Let's consider the three-phase system shown in Fig. 11.20. Calculate the real power loss in the line resistance for VL = 500 kV rms and 50 kV rms.

Question

Let's consider the three-phase system shown in Fig. 11.20. Calculate the real power loss in the line resistance for [latex]V_L[/latex] = 500 kV rms and 50 kV rms.

Accepted Answer

For [latex]V_{L}=500 \mathrm{kV} \mathrm{rms}, I_{L}=\frac{S_{ ext {load }}}{\sqrt{3} V_{L}}=\frac{1000}{\sqrt{3}(500)}=1.155 \mathrm{kA} \mathrm{rms}[/latex] and the real power losses in the line are [latex]P_{ ext {line }}=3 I_{L}^{2} R_{ ext {line }}=3(1.155)^{2}(0.1)=0.4 \mathrm{MW}[/latex].

For [latex]V_{L}=50 \mathrm{kV} \mathrm{rms}, I_{L}=\frac{1000}{\sqrt{3}(50)}=11.55 \mathrm{kA} \mathrm{rms}[/latex] and
[latex]P_{ ext {line }}=3 I_{L}^{2} R_{ ext {line }}=3(11.55)^{2}(0.1)=40 \mathrm{MW}[/latex]

The line losses at 50 kV rms are 100 times larger than those at 500 kV rms. This example illustrates that power transmission at higher voltages is more efficient because of the reduced losses. The transformer discussed in Chapter 10 allows voltage levels in ac systems to be changed easily. Electric generators at power plants generate line voltages up to 25 kV. Transformers are utilized to step up this voltage for transmission from the plants to the load centers.

Question 11.9: Let's consider the three-phase system shown in Fig. 11.20. C...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions