STUDENT HANDOUT PACKAGE 1 TRANSMISSION CIRCUIT CALCULATIONS (Short Lines) Resistance of Transmission Lines Table 1 provides the dc resistance values for the common transmission line cables. The physical formula for resistance shows how conductor resistance can be changed. π π = [ ππ × β ] π΄π΄ Where: β = ππππππππππβ ππππ ππππππππππππππππππ ππππ ππππππππ π΄π΄ = πΆπΆ. ππ. π΄π΄. ππππ ππππππππππππππππππ ππππ ππππππππππππππππ ππππππππ ρ (ππβππ) = ππππππππππππππππ ππππππππππππππππππππ ππππ π‘π‘βππ ππππππππππππππππππ ππππππππππππππππ Material ππππππππππππ πΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆ πΊπΊπΊπΊπΊπΊπΊπΊ π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄ ππππππππππππ Table 1 Rho π΄π΄ − ππππππππ/ππππ 9.546 10.09 13.32 15.94 41.69 In practical applications, we tend to use only Aluminium in HV Transmission lines as it is better in terms of weight, cost, and flexibility than copper, gold, and silver. This means our rho ρ will not be a variable. The equation then shows that if you increase the length β of the cable then R will increase. If you increase the cable cross sectional area CSA, the R will drop. π΅π΅ ↑ ππ ↑ ππ ↑ ππ ↓ The transmission line usually operates using ac. This causes an effect on the conductor that causes less current to flow in the conductor center and more to flow towards the outside. This is called the “skin effect”. Industry Tables exist to relate the RAC to the RDC. Unless you are told to use an industry table, for this course, we will use the following formula to relate RAC to RDC. ππ ππππ = ππ. ππππ × ππ ππππ 1/10 Example 3.1 A transmission line has been increased in length by 50% and the 500 kcmil (500 MCM) conductors uprated to 795 kcmil. Calculate the % change in resistance for the line. Remember that π π = πΉπΉππ = πΉπΉππ = [ ππππ × π΅π΅ππ ] π¨π¨ππ [ππ×β] π΄π΄ and that we have changed both β and A. [ ππππ × π΅π΅ππ ] π¨π¨ππ π΅π΅ππ = ππ. πππ΅π΅ππ π¨π¨ππ = {ππππππ / ππππππ} π¨π¨ππ ππππ = ππππ πΉπΉππ = πΉπΉππ = [ππππ × π΅π΅ππ ] π¨π¨ππ {ππππ × ππ. πππ΅π΅ππ } ππ. ππ {ππππ × π΅π΅ππ } ππ. ππππππ {ππππ × π΅π΅ππ } = = ππππππ ππ. ππππ π¨π¨ππ π¨π¨ππ οΏ½ οΏ½ π¨π¨ ππππππ ππ ππππππππππ πΉπΉππ = [ ππππ × π΅π΅ππ] = ππ. ππππππ πΉπΉππ π¨π¨ππ The above indicates the line resistance will drop 100 - 94.3 = 5.7% Temperature also affects conductor resistance, the higher the operating temperature the higher the conductor resistance. The change in the resistance of a material due to temperature is called the temperature coefficient α. The following formula can be used to determine the resistance of a material at any temperature if its resistance is known at any other temperature. π π ππ1 1 + ππ0 ππ1 = π π ππ2 1 + ππ0 ππ2 RT1 = Resistance at temperature T1 RT2 = Resistance at temperature T2 α0 = Temperature coefficient of the conductor material at 0° C 2/10 Temperature Coefficients of Common Materials Material Aluminium Copper Gold Lead Nichrome Silver Tungsten Carbon Nickel Example 3.2 Temp Coeff αo β¦/°πΆπΆ @ 0°πΆπΆ 0.00424 0.00390 0.00365 0.00466 0.00044 0.00041 0.00495 −0.000495 0.00680 Find the percent change of resistance of Drake conductor operated at 60°C from that given in your course data handout. Table 1-2 gives us the dc resistance as 0.07191 β¦ / km @ 20°C. Therefore T1 = 20°C and RT1 = 0.07191 β¦. We have made T2 = 60° C and from the above table we have the αo value for Aluminum to be αo = 0.00424 π π ππ1 1 + ππ0 ππ1 0.7191 1 + 0.00424 × 20 = = = π π ππ2 1 + ππ0 ππ2 π π ππ2 1 + 0.00424 × 60 π π ππ2 = 0.0902 = 0.08315β¦ 1.085 As expected, the resistance rose from 0.07191β¦/km at 20° C to 0.08315β¦/km at 60° C. [0.08315 − 0.07191] % ππβππππππππ = οΏ½ οΏ½ × 100 = 15.6% ππππππππ 0.07191 KEY TRANSMISSION LINE PARAMETER POINTS Industry Short Cuts Line Type XL (β¦/km) XC (β¦km) Aerial (OHL) non-bundled 0.40 400,000 Underground Cable 0.08 4,000 π π π΄π΄π΄π΄ = 1.04 × π π π·π·π·π· π π ππ1 1 + ππ0 ππ1 = π π ππ2 1 + ππ0 ππ2 3/10 TRANSMISSION CIRCUIT CALCULATIONS Once the transmission line parameters have been determined, the line may be simulated by its electrically equivalent circuit. Phase values are used and assuming a balanced load, a single line representation is used for calculations. Basic Transmission Line Equivalent Circuit LOAD 2 XC 2 XC G XL RAC I Sending End Receiving End The Short Transmission Line A short line is generally regarded as a line up to 80km in length or all lines of voltage less than 40kV. Capacitance can be neglected, and the short line is simulated by resistance and inductive reactance in series. The short line model is used for short transmission feeders and distributors. Short Line Equivalent Circuit: RAC ES XL VLINE VLOAD LOAD I Receiving End Sending End Applying Kirchhoff’s voltage to the circuit: πΈπΈππ = πππΏπΏπΏπΏπΏπΏπΏπΏ + πππΏπΏπΏπΏπΏπΏπΏπΏ ππβππππππ πππΏπΏπΏπΏπΏπΏπΏπΏ = πΌπΌ × πππΏπΏπΏπΏπΏπΏπΏπΏ ππππππππππππ πππΏπΏπΏπΏπΏπΏπΏπΏ = πΌπΌπ π π΄π΄π΄π΄ + πΌπΌπππΏπΏ 4/10 1) Lagging Power Factor Load: The following Vector Diagram shows a load with a lagging power factor, the most common type of load as the majority of load on a power system is inductive (lagging power factor load). Note that for a lagging power factor load, VR is much smaller than ES. ES VR θL IL ZL I L XL I L RL IL As you can see, the Sending End Voltage π¬π¬πΊπΊ must be greater than the Receiving End Voltage π¬π¬πΉπΉ . The following approximate formula can be used to determine the sending end voltage in a short line situation. π¬π¬πΊπΊ = π½π½πΉπΉ + π°π°π³π³π³π³π³π³π³π³ [πΉπΉπ³π³π³π³π³π³π³π³ πππππππππ³π³π³π³π³π³π³π³ − πΏπΏπ³π³π³π³π³π³π³π³ πππππππππ³π³π³π³π³π³π³π³ ] Where: πΉπΉπ³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ π π π π π π π π π π π π π π π π π π π π πΏπΏπ³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ πΌπΌπΌπΌπΌπΌπΌπΌπππ‘π‘π‘π‘π‘π‘π‘π‘ π π π π π π π π π π π π π π π π π π π½π½π³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ ππππππππππ πΉπΉπΉπΉπΉπΉπΉπΉπΉπΉπΉπΉ π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄ π½π½πΉπΉ = π π π π π π π π π π π π π π π π π π πΈπΈπΈπΈπΈπΈ ππππππππππππππ π¬π¬πΊπΊ = ππππππππππππππ πΈπΈπΈπΈπΈπΈ ππππππππππππππ Notes: the phase angle between π¬π¬πΊπΊ and π½π½πΉπΉ is Example A three-phase transmission line has a sending end voltage (π¬π¬πΊπΊ ) of 138kV. Calculate the voltage at the receiving end for a line current of 628A with a load power factor of 0.9 lagging. The line resistance (πΉπΉπ³π³ ) = 8.93β¦ and the line inductive reactance (πΏπΏπ³π³ ) = 29.97β¦. πππΏπΏ = cos −1 (0.9) = −25.84° Convert π¬π¬πΊπΊ to a phase value = 138 √3 = 79.674 kV π¬π¬πΊπΊ = π½π½πΉπΉ + IπΏπΏπΏπΏπΏπΏπΏπΏ [R πΏπΏπΏπΏπΏπΏπΏπΏ cosθπΏπΏπΏπΏπΏπΏπΏπΏ − XπΏπΏπΏπΏπΏπΏπΏπΏ sinθπΏπΏπΏπΏπΏπΏπΏπΏ ] π½π½πΉπΉ = π¬π¬πΊπΊ − IπΏπΏπΏπΏπΏπΏπΏπΏ [R πΏπΏπΏπΏπΏπΏπΏπΏ cosθπΏπΏπΏπΏπΏπΏπΏπΏ − XπΏπΏπΏπΏπΏπΏπΏπΏ sinθπΏπΏπΏπΏπΏπΏπΏπΏ ] π½π½πΉπΉ = 79,674 − 628 [8.93 cos 25.84 − 29.97 sin − 25.84] = 79.674 – 628 [8.04 – (−13.06)] π½π½πΉπΉ = 79,674 − 628 [8.04 + 13.06] = 79674 − 628 [21.06] = 79,674 − 13,226 π½π½πΉπΉ = 66.45kV Convert to a line value Vπ π = 66.45kV x √3 = 115.1kV Note that with a lagging power factor (inductive load), the receiving end voltage is much lower than the sending end voltage. 5/10 2) Unity Power factor Load Voltage regulation will depend on the magnitude and phase angle of the load. As the load becomes less lagging (inductive), the regulation gets better This can clearly be seen in the following phasor diagram for a load with a unity power factor. Remember, at unity power factor, θL = 0°. Note that for a unity power factor load, ER is smaller than ES. ES IL I L XL VR I L RL π¬π¬πΊπΊ = π½π½πΉπΉ + π°π°π³π³π³π³π³π³π³π³ [πΉπΉπ³π³π³π³π³π³π³π³ πππππππππ³π³π³π³π³π³π³π³ − πΏπΏπ³π³π³π³π³π³π³π³ πππππππππ³π³π³π³π³π³π³π³ ] Where: πΉπΉπ³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ π π π π π π π π π π π π π π π π π π π π πΏπΏπ³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ πΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌ π π π π π π π π π π π π π π π π π π π½π½π³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ ππππππππππ πΉπΉπΉπΉπΉπΉπΉπΉπΉπΉπΉπΉ π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄ π½π½πΉπΉ = π π π π π π π π π π π π π π π π π π πΈπΈπΈπΈπΈπΈ ππππππππππππππ π¬π¬πΊπΊ = ππππππππππππππ πΈπΈπΈπΈπΈπΈ ππππππππππππππ Example A Transmission line has a sending end voltage π¬π¬πΊπΊ of 138kV. Calculate the voltage at the receiving end for a load of 628A with a unity power factor. The line resistance (πΉπΉπ³π³ ) = 8.93β¦ and the line inductive reactance (πΏπΏπ³π³ ) = 29.97β¦. πππΏπΏ = cos −1 (1) = 0° Convert π¬π¬πΊπΊ to a phase value = 138 √3 = 79.674 kV π¬π¬πΊπΊ = π½π½πΉπΉ + IπΏπΏπΏπΏπΏπΏπΏπΏ [R πΏπΏπΏπΏπΏπΏπΏπΏ cosθπΏπΏπΏπΏπΏπΏπΏπΏ − XπΏπΏπΏπΏπΏπΏπΏπΏ sinθπΏπΏπΏπΏπΏπΏπΏπΏ ] π½π½πΉπΉ = π¬π¬πΊπΊ − IπΏπΏπΏπΏπΏπΏπΏπΏ [R πΏπΏπΏπΏπΏπΏπΏπΏ cosθπΏπΏπΏπΏπΏπΏπΏπΏ − XπΏπΏπΏπΏπΏπΏπΏπΏ sinθπΏπΏπΏπΏπΏπΏπΏπΏ ] π½π½πΉπΉ = 79,674 − 628 [8.93 cos 0 − 29.97 sin 0] = 79,674 – 5,608 π½π½πΉπΉ = 74.1kV Convert to a line value Vπ π = 74.1kV x √3 = 128.3kV Note that with a unity power factor (resistive load), the receiving end voltage is lower than the sending end voltage. 6/10 3) Leading Power factor Load The leading power factor current results in smaller sending end voltage than receiving end voltage making the voltage regulation negative. This may occur under light load conditions if capacitors are left in the system. The load current will now lead the load voltage and the following phasor diagram results. ES I L XL IL IL ZL θL I L RL VR π¬π¬πΊπΊ = π½π½πΉπΉ + π°π°π³π³π³π³π³π³π³π³ [πΉπΉπ³π³π³π³π³π³π³π³ πππππππππ³π³π³π³π³π³π³π³ − πΏπΏπ³π³π³π³π³π³π³π³ πππππππππ³π³π³π³π³π³π³π³ ] Where: πΉπΉπ³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ π π π π π π π π π π π π π π π π π π π π πΏπΏπ³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ πΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌπΌ π π π π π π π π π π π π π π π π π π π½π½π³π³ = πΏπΏπΏπΏπΏπΏπΏπΏ ππππππππππ πΉπΉπΉπΉπΉπΉπΉπΉπΉπΉπΉπΉ π΄π΄π΄π΄π΄π΄π΄π΄π΄π΄ π½π½πΉπΉ = π π π π π π π π π π π π π π π π π π πΈπΈπΈπΈπΈπΈ ππππππππππππππ π¬π¬πΊπΊ = ππππππππππππππ πΈπΈπΈπΈπΈπΈ ππππππππππππππ Example A Transmission line has a sending end voltage (π¬π¬πΊπΊ ) of 138kV. Calculate the voltage at the receiving end for a load of 62.8A with a power factor of 0.9 leading. The line resistance (πΉπΉπ³π³ ) = 8.93β¦ and the line inductive reactance (πΏπΏπ³π³ ) = 29.97β¦. πππΏπΏ = cos −1 (0.9) = 25.84° Convert π¬π¬πΊπΊ to a phase value = 138 √3 = 79.674 kV π¬π¬πΊπΊ = π½π½πΉπΉ + IπΏπΏπΏπΏπΏπΏπΏπΏ [R πΏπΏπΏπΏπΏπΏπΏπΏ cosθπΏπΏπΏπΏπΏπΏπΏπΏ − XπΏπΏπΏπΏπΏπΏπΏπΏ sinθπΏπΏπΏπΏπΏπΏπΏπΏ ] π½π½πΉπΉ = π¬π¬πΊπΊ − IπΏπΏπΏπΏπΏπΏπΏπΏ [R πΏπΏπΏπΏπΏπΏπΏπΏ cosθπΏπΏπΏπΏπΏπΏπΏπΏ − XπΏπΏπΏπΏπΏπΏπΏπΏ sinθπΏπΏπΏπΏπΏπΏπΏπΏ ] π½π½πΉπΉ = 79,674 − 62.8 [8.93 cos 25.84 − 29.97 sin 25.84] π½π½πΉπΉ = 79,674 − 62.8 [8.04 − 13.06] = 79674 − 62.8 [−5.06] = 79,674 + 318 π½π½πΉπΉ = 79,992V Convert to a line value Vπ π = 79,992V x √3 = 138.55kV Note that with a leading power factor (capacitive load), the receiving end voltage is higher than the sending end voltage 7/10 Short Line Voltage Regulation [VREG] The vector diagram shows that for normal lagging load conditions, the sending end voltage must be larger than the receiving end voltage to account for voltage drops along the transmission line. This difference in voltage is called the “voltage regulation” of the transmission line. Voltage regulation is defined as the percentage change in voltage at the receiving end when the load is removed. For the short line, when the load is disconnected, the receiving end voltage will rise to a value equal to the sending end voltage. ππππππ − πππΉπΉπΉπΉ πΈπΈππ − πππ π οΏ½ × 100 = οΏ½ οΏ½ × 100 πππΉπΉπΉπΉ πππ π % πππ π π π π π = οΏ½ Let us calculate the VREG for the 3 short line load possibilities; the lagging pf, the unity pf and the leading pf. From the previous 3 examples we can get the following information: Lagging pf Unity pf Leading pf πΈπΈππ = 138ππππ πππ π = 115.1ππππ πΈπΈππ = 138ππππ πππ π = 138.6ππππ πΈπΈππ = 138ππππ πππ π = 128.3ππππ We can now calculate the VREG for the 3 types of loads that can be found on the short line Lagging pf ES = 138kV VR = 115.1kV πΈπΈππ − πππ π οΏ½ × 100 πππ π % πππ π π π π π = οΏ½ (138 − 115) οΏ½ × 100 % πππ π π π π π = οΏ½ 115 23 οΏ½ × 100 115 % πππ π π π π π = οΏ½ % πππ π π π π π = 20% Unity pf % πππ π π π π π ES = 138kV VR = 128.3kV πΈπΈππ − πππ π =οΏ½ οΏ½ × 100 πππ π (138 − 128.3) οΏ½ × 100 % πππ π π π π π = οΏ½ 128.3 9.7 % πππ π π π π π = οΏ½ οΏ½ × 100 128.3 % πππ π π π π π = 7.56% Leading pf ES = 138kV VR = 138.6kV πΈπΈππ − πππ π οΏ½ × 100 πππ π % πππ π π π π π = οΏ½ (138 − 138.6) οΏ½ × 100 % πππ π π π π π = οΏ½ 138.6 −0.6 οΏ½ × 100 138.6 % πππ π π π π π = οΏ½ % πππ π π π π π = −0.43% 8/10 Transmission Line Efficiency The current flowing through the resistance of the transmission lines will cause a power loss which is dissipated as heat. π·π·π³π³π³π³π³π³π³π³ = (π°π°π³π³ ππ πΉπΉ) × ππ Efficiency is defined as: π·π·πΆπΆπΆπΆπΆπΆ π·π·πΆπΆπΆπΆπΆπΆ = π·π·π°π°π°π° π·π·πΆπΆπΆπΆπΆπΆ + π·π·π³π³π³π³π³π³π³π³ Example A section of transmission line running from the Stony Brook to Glenwood substations is 38 miles long and consists of 397.5 26/7 MCM ACSR conductors. If the load on the Glenwood substation is 43MVA with a power factor of 0.987 lagging, calculate the voltage required at the Stony Brook station to maintain 138kV at Glenwood. Determine the line regulation and transmission efficiency for this section of line. From Table 1, π π π·π·π·π· @ 20β = 0.1438β¦/km We apply our DC to AC conversion factor of 1.04 and we get: π π π΄π΄π΄π΄ @ 20°πΆπΆ = π π π·π·π·π· × 1.04 = 0.1438β¦/ππππ × 1.04 = 0.14955 β¦ / ππππ ππππππ ππππππππ π π = 0.14955 β¦/ππππ × 61ππππ = 9.122 β¦ This is a short line, so we next have to calculate the XL. Using our approximation figure for XL, we have the following: ππππππ ππππππππ πππΏπΏ = 0.4 β¦/ππππ × 61ππππ = 24.4 β¦ This gives us the following equivalent circuit ES XL 24.4 β¦ VLINE Stony Brook ER 138kV LOAD IL RAC 9.122 β¦ Glenwood 9/10 Because the load is lagging, we use the following equation πΈπΈππ = πππ π + πΌπΌπΏπΏπΏπΏπΏπΏπΏπΏ [π π πΏπΏπΏπΏπΏπΏπΏπΏ πππππππππΏπΏπΏπΏπΏπΏπΏπΏ − πππΏπΏπΏπΏπΏπΏπΏπΏ π π π π π π πππΏπΏπΏπΏπΏπΏπΏπΏ ] We can use the pf to find θ. ππ = ππππππ −1 ππππ = ππππππ −1 0.987 = −9.24° We also have to change ER to a phase value. ππβππππππ πππ π = πΏπΏπΏπΏπΏπΏπΏπΏ πππ π √3 = 138ππππ √3 = 79.67ππππ Lastly, we must calculate the load current IL. πΌπΌπΏπΏ = ππ √3 πππ π = 43,000ππππππ √3π₯π₯ 138ππππ = 180π΄π΄ We now have all the information we require to solve the equation πΈπΈππ = πππ π + πΌπΌπΏπΏπΏπΏπΏπΏπΏπΏ [π π πΏπΏπΏπΏπΏπΏπΏπΏ πππππππππΏπΏπΏπΏπΏπΏπΏπΏ − πππΏπΏπΏπΏπΏπΏπΏπΏ π π π π π π πππΏπΏπΏπΏπΏπΏπΏπΏ ] πΈπΈππ = 79.67ππ + 180 [9.122 ππππππ9.24 − 24.4 π π π π π π − 9.24] πΈπΈππ = 79.67ππ + 180 [9.0 + 3.9] = 79.67ππ + 2.32ππ πΈπΈππ = 82 ππππ (ππβππππππ) πΈπΈππ = 82 ππππ π₯π₯ √3 = 142 ππππ (πΏπΏπΏπΏπΏπΏπΏπΏ) We can now calculate the VREG of the circuit. (πΈπΈππ − πΈπΈπ π ) (ππππππ − πππΉπΉπΉπΉ ) οΏ½ × 100 = οΏ½ οΏ½ × 100 % πππ π π π π π = οΏ½ πππΉπΉπΉπΉ πΈπΈπ π 142 − 138 οΏ½ × 100 138 % πππ π π π π π = οΏ½ 4 % πππ π π π π π = οΏ½ οΏ½ × 100 138 % πππ π π π π π = 2.9% Now calculate the Line efficiency in terms of Power πππΏπΏπΏπΏπΏπΏπΏπΏ = 3 × πΌπΌπΏπΏ 2 × π π = 3 × 1802 × 9.122 = 0.886 ππππ ππππππππ = 43 ππππππ × ππππ = 43 × 0.987 = 42.44ππππ πππΏπΏπΏπΏπΏπΏπΏπΏ 42.44 πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ = οΏ½ οΏ½ × 100 = οΏ½ οΏ½ × 100 πππΏπΏπΏπΏπΏπΏπΏπΏ + πππΏπΏπΏπΏπΏπΏπΏπΏ 42.44 + 0.886 42.44 οΏ½ × 100 43.33 πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ = οΏ½ πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ = 98% 10/10