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oldgrowth
Joined: 27 Jun 2005
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 Posted: Tue Dec 28, 2010 2:29 am Post subject: |
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| rogerbum wrote: | Dalton,
Heating elements can be designed to run on a variety of different voltages. The amount of heat generated is related to the power dissipated in the heating element. Power in Watts (W) is calculated as current (I) times voltage (V) or
W=V*I .
Since a heating element is basically a resistor, the resistance (R) of the heating element determines how much current passes through the element. The relationship (Ohm's law) is
V=I*R (or I=V/R)
If you combine these two equations by replacing I in the power equation with V/R, you get
W = V*V/R = V^2/R (Voltage squared over resistance).
So if a manufacturer designs a water heater to use a certain amount of power at 120V, the same water heater run at 240V (double the voltage) would use 4 times as much power. Hence, to get the unit to operate at 240V and use the same amount of power, the resistance of the 240V element needs to be 4 times higher than the resistance of the 120V element.
Conversely, if you put an element that was designed to use a given amount of power at 240V into a system that is powered by 120V, it will use 1/4 as much power at 120V as it would at 240V. E.g. an element designed to draw 1500W of at 240V will only draw about 375W (3.1A) at 120V. Thus it can easily be powered from a lower power generator (it just takes about 4 times as much time to heat the water). |
Roger – I believe there is a slight flaw in you last paragraph (it just takes about 4 times as much time to heat the water). It seems as though something is missing here. If that were true, then we could use one watt of energy and given enough time have all the hot water we want.
Using a little logic/common sense because I don’t have training in this area:
When we cut the voltage in half we have to decrease the resistance by one fourth in the heating element in order to create the same amount of energy/heat.
Now let’s assume the 240v element is designed to operate at 440 degrees in order to heat the water to its desired temperature in a reasonable amount of time. A thermostat will shut it off and let’s assume that is at 140 degrees. Now using a 240v heating element with 120 volts will give one fourth the amount of energy/heat that 240 volts does. One fourth of 440 is 110. A heating element operating at 110 degrees will never heat the water to the desired cutoff temperature of 140 degrees.
Now my thought process could be wrong, so if it is tell me where.
Dave 
www.tolandmarine.com |
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TyBoo
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| Posted: Tue Dec 28, 2010 2:40 am Post subject: |
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Just to confuse things further, why would one need a "hot" water heater in the first place?
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Hooked on Powell
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| Posted: Tue Dec 28, 2010 7:28 am Post subject: |
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Greg & Roger,
Thanks for your help. I know I'm not the brightest bulb on the string when it comes to mechanical/electrical so your patience is much appreciated. You guys are the best!
Looks like changing out the water heater is on my list of boat projects now.
Greg, if you do come across the specs and supplier for the heating element please let us know.
Dalton |
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Pat Anderson
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| Posted: Tue Dec 28, 2010 9:25 am Post subject: |
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ZING! Right over their heads!
| TyBoo wrote: | | Just to confuse things further, why would one need a "hot" water heater in the first place? |
_________________
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rogerbum
Joined: 21 Nov 2004
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| Posted: Tue Dec 28, 2010 9:44 am Post subject: |
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| oldgrowth wrote: |
Roger – I believe there is a slight flaw in you last paragraph (it just takes about 4 times as much time to heat the water). It seems as though something is missing here. If that were true, then we could use one watt of energy and given enough time have all the hot water we want.
Using a little logic/common sense because I don’t have training in this area:
When we cut the voltage in half we have to decrease the resistance by one fourth in the heating element in order to create the same amount of energy/heat.
Now let’s assume the 240v element is designed to operate at 440 degrees in order to heat the water to its desired temperature in a reasonable amount of time. A thermostat will shut it off and let’s assume that is at 140 degrees. Now using a 240v heating element with 120 volts will give one fourth the amount of energy/heat that 240 volts does. One fourth of 440 is 110. A heating element operating at 110 degrees will never heat the water to the desired cutoff temperature of 140 degrees.
Now my thought process could be wrong, so if it is tell me where.
Dave
www.tolandmarine.com |
Dave,
You are both correct and incorrect so let me explain. You are correct in stating that there's something missing from my statement that it will take approximately 4 times as long to heat the water. It will take a little longer due to heat loss from the water heater. If there was no heat loss (e.g. the impossible "perfect" insulation), it would take 4 times longer. But there will be some heat loss so it will take a bit longer (depending on the percentage of heat lost relative to what goes in). Also, there's another complicating factor which is that the rate of heat transfer from the element to the water is related to the temperature difference between the element and the water and the temperature difference will be lower for the lower power element. The rest of your logic above is wrong. In a really well insulated system we could in fact heat water with 1W - it would take a really long time, and it would probably cost a hell of a lot of money to create the system with such fantastic insulation (probably a NASA like project).
The heating elements are not designed to operate at a certain temperature - they are simply resistive elements and they will continue to get hotter as long as current flows through them and there is no heat loss. E.g. the logic about the temperature of the heating element going down by a factor of 4 when the power goes down by a factor of 4 is completely screwed. The temperature at which a heating element in a water heater will operate is for the most part limited by the temperature of the water and the rate of thermal conductivity between the element and the water.
E.g. you can think of the cold body of water as "sucking away" heat/temperature from the heating element. The heating element will be above the temperature of the water but not by too much since the water serves as a heat sink for the heating element. Since water boils at 212F, the temperature of the heating element should never get too far above 212F until the tank goes dry. Also the temperature of the heating element is probably limited by an independent thermostat as a safety feature (at least that's how I'd design a water heater - I don't really know if this is the case). How much is "too much" and "too far" in the above sentences? I don't know since I don't really know the rate of thermal conductivity between the element and the water and even if I did it would require me to go back and look up some long forgotten physics to do the correct calculation.
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Alyssa Jean
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| Posted: Tue Dec 28, 2010 9:46 am Post subject: |
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It rightfully should be named a "cold" water heater.
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rogerbum
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| Posted: Tue Dec 28, 2010 9:46 am Post subject: |
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| TyBoo wrote: | | Just to confuse things further, why would one need a "hot" water heater in the first place? |
Ask George Carlin. It's the same reason why we put things in a pre-heated oven. |
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Aurelia
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| Posted: Tue Dec 28, 2010 11:10 am Post subject: |
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Not sure if George answers emails anymore.
Dalton,
Here is that element I ordered, you could likely find it elsewhere if you look.
Voltage 240
Hz 60
Watts 2500
Style Screw In
Insert Length (In.) 7-7/16
Plating Zinc
Sheath Material Copper
Includes 1" NPSM Screw Plug And Type AG-1 Rubber Gasket
http://www.grainger.com/Grainger/wwg/search.shtml?searchQuery=2E300&op=search&Ntt=2E300&N=0&sst=subset
Greg
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Gig Harbor
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oldgrowth
Joined: 27 Jun 2005
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| Posted: Tue Dec 28, 2010 2:48 pm Post subject: |
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| rogerbum wrote: | | oldgrowth wrote: |
Roger – I believe there is a slight flaw in you last paragraph (it just takes about 4 times as much time to heat the water). It seems as though something is missing here. If that were true, then we could use one watt of energy and given enough time have all the hot water we want.
Using a little logic/common sense because I don’t have training in this area:
When we cut the voltage in half we have to decrease the resistance by one fourth in the heating element in order to create the same amount of energy/heat.
Now let’s assume the 240v element is designed to operate at 440 degrees in order to heat the water to its desired temperature in a reasonable amount of time. A thermostat will shut it off and let’s assume that is at 140 degrees. Now using a 240v heating element with 120 volts will give one fourth the amount of energy/heat that 240 volts does. One fourth of 440 is 110. A heating element operating at 110 degrees will never heat the water to the desired cutoff temperature of 140 degrees.
Now my thought process could be wrong, so if it is tell me where.
Dave
www.tolandmarine.com |
Dave,
You are both correct and incorrect so let me explain. You are correct in stating that there's something missing from my statement that it will take approximately 4 times as long to heat the water. It will take a little longer due to heat loss from the water heater. If there was no heat loss (e.g. the impossible "perfect" insulation), it would take 4 times longer. But there will be some heat loss so it will take a bit longer (depending on the percentage of heat lost relative to what goes in). Also, there's another complicating factor which is that the rate of heat transfer from the element to the water is related to the temperature difference between the element and the water and the temperature difference will be lower for the lower power element. The rest of your logic above is wrong. In a really well insulated system we could in fact heat water with 1W - it would take a really long time, and it would probably cost a hell of a lot of money to create the system with such fantastic insulation (probably a NASA like project).
The heating elements are not designed to operate at a certain temperature - they are simply resistive elements and they will continue to get hotter as long as current flows through them and there is no heat loss. E.g. the logic about the temperature of the heating element going down by a factor of 4 when the power goes down by a factor of 4 is completely screwed. The temperature at which a heating element in a water heater will operate is for the most part limited by the temperature of the water and the rate of thermal conductivity between the element and the water.
E.g. you can think of the cold body of water as "sucking away" heat/temperature from the heating element. The heating element will be above the temperature of the water but not by too much since the water serves as a heat sink for the heating element. Since water boils at 212F, the temperature of the heating element should never get too far above 212F until the tank goes dry. Also the temperature of the heating element is probably limited by an independent thermostat as a safety feature (at least that's how I'd design a water heater - I don't really know if this is the case). How much is "too much" and "too far" in the above sentences? I don't know since I don't really know the rate of thermal conductivity between the element and the water and even if I did it would require me to go back and look up some long forgotten physics to do the correct calculation. |
Roger – thanks for your explanation, it almost makes sense to me. I figured you would talk about the heat loss from the rate of heat transfer from the element to the water in relation to the temperature difference between the element and the water.
I also know a heating element is not per say designed to operate at certain temperatures, but I believe it has a certain optimal temperature it reaches with a given size, material, resistance and voltage. Then any increase in temperature above that is negligible as current continues to flow through it. By increasing or decreasing the voltage with the resistance, size, and material being fixed, seems to me, will change the temperature of the element.
I do have some lingering questions though based on my observations.
1. If the heating element in a water heater continues to get hotter as long as there is current, why does a heating element on a portable space heater not continue to get hotter? They seem to reach a certain redness/brightness, then not get any hotter. If you increase the voltage or reduce it, the element changes its color/heat output.
2. Now look at a light bulb with an old style dimmer switch on it and the light produces less heat/brightness as the voltage is reduced. The element is in a vacuum so you cannot attribute the heat loss to air sucking the heat away.
3. If your statement that a heating element will continue to get hotter as long as current flows through it, why doesn’t the filament/element in the above light bulb not burn out when left on at full power?
4. Dosen’t the light bulb and space heater operate on the same principals as a water heater?
The reasons for these questions are so someone does not spend money on something that does not have a chance of working.
Dave
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Aurelia
Joined: 21 Aug 2009
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| Posted: Tue Dec 28, 2010 5:13 pm Post subject: |
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Not sure what temp the element gets up to or even exactly what the thermostat is currently set to on the tank (factory setting for now)...
47 minutes after running the generator and listening for a throttle change, it stepped back down to the lowest setting running with the ECO throttle switched to ON. At that time, there were no other appliances or accessories operating so I am making the assumption that the thermostat kicked in and switched off the heat. Now I would be happier with 27 minutes but if I can still run chargers for the batteries at the same time, it is worth running a little longer. If it took 2 hours I would call it a mis-judgment and throw it as far as I could.
Sure enough there was enough hot water to run a shower at that time and without mixing cold into the flow, I could not comfortably leave my hand in the stream. So 625 watts into a 240volt, 2500 watt element seems to do the trick in this configuration. I did think about finding a similar element for the factory tank but I guessed that 6 gallons would take too long to heat with that little power input. I think I made a good choice. |
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rogerbum
Joined: 21 Nov 2004
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| Posted: Tue Dec 28, 2010 6:13 pm Post subject: |
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| oldgrowth wrote: |
Roger – thanks for your explanation, it almost makes sense to me. I figured you would talk about the heat loss from the rate of heat transfer from the element to the water in relation to the temperature difference between the element and the water.
I also know a heating element is not per say designed to operate at certain temperatures, but I believe it has a certain optimal temperature it reaches with a given size, material, resistance and voltage. Then any increase in temperature above that is negligible as current continues to flow through it. By increasing or decreasing the voltage with the resistance, size, and material being fixed, seems to me, will change the temperature of the element.
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That's mostly true since the heating element is connected to a heat sink (e.g. the water tank) and there's some constant heat loss. However, once the heat sink (water) get's warmer, the element temperature will continue to rise for a given current. The element temperature is a combination of the power that's going in and the power that is flowing out. The power that's flowing out is related to the temperature difference of the element and the heat sink to which it's connected. So as the heat sink gets warmer, difference in temp decreases, the power loss is less and the element gets warmer (creating a bigger temperature difference and more heat flow). In practice, these factors keep in balance so that the temperature of the whole system rises smoothly rather than discontinuous explanation I provided. Eventually, the water gets warm enough and the element is turned off by the thermostat. Otherwise, the temperature of both the water and heating element would continue to rise until either the water boiled off or other heat losses (through the pipes and insulation) limit the temperature.
| oldgrowth wrote: |
I do have some lingering questions though based on my observations.
1. If the heating element in a water heater continues to get hotter as long as there is current, why does a heating element on a portable space heater not continue to get hotter? They seem to reach a certain redness/brightness, then not get any hotter. If you increase the voltage or reduce it, the element changes its color/heat output.
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Assuming you are not talking about a space heater with a fan - now you've hit on another source of hit loss (one I left out in my simplified explanation) - radiative heat loss. First, if you are talking about a space heater with a fan, the heat loss is mostly due to transfer to the air which is flowing over the heating element. In such a case, the temperature the element obtains will be limited by the combination of power in (from the electricity) and power out (warmed air moving across the element). But on a radiant heater, the heat loss is not so much due to movement of the air but due to radiative heat loss.
What is radiative heat loss you ask? It's light (both visible and invisible - most invisible infrared light) leaving the element. In this case the primary source of heat loss is that generated by photons (light) leaving the element. On elements where radiative heat loss is the dominant source of heat loss, the temperature of the element will rise with increase current/power and the temperature rise is directly reflected in the color of the element (orange, red, white). In such a case, the temperature will be still be limited by a combination of the power in vs power out, it's just that the power out is dominated by radiative heat loss as opposed to conductive heat loss.
| oldgrowth wrote: |
2. Now look at a light bulb with an old style dimmer switch on it and the light produces less heat/brightness as the voltage is reduced. The element is in a vacuum so you cannot attribute the heat loss to air sucking the heat away.
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A light bulb loses heat primarily through radiative loss. That's the heat you feel when you put your hand near it. It's essentially the same as a radiative space heater. In fact some people use light bulbs as space heaters (common in chicken coops and some well insulated dog houses).
| oldgrowth wrote: |
3. If your statement that a heating element will continue to get hotter as long as current flows through it, why doesn’t the filament/element in the above light bulb not burn out when left on at full power?
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Because of radiative heat loss and because the element is designed with a high enough resistance that the power in vs. radiative loss out is such that the temperature will not melt the element. However, hook any light bulb up to a high enough voltage and you can drive the temperature up to a point that it will melt/burn out the element. It's in a vacuum since that prevents oxidation of the element by air which would burn out the element faster.
| oldgrowth wrote: |
4. Dosen’t the light bulb and space heater operate on the same principals as a water heater?
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Yes in terms of power in, no in terms of the dominant source of heat loss from the element. Heating elements in space heaters operate at around 1800°F and elements in incandescent light bulbs operate at 3100–5400°F. At such very high temperatures and with no direct thermal connection to a heat sink, radiative heat loss dominates. The heating element in a water heater probably operates at a much lower temperature and has a direct thermal connection to a heat sink (the water tank). |
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oldgrowth
Joined: 27 Jun 2005
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| Posted: Tue Dec 28, 2010 9:19 pm Post subject: |
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Roger – I kind of followed most of your explanation but I did not understand the explanation for the following.
| rogerbum wrote: | | oldgrowth wrote: |
2. Now look at a light bulb with an old style dimmer switch on it and the light produces less heat/brightness as the voltage is reduced. The element is in a vacuum so you cannot attribute the heat loss to air sucking the heat away.
|
A light bulb loses heat primarily through radiative loss. That's the heat you feel when you put your hand near it. It's essentially the same as a radiative space heater. In fact some people use light bulbs as space heaters (common in chicken coops and some well insulated dog houses).
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In the light bulb example above when the voltage drops, all the other factors drop except the resistance of the element. If the voltage drops low enough, it will not give off any perceptible heat or light. It seems to me it would be the same for the heating element, regardless of its application.
After a little research I came up with the following.
Using the following formula to find the resistance of a heating element.
Resistance = Voltage squared over Wattage.
We know volts = 240 and watts = 1200. The current is 5 amps.
240 squared divided by 1200 = 48 ohms - the resistance of the heating element.
To find watts from a known voltage and resistance use the following formula.
Watts = volts squared divided by resistance.
Reduce the voltage to 120 volts, resistance remains the same because we still have the same element.
120 squared divided by 48 = 300. Current is 2.5 amps (see formula below)
Reduce the voltage to 12 volts
12 squared divided by 48 = 3 watts. Current is .25 amps (see formula below)
Knowing the resistance of the element we can figure the amperage
Amps = voltage divided by resistance.
240 divided by 48 = 5 amps
120 divided by 48 = 2.5 amps
12 divided by 48 = .25 amps
Looking at the above information, I would guess there has to be some internal lose in addition to the losses Roger identified. I don’t think 12 volts will ever under any condition heat the element any perceptible amount.
I believe using a 240v heating element with 120 volts will end up using more power than using it with 240 volts because of the internal loss as well as the loses Roger identified. It seems to me, using an element that draws the most power a battery or generator is capable of producing would be the most efficient.
To figure the most efficient you need the least resistance you can get by with in the heating element.
For a 1000 watt generator look for an operating draw of 900 watts.
120v squared divided by 900 watts = 16 ohms. So you need a heating element with 16 ohms of resistance.
120 divided by 16 = 7.5 amps
That would be the most efficient for a 1000 watt generator.
I don’t know how to measure heat other than BTU and 1 watt = 3.4121 BTU per hour.
Using the 240v element with 120 volts = 1,023 BTU per hour.
Using a 900 watt 120 volt heating element produces 3,070 BTU.
Roger - am I wrong somewhere here.
Dave
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Alyssa Jean
Joined: 02 Nov 2003
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| Posted: Tue Dec 28, 2010 9:57 pm Post subject: |
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| You guys are amazing. |
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TyBoo
Joined: 23 Oct 2003
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| Posted: Tue Dec 28, 2010 10:40 pm Post subject: |
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| Anna Leigh wrote: | | You guys are amazing. |
Yep they are. I'm looking forward to Joe posting the transflabulator video again. |
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rogerbum
Joined: 21 Nov 2004
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| Posted: Wed Dec 29, 2010 6:27 am Post subject: |
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| oldgrowth wrote: | Roger – I kind of followed most of your explanation but I did not understand the explanation for the following.
| rogerbum wrote: | | oldgrowth wrote: |
2. Now look at a light bulb with an old style dimmer switch on it and the light produces less heat/brightness as the voltage is reduced. The element is in a vacuum so you cannot attribute the heat loss to air sucking the heat away.
|
A light bulb loses heat primarily through radiative loss. That's the heat you feel when you put your hand near it. It's essentially the same as a radiative space heater. In fact some people use light bulbs as space heaters (common in chicken coops and some well insulated dog houses).
|
In the light bulb example above when the voltage drops, all the other factors drop except the resistance of the element. If the voltage drops low enough, it will not give off any perceptible heat or light. It seems to me it would be the same for the heating element, regardless of its application.
<remainder clipped>
Dave
www.tolandmarine.com |
Dave,
I general all of what you wrote this time is correct. I will point out that that lack of any perceptible heat or light is not equal to the lack of actual heat or light. What's really at issue is the dominant source of heat loss from the element (heating element or light bulb). The power in and the resistance behavior is essentially the same it's just that how they lose the heat is different. However, even a light bulb will lose some heat through conduction and a water heater will lose some heat through radiative losses. It just that at normal operating temperatures for each device (high for a light bulb element and lower for a water heater element), the dominant sources of heat loss are different.
You are correct that the most efficient water heater element would be operated at close to max available current if the definition of most efficient is to get the maximum rise in temperature of the water per unit of power. The reason for this is that since heat energy is always leaking out of the water heater (through conduction and radiative loss), the quicker you get the water to temp, the less heat is lost to other places. However, as a practical matter water heater elements are only available in certain watt ratings for certain voltages so one can't necessarily tune the resistance of the element to maximize efficiency. Also, my guess is that since the water heaters are fairly well insulated, the loss is relatively minimal. I know that when I had the 6 gallon water heater on for an hour or so in my Tomcat, the water was still quite warm 6-8 hours after I turned the water heater off. [/i] |
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