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Homework Help: Factory Power Demand Costs

  1. Jul 28, 2016 #1
    1. The problem statement, all variables and given/known data
    A factory has an average demand of 520 000 units per week. The
    maximum demand is 25 MVA at 0.8 power factor and the minimum
    power factor of 0.6 occurs when the demand is 11 MVA.
    The factory is charged at 2.5 pence per unit with a surcharge of 0.2 pence
    per unit for each 500 kW by which the maximum demand exceeds
    18 MW and a further surcharge of 3% (of charge, plus surcharge) for
    every increment of 0.05 by which the minimum power factor falls
    below 0.8.
    There is a large drive which operates continuously and is powered by an
    induction motor with draws 2 MW at a power factor of 0.8 lagging.
    This motor is replaced by a synchronous motor which draws the same
    power but runs at a power factor of 0.8 leading.
    Assuming the maximum demand penalty does not change

    2. Relevant equations
    (i) Show that the maximum demand power is 20 MW.
    (ii) Show that the total weekly charge for the factory is £19 219.20.
    (iii) Calculate the new power factor and reactive penalty charge when
    demand is 11 MVA.
    (iv) Show that the new total weekly cost is £17 644.50.
    (v) If the synchronous motor costs £250 000, calculate the time required
    to recover the cost of the motor.

    3. The attempt at a solution
    (i) So this one seems obvious,
    VAcosø = 25*10^10 * 0.8 = 20 MW

    Unit cost, (520000 * 2.5)/100 = £13,000

    I’m at a loss at to what to do next, any advice on next steps would be very welcome.
  2. jcsd
  3. Aug 3, 2016 #2
    So finally getting some time to come back to this, for (ii) think there are essentially 3 costs to calculate,

    Pence per unit charge:

    Unit cost, (520000 * 2.5)/100 = £13,000

    Max demand surcharge:

    (18*10^6)/1000 = 18000 units

    18000*2.5 = 45000/100 = £450.00

    (2*10^6)/1000 = 2000 UNITS

    2000*2.7 = 5400/100 = £54.00

    So total charges so far are £13,504.00

    Reactive penalty charge:

    3% of £13,504.00 = £405.12

    And this seems to be as far as I can get, am I on the correct lines so far?
  4. Aug 7, 2016 #3
    Hi everybody, I don't seem to be receiving any responses to my post, could anybody advise as to what I might need to do to generate some help.

  5. Aug 7, 2016 #4


    User Avatar

    Staff: Mentor

    First, I think anyone whose native language is not English should forget this exercise---go and spend your time on some other problem.

    I have arrived at the answer they want, so apparently that says I've managed to solve the riddle? :smile:
    Can you show how to calculate this surcharge.
  6. Aug 7, 2016 #5
    Thanks for the response, so for the max demand surcharge I have taken this to be an additional two-part cost,

    Part one, max demand of 18MW will be at standard charge of 2.5pence per unit, so to convert 18MW to units, (18*10^6)/1000 = 18000 units

    18000 units at 2.5pence per unit then converted to pounds, 18000*2.5 = 45000/100 = £450.00

    Part two, additional surcharge above 18MW. Max demand from part (i) is 20MW so 20MW-18MW leaves 2MW at surcharge cost.

    2MW to units, (2*10^6)/1000 = 2000 UNITS

    2000 units at surcharge cost of 2.7pence per unit(2.5 + 0.2) then converted to pounds gives me, 2000*2.7 = 5400/100 = £54.00

    So my total max demand surcharge cost is £450.00+£54.00 = £504.00.
  7. Aug 8, 2016 #6


    User Avatar

    Staff: Mentor

    Interesting. Somehow you understand that 1kW equals 1 unit of demand power? Is this something I should know? In any case, it isn't necessary to know (or to assume) this, to do the calculations.

    A surcharge is a charge on top of the base charge, so in calculating the surcharge, you'll use the 0.2p figure but not the 2.5p.

    I said ##max\ demand\ surcharge = 520,000 \times \left(\dfrac{20\ MW - 18\ MW}{500kW}\right) \times 0.2p##

    As stated in words here:
    I'm applying the surcharge to all units, not just those in excess of 18MW. The fact that this gives the textbook's answer seems to confirm it.

    Do you follow this, so far?

    So, adding the base charge on average demand together with this max demand surcharge you arrive at what combined cost?
  8. Aug 8, 2016 #7


    User Avatar

    Staff: Mentor

    This would be energy, in kWhr, do you think? I can't see how it could be anything else.
  9. Aug 8, 2016 #8
    So I understood 1 unit of billed electricity to = 1000W, so 1KW consumed in an hour would = 1KWhr

    Your calculation makes obvious sense now, the surcharge costs calculates to £4160.00

    3% of this cost combined with the unit charge gives me £514.80 * (0.8 - 0.6)/0.05 = £2059.20

    All three costs combined gives me the target of £19,219.20.

    Thanks for the help so far.

    Now for part (iii)
  10. Aug 8, 2016 #9


    User Avatar

    Staff: Mentor

    What do you understand part(iii) to be about?
  11. Aug 9, 2016 #10
    So my first attempt is clearly well off, but I think I need to find the corrected reactive power in order to calculate the new P.F.

    cosø = MW/MVA

    Active power of load MVAcosø = 11*10^6 * 0.6 = 6.6MW

    Reactive power of load MVAsinø = 11*10^6 * 0.8 = 8.8MVAr lagging

    Active power of synchronous motor = 2MW

    Reactive power of motor = 2/0.8 = 2.5MVA = MVAsinø = 2.5*10^6 * 0.6 = 1.5KVAr leading

    Corrected reactive power = 8.8 – 1.5 = 7.3KVAr lagging

    MVA = Root 6.6^2 + 7.3^2

    =9.84 MVA

    New calculated P.F is not correct with these new outputs.
  12. Aug 9, 2016 #11


    User Avatar

    Staff: Mentor

    Are you saying the textbook tells you what the pf should be, for this part?

    My attempts seem to have reached an impasse...
  13. Aug 10, 2016 #12
    No apologies, I did not complete the equation based on my my findings,

    cosø = MW/MVA

    6.6/9.84 = 0.67
  14. Aug 10, 2016 #13


    User Avatar

    Staff: Mentor

    I tried a couple of interpretations, but failed to divine their precise thinking.

    What you could do is take their new costing and work backwards to determine the pf this would be based on. While the answer is rounded to the next 0.05, at least you'd be able to see roughly what you're aiming for.
  15. Aug 11, 2016 #14


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    Staff: Mentor

    My final thoughts...

    When using the induction motor, at 11MVA the pf was 0.6, so load is 6.6 - j8.8 MVA
    replacing with the synchronous motor, at this same real power, the load becomes 6.6 - j6.4 MVA,
    this is 9.19 MVA at pf of 0.718, and means the factory will now incur 2 units of power factor penalty.

    520000p × 3.3 × (1 + 2 × 0.03) = £18,189.60
    This is not in agreement with the cost figure provided in the question.
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