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Example Cable Fault Calculation

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Example Cable Fault Calculation - version

This calculation provides an example of a cable fault calculation

Cable Inputs

The cable is assumed to be XLPE single core copper, 185 mm, installed flat, spaced one diameter apart and 52 m in length.

Under the installation conditions the positive sequence impedance, Z1 and zero sequence impedance Z0 are calculatd and found to be:

Z1 = 0.007589471899785190 + 0.0072621169251118863i;
Z0 = 0.024855569961970723 + 0.0054850851667301833i;

% The system operation voltage is 400 V three phase, and the source
% fault levels at the cable are 10 kA at 0.25 pf phase fault and
% 9.8 ka, 0.25 pf earth fault.
U = 400/sqrt(3);            % we use the phase voltage for calculations
Ik3 = 10000; pfk3 = 0.25;   % phase fault level
Ik0 = 9800; pfk0 = 0.25;    % earth fault level

% To be able to use the fault levels, these need to be converted to complex
% form and then the impedance found. The current will be lagging the
% voltage,% which gives
Ik3 = Ik3 * pfk3 - Ik3 * sin(acos(pfk3))* 1i;   % convert to complex
Ze3 = U/Ik3;                                    % convert to impedance
Ik0 = Ik0 * pfk0 - Ik0 * sin(acos(pfk0))* 1i;   % convert to complex
Ze0 = U/Ik0;
disp(['Phase fault source impedance = ', num2str(Ze3, '%0.4f' ),' ohm'] );
disp(['Earth fault source impedance = ', num2str(Ze0, '%0.4f' ),' ohm'] );
Phase fault source impedance = 0.0058+0.0224i ohm
Earth fault source impedance = 0.0059+0.0228i ohm

Load End Fault Levels

To calculate the load end fault levels we use the total impedance (source + cable). In addition, IEC 60609 requires the user of a C factor, which for low voltage cables is Cmax=1.05, Cmin=0.95. For the phase fault we have:

Cmax=1.05; Cmin=0.95;
Zt3 = Ze3 + Z1;
If3 = Cmax*U/Zt3;
disp(['Phase fault load end, maximum = ', num2str(abs(If3)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If3)), '%0.2f' ),' pf)'] );
If3 = Cmin*U/Zt3;
disp(['Phase fault load end, minimum = ', num2str(abs(If3)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If3)), '%0.2f' ),' pf)'] );

%
% And for the earth fault:
Zt0 = Ze0 + Z1+ Z0;
If0 = Cmax*U/Zt0;
disp(['Earth fault load end, maximum = ', num2str(abs(If0)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If0)), '%0.2f' ),' pf)'] );
If0 = Cmin*U/Zt0;
disp(['Earth fault load end, minimum = ', num2str(abs(If0)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If0)), '%0.2f' ),' pf)'] );
Phase fault load end, maximum = 7.46 kA(@ 0.41 pf)
Phase fault load end, minimum = 6.75 kA(@ 0.41 pf)
Earth fault load end, maximum = 4.64 kA(@ 0.73 pf)
Earth fault load end, minimum = 4.20 kA(@ 0.73 pf)

BS 7671 Earth Fault Loop Impedance

BS 7671 regulation 411.4.5 has a requirement to use a minimum voltage factor Cmin in the calculation of earth fault loop impedance. The value to use for Cmin is given as 0.95.ce

Cmin = 0.95;
If0 = Cmin*U/Zt0;
disp(['Earth fault load end = ', num2str(abs(If0)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If0)), '%0.2f' ),' pf) - considering Cmin'] );
disp(['Earth fault loop impedance = ', num2str(abs(Zt0), '%0.4f' ),' ohm'] );
Earth fault load end = 4.20 kA(@ 0.73 pf) - considering Cmin
Earth fault loop impedance = 0.0523 ohm

Additional External Conductor

Also consider the case where an additional external conductor is added into the circuit to reduce the overall earth fault look impedance. Select a 50 mm2 copper conductor, with an impedance of 0.495e-3 + j0.135e-3 ohm/meter. Assume the CPC to 50 m in length:

Z2 = ( 0.495e-3 + 0.135e-3)*50;
%
% Overall earth fault loop impedance Z2, is that of the external CPC in
% parallel with the Z0 due to the cable armouring:
Z2 = (Z2*Z0) / (Z2 + Z0);
Zs = Ze0 + Z1+ Z2;
If0 = Cmax*U/Zs;
disp(['Earth fault load end, maximum = ', num2str(abs(If0)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If0)), '%0.2f' ),' pf)'] );
If0 = Cmin*U/Zs;
disp(['Earth fault load end, minimum = ', num2str(abs(If0)/1000, '%0.2f' ),' kA', ...
        '(@ ', num2str(cos(angle(If0)), '%0.2f' ),' pf)'] );
disp(['Earth fault loop impedance = ', num2str(abs(Z2), '%0.4f' ),' ohm'] );
Earth fault load end, maximum = 5.77 kA(@ 0.65 pf)
Earth fault load end, minimum = 5.22 kA(@ 0.65 pf)
Earth fault loop impedance = 0.0142 ohm

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