- #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 in to 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 in to 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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