Question

Question: For the distribution substation given in Problem 14.9, assume that each of the four circuit breakers on the 12.5-kV side of the distribution substation transformers has a maximum continuous current of 2000A/phase during both normal and emergency conditions. Determine the summer allowable substation rating under the single-contingency loss of one transformer, based on not exceeding the maximum continuous current of these circuit breakers at 12.5-kV operating voltage. Assume a 5% reduction for unequal transformer loadings. Comparing the results of this problem with Problem 14.9, what limits the substation allowable rating, the circuit breakers or the transformers?

Short answer

Answer

The circuit breaker limits the substation allowable rating.

Step-by-step solution

Determine maximum power 

Write the expression for the maximum power through the circuit breaker by considering the maximum continuous current of .

${\mathbf{P}}_{\mathbf{max}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{L}}{\mathbf{I}}_{\mathbf{L}}$ …… (1)

Substitute 12.5 KV for V_L and$3\times 2000A$ for L_L .

$$\begin{gathered} P_{\max }=\sqrt{3}\left(12.5k\right)\left(2000\times 3\right) \\ =129.9\times {10}^{3}kVA \\ \approx 130MVA \end{gathered}$$

Write the expression for the reduced circuit breaker maximum loading due to unequal transformer loadings which is times the maximum loading of the circuit breaker.

$$\begin{gathered} P_{\max \_reduced}=0.95\times 130 \\ =123.5MVA \end{gathered}$$

Determine maximum power

Assume single-contingency loss of the one transformer for two hour emergency.

Consider that the 30 MVA transformers are also being repaired.

Assume balanced loading on the three transformers because the emergency only lasts for two hours.

The reduction is 10% for unequal transformer loading.

Write the expression for output of the transformers TR2, TR3 and TR4.

90% of 33.3 MVA

$0.90\times 33.3MVA=29.97MVA$

Write the expression for the total rating of four transformers.

$3\left(29.97\right)=89.91MVA$

The normal summer loading is 150% of the total loading.

$1.5\times 89.91MVA=134.86MVA$

Thus the summer allowable substation rating is134.86 MVA.

As the rating of circuit breaker of 123.5 MVA. is less than the summer allowable substation rating is134.86 MVA. .

Thus, the circuit breaker limits the substation allowed rating to 123.5 MVA.