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Murray Loop Test and Locating Faults

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Hi there,
Can anyone answer the following questions please?

Q.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. Ra = 2 ohms, Rb = 1 ohm 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?

Q.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,
Rb/(Rb + Ra), is 1/3

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

Many thanks in advance.

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https://brainmass.com/engineering/electrical-engineering/murray-loop-test-locating-faults-516389

Solution Preview

Please refer to the attachment.

Q.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. Ra = 2 ohms, Rb = 1 ohm 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?

Solution: (a) Murray loop test is based on the Wheatstone bridge principle.
Wheatstone bridge:

Ra Rc
Ammeter

Rb Rd

V

Wheatstone principle: When current through the ammeter is zero (balanced Wheatstone bridge), various resistors are related as follows: Ra/Rb = Rc/Rd
Murray loop test:
Length L = 300m Three core cable
A B
C Rc D
Ra E E Rd G F
x m (300-x) m
Earth fault
...

Solution Summary

The expert examines Murray loop tests and locating faults. A cable measurement accuracy is determined.

$2.19
See Also This Related BrainMass Solution

Varley bridge/ Wheatstone bridge

The Varley loop test is a method of fault location in telecom and pilot wire circuits where fairly high circuit resistances are encountered. This solution determines the distance to the fault.

3. A Varley Bridge is connected to a faulty three-core copper cable by two identical copper leads of resistance R1.
(a) Show that for the initial reading (connection to earth);
2Rx = 2R, — Ri (1)
where R, is the resistance of the cable core
Ri is the initial reading of the bridge
Rx is the cable resistance to the fault from the bridge
and for the final reading:
2R, = R f — 2R1 (2)
where R1 is a lead resistance
Rf is the final reading resistance.

Then by substituting (2) in (1) and rearranging the equation, show:

(see the PDF for correct calculations)

(b) By multiplying the rhs brackets and collecting terms, show the effect of the leads is given by:

(see the PDF for correct calculations)

i.e. = effect with no leads — ratio of initial and final readings
x lead resistance

(c) Determine the distance to the fault by modifying the expression in (b)

(see the PDF for correct calculations)

where x is the cable distance to the fault
and L is the length of a cable core.
Derive an expression for x, the distance to the fault.

(d) Using R = pi, A
where p is the resistivity,
L is the length
and A is the cross-sectional area of a cable core

(see the PDF for correct calculations)

and by substituting for R1 show that the distance to the fault is given by:

(see the PDF for correct calculations)

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

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