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Water Heater issue (NOT Homework)

  1. Mar 11, 2010 #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: Mar 11, 2010
  2. jcsd
  3. Apr 14, 2010 #2
  4. Apr 14, 2010 #3
    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,
     
  5. Apr 15, 2010 #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?

    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: Apr 15, 2010
  6. Apr 15, 2010 #5
    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.
     
  7. Apr 15, 2010 #6

    lanedance

    User Avatar
    Homework Helper

    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...
     
  8. Apr 15, 2010 #7

    lanedance

    User Avatar
    Homework Helper

    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
     
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