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Unformatted text preview: 2.25 Skin effect b.
What is the skin depth for an iron wire carrying a current at 60 Hz? The resistivity of iron at 27 °C is 97 n!) m. Assume that its relative permeability is pm 700. How does
this compare with the copper wire? Discuss why copper is preferred over iron for power transmission even though the iron is nearly 100
times cheaper than copper. Solution
b. What is the skin de th for an iron wire 0 'n a current at 60 Hz? The conductivity is 1/ ,0. The relative permeability (yr) for Iron is 700, thus ,upe = 700/10. The angular
frequency is a) = 279’= 272(60 Hz). Using these values in the equation for skin depth (6): 1 5pc =ﬁ— (+1)
_a) _
2 a pluFe
1 (700 4: 10‘7H ‘1 +1 (+1)
—(27r60s’1) ﬁx m ( ) 97x10—9Qm
( ) 6= 0.000765 m or 0.765 mm (+1)
Thus the skin depth is 0.765 mm, about 11 times less than that for copper. 2 2.1 The resistivity of iron at 27 °C is 97 n9 m. Assume that its relative permeabilig is gr} 700. How does
this compare with the copper wire? To calculate the resistance we need the cross sectional area for conduction. The material cross sectional
area is 7zr2 where r is the radius of the wire. But the current ﬂow is within depth 6. We deduct the area of
the core, 7r(p — ($2, from the overall area, zzrz, to obtain the cross sectional area for conduction. Comparison of Cu and Fe based on solid core wires: The resistance per unit length of the solid core Fe wire (RFC) is: R zpFe Fe
AFe pr
R =——6 +1
Fe 72TFez_7z(rFe_6Fe)2( ) The resistance per unit length of solid core Cu wire is: R —pcu=¢ pcu Cu _ 2 2 2
AC" 727“Cu _ 7z(rCu _ 6Cu ) ZWCU 6Cu — 72'6Cu If we equate these two resistances, we can make a comparison between Fe and Cu: RFC = RCu pFe pCu 2 — 2
27sze6Fe _ 71.6% 2727«CuaCu _ 7I6Cu 27:,0CurFeé‘Fe = pFe (2727016Cu —7r5Cu2) (Neglect the 6&2 term which is small) 2 pFe szu6Cu _7[§Cu rFe= 272IIOCué‘Fe We can assume a value for rcu for calculation purposes, rCu = 10 mm. The resistivity pen is given as 17
n9 m and the skin depth of Cu is known to be do“ = 8.47 mm. The resistivity of Fe is given as ppe = 97 119 m and its skin depth was just calculated to be 5pc = 0.765 m. We can substitute these values into
the above equation to determine the radius of Fe wire that would be equivalent to 10 mm diameter Cu wire.
r _ 97x109 Q m 27r(0.010 m)(0.00847 m)— 7r(0.00847 m)2 (+1)
Fe 27r 17x 109 Q m 0.000765 m) m = 0.364 m or 36.4rcu = rre (+1) 2.2 Discuss wh co er is referred over iron for ower transmission even thou the iron is nearl 100
times cheaper than copper. Now compare the volume (V) of Fe per unit length to the volume of Cu per unit length: : sze2(1m)= (0.364 m)2
Cu ﬂrc“2(lm) (0.010 my VFe
V = 1325 Even though Fe costs 100 times less than Cu, we need about 1300 times the volume of Cu (+1) if Fe is
used. The cost disadvantage is 13 times in addition to weight disadvantage. 2.3 PROBLEMS PROBLEMS 1w gﬁ4
M @H/chys'f : g
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@ 0,1.5:(ﬂe T')=(/re 5):; X o'z/wsw X 1 /Z' ? Ml,”
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.3 amn‘ be N0! (56/ n5 /e(r7 {ﬁll/Eva A an {'i/h( [3 3. ;_$ ® (W 42 W WW as recrysn/WAM (Saws af *‘IOL
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This note was uploaded on 11/05/2011 for the course MSE 302 taught by Professor Norton during the Spring '11 term at Western Washington.
 Spring '11
 Norton

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