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It is very often claimed (by both government energy efficiency experts and water-heater sellers/service people) that calcified heating elements in electrical water heaters lead to much higher energy bills for water heating. Numbers often wildly ranging from 30% to over 400% percent increase in electrical power bills. Thus, it is important that water heaters get serviced yearly to clean out the sedimentation.

Now, I can see why calcification on and around heating element is bad, as it would be acting as thermal insulation between itself and water, which would:

  1. prolong the time needed for heating element to heat the full tank of water (due to slower transfer of heat)

  2. reduce life of heating element due to it overheating (as thermostat shutoff would be delayed due to previous point.)

However, I do not see why it would cause any increase in energy usage?

Is it true that calcified water heater elements would use up more electrical energy?

Edit1: example claims

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    Welcome to Skeptics! You say this is often claimed. I did a 2-minute search and didn't find any examples (except in reference to gas hot-water heaters). Could you please add a couple of links to examples (and quote from them)? Thanks. – Oddthinking Nov 15 '14 at 7:04
  • Perhaps it makes the pump less efficient, so it needs to be on longer to get the same effect. – ChrisW Nov 15 '14 at 15:58
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    @ChrisW there is no pump - positive pressure from cold water intake is what is pushing hot water out. There are only resistive heaters inside that consume electricity. – Matija Nalis Nov 15 '14 at 23:18
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    @Andy no, I don't. It does take longer to heat a full tank, but the full tank also stays warm longer, as the water continues to heat even when power is turned off - see law of conservation of energy as in practice used in for example Storage heater. So if conservation of energy applies (as it should) it must even out (so it doesn't matter if you heat 10 times for 1 hour each, or 5 times for 2 hours each) – Matija Nalis Nov 25 '14 at 1:02
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    @Andy I'm aware of heat loss from the tank to the surroundings. But it is (IMHO) function of temperature difference (water vs surroundings), and coefficient and depth of tank wall insulator material. It should not depend on the heating element (not even on it's existence - you could pour hot water to the tank, and its insulation would still work the same). It is also usually in specs, and is negligible <2kWh/day. As for efficiency, I'm listening - how could resistive heater be anything but 100% efficient? I'm indeed expecting any waste to be "waste heat", which is main purpose in our case. – Matija Nalis Nov 25 '14 at 1:35
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Calcification's effect on water heater efficiency depends on the design of the water heater and the type of calcification. The question referred specifically to electric water heaters, however, some of the links were referring to gas water heaters. In a gas water heater, the burner is usually located under and outside of the tank and heats a heat exchanger on the bottom of the tank that transfers heat to the water in the tank. In this configuration, sediment at the bottom of the tank can significantly effect efficiency because it insulates and prevents some of the heat from the burner from being transferred into the water (the heat instead is transferred to the air around the bottom of the burner compartment). Some of the links cited are referring to this effect in gas water heaters, not an electic water heater.

In an electric water heater, the element(s) generally are located inside the tank some distance from the bottom of the tank. Sediment on the bottom of the tank has little effect on efficiency, in fact, the sediment may actually provide additional insulation for the small amount of heat that can be lost out of the bottom of the tank. If there is so much sediment at the bottom of the tank that it builds up high enough to cover the heating element, the element will quickly overheat and burn out because the large amount of heat it produces will not be transferred quickly to the water (it will be concentrated in the sediment around the element instead).

Calcification can form on the heating element itself if the anode is not properly maintained. In this case, the minerals can act as an insulator around the element, however, the insulating effect is minimal. Limestone actually has a higher thermal conductivity than water itself (although water in the liquid state transfers heat more quickly because of convection). The tiny insulating effect of the calcification around the element will cause the element to operate at a slightly higher temperature. This has multiple, but insignificant effects. First, the higher temperature will cause that heat to transfer more quickly through the calcification (although not as fast as it would if the element were in direct contact with the water). Second, the higher temperature will cause more heat to conduct through the flange and the electrical supply wires connected to the element assembly. Some of that heat will eventually be lost through the wire's insulation and into the the ambient air around the wire. Third, the extra heat in the wire will slightly increase the resistivity of the wire as a conductor (assuming copper or aluminum wire). That extra resistance will cause a tiny increase in heat loss from the wire. There is another very insignificant effect in that there will be a little more of a time delay in the heat from the element eventually being "seen" by the thermostat. This increase in the hysteresis of the system will mean that the water temperature will cycle to a very slightly higher temperature and thus there will be a very slight increase of general heat loss through the tank's insulation. (This is the same effect as setting the temperature a tiny bit higher: The higher the water temperature, the more heat is lost through the walls of the tank.)

All of those effects of direct calcification on the element are insignificant compared to the amount of heat involved in heating a hot water heater, so it is safe to say that direct calcification on the electric heating element has no significant effect on efficiency. The website that claimed that there could be a 40% loss of efficiency was also selling some kind of product to prevent it, so I would question the validity of the claims from that website.

  • Welcome to Skeptics! Please provide some references to support your claims. – Larian LeQuella Nov 26 '14 at 6:32
  • Thanks for the answer, Heidi! I understand that the market might make lying profitable (regarding 40% loss claim for 6.4mm), but I find similar results even in sources which should not be suspectable - for example, newspapers (not in ads) and Croatian (UNDP+EU sponsored) energy efficiency foundantion (which does not sell anything nor is supported by commercial entities) also claims 10% energy loss for 1mm lime on heating element. – Matija Nalis Nov 27 '14 at 1:40
  • @MatijaNalis Thank you for the comment. Looking over some of the sources, it appears that some of them may have used the questionable source and simply repeated that 40% number. Given the physics, the 405 – Heidi Guttenheim Nov 27 '14 at 4:41
  • @MatijaNalis Thank you for the comment. Looking over some of the sources, it appears that some of them may have used the questionable source and simply repeated that 40% number. Given the physics, the 40% number is certainly an extrodinary claim and, lacking any support for the claim, should be questioned. Also, round numbers like 10%, 30%, etc., suggest that the numbers are not based on empirical evidence (although it is possible to arrive at a round number analytically or experimentally). Sources that explain how the number was arrived at are usually more creditible. – Heidi Guttenheim Nov 27 '14 at 4:51
  • The gold standard is an exeriment conducted using a standard protocol, a control, and a large number of data points; and published in a peer-reviewed journal. We probably are not going to find such a source for this question. – Heidi Guttenheim Nov 27 '14 at 4:58

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