- #1
everseeker
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Hi.
Homeowner here who is looking at his brand new hot water setup...
2 tanks. 50 gallons each, with a 4500 watt electrical heater in each one
Tank 1 output feeds to tank 2 input.
Outside water is fairly constant 50 degrees F
I am curious: how long can I take a 3.5GPM shower? (output 120<x<130)
Looked up the following formula:
KW input = (GPH X °Rise X 8.33) / (Thermal Efficiency X 3412 BTU/KW)
Thermal Eff for electric heaters is .98 For Gas heaters is = .75)
1GPM= 60GPH, shower = 3.5GPM, or 210 GPH
temp in = 50 Desired temp out = 130 Delta = 80
So, mangle the formula a bit...
4500=X*80*8.33/.98*3412
4500=X*666.4/3343.76
4500/X=666.4/3343.76
4500/X=.1993
1/x=.1993/4500
1/x=4.4288E-5
X=22579.412
do not believe... plugging into crosscheck...
22500*80*8.33/.98*3412=...oh, = 4500KW... heh
Restarting from 4500/X=.1993
4.5/X=.1993
1/X=.1993/4.5
1/x=0.044288
x=22.57941
So, 22.6 GPH
and, for tank 2
4500=X*10*8.33/.98*3412
4.500/X=(83.3/3343.76
1/x= 0.0249/4.5
X=180.635
+ the 22 GPH
202GPH
210 = shower
(loss of 8 GPH)
(50 gallons + 22 gallons from first one =72/8=
9 hours. seems nice... too nice.
-------------
Try another way...
For a continuous shower at 3.5 GPM:
X=(210*80*8.33/.98*3412
X=41.8523 KW
or, 41,850 Watts!
ok, so the numbers simply do NOT add up
----------
Got to thinking... the numbers are kinda more complex then I am thinking... as tank 1 feeds tank 2, the temp rise is constant at first, then begins to change , because the draw from tank 1 exceeds 22 GPH
I am envisioning a couple curves... this smacks of calculus, which I have not had in over 20 years... Thoughts anyone?
Homeowner here who is looking at his brand new hot water setup...
2 tanks. 50 gallons each, with a 4500 watt electrical heater in each one
Tank 1 output feeds to tank 2 input.
Outside water is fairly constant 50 degrees F
I am curious: how long can I take a 3.5GPM shower? (output 120<x<130)
Looked up the following formula:
KW input = (GPH X °Rise X 8.33) / (Thermal Efficiency X 3412 BTU/KW)
Thermal Eff for electric heaters is .98 For Gas heaters is = .75)
1GPM= 60GPH, shower = 3.5GPM, or 210 GPH
temp in = 50 Desired temp out = 130 Delta = 80
So, mangle the formula a bit...
4500=X*80*8.33/.98*3412
4500=X*666.4/3343.76
4500/X=666.4/3343.76
4500/X=.1993
1/x=.1993/4500
1/x=4.4288E-5
X=22579.412
do not believe... plugging into crosscheck...
22500*80*8.33/.98*3412=...oh, = 4500KW... heh
Restarting from 4500/X=.1993
4.5/X=.1993
1/X=.1993/4.5
1/x=0.044288
x=22.57941
So, 22.6 GPH
and, for tank 2
4500=X*10*8.33/.98*3412
4.500/X=(83.3/3343.76
1/x= 0.0249/4.5
X=180.635
+ the 22 GPH
202GPH
210 = shower
(loss of 8 GPH)
(50 gallons + 22 gallons from first one =72/8=
9 hours. seems nice... too nice.
-------------
Try another way...
For a continuous shower at 3.5 GPM:
X=(210*80*8.33/.98*3412
X=41.8523 KW
or, 41,850 Watts!
ok, so the numbers simply do NOT add up
----------
Got to thinking... the numbers are kinda more complex then I am thinking... as tank 1 feeds tank 2, the temp rise is constant at first, then begins to change , because the draw from tank 1 exceeds 22 GPH
I am envisioning a couple curves... this smacks of calculus, which I have not had in over 20 years... Thoughts anyone?
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