Gary said:
Looks like the (14 hour?) stratification time constant is a lot less than
the 180x0.5^2 = 45 hours I figured. Maybe the drum wall conductance... If
it's 1/16" thick and 2Pi feet around, that's like a 0.39 ft^2 vertical
steel rod conductor 1.5' long with G = 25x0.39/1.5 = 6.5 Btu/h-F between
two 225 Btu/F caps with RC = 225/2/6.5 = 17.2 hours. And with finite drum
insulation, some warm water will flow down near the walls and back up in
the middle. It might be interesting to try this again with a plastic drum
or kerfed insulation board and a plastic film liner inside the drum wall.
We might make 7 kerfs 8.26" apart in a 35"x66" piece of 1" foil-faced
foamboard, as below, and slip a 1x2 into the 0.78" gap between the drum
wall and the middle of each side of the foamboard to maintain the gap,
which adds some R-value...
20 PI=4*ATN(1)
30 RO=22.5/2'inner drum radius (inches)
40 N=8'insulation board number of sides
50 T=.5'insulation board thickness (inches)
60 THETA=2*PI/N'angle per side (radians)
70 RIMAX=RO-T*COS(THETA/2)'outer board radius (inches)
80 K=2*RIMAX*COS(PI/2-THETA/2)'distance between kerfs (inches)
90 L=N*K'foamboard length (inches)
100 LC=2.54*L'foamboard length (cm)
110 DIMIN=2*SQR(RIMAX^2-(K/2)^2)
120 BOARDGAP=RO-(DIMIN/2+T)
130 PRINT N,L,LC,DIMIN,BOARDGAP
140 D=2.54*L/N'kerf spacing (cm)
150 FOR K= 0 TO N-1
160 PRINT K,K*D
170 NEXT K
# sides foamboard length foamboard id drum-board gap
8 66.05458 (in) 167.7786 (cm) 19.93374 (in) .7831316 (in)
kerf # distance from end
0 0 (cm)
1 20.97233
2 41.94466
3 62.91698
4 83.88931
5 104.8616
6 125.834
7 146.8063
Nick