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  • Full Load Amps

    I'm designing the electrical feed to a well. I need to size the transformer & cable feeding the main switchboard based on the KVA. I also need to run my voltage drop calcs based on the KW.

    I have a single line consisting of multiple pumps, a 100A panel for misc power needs (lighting, recaptacles, etc...) & a unit heater.

    I have the HP of the pumps & a panel schedule for the 100A panel. But the only info I have for the heater is 19 FLA. I don't know the BTU's or even the model number to look up more info.

    Is there any way to determing the KVA or KW based solely on FLA? Or will I need to keep digging for more info?

    Thanks!

  • #2
    Re: Full Load Amps

    Originally posted by twinspa
    I'm designing the electrical feed to a well. I need to size the transformer & cable feeding the main switchboard based on the KVA. I also need to run my voltage drop calcs based on the KW.

    I have a single line consisting of multiple pumps, a 100A panel for misc power needs (lighting, recaptacles, etc...) & a unit heater.

    I have the HP of the pumps & a panel schedule for the 100A panel. But the only info I have for the heater is 19 FLA. I don't know the BTU's or even the model number to look up more info.

    Is there any way to determing the KVA or KW based solely on FLA? Or will I need to keep digging for more info?

    Thanks!
    The only other thing you need to know is the rated voltage; I don't know if you are using 115 V, 230 V or some other voltage rated components. Power = Voltage * Amperage, or P = V*I.

    However, for ac circuits, there is an additional complication. The voltage and the current may not be exactly in phase; this is called "power factor." If the electrical phase angle between current and voltage is theta (I can't make the Greek letter here), then P = V*I*cos (theta). But for many purposes, the product of volts and amps is used, and called VA. If divided by 1000, then kVA. NOTE: On a heater, current and voltage should be in phase, Power factor mostly applies to motor loads.

    Actually just the current, and the resistance of the wires suffices to calculate the voltage drop, but what is allowed may depend on the nominal voltage.

    Comment


    • #3
      Re: Full Load Amps

      Originally posted by Unregistered
      The only other thing you need to know is the rated voltage; I don't know if you are using 115 V, 230 V or some other voltage rated components. Power = Voltage * Amperage, or P = V*I.

      However, for ac circuits, there is an additional complication. The voltage and the current may not be exactly in phase; this is called "power factor." If the electrical phase angle between current and voltage is theta (I can't make the Greek letter here), then P = V*I*cos (theta). But for many purposes, the product of volts and amps is used, and called VA. If divided by 1000, then kVA. NOTE: On a heater, current and voltage should be in phase, Power factor mostly applies to motor loads.

      Actually just the current, and the resistance of the wires suffices to calculate the voltage drop, but what is allowed may depend on the nominal voltage.
      Thanks for the reply...

      I ended up contacting the EE who produced the plan. He informed me that he has since been told that the actual load is 13amps.

      So applying that to the formula above (using 3 phase 480V) I get 480V*13A=6.24kVA

      He also gave me the model of the heater as Reznor EGEA 10. So I looked up the specs on that & they list 34 MBH which converts to 9.96kW. Using 85% power factor(SCE power company standard for commercial applications) I get 11.72kVA.

      Considering that the total HP of the pumps involved at this well site is 450, this heater will be practically inconsequential, but I wanted to give an accurate number to SCE when I submit.

      Thanks again for the response.

      Comment


      • #4
        Re: Full Load Amps

        So applying that to the formula above (using 3 phase 480V) I get 480V*13A=6.24kVA
        Your calculation is wrong for finding power in 3-phase systems. There is a multiplying factor of either 3^(1/2) or 3 depending on if it is Delta or Wye. Heating elements are Wye configured, so therefore your power is 3^(1/2)*13A*480= 10.81 KW. There is no PF for heating elements since they are purely resistive, therefore my calculation will suffice.

        Comment


        • #5
          Re: Full Load Amps

          a 7.5k mini power panel
          The panel is 480/120/240v 1 phase. I need the load to add to existing MCC

          Comment


          • #6
            Re: Full Load Amps

            Originally posted by Unregistered View Post
            a 7.5k mini power panel
            The panel is 480/120/240v 1 phase. I need the load to add to existing MCC
            Do you mean 7.5 kW?

            It will depend on the voltage. Divide 7500 W by rms voltage
            120 V: 62.5 A
            or 240 V: 31.25 A
            or 480 V: 15.6 A
            Each alone would be 7500 W

            If you run loads at different voltages, multiply each voltage by the load in amps at that voltage, add the wattages. Must be less than 7500 W

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