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Energy and power costs (fill in the table)

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    So i have an idea of what im supposed to do, but i dont know if im on the right track, Any help would be appreciated! Thanks in advance :)
    upload_2016-4-3_1-45-41.png

    2. Relevant equations
    relevant equations are listed below with each question

    3. The attempt at a solution

    d) yearly energy cost

    ok so to calculate the yearly energy cost i would use the equation Y = Energy x cost
    1. Y = (448 kWh/yr) x ($0.06) = $26.9 /year
    2. Y = (393 kWh/yr) x ($0.06) = $23.6/year
    3. Y = (420 kWh/yr) x ($0.06) = $25.2 /year
    4. Y = (362 kWh/yr) x ($0.06) = $21.7 /year
    5. Y = (301 kWh/yr) x ($0.06) = $18.1/year
    6. Y = (153 kWh/yr) x ($0.06) = $9.18/year

    e) energy used in one week

    so to calculate the energy each appliance uses in one week, i would use the equation
    energy = Power x time
    since we are only solving for the energy used in 1 week, the total time is 168 hours
    the only problem is how do i calculate the power in watts, do i use the energuide rating (in part b) and convert it into watts, so i can solve for the energy consumption in one week? im not even sure if my equation is right,
    for ex:
    1. E = 448 kWh x 168 hours (does 448 kWh have to be converted to Watts)
    can someone please verify and explain what to do here?

    f)
    when they say 'estimate' do they mean make a guess for how many hours a week each appliance is used?

    g) power for one week
    ok so i know that Power = Work/time
    it can also be Power = Energy/time
    so i would take the energy used in one week, for each appliance (calculated in part e), and divide it by the total time which is 168 hours. is this correct?



     
  2. jcsd
  3. Apr 3, 2016 #2

    billy_joule

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    kWhr is already a unit of energy.
    Like you said:
    Energy = power * time

    I assume you know kW is a unit of power and hour is a unit of time so if we put that all together:

    kW* hr = Power* time = Energy

    (if you can't figure out how to convert kWhr to Joules, try googling it)

    So if 448 kWh of energy are used in a year, how much of that is used in one week of that year?

    Yes.
    While a fridge is on all the time, the compressor motor is only on when the thermostat tells it to. You've probably heard your own the fridge turn off and on seemingly randomly. The proportion of time it's running is called the 'duty cycle'.

    This question is phrased poorly.
    It's unclear weather they want the average power consumption (based on b) or e) ) or the actual power consumption based on f).
    The latter should match the actual power consumption listed on the appliance by the manufacturer. The former will be a small fraction of that value and offer no new insight.
    So, I assume they want the appliance power that corresponds to your estimated duty cycle.
     
  4. Apr 3, 2016 #3
    for part e) , i understand that kW is a unit of power and hr is a unit of time, but i dont get what exactly im supposed to do to solve for the energy.
    so you mentioned converting 448 to joules, why would i need to do that? wouldnt that give me work instead of power?
    or since 448 kWh is the energy used in 1 year, would i have to divide to solve for the energy used in one week? and by what?
     
  5. Apr 3, 2016 #4

    billy_joule

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    You don't need to convert kWhr to joules. I thought there was confusion on what kWhr were so was just explaining that.
    Well, how many weeks are there in a year?
     
  6. Apr 3, 2016 #5
    there are 52 weeks in a year.
     
  7. Apr 3, 2016 #6

    billy_joule

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    Right, so if X amount of something is used in a year what fraction of X is used in a week?
    Eg if you get paid $60,000/year how much do you get paid per week?
     
  8. Apr 3, 2016 #7
    do i divide $60 000 by 52 weeks?
    $60 000 / 52 weeks = $1154 /week
     
  9. Apr 4, 2016 #8

    billy_joule

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    That's right. If you're working with unit conversions, laying it out in a consistent way makes it much easier to understand and solve:

    [tex]( \frac{$60,000}{ year} )( \frac{year}{ 52 ~ weeks}) = (\frac{$60,000}{ 1 } )( \frac{1 }{ 52 ~ weeks} )= \frac{$60,000}{ 52 ~weeks} =\frac{ $1154}{week} [/tex]
     
  10. Apr 4, 2016 #9
    so for part e)

    1. 448 kWh / 52 weeks = 8.62 kWh / week

    2. 393 kWh / 52 weeks = 7.56 kWh/week

    3. 420 kWh /52 weeks = 8.1 kWh/week

    4.362 kWh/52 weeks = 6.96 kWh/week

    5. 301 kWh/52 weeks = 5.79 kWh/week

    6. 153 kWh/52 weeks = 2.94 kWh/week
     
  11. Apr 4, 2016 #10
    f) Estimate of hours in use per week
    is this accurate?
    1. 90 hours/week

    2. 75 hours/week

    3. 80 hours/week

    4. 76 hours/week

    5. 21 hours/week

    6. 14 hours/week
     
  12. Apr 4, 2016 #11
    ok so i re-did part f) and then answered part g). is this correct?
    f) Estimate of hours in use per week

    1. 84 hours/week

    2. 80 hours/week

    3. 83 hours/week

    4. 76 hours/week

    5. 6 hours/week

    6. 5 hours/week

    g) Power for one week (W)

    To calculate the power of each appliance weekly, the following equation can be used:

    P = E / t (weekly)

    1. P = 8.62 kWh / 84 hours = 0.10 kW

    2. P = 7.56 kWh/ 80 hours = 0.09 kW

    3. P = 8.1 kWh / 83 hours = 0.098 kW

    4. P = 6.96 kWh/ 76 hours = 0.092 kW

    5. P = 5.79 kWh/6 hours = 0.965 kW

    6. P = 2.94 kWh/5 hours = 0.588 kW
     
  13. Apr 5, 2016 #12

    billy_joule

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    That looks good. I think your power values for the fridges & freezers are a little low ( which you'd only know if you're familiar with domestic refrigeration power consumption), but they're in the ballpark and how accurately you can guess a refrigerators duty cycle isn't really the point of the exercise.
    Good work.
     
  14. Apr 5, 2016 #13
    ok thanks for your help :smile:
     
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