- Blavier Test Definition: The Blavier Test is used to find earth fault locations in underground cables by measuring resistance at different points.
- Murray Loop Test Process: This test uses a Wheatstone Bridge to locate faults by comparing resistances in known cable lengths.
- Varley Loop Test Definition: The Varley Loop Test finds underground cable faults using a Wheatstone Bridge without relying on cable lengths, by adjusting and measuring variable resistors.
- Varley Loop Test Considerations: Uniform cable sections and lower current are needed to ensure accurate results due to temperature effects.
- Fisher Loop Test Definition: The Fisher Loop Test uses two healthy cables of the same length and cross-sectional area as the faulty one to locate faults by comparing bridge connections.
Blavier’s Test is used to find the earth fault location in an underground cable. The cable ends are called the sending end and the far end. In this test, the sending end must be open and isolated. First, measure the resistance between the sending end and the earth point with the far end isolated. Then, measure it again with the far end grounded.
Suppose we get resistance values R1 and R2 from these two measurements. At the fault location, the conductor is shorted to ground. This short circuit may have some resistance, denoted as g.
In Blavier’s test, the total line resistance is denoted as L. The resistance from the sending end to the fault end is x, and the resistance from the fault end to the far end is y.
So, the total resistance L is equals to the addition of x and y resistances.
Now, the total resistance of the x and g loop is nothing but R1 – the conductor resistance between sending end and earth by keeping far end open.
The total resistance of the entire loop of the above circuit is nothing but R2 – the conductor resistance between sending end and earth by keeping far end earthed.
By solving the above three equation and eliminating g and y;
This expression gives the resistance from the sending end to the fault location. The corresponding distance is calculated by known resistance per unit length of the cable. A practical difficulty in Blavier’s test is that the resistance to ground g is variable, being influenced by the amount of moisture present in the cable and the action of the current at the fault condition. Also, the resistance g may be so high that it exerts very little shunting action when y is placed in parallel with it by grounding the far end of the line.
Murray Loop Test
This test is used to find the fault location in an underground cable by making one Wheatstone Bridge in it and by comparing the resistance we shall find out the fault location. But we should use the known length of the cables in this experiment. The necessary connection of the Murray loop test is shown in figure 2 and 3. The figure 2 shows that the circuit connection for finding the fault location when the ground fault occurs and the figure 3 shows that the circuit connections for finding the fault location when the short circuit fault occurs.
In this test, the faulty cable is connected to a sound cable using a low resistance wire. This wire should not affect the total resistance of the cable and must allow the loop current to circulate through the bridge circuits without loss.
The variable resistors R1 and R2 are forming the ratio arms. Balance of the bridge is achieved by adjusting the variable resistors. G is the galvanometer to indicate the balance. [R3 + RX] is the total loop resistance formed by the sound cable and the faulty cable. At the balance condition,
When the cross section area of the both sound cable and faulty cable are equal, then the resistance of the conductors are directly proportional to their lengths. So, if LX represents the length between test end to the fault end of the faulty cable and if L represents the total length of the both cables, then the expression for LX is as follows;
The above test is only valid when the lengths of the cables are known. In Murray Loop Test, the fault resistance is fixed and it may not be varied. Also it is difficult to set the bridge as balance. Thus, the determination of the fault position is not accurate. Then the current circulation through the cable would cause temperature rises due to high voltage or high current. If the resistance varies according to the temperature, then the balance collapses. So, we need to apply less voltage or less current to this circuit.
Varley Loop Test
This test is used to find the fault location in an underground cable by making one Wheatstone Bridge in it and by comparing the resistance we shall find out the fault location instead of calculating it from the known lengths of the cable. The necessary connection of the Varley loop test is shown in figure 4 and 5. The figure 4 shows that the circuit connection for finding the fault location when the ground fault occurs and the figure 5 shows that the circuit connections for finding the fault location when the short circuit fault occurs.
In this test, the faulty cable is connected with sound cable by a low resistance wire, because that resistance should not affect the total resistance of the cable and it should be able to circulate the loop current to the bridge circuits without loss. A single pole double through switch ‘S’ is used in this circuit. There would be a variable resistor ’ which is used to balance the bridge circuit during the working period.
If the switch S is in position 1, then we need to adjust the variable resistance R to balance the circuit. Let us assume that the present R value as RS1. At this position, the expressions are as follows;
This expression gives the value of [R3 + RX], if the value of R1, R2 and RS1 are known.
If the switch S is in position 2, then again we need to adjust the variable resistance R to balance the bridge circuit. Let us assume that the new R value as RS2. At this position, the expressions are as follows;
By solving the equation (1) and (2),
Therefore, the unknown resistance RX is,
Varley Loop Test is valid only when the cable sections are uniform throughout the loop. The current flowing through the cable would cause the temperature effect. Due to this temperature effect, the resistance of the cable would change. Thus, we need to apply less current to this circuit to carry out the experiment.
Fisher Loop Test
In this Fisher Loop Test, there must be two healthy sound cables which must have the same length and same cross sectional area as the faulty cable. As per the Fig.6 and 7 circuit diagram, all the three cables are connected by a low resistance wire.
In the Fig.6 circuit connection, the bridge connection is connected to ground. Now, the bridge arms are RA, RB, RX and [RS1 + RY]. In the Fig.7 circuit connection, the bridge connection is connected to ‘Sound Cable 2’.
Now, the bridge arms are RA‘, RB‘, RS2 and [RX + RY]. Here [RS1 = RS2]. Two balancing are necessary as per the two different circuits. Let, for the first balance, the expressions are as follows;
For the second balance, the expressions are as follows;
From the expression (1) and (2),
In this two circuits, if the bridge arm resistors are equal (or) if [(RA + RB) = (RA‘ + RB‘)], then the expression (3) can be modified as,
So, when the resistance per unit length of the conductor is uniform in all conditions, then the fault location LX is as follows;
Here ‘L’ if the total length of the faulty cable. But practically, this is not possible. There would be fractional changes in the bridge arms. Thus, the fault location LX is as follows;
This is about the working principle of “Fisher Loop Test”.
