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  • Cable/conductor size determination.

    aj_onduty

    Member

    Updated: Oct 21, 2024
    Views: 1.3K
    :smile: Greetings.
    Hi, Can anyone help me by explaining a method to determine conductor sizes. I am basically talkingg about HT cables. I am being confused by the idea. If u can, please give me a good explanation about the basics, and also the practical factors affecting the conductor size determination.
    Sorry for taking your time. I will be really thankful if u will answer this query. :smile:
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  • Voltaire

    Member • Aug 26, 2010

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  • aj_onduty

    Member • Aug 26, 2010

    Thank you Mr. Voltaire for your answer. You gave me a good amount of information.
    So, I understand that there are many factors are to be considered to determine the cable size. Can anyone give me a small explanation, an insight of what I am trying to dig into? Can any one at least name the factors we have to take into account to calculate cable sizes? I would be very thankful if anyone would help. Its very important for me. Kindly help me.
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  • Voltaire

    Member • Aug 27, 2010

    Use the analogy of a garden hose: the voltage equates to pressure and the water flow rate through the pipe equates to the current. The higher the water flow rate the higher the friction losses in the pipe (because the friction is a function of velocity squared the friction losses are 4 about times higher at 2m/s than at 1m/s). Electricity is a quantum packet. What this means, in simple terms, is that we can convert current (flow) into voltage (pressure) and vice versa. Do you remember that V = I x R and P = VI = I[sup]2[/sup] x R = V[sup]2[/sup]/R? Consider a 12V, 1A electric current (12W) flowing through a resistor. We can convert that to a 6V stream with a simple transformer. Since the circuit power is 12W, VI = 12W for case 2 as well, hence 12 = 6 x I and hence I = 2A i.e. when the voltage doubles then the current halves and vice versa.
    Case 1: V = 12V, P = 12W, I = 1A
    - P = V[sup]2[/sup]/R hence R = 144/12 = 12Ω
    Case 2: V = 6V, P = 12W, I = 2A
    - P = V[sup]2[/sup]/R hence R = 36/12 = 3Ω

    What this shows is that for a large current only a small resistance is needed to dissipate the entire power supplied while for a small current, the same resistance would cause only a small voltage drop for the same wattage viz. if a 3Ω resistor was used in the 12V circuit then dV = IR = 3V.

    Every conductor has a specific resistance and resistivity. That is: how much and to what extent does it resist the flow of electrons? The calculations for resistance is:
    R = r â„“/A
    Where
    R = resistance in ohm
    â„“ = length in metres
    A = area in metres squared
    r = specific resistivity of the solid conductor in ohm.metres squared per metre (usually given as W.mm[sup]2[/sup]/m)

    The dimensional analysis is R = W.m[sup]2[/sup]/m x m x 1/m[sup]2[/sup] = W

    The formula R = r â„“/A tells you that an increase in the area of a conductor would decrease the resistance and hence the voltage drop in a given conductor for a given power flow. You also saw that a low amp current of the same power has a lower voltage drop than a high amp current. When someone wants to transfer a huge amount of power then the cables can become very large in diameter and very costly.
    This is why energy utility companies transfer power cross-country at 132,000 volts because then they can use smaller diameter cables for the same wattage 😀
    =======================================================
    Just to satisfy your curiosity:
    The specific resistivity of the solid r is a function of the scattering of electrons. In superconductors the electrons flow in streamline paths and hence there is near zero resistance. In ordinary conductors such as copper and aluminum the electrons bounce off the walls and off each other and this causes an increase in potential energy i.e. loss of kinetic energy = loss of 'pressure' (as in our garden pipe analogy) i.e. loss of voltage or voltage drop. The value for r varies with temperature due to collisions between electrons and phonons. Phonons can be seen as quasiparticles (not real or virtual particles) that increase in quantity as lattice vibrations increase in the conductor. Since vibrations increase when temperature increases the conductor becomes less conductive (more resistive) at higher temperatures because there are more collisions between electrons and phonons.

    The specific conductivity is given by (see <a href="https://physics.info/electric-resistance/" target="_blank" rel="nofollow noopener noreferrer">Electric Resistance – The Physics Hypertextbook</a>)
    σ= (ne[sup]2[/sup] ℓ)/(m[sub]e[/sub] v[sub]rms[/sub])
    Where …
    σ = electrical conductivity
    n = density of free electrons
    e = charge of an electron
    m[sub]e[/sub] = mass of an electron
    v[sub]rms[/sub] = root-mean-square speed of electrons
    â„“ = mean free path length

    Resistivity is the inverse of conductivity and hence ρ = σ[sup]-1[/sup] so that ρ= (m[sub]e[/sub] v[sub]rms[/sub])/(ne[sup]2[/sup] ℓ)
    =======================================================
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  • bill190

    Member • Aug 27, 2010

    Also the ambient temperature is a factor as well as if many wires are tightly bundled together (heat build-up) or run loose with good air circulation as well as the rating of the insulation used.

    The word for that is "derating". Search for that word and here is one page on this...
    #-Link-Snipped-#

    (You might want to use a larger conductor under certain circumstances...)
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  • aj_onduty

    Member • Aug 27, 2010

    @Voltaire
    Sir, You gave me a good load of information, but I want to bring to your notice one fact that I am feeling that what all information you have given me is theoretical. I am looking for a it more practical point of views and also all the external factors you will face while you are actually determining the wire size for large power systems. Its true that we have resistivity as a main factor, but I am getting a feeling that I cannot simply use the things you have mentioned as the only factors to design a grid or determine wire sizes for transmission lines. Do you think the same way or is it that its just my feeling or should I use these factors only? Kindly let me know and if you think this is insufficient, kindly let me know that.
    Again, I thank you very, very much for your efforts.
    Expecting your answer on this,
    -Ajit
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  • Voltaire

    Member • Aug 28, 2010

    If you want to design a power distribution grid without doing the math then good luck!
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  • bill190

    Member • Aug 28, 2010

    I searched google.com for the words...
    power distribution calculations
    ...and came up with a lot of information including design software. Here is that search...
    <a href="https://www.google.com/#hl=en&source=hp&q=power+distribution+calculations&aq=0sx&aqi=g-sx1&aql=&oq=power+distrubution+calcu&gs_rfai=C0Tkrl5F5TL3bDqXgoAS_krGrBAAAAKoEBU_QwcmN&pbx=1&fp=ac77ff06aef7355b" target="_blank" rel="nofollow noopener noreferrer">Google</a>

    Another search is using the words...
    electrical calculations
    ...here is that search...
    <a href="https://www.google.com/#hl=en&safe=off&q=electrical+calculations&aq=f&aqi=g-c10&aql=&oq=electrical+calculations&gs_rfai=&pbx=1&fp=d575273b06deb3e2" target="_blank" rel="nofollow noopener noreferrer">Google</a>
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  • aj_onduty

    Member • Aug 28, 2010

    Voltaire
    If you want to design a power distribution grid without doing the math then good luck!
    sir.i never meant offence.i am really sorry that i made you feel bad.i apologise for this.as i have logged in through my mobile,i cannot type all the things that i meant right now.again,i am extremely sorry and i never meant that the data you gave wasnt helpful.
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  • aj_onduty

    Member • Aug 29, 2010

    What I wanted to know is that, is there any other factors(apart from calculation of resistance, resistivity) such as environmental factors which we have to take under our consideration to determine the conductor size. I am sorry again Voltaire sir, I am still feeling very bad for making you feel bad. Really sorry.
    I was just being desperate. The problem is that my seniors are asking me these factors and I have no other source for help. Very very sorry for being rude to you sir.
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  • aj_onduty

    Member • Aug 30, 2010

    Am I asking too much? Anyone, please.
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  • Voltaire

    Member • Aug 31, 2010

    Ajit
    Looking at the above posts, what do you think are the influencing factors?
    Secondly: what is your brief? Do you, for example, need to consider tensile forces?
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  • aj_onduty

    Member • Aug 31, 2010

    Voltaire
    Ajit
    Looking at the above posts, what do you think are the influencing factors?
    Secondly: what is your brief? Do you, for example, need to consider tensile forces?
    Firstly, The environmental factors. For example,
    >>the place where cables are going to be, overhead or underground.
    >>If overhead, in hazardous environment or climatically challenged place.
    >>Corona, heating up, and other factors.
    In short, I want someone who has designed some transmission lines and cables and has faced the challenges while doing it. I would like him/her to share the experriences the person had while working. Not just that person, but someone experienced enough to speak.
    Kindly let me know if my question is genuine enough.
    Thank you.

    P.S.-Am I asking too much?
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  • Voltaire

    Member • Sep 2, 2010

    Ok fellas
    Ir. Rhett Kelly is a professional electrical engineer and the Chief Engineer of a leading power utility supplier. In case some of you do not know; Chief Engineer is the highest job title for any engineer in any discipline.
    He responds as follows:
    =======================================
    The question is unclear and non specific.

    A brief summary of issues to consider for cable conductor size determination (from an electrical perspective):

    (note as this is intended for students in a foreign country, answers are not provided, but rather the questions/issues to consider)

    1. required load (in MVA or Amperes) and load profile (i.e. load factor)
    2. prospective short circuit current (symmetrical and earth fault) and duration (typically 1 second). The duration is determined, in general, by the protection settings applied and type of protection functions.
    3. allowable voltage drop (where applicable)
    4. technical losses (IxIxR) associated with a particular conductor size over a 25 year (typical) period brought back to net present value (often warranting a larger conductor / cable size than the load and system requires).
    5. planned spare capacity for emergency/contingency scenarios

    To determine the current ratings of insulated cables, refer to IEC 60287 (for 100% load factor) and IEC 60853 (for cyclic rating factors) for detailed equations or consult manufacturers catalogues. The short circuit ratings of cables can be calculated assuming adiabatic conditions using equations based on the initial and final conductor temperatures, the conductor material, the conductor cross section and the short-circuit duration (typically 1 sec).

    The current rating of a cable is affected by the following:

    1. the maximum continuous conductor temperature allowed - this depends on the insulation material
    2. the maximum short circuit conductor temperature allowed - this also depends on the insulation material
    3. the installation conditions - namely:
    -free-air above ground vs directly buried underground cable vs cable installation in pipe ducts vs cable installation in air within ventilated tunnels)
    -ambient air temperature (annual average) for installations in air
    -soil temperature (annual average) for direct buried cables or cables in pipe ducts in the ground
    -spacing between cables and/or other circuits (i.e. presence of external thermal sources in proximity) for direct buried cables
    -depth of burial for direct buried cables
    -soil thermal resistivity for direct buried cables
    -for single core cables - the configuration (flat vs trefoil formation)
    -for single core cables - the type/method of earthing applied to earth the cable metallic screen/sheath (also referred to as special bonding to limit / eliminate sheath circulating currents)

    Regards,

    Rhett Kelly Pr. Eng.
    Chief Engineer - Technology Development
    Power Plant Technologies
    Eskom Corporate Services
    Johannesburg, South Africa
    #-Link-Snipped-#
    =======================================
    Please note that Ir. Kelly echoes my gripe viz.
    The question is unclear and non specific.
    PS a BIG thanks also to Dr. Mark van der Riet
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  • aj_onduty

    Member • Sep 2, 2010

    @Voltaire.
    Sir,
    I never expected such a precise satisfaction of my curiosity. I thank you from the bottom of my heart for the efforts you have taken behind answering this question. This answer is really going to help me a lot. And let this answer come of good use to other CEans here.
    Again and yet again, I thank you for answering this question.
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  • chuackl

    Member • Sep 2, 2010

    Well, the above information provided above are very helpful to all CEans, thank you voltaire😁
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  • Kaustubh Katdare

    Administrator • Sep 2, 2010

    CEan - Voltaire is taking CE to a newer height! Unbelievable!
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  • aj_onduty

    Member • Sep 2, 2010

    Sure BigK. I was (seriously) searching for a precise answer for about a month, and I couldn't find it in books, I didn't get it in Google, and at last I thought I won't find it. But look at this beauty.Hats off Voltaire.
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  • Voltaire

    Member • Sep 3, 2010

    Fellas
    Thanks anyways but the people who really deserve your gratitude are Dr. Mark van der Riet, Ir. Rhett Kelly and Ir. Barry MacColl of ESKOM South Africa
    I have conveyed your appreciation to them and I'm sure that they would welcome it
    Voltaire
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  • chuackl

    Member • Sep 3, 2010

    Well if there is no Voltaire in CE forum the answer also won't be publish at here am i right?😀 So you deserve to be thank also and i remember last time you did approach my question to a professor also. At the moment i see the reply i can't think of anything thing else other than keep in mind saying thank you to you.
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  • aj_onduty

    Member • Sep 3, 2010

    To be frank, I am proud to be in a forum where experts' advices and views are posted. Really amazing.
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