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Transmission line Voltage loss

  1. May 6, 2015 #1
    Ok I do not know if I am right about this but :
    if you have an AC 120 V 60Hz generator connected to a wire of Length L(1000KM) and Resistance R going to a motor down a well, to know if you will have Voltage loss you do

    λ = C / F
    C is the speed of Light = 299 792 458 m/s
    F is the 60Hz
    λ = 299 792 458 (m/s) / 60 Hz = 4996540.966666667 m
    = 1000 Km / 4996540.966666667m = 0.20013845711889121639354535604153 = 20%
    20% < 10% so there will be voltage loss.
    and the end Voltage( Voltage to the Motor) will be Ve
    Ve = 120 V Cos( 2 pi (60) 1000E3/ 299 792 458 ) = 120V ( 0.99975916026658575202469779578012) =
    Ve = 119.97109923199029024296373549361 V
    is that right ?
    it seems small ...

    Also should R and C play a part in the Voltage loss? I feel like I am going something wrong.
    You loss Voltage because of power/ heat in the wire right ? so I2R?
    so I feel like I should use R to get Ve ( Voltage at the motor)
     
  2. jcsd
  3. May 6, 2015 #2

    jim hardy

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    you postulate a 1/5 wavelength transmission line.

    Indeed transmission line effects do happen on extremely long lines.

    try this link starting about page 23.

    https://ece.uwaterloo.ca/~raelshat/COURSES_dr/eleb7/NewNotes/Power%20System%20Components_Part1.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. May 6, 2015 #3

    Jim Hardy,

    I looked at your link, but It is not answering my questions.
    did I solve for the Ve ( Voltage at the motor) right ? if not how should I have?
    also should I use R and I so solve for Ve?
    R is The distributed resistance e1e1d3d40573127e9ee0480caf1283d6.png of the conductors is represented by a series resistor (expressed in ohms per unit length).
     
    Last edited by a moderator: May 7, 2017
  5. May 6, 2015 #4

    berkeman

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    Why are you running 120Vrms for 1000km? That is a very lossy setup. And it shouldn't have much to do with the wavelength, etc. It will be dominated by IR losses
     
  6. May 6, 2015 #5
    1000Km is just an example ...
    in the real world the oilwell are no more than 743.2m (9000 feet). But the motor would be at the bottom of the well connected to a pump.

    anyhow,
    did I do Ve right ?
     
  7. May 6, 2015 #6

    berkeman

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    I don't think so, but I didn't really follow what you were trying to do. Once you mentioned the speed of light, I kind of zoned out reading the post... :woot:

    To calculate the voltage loss across the cable, you need the total resistance of the cable, and the current flowing in the cable. If you know the wire gauge, you can use Google to find the resistivity in Ohms per thousand feet or similar dimensions. For a 1000m run of cable, you will have 2000m of wire (out and back). From the resistivity and cable length, you get your total resistance.

    From the motor power, you get the input current: P = Vrms * Irms, assuming a reasonable power factor for the motor. Use that Irms * Rtotal to calculate how much Vrms loss you get across the cable. :smile:
     
  8. May 6, 2015 #7
    The motor run on 120V, I do not care to know the power, just to know if 120V is making it to the motor down the well.
    why did you say you zoned out reading the post after I mention the speed of light ?
    you need C to find λ
     
  9. May 6, 2015 #8

    berkeman

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    You don't need λ to find the resistive losses in your power cable. Transmission line effects do not enter into this calculation, as long as the motor is consuming the power that is sent to it over the power cable. Think of it as a well-matched load termination -- there will be very little reflected power, even if your power cable were near λ long.

    I don't care if you don't care to know the power. :smile: You need the Irms that is being fed to the motor to calculate the Vrms loss.

    So for example, if the motor is 10hp, that's about 7500W, which is 7500W / 120Vrms = 62.5Arms. (that's a lot of current)

    If you use 00 AWG cable (0.079 Ohms per 1000 feet), and the cable is 2000 feet long, that is 4000 feet of wire, which is 4 * 0.079 = 0.316 Ohms. So your V = I*R drop is about 62.5Arms * 0.316 Ohms = 20V. So your 120V starting voltage becomes about 100Vrms at the motor for a 2000 foot long 00 AWG power cable.

    What is your motor power? What gauge are you planning on using for your power cable?


    NOTE -- I edited my post to fix a math error, in case you saw the longer version of it before I deleted it and fixed it.
     
  10. May 6, 2015 #9
    WHy 4 * 0.079 = 0.316 Ohms ? and not 4000 * 0.079 ?
    and why do I not need λ?
    I am sorry, I thought λ is used for transmission lines and that this was a transmission line ...
    is it not a transmission line ?
     
  11. May 6, 2015 #10

    berkeman

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    That's the math error that made me delete and fix my post. The resistivity is 0.079 Ohms per 1000 feet, not per foot. :smile:

    No, transmission line effects do not come into play here. The wavelength in free space at 60Hz is about 300,000,000m/s / 60Hz = 5 million meters. That matters for long power transmission lines, but not for 1000m of cable. And again, your motor is absorbing most of the power sent to it over the power cable, so there is not much lost in reflection from the motor. Just stick with calculating the I*R losses in the cable for this situation. :smile:
     
  12. May 6, 2015 #11

    jim hardy

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    what does that mean ? I too tuned out because to me it's nonsensical unless you are trying to treat it like a "transmission line" as in radio and TV.
    And your inequality sign points the wrong way ? 20% is > 10% .

    At line lengths approaching 1/4 wavelength , that treatment becomes appropriate. Yours is 1/5 wavelength. So i gave you a link to that treatment of 60 hz power transmission lines. It's a real phenomenon.

    A question well stated is half answered.
     
  13. May 6, 2015 #12

    berkeman

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    Oops, did I make another math error?
     
  14. May 6, 2015 #13

    jim hardy

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    more likely i made it.

    Let's see c= 3e8 m/sec
    divided by 60hz

    equal 5e6 meters = 5,000km
    and he said 1,000km ?
    i think he's at 1/5 wavelength. That's why i answered the wrong question.
    another "Red Herring" . No harm intended.

    Power system guys do have to take into account that radio wave like behavior even at their measly 60 hz .
    It's un-economic to let energy radiate into space because a long transmission line is acting like an antenna.
    I think that's part of the reason for "transposing" the phases on long lines, to make them sort of a twisted pair.

    Next time you drive along a long power line, like a hundred mile one, watch the lines and you'll see them roll the phases every few miles - i think like twenty. But don't take that number as an authoritative estimate, just from my rusty old memory .
    You can see it along FPL's 500 KV line parallel to US27 between Clewiston and Miami. And that's a nice drive.

    In the days of analog telephone over copper wire pairs on poles we rolled the pair according to the frequency of the telephone carriers. A 20 khz line carrying five conversations would be rolled at every pole, a straight voice line only every ten or twenty poles.


    old jim
     
  15. May 6, 2015 #14

    berkeman

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    Oops, I had 1000m in my head, not 1000km. You are correct.

    So to the OP, you will get about 0.1Vrms at the end of your 1000km power cable. If you really want to go that far, you need to do what the power companies do, and use step-up and step-down transformers to minimize IR losses in your long power cable. So this is for deep drilling applications?
     
  16. May 6, 2015 #15

    davenn

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    I don't know of anything that drills 1000 km deep
    at that depth you are well into the Earth's molten mantle

    I suspect he really meant 1000m (1km) ( rereading he did say just an example)
    a note to @btb4198 it is better to keep things realistic with your posed questions/problems :smile:


    deepest depths drilled are as listed from wiki .....

    The Kola Superdeep Borehole (Russian: Кольская сверхглубокая скважина, Kolskaya sverkhglubokaya skvazhina) is the result of a scientific drilling project of the Soviet Union in the Pechengsky District, on the Kola Peninsula. The project attempted to drill as deep as possible into the Earth's crust. Drilling began on 24 May 1970 using the Uralmash-4E, and later the http://en.wikipedia.org/w/index.php?title=Uralmash-15000&action=edit&redlink=1 [Broken] series drilling rig. A number of boreholes were drilled by branching from a central hole. The deepest, SG-3, reached 12,262 metres (40,230 ft) in 1989 and still is the deepest artificial point on Earth.[1]

    In terms of true depth, it is the deepest borehole in the world. For two decades it was also the world's longest borehole, in terms of measured depth along the well bore, until surpassed in 2008 by the 12,289-metre-long (40,318 ft) Al Shaheen oil well in Qatar, and in 2011 by 12,345-metre-long (40,502 ft) Sakhalin-I Odoptu OP-11 Well (offshore the Russian island Sakhalin).[2]

    so a bit over 12km a VERY long way from 1000 km


    Dave
     
    Last edited by a moderator: May 7, 2017
  17. May 6, 2015 #16

    berkeman

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    Thanks Dave! :smile:
     
  18. May 6, 2015 #17
    The Well Is only about 1000m (3200Ft) deep that was my bad. I have a FVD (Altivar 71) that powers a pumper: 120V and 4 HP.

    so when I buy the wire I can get the R' (Ohms per 1000 feet) ?

    Also Transmission line seem to only be use for high F and very very long distance...
    but I should just use
    V= IR
    so I would have
    P = iv
    I = 2982.799488 W / 120 V = 24.8566624A
    so using 00 AWG cable (0.079 Ohms per 1000 feet), 0.079 * (3.28084) * 2 = 0.51837272 ohms
    V ( Voltage Drop) = 24.8566624( 0.51837272 ) = 12.88499579289728V

    um... is that wrong ?
     
    Last edited: May 6, 2015
  19. May 6, 2015 #18

    jim hardy

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    it'd sure be easier for me to read if you rounded to three digits, slide rule accuracy.

    It looks okay so far as voltage drop for the current you calculated.
    Glad to see you doubled resistance to account for the round trip , it's common to forget that .

    You'll want to get the actual current for the pump motor. You just divided 4hp's worth of watts by 120 volts and that's very optimistic..
    The motor has a "power factor" so it'll draw probably 25% more current than you figured
    and since most motors will will draw a lot more current when starting than when running
    your wire must deliver enough voltage for the pump motor to start.

    i tried to find the manual for Altivar 71 VFD but only could get to a DC braking unit.
    so I dont know how the VFD behaves. I've never used one. Surely its manual tells how to size the cables.

    When you don't have a wire table handy
    here's a handy "Rule of Thumb":
    #10 wire is 1 ohm per thousand feet and that's easy to remember
    every three gages that doubles
    so every gage changes by cube root of two
    so #13(an unusual size) is two ohms/thousand
    and #16 is 4 ohms/thousand
    and #7 is a half ohm/thousand
    #4 a quarter

    here's a table, any search engine will turn one up for you
    http://en.wikipedia.org/wiki/American_wire_gauge
    http://hyperphysics.phy-astr.gsu.edu/hbase/tables/wirega.html

    as you see the "Rule of Thumb" gives a reasonable approximation

    hang in there - it becomes easier as your familiarity grows.

    old jim
     
    Last edited: May 7, 2015
  20. May 7, 2015 #19
    is it called Voltage drop or Resistive Losses ?
     
  21. May 7, 2015 #20

    nsaspook

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    We have wire Resistive Loss as a constant and a Voltage Drop on that wire thats a function of the voltage/current ratio for the power delivered at the load. Higher source voltage for the same power at the load (resistance increase for lower wire currents) gives a smaller 'Voltage drop' as a percentage of the total source voltage delivered to the load.
    If we were to increase voltage while maintaining the same wire current by increasing load resistance to keep the delivered power the same, we see the same effect of reduction in 'Voltage Drop' as a percentage of the total source voltage delivered to the load.
     
    Last edited: May 7, 2015
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