INSULATE HOT WATER PIPE
I believe that all hot water pipe should be insulated. The house I live in now has no
basement, so standard practice in USA is to run the hot water pipe under the slab. With
this configuration, five minutes after running hot water, the water in the pipe is cold
again, particularly in the winter. I will show my calculation later, when I get more
organized. The upper bath had to run two gallons of water to get hot again.
The house is two story with the second floor supported by trusses, so I re-routed the hot
water pipe between the floors. That is between the ceiling of the first floor and the floor
of the second floor. Below is my crude drawing. Note that I used this drawing for
planning many times, including editing, and behind the wall explorations.
I used steam pipe insulation, one and one half inches of fiberglass. The pipe is three
quarter inch copper.
The results are good. The water is still warm after fifteen minutes and is fully hot five
minues after taking a shower. This was one of my criteria. (note the sophistication of
developing critiera to determine final project success before project is undertaken.) I
wrote up the whole thing and it appears below.
I need to do a test to determine how close my calculations were. At this time I believe
the calculations are close to correct, but no definate measuring test has yet been done.
ORIGINAL CALCULATIONS BELOW
Calculations for Heat Loss in Super Insulated Pipe
The heat flow through pipe insulation with outer diameter do and inner diameter (of the
insulation--equal to outer diameter of the pipe) di is
where is the length of the pipe, k is the thermal conductivity of the insulation material,
and ΔT is the temperature difference between the inner and outer 2 pi is 6.2832
Time depends on the k value. k = 0.045 watts/m/C = 0.026 BTU/ft./hr./F
1 W = 1 J/sec and 1 calorie = 4.186 joules and 1 calorie = 1 cc water 1 degree C
Criterion for ¾ inch copper pipe in 60 degree ambient air
Charged at 120 degrees = 55 C (ambient is 15 C)
Lose only 10% over 15 minutes or to 108 degrees 47.5 C
Retain warmth of 85 degrees after 3 hours = 33 C
Q is the amount of energy transferred
/ 0.693 for do = 2.25” and di = 0.75” (0.75 thick insulation) or do = 57mm di = 19 mm
Q = 10.29 watts per meter degree Centigrade - using k = 0.045 for 0.75 inch insulation
Q = 10.29 watts (10.29/4.186 joule seconds) = 2.458 calories
one meter of pipe contains (1.905cm/2)2pi X 100 = 285 cc
285 cc of water at 55 C = 15,675 calories of energy
minus 2.458 calories per second at (15 min X 60) = 900 seconds
15,675 – (900 X 2.458 calories) = 13,463 calories remain
13,463 / 285 cc = 47.24 degrees
(258 X 55 – 900 x 2.458) / 285 = 46.4 degrees DOES NOT MEET CRITERIA
Q = 7.027 watts per meter degree Centigrade - using k = 0.045 for 1.5 inch insulation
Q = 7.027 watts (7.027/4.186 joule seconds) = 1.679 calories lost
285 cc of water at 55 C = 15,675 calories of energy
minus 1.679 calories per second at (15 min X 60) = 900 seconds
15,675 – (900 X 1.679 calories) = 14,163 calories remain
14,163 / 285 cc = 49.7 degrees DOES MEET 15 min. CRITERIA
LATEST BEST NUMBERS LATEST BEST NUMBERS
Re-Revised Criterion for ¾ inch copper pipe in 60 degree ambient air
Charged at 115 degrees = 52 C (ambient is 21 C)
Lose only 10% over 15 minutes or to 108 degrees 47.5 C
Retain warmth of 85 degrees after 3 hours = 33 C
Q = 5.97 watts per meter degree Centigrade - using k = 0.045 for 1.5 inch insulation and
120 F for water and 70 F for belly area
Q = 5.97 watts (5.97/4.186 joule seconds) = 1.43 calories lost per second per meter
285 cc of water at 55 C = 15,675 calories of energy
minus 1.43 calories per second at (15 min X 60) = 900 seconds
15,675 – (900 X 1.43 calories) = 14,388 calories remain
14,388 / 285 cc = 50.5 C, 113 F DOES MEET Revised 15 min. CRITERIA
Q = 5.21 watts per meter degree Centigrade - using k = 0.045 for 2.0 inch insulation
Q = 5.21 watts (5.21/4.186 joule seconds) = 1.24 calories lost
285 cc of water at 55 C = 15,675 calories of energy
minus 1.24 calories per second at (15 min X 60) = 900 seconds
15,675 – (900 X 1.24 calories) = 14,559 calories remain
14,559 / 285 cc = 51.1 c, 114 F, ONLY 1.1 DEGREE BETTER – STILL AFTER ONLY 15
MINUTES
Numbers for ¾ inch copper pipe in 70 degree ambient air no insulation
Q = 5.21 watts per meter degree Centigrade - using k = 401.0 for 0.0156 inch copper
Q = 555,723 watts (555,723/4.186 joule seconds) = 132,757 calories lost
285 cc of water at 55 C = 15,675 calories of energy
minus 132,757 calories per second at (15 min X 60) = 900 seconds
NOT CORRECT – NEED TO INTEGRATE – ALSO NEED SKIN OF AIR
NEGLECT INSULATION VALUE OF COPPER
Second Criterion for ¾ inch copper pipe in 70 degree ambient air
Retain warmth of 85 degrees after 3 hours = 33 C
Q = 4.92 watts per meter degree Centigrade - using k = 0.045 for 1.5 inch insulation with
temp delta for 100 F to 60 F = 27.5 C
285 cc of water at 55 C = 15,675 calories of energy
Q = 4.92 watts (4.92/4.186 joule seconds) = 1.18 calories lost
minus 1.18 calories per second at (3 hr. X 3600 ) = 12,749 seconds
15,675 – (12,749 X 1.18 calories) = 631 calories remain
631 / 285 cc = 2.2 degrees NOT CORRECT, NEED TO INTEGRATE
INTEGRATION SUBSTITUTE
First 15 min. Q = 7.027 watts per meter degree Centigrade - using k = 0.045 for 1.5 inch
insulation
Q = 7.027 watts (7.027/4.186 joule seconds) = 1.679 calories lost
285 cc of water at 55 C = 15,675 calories of energy
minus 1.679 calories per second at (15 min X 60) = 900 seconds
15,675 – (900 X 1.679 calories) = 14,163 calories remain
14,163 / 285 cc = 49.7 degrees
SECOND FIFTEEN MINUTES
49.7 – 16 = 33.7 (delta T)
Q = 5.92 watts (5.92/4.186 joule seconds) = 1.414 calories lost
14,163 – (900 X 1.414 calories) = 12,890 calories remain
12,890 / 285 cc = 45.2 degrees
THIRD FIFTEEN MINUTES
45.2 – 16 = 29.2 (delta T)
Q = 5.13 watts (5.13/4.186 joule seconds) = 1.226 calories lost
12,890 – (900 X 1.414 calories) = 11,617 calories remain
11,617 / 285 cc = 40.8 degrees
FORTH FIFTEEN MINUTES
40.8 – 16 = 24.8 (delta T)
Q = 4.36 watts (4.36/4.186 joule seconds) = 1.041 calories lost
11,617 – (900 X 1.041 calories) = 10,680 calories remain
10,680 / 285 cc = 37.5 degrees (AFTER ONE HOUR)
FIFTH FIFTEEN MINUTES
37.5 – 16 = 21.5 (delta T)
Q = 3.78 watts (3.78/4.186 joule seconds) = 0.903 calories lost
10,680 – (900 X 0.903 calories) = 9867 calories remain
9867 / 285 cc = 34.6 degrees
SIXTH FIFTEEN MINUTES
34.6 – 16 = 18.6 (delta T)
Q = 3.28 watts (3.28/4.186 joule seconds) = 0.784 calories lost
9867 – (900 X 0.676 calories) = 9161 calories remain
9161 / 285 cc = 32.1 degrees
SEVENTH FIFTEEN MINUTES
32.1 – 16 = 16.1 (delta T)
Q = 2.83 watts (2.83/4.186 joule seconds) = 0.784 calories lost
9161 – (900 X 0.784 calories) = 8553 calories remain
8553 / 285 cc = 30 degrees
EIGHTH FIFTEEN MINUTES
30 – 16 = 14 (delta T)
Q = 2.46 watts (2.46/4.186 joule seconds) = 0.588 calories lost
8553 – (900 X 0.588 calories) = 8024 calories remain
8024 / 285 cc = 28 degrees AFTER TWO HOURS
DOES NOT MEET CRITERIA (33)
Q = 4.144 watts /ft. F for 0.75 inch insulation or 7.027 watts per ft for 1.5 inch insulation
Using k = 0.0387 kcal/m/hour Q = 14.03 E3 calories per hour = 14,030
so that aint right.
SOFTWARE VALUES = insulation 0.75 inch = 9 to 11 BTU per linear ft per hour
insulation 1.5 inch = 6.5 to 8 BTU per liner ft per hour
How many pounds of water in a meter? 285 cc per meter of ¾ inch pipe = 2.6 pounds
So contents of one foot = 2.6 / 3.281 = 0.79 pounds
So BTU contents of one foot of pipe at 120 F is 120 – 60 X 0.79 = 69.5 BTU
So after fifteen minutes at 0.75 inch insulation 69.5 – (10 / 4) = 67
So temperature of water with 67 BTU is (67 + 32)/0.79 = 99 DOESN’T MEET CRITERIA
Using 1.5 inch insulation 69.5 – (7 / 4) = 67.75
So temperature of water with 67.75 BTU is 67.75 + 32 = 99.75 DOESN’T MEET CRITERIA
Reduce criteria to 10 minutes 69.5 – 7/6 = 68.33 temp is 68.33 + 32 = 100.33
Charged at 120 degrees = 55 C (ambient is 15 C)
Lose only 10% over 15 minutes or to 108 degrees 47.5 C
Retain warmth of 85 degrees after 3 hours = 33 C
Fiberglass R = 4.35 per inch = k =1/4.35 = 0.23 English units
Polyethylene =
1 BTU = 1 pound water 1 degree F
1 BTU/hr = 0.293 W and 1 W = 3.413 BTU/hr
3 W/ft heat loss with ½ inch insulation = about 9 W/m (for ¾ inch pipe)
1.5 W/ft heat loss with 1.5 inch insulation = 4,5 W/m (for ¾ inch pipe)
Not insulated = 30 W/m or 30 BTU hr / ft
7 times improvement with 1.5 inch
3.3 times improvement with 0.5 inch
1 gallon of water is 8.6 pounds so 1 pound is 0.116 gallon
0.116 X 3786 cc is 439 cc