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Jonathan Ball
 
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Default Power cost of idle electric water heater

Jonathan Ball wrote:

wrote:

wrote:


A larger tank should help, given a smaller surface to lose heat from...




Larger tanks have less surface?



Imagine you have a cube tank that's one cubic foot. Its surface area is
six square feet.


I meant to specify it's 1 x 1 x 1.

Now you take another identical tank, cut the bottom
off it, cut the top off the first, and weld them together. You've
doubled your volume to two cubic feet, but you've only increased your
surface area by four square feet: two 1-foot cubes stacked on top of
one another have an *external* surface area of 10 square feet, not 12.

If you have a cylindrical tank, the volume is given by

h * (? * r * r)

where r * r = 'r squared' (no exponents in plain text font) and h is the
height.

The surface area is two times the surface area of the circle on the end,
plus the surface area of the sides:

(2 * (? * r * r)) + (2 * ? * r * h)

where the surface area of the sides is given by the circumference of the
cylinder's base, 2 * ? * r, times the height.

If you now double the height of your cylinder, which obviously doubles
the volume, you can see that only the expression to the right of the '+'
sign in the formula for the surface area is increased; 'h' doesn't
appear in the term to the left of the '+', meaning that as height
increases, that term is constant.