Reference no: EM13967361

Problem 1. (a) Using the Murray Loop test, determine the distance to an earth fault on one of the cores of a uniform three-core underground cable. R_a = 2 ohms, R_b = 1Ω and the cable length is 300 m.

(b) If the cable measurement is accurate to ±1% what length of excavation would be required to locate the fault?

Problem 2. A cable run consists of 100 m of 120 mm² three-core cable jointed to 100 m of 240 mm² cable. The ratio of the potentiometer resistances in a Murray Loop test, R_b/(R_b + R_a), is 1/3.

For an earth on one core of the cables, determine the location of the fault.

Problem 3. A Varley Bridge is connected to a faulty three-core copper cable by two identical copper leads of resistance R_l.

(a) Show that for the initial reading (connection to earth);

2R_x = 2R_c - R_i .............................. (1)

where R_c is the resistance of the cable core

R_i is the initial reading of the bridge

R_x is the cable resistance to the fault from the bridge

and for the final reading:

2R_c = R_f - 2 R_l .............................. (2)

where R_l is a lead resistance

R_f is the final reading resistance.

(d) Using R = ρL/A

where ρ is the resistivity,

L is the length

and A is the cross-sectional area of a cable core

show that

R_l = R_cA_cL_c/A_lL_c      (where L_c = L)

and by substituting for Rl show that the distance to the fault is given by:

x = [(R_f - R_i)/R_f - R_i/R_f(A_cL_i/A_lL)] L

(e) Determine the distance to a fault on a 200 m, 120 mm² copper cable if copper 10 mm² test leads of length 10 m are used and R_i/R_f = 0.2.

Problem 4. FIGURE 1 shows a Wheatstone Bridge connection used to determine the location of a fault on a cable in which two cores are of different cross-sectional area (the Murray Fisher method). Ignoring test lead resistance, show that:

(a) For the first reading position, the resistance to the fault, R_x, is given by

R_x = R_b1/(R_a1 + R_b1).(R_c1 + R_c2)

(b) For the second reading position and substituting for Rc₁, the distance to the fault, x, is given by:

x = R_b1/(R_a1 + R_b1) x (R_a2 + R_b2)/R_b2 x L

x = R_b1/ R_b2[R_a2 + R_b2/R_a1 + R_b1]L

(c) If the cable tested is 100 m in length, R is 1.5 ohms, R_b is 1.0 ohm and R_a2 /R_b2 = 0.8, calculate the distance to the fault.