Water Heater issue (NOT Homework)

  • Thread starter everseeker
  • Start date
  • #1
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?
 
Last edited:

Answers and Replies

  • #3
144
0
You should put up some relevant units, and maybe some of the names of the equations you're using. I look at this and see a mess of information.
I know that you should be using Specific heat of water to raise it to desired temperature,
to get joules used etc,
 
  • #4
OK, Condensed a bit...

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)

Found this on the internet:
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)

----------
Cut out all the attempts I made to get an answer
----------

Thoughts anyone?

You should put up some relevant units, and maybe some of the names of the equations you're using. I look at this and see a mess of information.
I know that you should be using Specific heat of water to raise it to desired temperature,
to get joules used etc,

Well, since this is a real world question, I used real world Units...
BTU= British Thermal Unit (Delve in here if you want to see where all the Joules/Specific heat etc... are hiding)
GPM=Gallons per minute
GPH=gallons per hour
Temp in Degrees F
Power in Watts
3.5GPM = limit set in US on max flow allowed in a shower



The "Formula"s name is
Found on the internet in an article on sizing water heaters... It's used in the trade to calculate the KW input needed for large facilities
If you need a name...
Will calling it "Joe" help?




In the end, I am looking to answer: How long can I take a shower before it gets cold (cold = 99 degrees F)
 
Last edited:
  • #5
144
0
Well certain formula's only apply in certain situations. In thermodynamics, there are hundreds of equations that apply to only specific situations, like constant pressure, constant volume, etc. So if you're using a thermal equation that has specific conditions, and you're just ignoring them, equation "joe" probably isn't going to work. But then again, this is a homework forum, and this is also a calculus and beyond forum. Your question applies to NONE of these.
 
  • #6
lanedance
Homework Helper
3,304
2
Homework starts at home... ;)

It depends on how a tank fills, but i woud start by analysing a single tank under 3.5GPM flow rates (i'm not great with field units by the way)

so assume
- water flows out at tank temp at 3.5GPM
- water flows in at outside temp at 3.5GPM
- energy is supplied to the tank at 0.98*4500W

T(0) = initial tank temperture

by setting up an energy balance, you should be able to determine the tank temp as a function of time T(t)

once you have done it for one, expand the problem to both tanks

you could do this and i think you'll come up with a reasonably simple DE (differential equation), or if you just want an approximate answer, why not just set up a spreadsheet that calculates the tank temp after evry minute say...
 
  • #7
lanedance
Homework Helper
3,304
2
as mr tt mentioned, the only real constant you need is specfic heat capacity of water (easy to find) as you already have rate & input power
 

Related Threads on Water Heater issue (NOT Homework)

  • Last Post
Replies
3
Views
784
  • Last Post
Replies
7
Views
990
  • Last Post
Replies
1
Views
644
  • Last Post
2
Replies
26
Views
3K
Replies
1
Views
1K
Replies
8
Views
2K
  • Last Post
Replies
2
Views
794
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
966
Top