Chapter 4.2 Resistance PDF

Module 4 Chapter 4.2: Resistance All Rights Reserved 1 Resistance DC resistance of a conductor at temp. T is: Rdc,T = ƿT l / A Where: Ω (4.2.1) ƿT conductor resistivity at temperature T l = conductor length A = conductor cross-sectional area Area: Where: cmil = circular mils 1 cmil = π/4 sq....

Module 4 Chapter 4.2: Resistance All Rights Reserved 1 Resistance DC resistance of a conductor at temp. T is: Rdc,T = ƿT l / A Where: Ω (4.2.1) ƿT conductor resistivity at temperature T l = conductor length A = conductor cross-sectional area Area: Where: cmil = circular mils 1 cmil = π/4 sq.mil 1 cmil = m2 2 Resistance TABLE 4.2 Comparison of SI and English units for calculating conductor resistance Quantity Resistivity Length Cross-sectional area dc resistance Symbol SI Units English Units r l A Wm m m2 W-cmil/ft ft cmil W W Rdc ¼ rl A TABLE 4.3 % Conductivity, resistivity, and temperature constant of conductor metals T r20 o C Resistivity at 20 oC Material Copper: Annealed Hard-drawn Aluminum Hard-drawn Brass Iron Silver Sodium Steel Temperature Constant % Conductivity Wm x 10-8 W-cmil/ft 100% 97.3% 1.72 1.77 10.37 10.66 234.5 241.5 61% 20–27% 17.2% 108% 40% 2–14% 2.83 6.4–8.4 10 1.59 4.3 12–88 17.00 38–51 60 9.6 26 72–530 228.1 480 180 243 207 180–980 o C 3 Resistance Conductor Resistance depends on: 1. 2. 3. 4. Spiraling Temperature Frequency (skin effect) Current magnitude – magnetic conductors 4 Resistance 1. Spiraling: Alternate layers of strands are spiraled in opposite directions to hold the strands together. Spiraling makes the strands 1 or 2% longer than the actual conductor length. As a result, the dc resistance of a stranded conductor is 1 or 2% larger than that calculated from (4.2.1) for a specified conductor length. 5 Resistance 2. Temperature: Resistivity of conductor materials varies linearly over normal operating temperatures: Where: ƿT2 and ƿT1 - resistivities at temperatures T2 and T1 oC, T - is a temperature constant that depends on the conductor material, and is listed in Table 4.3. AC Resistance or Effective Resistance of a conductor Where: ƿLoss = Conductor Real Power Loss; I = rms conductor current For DC, Current distribution is uniform throughout the conductor cross section 6 Resistance 3. Skin Effect: For AC, the current distribution is non-uniform. As frequency increases, the current in a solid cylindrical conductor tends to crowd toward the conductor surface, with smaller current density at the conductor center. This phenomenon is called skin effect. 4. Current magnitude—magnetic conductors: For magnetic conductors, such as steel conductors used for shield wires, resistance depends on current magnitude. The internal flux linkages, and therefore the iron or magnetic losses, depend on the current magnitude. For ACSR conductors, the steel core has a relatively high resistivity compared to the aluminum strands, and therefore the effect of current magnitude on ACSR conductor resistance is small. Tables on magnetic conductors list resistance at two current levels. (see appendix Table A.4, Page 924). 7 Example 1a: A 4/O copper conductor has 12 strands. Strand diameter is: 0.1328 in (0.3373 cm) Find: a) Total copper cross-sectional area (A) in mm2 Example: Solution a) Total copper cross-sectional area (A) in mm2 d = 0.3373 cm = 3.373 mm A = π r2 in general For 12 strands: A = 12 π r2 = 12 π (d/2)2 = 12 π (3.373 mm / 2)2 = 107.23 mm2 Example 1b: A 4/O copper conductor has 12 strands. Strand diameter is: 0.1328 in (0.3373 cm) Find: b) Rdc at 50o C with length of 1 km. Use hard-drawn copper, assume a 2% increase in resistance due to Spiraling (see appendix table A.3, Page 923; and also table 4.3, page 179) TABLE 4.3 % Conductivity, resistivity, and temperature constant of conductor metals T r20 o C Resistivity at 20 oC Material Copper: Annealed Hard-drawn Aluminum Hard-drawn Brass Iron Silver Sodium Steel Temperature Constant % Conductivity Wm x 10-8 W-cmil/ft oC 100% 97.3% 1.72 1.77 10.37 10.66 234.5 241.5 61% 20–27% 17.2% 108% 40% 2–14% 2.83 6.4–8.4 10 1.59 4.3 12–88 17.00 38–51 60 9.6 26 72–530 228.1 480 180 243 207 180–980 Example: Solution b) Rdc at 50o C with length of 1 km. Use equation (4.2.1) to solve: Rdc,T = ƿT l / A Ω From (4.2.3) we have: (4.2.1) Where T2 = 50oC and T1 = 20oC (from Table 4.3) P50oC = 1.77 x 10-8 ( 50 + 241.5 / 20 + 241.5) = 1.973x10-8 Ωm Rdc,T = ƿT l / A Ω Rdc, 50oC = [(1.973x10-8 (4.2.1) Ωm ) x (103 x 1.02)m] / 107.23 mm2 = 0.1877 Ω / km