To address a few of the issues that have become entwined on this thread: <div>* I believe the purpose of 190 degree instant hot water heaters is for making tea, instant oatmeal, etc. Dialing it back to 140 degrees is not the point, as these devices would be installed in addition to a domestic hot water tap.</div>
<div>* I believe this list discussed the efficiency of different methods of boiling water, and (to my surprise) microwave was not the most efficient; I think the gas burner won that contest, but an electric kettle was a strong contender; see this blog post for further discussion:</div>
<div><a href="http://blog.plotwatt.com/2009/08/best-way-to-boil-water.html">http://blog.plotwatt.com/2009/08/best-way-to-boil-water.html</a></div><div>* Speireag was addressing the key issue: it's not the relative efficiency of heating the water you use, it's the standby losses, which can be calculated using the surface area and R-value of the tank:</div>
<div><br></div><div>1 gallon = 231 cubic inches, so 1/2 gallon = 116 cubic inches</div><div><br></div><div>The minimum surface area (best case) would be a sphere: V = 4/3 pi r^3</div><div>116 / 4.2 = r^3</div><div>r = 3</div>
<div>surface area = 4 pi r^2 = 113 sq in = 0.79 sq ft (which is more compatible with the units for R-value)</div><div><br></div><div>The R-value is a bit harder to guess, but if they managed to fit in 2" of foam that would give you R-10 or R-12; let's continue to take the best-case scenario and pick R-12.</div>
<div><br></div><div>Heat flow is calculated by U*A*delta-T, where U = 1/R-value, A is area in square feet, and delta-T is the difference in temperature between the inside and outside; the result is in units of BTU/hour.</div>
<div>U = 1/12</div><div>A = 0.79 sq ft</div><div>delta-T = 190-70 = 120 deg F (again, best-case scenario with a relatively warm house, or at least a warm under-sink cabinet with this heater in it!)</div><div><br></div><div>
(0.79 * 120) / 12 = 7.9 BTU/hr</div><div>There are 8,760 hrs/year, so the annual heat loss is 7.9 * 8,760 = 69,000 BTU</div><div>There are 3,412 BTU in a kWh, so we can convert to native units for electric energy billing: </div>
<div>69,000 / 3,412 = 20 kWh</div><div><br></div><div>At $0.10 - $0.15 / kWh for electricity (perhaps less in some areas) that comes to $2 or $3 per year for best-case assumptions.</div><div><br></div><div>Realistically, you might have 20% more surface area for a box-shaped tank (much more likely) and there might only be R-5 insulation (double the heat loss), so that would be closer to $5-7/year; let's call it $6/year. </div>
<div><br></div><div>On the other hand, about 1/3 of the year is heating season (in the northeast, at least), so about $2 of that is actually not simply waste, but is displacing heat that you would be buying anyway. Using the fuel cost calculator that I built for BuildingGreen:</div>
<div><a href="http://www.buildinggreen.com/calc/fuel_cost.cfm">http://www.buildinggreen.com/calc/fuel_cost.cfm</a></div><div>we see that electric resistance heat costs about twice as much as heat from a natural gas-fired system, so we would have spent $1 rather than the $2 we paid for the heat given off by hot-water unit, so only $5 of the $6/year is truly waste (the other $1 was useful work). </div>
<div><br></div><div>If you want to go further, part of the year is heating season, so you will pay 1/3 again for any waste heat from electric sources during that time, assuming your air-conditioner has a COP of 3. This is a small part of the year where I am (Burlington, VT) but may be more important in other parts of the country. For example, if 1/5 of the year is heating season, the $1 of waste heat costs $1.33 by the time you remove that heat with mechanical cooling (A/C).</div>
<div><br></div><div>Just to check our assumption that the waste heat is really more significant than the heat that's used to make tea water, let's assume that you use 2 cups of hot water per day and 3 on weekends for 16 cups = 1 gallon / week or 52 gallons of hot water per year. Assuming that you are raising the temperature from 50 to 190 degrees F, you have used 52 * (190 - 50) = 7,300 BTU / year or 7,300 / 3,412 = 2.1 kWh / year versus 20 to 40 kWh / year of waste heat (depending on your level of insulation). As a side note, if you typically heat twice as much water as you need in the kettle, you are wasting another 2 kWh / year, whereas the instant hot water unit would allow you to dispense the right amount every time.</div>
<div><br></div><div>My conclusion: the 2 kWh / year used to heat the water you use is actually moderately efficient compared to the alternatives, but the 20 to 40 kWh / year of standby losses is almost pure waste. If you really want the convenience then make sure you have adequate insulation, adding more until the outside no longer feels warm; that could save you $2 or $3 per year with a fairly small amount of insulation. </div>
<div><br></div><div>If you're still reading, I'll reward you with a screen-shot of my energy monitor showing breakfast at my house: 2 cups of coffee and some toast, using an electric kettle and toaster. Note the green dashed line is the fridge, which runs about half the time or a little more than that when we're cooking.</div>
<div><br></div><div><img src="cid:ii_133de1d78cbea716" alt="Picture 1.png" title="Picture 1.png"><br></div><div><br></div><div>Happy energy calculating,</div><div>Ethan</div><div><br><div class="gmail_quote">On Fri, Nov 25, 2011 at 12:58 PM, Speireag Alden <span dir="ltr"><<a href="mailto:speireag@gmail.com">speireag@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">Sgrìobh Ben:<br>
<div class="im"><br>
> My house came with a small "Instant hot water dispenser" to keep 1/2<br>
> gallon of water at 190 degrees f. I've had it unplugged since day one<br>
> because i thought it was wasteful expensive to run. My wife wants to<br>
> use it this winter and asked me how much it would cost us. Good<br>
> question. Can anyone give me an estimate on how much it costs to<br>
> run--versus a kettle or a microwave?<br>
<br>
</div> If you are using it in a space which you must still heat after you use it, and that you are using electricity to heat that space, it costs you exactly nothing. Any "waste heat" simply heats the space it's in.<br>
<br>
If you are using something other than electricity to heat that space, then the effective cost is the difference between the "waste heat" generated by using it, and what it would have cost you to produce the "waste heat" using your space-heating method.<br>
<br>
Practically speaking, you will be using this device so little that you will not notice the difference unless you are one of those types who heats the house by throwing a ball for the dog.<br>
<br>
So, during the heating season, the device is essentially cost-free. All the heat it generates goes to heat the house anyway.<br>
<br>
In the non-heating-season, the equation is different and I bow to the more math-capable people here.<br>
<br>
But you asked about the winter.<br>
<br>
Hope this helps,<br>
<br>
-Speireag.<br>
<span class="HOEnZb"><font color="#888888"><br>
<br>
--<br>
Truth, in matters of religion, is simply the opinion that has survived.<br>
--Oscar Wilde<br>
</font></span><div class="HOEnZb"><div class="h5"><br>
<br>
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