ohm per meter as a compromise between the 0.2 of salt water and the 2-200 of drinking water). 128 kW's across this resistance equates to roughly 191 volts and 669 amps.

Note that all of the above calculations are very rough, use many simplifications, and were done by a guy whose a little rusty at solving these kinds of problems.
 

callie123

Gold Member
19
8
It depends. A 20 foot stream that's 10 feet deep and moving is a very difficult thing to heat up. Assuming the stream's water speed is roughly 7 ft/s (5 mph), that's over 1400 cubic feet of water moving through the stream every second, or almost 10,500 gallons. To raise the temperature of the stream by 5 degrees F takes 437,500 BTU's, or 128 kW's of power.

This is a lot of power, but not that much. It's equivalent to about 171 horsepower.

To continue, we now need the electrical resistance of the water. But here there's a problem. In a standard circuit, all components are of a small size with clear-cut boundaries. A resistor, for example, has clearly defined boundaries that make it relatively easy to calculate the resistance. A moving body of water, on the other hand, has no clear-cut boundaries except at where you place your electrodes. Basically, the current can run not only straight through the volume of water we just talked about to get to the electrode on the other side, but also up and then back down the stream, or down and then back up the stream.

The net result is that the resistance of the water is very low. Even our volume of water by itself has a resistance of only 0.286 ohms or so (using a resistivity of 1 ohm per meter as a compromise between the 0.2 of salt water and the 2-200 of drinking water). 128 kW's across this resistance equates to roughly 191 volts and 669 amps.

Note that all of the above calculations are very rough, use many simplifications, and were done by a guy whose a little rusty at solving these kinds of problems.
Thanks a lot :-)
 
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Should I tell them I had to look up the definition of "sloe"? Nah.
I'd heard of a drink called a sloe gin fizz years and years ago, but it wasn't until I was in former Yugoslavia in '74 that it dawned on my what a sloe was. They have a liqueur there called "slivovitz" that is made from a type of plum -- the sloe. The stuff was amazingly cheap, at 21 dinar a liter, about $1.05 at the time.
Had kind of a kerosene flavor to it.
 
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I'd heard of a drink called a sloe gin fizz years and years ago, but it wasn't until I was in former Yugoslavia in '74 that it dawned on my what a sloe was. They have a liqueur there called "slivovitz" that is made from a type of plum -- the sloe. The stuff was amazingly cheap, at 21 dinar a liter, about $1.05 at the time.
Had kind of a kerosene flavor to it.
I tried it once. Never again. You are being overly generous in your description of the flavor. :smile:
 
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By the way, Mark, what were you doing drinking kerosene? :oldlaugh:
 
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By the way, Mark, what were you doing drinking kerosene? :oldlaugh:
I was on a train in Yugoslavia with two British girls and we stopped off in Dubrovnik. At the time, locals would go to the train station and offer a spare room in their house to train travellers, as a way to make a little extra cash. The two British girls and I took one of the people up on his offer. Once we got situated, I went to the living room, and the man of the house offered me a small glass of slivovitz (or rijeka, which I think is the same stuff). We carried on a bit of a conversation, as I knew a fair amount of Russian and there are a lot of words that are the same in Croatian. I enjoyed talking with the fellow while his son was watching a Bugs Bunny cartoon on TV, dubbed in Croatian.
 
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Back to topic, is this for a fish fence ? Or a total-flow current meter ??

ps: Mum & Dad used to make Damson and/or Sloe gin in BIG Kilner jars: Layers of sugar, perforated fruit, more sugar, fill with a respectable but budget brand of gin. Store in dark all Autumn, invert weekly to mix. Eventually, all the sugar dissolved, and the fruit shrank as if sun-dried. A seriously wicked liqueur ensued...
 
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Prisoners are known to use a 'stinger' to heat their coffee.
Essentially it's a cut off power cable.
Plug it in, dunk it.
Heats fast quick.
 
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Prisoners are known to use a 'stinger' to heat their coffee.
Essentially it's a cut off power cable.
Plug it in, dunk it.
Heats fast quick.
That makes no sense. It would just blow a fuse. I think you are thinking about the thing that has a heating element, not just a bare wire.
 
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It would just blow a fuse.
That assumes there is a fuse to blow.

Years ago I was working onsite in Asia and the very low-rent office had power points that were just cut off power cables, like the electrician hadn't gotten around to connecting the actual socket and fascia. The local guys would wrap the ends around my laptop power prongs so I could work. They wore rubber gloves, but even being careful, sparks would fly. I asked about fuses and that caused great amusement.

I guess it was a pretty stupid question, given the circumstances :biggrin:
 
Coffee isn't a complete short. Water actually has a lot of resistance to it.
People always say water is a good conductor, but really there are much better liquids for conducting electricity.

Also, I think you have to make sure you don't ground out the coffee.
 
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It depends. A 20 foot stream that's 10 feet deep and moving is a very difficult thing to heat up. Assuming the stream's water speed is roughly 7 ft/s (5 mph), that's over 1400 cubic feet of water moving through the stream every second, or almost 10,500 gallons. To raise the temperature of the stream by 5 degrees F takes 437,500 BTU's, or 128 kW's of power.

This is a lot of power, but not that much. It's equivalent to about 171 horsepower.

To continue, we now need the electrical resistance of the water.
But we have it.
Because the current is specified in the OP. 75 milliamperes.
75 milliamperes producing 128 kW gives the voltage as 1700 kV.

Such a high-voltage, low-current leakage through some insulation... gives us the resistance. 23 MΩ.
What does the geometry need to be in order for the 23 MΩ resistor to support 128 kW thermal output?
Also: if the bulk of stream water is only heated by 5 F - less than 3 C - it does not mean that the water at the heating element is only heated by 3 C. It is heated much more, before the hot water and steam bubbles get mixed in the stream water.
The resistivity of water is high in cold fresh water. It drops slightly as the water is heated - then increases suddenly as the water boils, though the bubbles will travel up by buoyancy and be replaced by water - then drops a lot as the steam is heated to thousands of Celsius and ionized to plasma.

How do corona discharges work in water, where one electrode is an insulation leak and the other is bulk water? What are typical resistances of arc and corona discharges in fresh water?
 

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