Search results

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
  1. S

    Voltage Drop Example Problem

    Cable resistance is directly from Table 9 under alternating Current Resistance for uncoated wires. Not sure where you see effective Z in the answer? cos(13) = .97 so you are not using effective Z. I get the same answer using the table and accounting for a 4000 foot run.
  2. S

    NCEES #533

    Yes, that's correct
  3. S

    PE power 148

    This is the correct solution. When the transformer is at full load both copper loss and core loss occur. But core loss is negligible compared to the copper loss as there will be a huge surge of current passing through it as it is in short circuit. So its 12 hours at full load, then 12 hours at...
  4. S

    PE power 148

    If I have 10 transformers in my house, plugged in and running at no load all YEAR...….Will I not get a bill? I have no load at all on them. Yes, because those core losses are still drawing power
  5. S

    Percent Voltage Drop

    3Ø - 480: Vd-Value * sqrt(3) 480 1Ø - 277/480 Vd-Value 277
  6. S

    PE power 148

    Your cost is given in kWh. It's way easier to multiply what you're already given "kW * hours" for your kWh
  7. S

    PE power 148

    I think this is: Power losses cost for one day = 0.07* (5+3.5)*12 + 0.07*(3.5)*12 = $10.08/day Annual losses cost = 10.08 * 365day=3679.2 $/year The core losses are occurring with load and no load. What is the actual answer? I'm curious
  8. S

    Complex imaginary

    142.5 is your line current on primary just like 62.7 is your line on secondary side. I'm not sure what you're calculating. Secondary = a*I/1.732 = 62.7 isn't necessary at they told you the current. you're not solving for the secondary current. Taking the secondary current across to the...
  9. S

    Complex Imaginary Q#3

    I think you may be right
  10. S

    Complex imaginary

    They gave you a current of 62.7A on the secondary of T1. To get the primary current just transfer the current back the primary side of T1. 62.7 * ( 15000 / 6600) = 142.5 A
  11. S

    Power factor correction question - PPI Exam 1 Question 10

    The second chart has an error to me unless I'm not seeing something. 200 W & 150 VAR does not give 200VA. It gives 250VA-angle(36.9). 3uF is the capacitor rating in each phase. Total capacitor rating would be 9uF to get unity PF for 3 phases I think. I'm not sure delta or wye matter. 150VAR...
  12. S

    October 2019 Results Map

    There were definitely about 350+ Engineers in Denver taking the test with me in October
  13. S

    Free giveaway! - New Practice Exam for the PE by Electrical PE Review just released

    Please allow more copies. I’ve been trying for a while and I keep getting invalid
Back
Top