STUDENT HANDOUT PACKAGE 1

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. 𝑅𝑅 = [ 𝜌𝜌 × ℓ ] 𝐴𝐴 Where: ℓ = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐴𝐴 = 𝐶𝐶. 𝑆𝑆. 𝐴𝐴. 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ρ (𝑟𝑟ℎ𝑜𝑜) = 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 Material 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 Table 1 Rho 𝜴𝜴 − 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄/𝒇𝒇𝒇𝒇 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 ℓ of the cable then R will increase. If you increase the cable cross sectional area CSA, the R will drop. 𝓵𝓵 ↑ 𝐑𝐑 ↑ 𝐀𝐀 ↑ 𝐑𝐑 ↓ 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. 𝐑𝐑 𝐀𝐀𝐀𝐀 = 𝟏𝟏. 𝟎𝟎𝟎𝟎 × 𝐑𝐑 𝐃𝐃𝐃𝐃 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 𝑅𝑅 = 𝑹𝑹𝟏𝟏 = 𝑹𝑹𝟐𝟐 = [ 𝝆𝝆𝟏𝟏 × 𝓵𝓵𝟏𝟏 ] 𝑨𝑨𝟏𝟏 [𝜌𝜌×ℓ] 𝐴𝐴 and that we have changed both ℓ and A. [ 𝝆𝝆𝟐𝟐 × 𝓵𝓵𝟐𝟐 ] 𝑨𝑨𝟐𝟐 𝓵𝓵𝟐𝟐 = 𝟏𝟏. 𝟓𝟓𝓵𝓵𝟏𝟏 𝑨𝑨𝟐𝟐 = {𝟕𝟕𝟕𝟕𝟕𝟕 / 𝟓𝟓𝟓𝟓𝟓𝟓} 𝑨𝑨𝟏𝟏 𝝆𝝆𝟏𝟏 = 𝝆𝝆𝟐𝟐 𝑹𝑹𝟐𝟐 = 𝑹𝑹𝟐𝟐 = [𝝆𝝆𝟐𝟐 × 𝓵𝓵𝟐𝟐 ] 𝑨𝑨𝟐𝟐 {𝝆𝝆𝟏𝟏 × 𝟏𝟏. 𝟓𝟓𝓵𝓵𝟏𝟏 } 𝟏𝟏. 𝟓𝟓 {𝝆𝝆𝟏𝟏 × 𝓵𝓵𝟏𝟏 } 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗 {𝝆𝝆𝟏𝟏 × 𝓵𝓵𝟏𝟏 } = = 𝟕𝟕𝟕𝟕𝟕𝟕 𝟏𝟏. 𝟓𝟓𝟓𝟓 𝑨𝑨𝟏𝟏 𝑨𝑨𝟏𝟏 � � 𝑨𝑨 𝟓𝟓𝟓𝟓𝟓𝟓 𝟏𝟏 𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 𝑹𝑹𝟏𝟏 = [ 𝝆𝝆𝟏𝟏 × 𝓵𝓵𝟏𝟏] = 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗 𝑹𝑹𝟏𝟏 𝑨𝑨𝟏𝟏 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. 𝑅𝑅𝑇𝑇1 1 + 𝑎𝑎0 𝑇𝑇1 = 𝑅𝑅𝑇𝑇2 1 + 𝑎𝑎0 𝑇𝑇2 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 Ω/°𝐶𝐶 @ 0°𝐶𝐶 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 Ω / km @ 20°C. Therefore T1 = 20°C and RT1 = 0.07191 Ω. We have made T2 = 60° C and from the above table we have the αo value for Aluminum to be αo = 0.00424 𝑅𝑅𝑇𝑇1 1 + 𝑎𝑎0 𝑇𝑇1 0.7191 1 + 0.00424 × 20 = = = 𝑅𝑅𝑇𝑇2 1 + 𝑎𝑎0 𝑇𝑇2 𝑅𝑅𝑇𝑇2 1 + 0.00424 × 60 𝑅𝑅𝑇𝑇2 = 0.0902 = 0.08315Ω 1.085 As expected, the resistance rose from 0.07191Ω/km at 20° C to 0.08315Ω/km at 60° C. [0.08315 − 0.07191] % 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = � � × 100 = 15.6% 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 0.07191 KEY TRANSMISSION LINE PARAMETER POINTS Industry Short Cuts Line Type XL (Ω/km) XC (Ωkm) Aerial (OHL) non-bundled 0.40 400,000 Underground Cable 0.08 4,000 𝑅𝑅𝐴𝐴𝐴𝐴 = 1.04 × 𝑅𝑅𝐷𝐷𝐷𝐷 𝑅𝑅𝑇𝑇1 1 + 𝑎𝑎0 𝑇𝑇1 = 𝑅𝑅𝑇𝑇2 1 + 𝑎𝑎0 𝑇𝑇2 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: 𝐸𝐸𝑆𝑆 = 𝑉𝑉𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 + 𝑉𝑉𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑊𝑊ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 𝐼𝐼 × 𝑍𝑍𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑉𝑉𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 𝐼𝐼𝑅𝑅𝐴𝐴𝐴𝐴 + 𝐼𝐼𝑋𝑋𝐿𝐿 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 𝑬𝑬𝑺𝑺 must be greater than the Receiving End Voltage 𝑬𝑬𝑹𝑹 . The following approximate formula can be used to determine the sending end voltage in a short line situation. 𝑬𝑬𝑺𝑺 = 𝑽𝑽𝑹𝑹 + 𝑰𝑰𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 [𝑹𝑹𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 − 𝑿𝑿𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 ] Where: 𝑹𝑹𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑿𝑿𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝑐𝑐𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝜽𝜽𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑽𝑽𝑹𝑹 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝑬𝑬𝑺𝑺 = 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 Notes: the phase angle between 𝑬𝑬𝑺𝑺 and 𝑽𝑽𝑹𝑹 is Example A three-phase transmission line has a sending end voltage (𝑬𝑬𝑺𝑺 ) 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 (𝑹𝑹𝑳𝑳 ) = 8.93Ω and the line inductive reactance (𝑿𝑿𝑳𝑳 ) = 29.97Ω. 𝜃𝜃𝐿𝐿 = cos −1 (0.9) = −25.84° Convert 𝑬𝑬𝑺𝑺 to a phase value = 138 √3 = 79.674 kV 𝑬𝑬𝑺𝑺 = 𝑽𝑽𝑹𝑹 + I𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [R 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 cosθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − X𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 sinθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝑽𝑽𝑹𝑹 = 𝑬𝑬𝑺𝑺 − I𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [R 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 cosθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − X𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 sinθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝑽𝑽𝑹𝑹 = 79,674 − 628 [8.93 cos 25.84 − 29.97 sin − 25.84] = 79.674 – 628 [8.04 – (−13.06)] 𝑽𝑽𝑹𝑹 = 79,674 − 628 [8.04 + 13.06] = 79674 − 628 [21.06] = 79,674 − 13,226 𝑽𝑽𝑹𝑹 = 66.45kV Convert to a line value V𝑅𝑅 = 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 𝑬𝑬𝑺𝑺 = 𝑽𝑽𝑹𝑹 + 𝑰𝑰𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 [𝑹𝑹𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 − 𝑿𝑿𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 ] Where: 𝑹𝑹𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑿𝑿𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝜽𝜽𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑽𝑽𝑹𝑹 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝑬𝑬𝑺𝑺 = 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 Example A Transmission line has a sending end voltage 𝑬𝑬𝑺𝑺 of 138kV. Calculate the voltage at the receiving end for a load of 628A with a unity power factor. The line resistance (𝑹𝑹𝑳𝑳 ) = 8.93Ω and the line inductive reactance (𝑿𝑿𝑳𝑳 ) = 29.97Ω. 𝜃𝜃𝐿𝐿 = cos −1 (1) = 0° Convert 𝑬𝑬𝑺𝑺 to a phase value = 138 √3 = 79.674 kV 𝑬𝑬𝑺𝑺 = 𝑽𝑽𝑹𝑹 + I𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [R 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 cosθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − X𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 sinθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝑽𝑽𝑹𝑹 = 𝑬𝑬𝑺𝑺 − I𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [R 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 cosθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − X𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 sinθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝑽𝑽𝑹𝑹 = 79,674 − 628 [8.93 cos 0 − 29.97 sin 0] = 79,674 – 5,608 𝑽𝑽𝑹𝑹 = 74.1kV Convert to a line value V𝑅𝑅 = 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 𝑬𝑬𝑺𝑺 = 𝑽𝑽𝑹𝑹 + 𝑰𝑰𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 [𝑹𝑹𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 − 𝑿𝑿𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 ] Where: 𝑹𝑹𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑿𝑿𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝜽𝜽𝑳𝑳 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑽𝑽𝑹𝑹 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝑬𝑬𝑺𝑺 = 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐸𝐸𝐸𝐸𝐸𝐸 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 Example A Transmission line has a sending end voltage (𝑬𝑬𝑺𝑺 ) 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 (𝑹𝑹𝑳𝑳 ) = 8.93Ω and the line inductive reactance (𝑿𝑿𝑳𝑳 ) = 29.97Ω. 𝜃𝜃𝐿𝐿 = cos −1 (0.9) = 25.84° Convert 𝑬𝑬𝑺𝑺 to a phase value = 138 √3 = 79.674 kV 𝑬𝑬𝑺𝑺 = 𝑽𝑽𝑹𝑹 + I𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [R 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 cosθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − X𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 sinθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝑽𝑽𝑹𝑹 = 𝑬𝑬𝑺𝑺 − I𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [R 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 cosθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − X𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 sinθ𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝑽𝑽𝑹𝑹 = 79,674 − 62.8 [8.93 cos 25.84 − 29.97 sin 25.84] 𝑽𝑽𝑹𝑹 = 79,674 − 62.8 [8.04 − 13.06] = 79674 − 62.8 [−5.06] = 79,674 + 318 𝑽𝑽𝑹𝑹 = 79,992V Convert to a line value V𝑅𝑅 = 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. 𝑉𝑉𝑁𝑁𝑁𝑁 − 𝑉𝑉𝐹𝐹𝐹𝐹 𝐸𝐸𝑆𝑆 − 𝑉𝑉𝑅𝑅 � × 100 = � � × 100 𝑉𝑉𝐹𝐹𝐹𝐹 𝑉𝑉𝑅𝑅 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � 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 𝐸𝐸𝑆𝑆 = 138𝑘𝑘𝑘𝑘 𝑉𝑉𝑅𝑅 = 115.1𝑘𝑘𝑘𝑘 𝐸𝐸𝑆𝑆 = 138𝑘𝑘𝑘𝑘 𝑉𝑉𝑅𝑅 = 138.6𝑘𝑘𝑘𝑘 𝐸𝐸𝑆𝑆 = 138𝑘𝑘𝑘𝑘 𝑉𝑉𝑅𝑅 = 128.3𝑘𝑘𝑘𝑘 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 𝐸𝐸𝑆𝑆 − 𝑉𝑉𝑅𝑅 � × 100 𝑉𝑉𝑅𝑅 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � (138 − 115) � × 100 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � 115 23 � × 100 115 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 20% Unity pf % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 ES = 138kV VR = 128.3kV 𝐸𝐸𝑆𝑆 − 𝑉𝑉𝑅𝑅 =� � × 100 𝑉𝑉𝑅𝑅 (138 − 128.3) � × 100 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � 128.3 9.7 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � � × 100 128.3 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 7.56% Leading pf ES = 138kV VR = 138.6kV 𝐸𝐸𝑆𝑆 − 𝑉𝑉𝑅𝑅 � × 100 𝑉𝑉𝑅𝑅 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � (138 − 138.6) � × 100 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � 138.6 −0.6 � × 100 138.6 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = −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. 𝑷𝑷𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 = (𝑰𝑰𝑳𝑳 𝟐𝟐 𝑹𝑹) × 𝟑𝟑 Efficiency is defined as: 𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶 𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶 = 𝑷𝑷𝑰𝑰𝑰𝑰 𝑷𝑷𝑶𝑶𝑶𝑶𝑶𝑶 + 𝑷𝑷𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 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, 𝑅𝑅𝐷𝐷𝐷𝐷 @ 20℃ = 0.1438Ω/km We apply our DC to AC conversion factor of 1.04 and we get: 𝑅𝑅𝐴𝐴𝐴𝐴 @ 20°𝐶𝐶 = 𝑅𝑅𝐷𝐷𝐷𝐷 × 1.04 = 0.1438Ω/𝑘𝑘𝑘𝑘 × 1.04 = 0.14955 Ω / 𝑘𝑘𝑘𝑘 𝑂𝑂𝑂𝑂𝑂𝑂 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑅𝑅 = 0.14955 Ω/𝑘𝑘𝑘𝑘 × 61𝑘𝑘𝑘𝑘 = 9.122 Ω This is a short line, so we next have to calculate the XL. Using our approximation figure for XL, we have the following: 𝑂𝑂𝑂𝑂𝑂𝑂 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑋𝑋𝐿𝐿 = 0.4 Ω/𝑘𝑘𝑘𝑘 × 61𝑘𝑘𝑘𝑘 = 24.4 Ω This gives us the following equivalent circuit ES XL 24.4 Ω VLINE Stony Brook ER 138kV LOAD IL RAC 9.122 Ω Glenwood 9/10 Because the load is lagging, we use the following equation 𝐸𝐸𝑆𝑆 = 𝑉𝑉𝑅𝑅 + 𝐼𝐼𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [𝑅𝑅𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑐𝑐𝑐𝑐𝑐𝑐𝜃𝜃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − 𝑋𝑋𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] We can use the pf to find θ. 𝜃𝜃 = 𝑐𝑐𝑐𝑐𝑐𝑐 −1 𝑝𝑝𝑝𝑝 = 𝑐𝑐𝑐𝑐𝑐𝑐 −1 0.987 = −9.24° We also have to change ER to a phase value. 𝑃𝑃ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑉𝑉𝑅𝑅 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑉𝑉𝑅𝑅 √3 = 138𝑘𝑘𝑘𝑘 √3 = 79.67𝑘𝑘𝑘𝑘 Lastly, we must calculate the load current IL. 𝐼𝐼𝐿𝐿 = 𝑆𝑆 √3 𝑉𝑉𝑅𝑅 = 43,000𝑘𝑘𝑘𝑘𝑘𝑘 √3𝑥𝑥 138𝑘𝑘𝑘𝑘 = 180𝐴𝐴 We now have all the information we require to solve the equation 𝐸𝐸𝑆𝑆 = 𝑉𝑉𝑅𝑅 + 𝐼𝐼𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 [𝑅𝑅𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑐𝑐𝑐𝑐𝑐𝑐𝜃𝜃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 − 𝑋𝑋𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ] 𝐸𝐸𝑆𝑆 = 79.67𝑘𝑘 + 180 [9.122 𝑐𝑐𝑐𝑐𝑐𝑐9.24 − 24.4 𝑠𝑠𝑠𝑠𝑠𝑠 − 9.24] 𝐸𝐸𝑆𝑆 = 79.67𝑘𝑘 + 180 [9.0 + 3.9] = 79.67𝑘𝑘 + 2.32𝑘𝑘 𝐸𝐸𝑆𝑆 = 82 𝑘𝑘𝑘𝑘 (𝑃𝑃ℎ𝑎𝑎𝑎𝑎𝑎𝑎) 𝐸𝐸𝑆𝑆 = 82 𝑘𝑘𝑘𝑘 𝑥𝑥 √3 = 142 𝑘𝑘𝑘𝑘 (𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿) We can now calculate the VREG of the circuit. (𝐸𝐸𝑆𝑆 − 𝐸𝐸𝑅𝑅 ) (𝑉𝑉𝑁𝑁𝑁𝑁 − 𝑉𝑉𝐹𝐹𝐹𝐹 ) � × 100 = � � × 100 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � 𝑉𝑉𝐹𝐹𝐹𝐹 𝐸𝐸𝑅𝑅 142 − 138 � × 100 138 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � 4 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = � � × 100 138 % 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 2.9% Now calculate the Line efficiency in terms of Power 𝑃𝑃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 3 × 𝐼𝐼𝐿𝐿 2 × 𝑅𝑅 = 3 × 1802 × 9.122 = 0.886 𝑀𝑀𝑀𝑀 𝑃𝑃𝑂𝑂𝑂𝑂𝑂𝑂 = 43 𝑀𝑀𝑀𝑀𝑀𝑀 × 𝑝𝑝𝑝𝑝 = 43 × 0.987 = 42.44𝑀𝑀𝑀𝑀 𝑃𝑃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 42.44 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = � � × 100 = � � × 100 𝑃𝑃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 + 𝑃𝑃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 42.44 + 0.886 42.44 � × 100 43.33 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = � 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 98% 10/10

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