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 October 3rd, 2016, 06:57 AM #1 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,143 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions Transmission losses in electricity distribution networks Need some help with an electricity query... The formula for power losses in a wire is $\displaystyle P_L = I^2 R$ where I is the current passing through the wire and R is the resistance. If the power transmitted along the cable is P and the voltage level is V, then the losses can be written as $\displaystyle P_L = \frac{RP^2}{V^2}$ In an A.C. circuit, V is not constant, so can substitute $\displaystyle V = |V|\cos\phi$ where $\displaystyle \cos\phi$ is called the 'power factor' and is a parameter of the generator. The formula can therefore be rewritten as $\displaystyle P_L = \frac{R}{|V|^2 \cos^2 \phi } \cdot P^2 = B P^2$ where B is called the 'B-loss coefficient', which is $\displaystyle B = \frac{R}{|V|^2 \cos^2 \phi }$ Now... for the case of two generators connected to a single node (aka bus) via two separate wires, we can rewrite the loss formula to be $\displaystyle P_{L,i} = \frac{R_i}{|V_i|^2 \cos \phi_i }$ I would've thought that $\displaystyle P_L = P_{L,1} + P_{L,2}$ $\displaystyle P_L = \frac{R_1}{|V_1|^2 \cos \phi_1 } P_1^2 + \frac{R_2}{|V_2|^2 \cos \phi_2 } P_2^2$ or, if we define $\displaystyle B_i = \frac{R_i}{|V_i|^2 \cos^2 \phi_i }$ we get $\displaystyle P_L = B_1 P_1^2 + B_2 P_2^2$ However, according to an online journal article, (http://www.arpapress.com/volumes/vol...as_12_2_20.pdf) they seem to define B something like this (not sure about this one): $\displaystyle B_{ij} = \frac{R}{|V_i||V_j| \cos \phi_i \cos \phi_j}$ and then quote the following formula as the solution: $\displaystyle P_L = B_{11} P_1^2 + 2B_{12} P_1 P_2 + B_{22} P_2^2$ My questions are: 1. Why isn't the power loss just the sum of the power losses across each cable? What is the significance of the additional $\displaystyle 2 B_{12} P_1 P_2$ term? 2. What is R in the definition of $\displaystyle B_{ij}$? Is it the total resistance calculated from $\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$? If so, how come we need to know the total resistance of a circuit to calculate the power losses in a single wire? Is it because of the dependence of power injection across the other wire? Last edited by Benit13; October 3rd, 2016 at 07:02 AM. October 3rd, 2016, 07:35 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,622 Thanks: 2611 Math Focus: Mainly analysis and algebra I'm not certain of the configuration of your generators, but power-loss would indicate resistance, so in parallel the wouldn't sum directly. October 6th, 2016, 06:04 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,143 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions The plot thickens... The paper I linked basically copied a bunch of stuff from another paper: https://www.researchgate.net/publica...STEM_OPERATION In that paper, they define $\displaystyle B_{mn}$ as $\displaystyle B_{mn} = \frac{\cos(\delta_m - \delta_n)}{|V_m||V_n| pf_m pf_n} \sum_k N_{km} N_{kn} R_k$ and then go on to give an example calculation where they don't use this formula... So, they haven't made it clear how the power losses were defined in terms of $\displaystyle P_1$ and $\displaystyle P_2$ and then don't use their own mathematics in the example calculation. So sloppy. I can derive $\displaystyle P_L = B_{11} P_1^2 + 2B_{12} P_1 P_2 + B_{22} P_2^2$ if I instead define $\displaystyle B_{mn}$ like my last post, but then the meaning of I and R isn't clear... This material seems to crop up fairly often as references in other papers, including this lab script at a university (making students follow the exact same sloppy mathematics...): https://www.google.co.uk/url?sa=t&rc...8FMYjoh20f3uSQ and this paper: https://www.researchgate.net/publica..._Power_Systems However, that latter paper is the only one to actually address the shortcomings of the sloppiness in the original paper by making it clearer how the formula is used in their example calculation. In this one they define the B-loss coefficients in the same manner and then consistently use those to get a result. I find the whole thing confusing so I'm going to order a textbook. Hopefully that will shed some light on the problem. Oddly enough, if I ignore reactive power issues, I get power losses just being the sum of all individual $\displaystyle I^2R$ values for each transmission cable, so it's probably a fuss over nothing... October 6th, 2016, 06:54 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,143 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions Okay... Think I got it... Consider this diagram where electricity is being generated at 1 and 2 (with power output $\displaystyle P_1$ and $\displaystyle P_2$). 3 is an intermediate bus with no power injection and bus 4 can be considered as a sink. Code: 1 ------>-------3-------<------2 | | | \ / | | 4 Arrows show current / power flow. There is a resistor between 1 and 3 ($\displaystyle R_1$), a resistor between 2 and 3 ($\displaystyle R_2$) and a resistor between 3 and 4 ($\displaystyle R_D$) Power loss is generally given by $\displaystyle P_L = I^2R = \frac{RP^2}{|V|^2 \cos^2 \phi}$ So total power loss is $\displaystyle P_L = I_1^2R_1 + I_2^2R_2 + I_D^2R_D$ $\displaystyle = I_1^2R_1 + I_2^2R_2 + (I_1 + I_2)^2R_D$ $\displaystyle = I_1^2R_1 + I_2^2R_2 + (I_1^2 + 2I_1 I_2 + I_2^2)R_D$ $\displaystyle = I_1^2(R_1 + R_D) + 2 I_1 I_2 R_D + I_2^2(R_2 + R_D)$ So if the loss coefficient is defined as $\displaystyle B_{mn} = \frac{1}{|V_m| |V_n| \cos \phi_m \cos \phi_n} \sum_k R_k$ where m and n is over all power sources and k is the sum of all resistances that involve current from m and/or n. $\displaystyle P_L = B_{11} P_1^2 + 2B_{12}P_1 P_2 + B_{22} P_2^2$ occurs. So the middle term arises naturally because some parts of an electricity network involve current from multiple sources. 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