Question 7.10: Consider the 11-kV radial system shown in Figure 7.42. Assum...
Consider the 11-kV radial system shown in Figure 7.42. Assume that all loads have the same power factor. Determine relay settings to protect the system assuming relay type CO-7 (with characteristics shown in Figure 7.43) is used.
The load currents are calculated as follows:
I_{1}=\frac{4\times 10^{6} }{\sqrt{3}\left(11\times 10^{3} \right) } =209.95 AI_{2}=\frac{2.5\times 10^{6} }{\sqrt{3}\left(11\times 10^{3} \right) } =131.22 A
I_{3}=\frac{6.75\times 10^{6} }{\sqrt{3}\left(11\times 10^{3} \right) } =354.28 A
The normal currents through the sections are calculated as
I_{21}=I_{1}=209.95 A
I_{32}=I_{21}+I_{2}=341.16 A
I_{S}=I_{32}+I_{3}=695.44 A
With the current transformer ratios given, the normal relay currents are
i_{21}=\frac{209.92}{\frac{200}{5} } =5.25 Ai_{32}=\frac{341.16}{\frac{200}{5} } =8.53 A
i_{S}=\frac{695.44}{\frac{400}{5} } =8.69 A
We can now obtain the current tap settings (C.T.S.) or pickup current in such a manner that the relay does not trip under normal currents. For this type of relay, the current tap settings available are 4, 5, 6, 7, 8, 10, and 12 amperes. For position 1, the normal current in the relay is 5.25 A; we thus choose
\left(C.T.S.\right) _{1}=6 AFor position 2, the normal relay current is 8.53 A, and we choose
\left(C.T.S.\right) _{2}=10 ASimilarly for position 3,
\left(C.T.S.\right) _{3}=10 AObserve that we have chosen the nearest setting higher than the normal current.
The next task is to select the intentional delay indicated by the time dial setting (T.D.S.). We utilize the short-circuit currents calculated to coordinate the relays. The current in the relay at 1 on a short circuit at 1 is
Expressed as a multiple of the pickup or C.T.S. value, we have
\frac{i_{SC_{1} }}{\left(C.T.S.\right) _{1}} =\frac{62.5}{6} =10.42We choose the lowest T.D.S. for this relay for fastest action. Thus
\left(T.D.S.\right) _{1}=\frac{1}{2}By reference to the relay characteristic, we get the operating time for relay 1 for a fault at 1 as
T_{1_{1} } =0.15 sTo set the relay at 2 responding to a fault at 1, we allow 0.1 second for breaker operation and an error margin of 0.3 second in addition to
T_{1_{1} } . Thus,
The short circuit for a fault at 1 as a multiple of the C.T.S. at 2 is
\frac{i_{SC_{1} }}{\left(C.T.S.\right) _{2}} =\frac{62.5}{10} =6.25From the characteristics for 0.55-second operating time and 6.25 ratio, we get
\left(T.D.S.\right) _{1}\cong 2The final steps involve setting the relay at 3. For a fault at bus 2, theshort-circuit current is 3000 A, for which relay 2 responds in a time T_{22} obtained as follows:
\frac{i_{SC_{2} }}{\left(C.T.S.\right) _{2}} =\frac{3000}{\left\lgroup\frac{200}{5} \right\rgroup 10 }=7.5For the \left(T.D.S.\right) _{2} = 2, we get from the relay’s characteristic,
T_{22 }=0.50 sThus allowing the same margin for relay 3 to respond to a fault at 2, as for relay 2 responding to a fault at 1, we have
T_{32 }=T_{22 }+0.1+0.3
=0.90 s
The current in the relay expressed as a multiple of pickup is
\frac{i_{SC_{2} }}{\left(C.T.S.\right) _{3}} =\frac{3000}{\left\lgroup\frac{400}{5} \right\rgroup 10 }=3.75Thus for T_{3 } = 0.90, and the above ratio, we get from the relay’s characteristic
\left(T.D.S.\right) _{3}\cong 2.5We note here that our calculations did not account for load starting currents that can be as high as five to seven times rated values. In practice, this should be accounted for.