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Hope someone can help me out and point me the right direction to solving this.

I am trying to do a voltage drop calculation to see the VD % for the cable selected. The following are the values,

Current: 140A (I)
Resistance of Cable: 0.0001 Ohm/m (R)
Reactance of Cable: 0.0000704 Ohm/m (X)
PF: 1
Distance: 300m (L)
Voltage: 20 kV (20000V)

I am using the following voltage drop formula that I found in a very old school notes (can someone pls verify this formula)?,

Vd = (I x (R cos theta + X sin theta) x L) / 1000

If I input the values into the formula, my answer is 0.266V. This does not look right for me as I will get a weird VD %.

Can someone please assist? Thanks!

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  • \$\begingroup\$ Shouldn't theta = pf? \$\endgroup\$ – SunnyBoiz Dec 29 '17 at 8:43
  • \$\begingroup\$ @SunnyBoiz no, ‘Cos theta’ equals the power factor! \$\endgroup\$ – rfkortekaas Dec 29 '17 at 9:49
  • \$\begingroup\$ Ic! What does 'sin theta' equals to? (where do I obtain the value?) \$\endgroup\$ – SunnyBoiz Dec 30 '17 at 3:53
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I think that the 1000 in your formula is placed incorrect and probably used when the length is in kilometers. So basically for a one phase load the formula is:

$$V_d = 2I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L$$ Note: the first 2 is to get the result from source to load and the return as the return cable also influences the voltage drop

For a three phase system the formula is: $$V_d = \sqrt{3} I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L$$

Where:

\$ \begin{align} V_d&=\text{voltage drop in volts}\\ I&=\text{current in amperes}\\ R&=\text{conductive resistance in ohms/m}\\ X&=\text{conductor inductive reactance in ohms/m}\\ L&=\text{one way length of circuit in m (or km/1000 in your formula)}\\ \theta&=\text{phase angle of the load}\\ PF&=\cos(\theta)\\ \end{align}\\ \$

Answer

\$ \begin{align} PF&=1\\ \theta&=\arccos(PF)=0\\ \\ V_d&= 2I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot\cos(0) + 0.0000704\cdot\sin(0)\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot1 + 0.0000704\cdot0\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.0001\bigr)\cdot 300\\ V_d&= 2\cdot4.2\\ V_d&= 8.4\\ \\ \end{align}\\ \$

So this gives a voltage drop for one run of 4.2 V and for a double run 8.4 V.

Answer with different PF

Because a PF of 1 in an AC circuit is not a real world example I will show the influence of a PF of 0.8 on the voltage drop:

\$ \begin{align} PF&=0.8\\ \theta&=\arccos(PF)=36.8699°\\ \\ V_d&= 2I \bigl(R \cos(\theta) + X \sin(\theta)\bigr) L\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot\cos(36.8699°) + 0.0000704\cdot\sin(36.8699°)\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.0001 \cdot0.8 + 0.0000704\cdot0.6\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.00008+0.00004224\bigr)\cdot 300\\ V_d&= 2\cdot140\cdot\bigl(0.00012224\bigr)\cdot 300\\ V_d&= 2\cdot5.13408\\ V_d&= 10.26816\\ \\ \end{align}\\ \$

As a result of the reactive power the voltage drop will increase.

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  • \$\begingroup\$ Hi! Thanks for the response. After checking the school notes again, you are right! I misplaced my 1000 division and it is meant to be used for L. Can you enlighten how you obtain 4.524V in your calculation? I input the values into the formula but my answer is 266.112V. \$\endgroup\$ – SunnyBoiz Dec 29 '17 at 15:17
  • \$\begingroup\$ Hi rfkorekass, thanks for the steps. Can I ask one more thing? How come it is 2pi? I thought if the pf is given, in this case is 1, I just have to sub in to cos and sin respectively. And when I key it in my calculator it will be cos(inverse)1 and sin(inverse)1. Can you show me how to derive to 2pi? If this is a three phase load, I have to multiply sqrt 3? \$\endgroup\$ – SunnyBoiz Dec 30 '17 at 3:57
  • \$\begingroup\$ Thank you so much everyone. I am almost there to understanding the equation. One more thing, for the sin(theta) value, do I just simply sub in the arccos(PF) value of 36.86 into the equation? \$\endgroup\$ – SunnyBoiz Dec 31 '17 at 2:38
  • \$\begingroup\$ Yes, you need to replace \$\theta\$ with \$\arccos(PF)\$. As in the example \$PF=0.8\$ then \$\sin(\theta)=\sin(\arccos(0.8)=0.6\$. \$\endgroup\$ – rfkortekaas Dec 31 '17 at 7:20
  • \$\begingroup\$ Thank you! For this particular voltage drop calculation, the 2 is necessary? I thought the multiplication of 2 is only for single phase/DC. If this is a three phase, is there any addition I must add into equation? \$\endgroup\$ – SunnyBoiz Jan 1 '18 at 2:57
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Let's do this step by step. Shoving everything into one formula usually obscures what's happening.

0.0001ohms/s is quite big cable. 1mm2 cable is about 17mohms/m, so 0.1mohm/m means an area of 170mm2, which is a diameter of about 15mm. Is your cable that big?

140A in 0.1mohm gives 14mV voltage drop per metre.

300m of cable at 14mV/m is a total drop of 4.2v. Way above your figure, and we've not included the reactance yet, which will increase it.

What's that /1000 doing in your formula? Unless you understand why it's there, where it's come from, you might get answers that are 1000 times too small!

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  • \$\begingroup\$ Hi Neil. Yes the cable used in the example is a huge cable. I have no idea what is the purpose of the division of a thousand. Hence I will like someone to verify the above formula (afterall its taken from my old school notes). Couldn't find anything online. T_T \$\endgroup\$ – SunnyBoiz Dec 29 '17 at 8:46
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How the voltage drop formulas derived; Simply voltage drop of cable is the voltage occured due to the resistance (R) and inductance (X) characteristic of the cable itself which depend on load (amp) and total cable length (meter). Taken both R and X as impedance (Z).

V = IZ V cable drop = Icable x Z cable Phytagoras formula, Z²= R²+X² Z = R²/Z + X²/Z Apply trigonometric function; cos ϴ = R/Z sinϴ = X/Z Thus Z = R(cos ϴ) + X(sin ϴ)

For 1 phase 2 wire, we need to multiply Z by 2 since it is for live and neutral cable. Vd= I x Z x Length of 2 cable (or L x 2)

For 3 phase, we need to multiply by √3

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Assuming the voltage, current and power factor are receiving end quantities the magnitude difference in voltage obtained is 4.2V which is no surprise because, the cable impedance is very low.

Your formula is correct, the "/1000" is most probably because the required answer is in kiloVolts rather than in Volts. On practical transmission lines the voltage drop due to reactance is usually in the range of kilovolts.

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  • \$\begingroup\$ 4.2 is your final answer? I assume you forgot about reactance? \$\endgroup\$ – MCG Dec 29 '17 at 11:03
  • \$\begingroup\$ Please look at what power factor is given \$\endgroup\$ – spaul Dec 29 '17 at 11:04
  • \$\begingroup\$ Oh right..... which means you didn't answer the question then? \$\endgroup\$ – MCG Dec 29 '17 at 11:07
  • \$\begingroup\$ I've improved my ans \$\endgroup\$ – spaul Dec 29 '17 at 11:15
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    \$\begingroup\$ Look at the other 2 answers. They give explanations, all you did was say omit the 1000, and the formula is wrong. You did not explain how/why you came to this conclusion \$\endgroup\$ – MCG Dec 29 '17 at 11:25

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