Load Sharing by Two Alternators in Parallel Operation

Electronics & ElectricalElectronDigital Electronics

Consider two synchronous generators or alternators are operating in parallel and their load-frequency characteristics are shown in the figure.

Let

  • $𝑓_{1(𝑛𝑙)}$ = No load frequency of alternator 1

  • $𝑓_{2(𝑛𝑙)}$ = No load frequency of alternator 2

  • $𝑓_{1(𝑓𝑙)}$ = Full load frequency of alternator 1

  • $𝑓_{2(𝑓𝑙)}$ = Full load frequency of alternator 2

  • 𝑓 = Common operating frequency when the two alternators are operating in parallel

  • $𝑊_{1}$ = Full load power rating of alternator 1

  • $𝑊_{2}$ = Full load power rating of alternator 2

  • $𝑃_{1}$ = Power shared by alternator 1

  • $𝑃_{2}$ = Power shared by alternator 2

  • 𝑃 = Total power delivered by the two alternators

For Alternator 1:

$$\mathrm{Drop\:in\:frequency\:from\:no\:load\:to \:full\:load = 𝑓_{1(𝑛𝑙)} − 𝑓_{1(𝑓𝑙)}}$$

$$\mathrm{Drop\:in\:frequency\:per\:unit\:power \:rating =\frac{𝑓_{1(𝑛𝑙)}− 𝑓_{1(𝑓𝑙)}}{𝑊_{1}}}$$

$$\mathrm{∴\:Drop\:in\:frequency\:for\:a\:load\:of \:power\:𝑃_{1} =\frac{𝑓_{1(𝑛𝑙)} − 𝑓_{1(𝑓𝑙)}}{𝑊_{1}}× 𝑃_{1} … (1)}$$

Therefore, the operating frequency of the alternator 1 would be,

$$\mathrm{𝑓_{1} = No\:load\:frequency − Drop\:in\: frequency}$$

$$\mathrm{\Rightarrow\:𝑓_{1} = 𝑓_{1(𝑛𝑙)} − \left( \frac{𝑓_{1(𝑛𝑙)} − 𝑓_{1(𝑓𝑙)}}{𝑊_{1}}× 𝑃_{1}\right)… (2)}$$

For Alternator 2:

$$\mathrm{Drop\:in\:frequency\:from\:no\:load\:to \:full\:load = 𝑓_{2(𝑛𝑙)} − 𝑓_{2(𝑓𝑙)}}$$

$$\mathrm{Drop\:in\:frequency\:per\:unit\:power \:rating =\frac{𝑓_{2(𝑛𝑙)} − 𝑓_{2(𝑓𝑙)}}{𝑊_{2}}}$$

$$\mathrm{∴\:Drop\:in\:frequency\:for\:a\:load\:of\:power\:𝑃_{2} =\frac{𝑓_{2(𝑛𝑙)} − 𝑓_{2(𝑓𝑙)}}{𝑊_{2}}× 𝑃_{2}… (3)}$$

Therefore, the operating frequency of the alternator 2 would be,

$$\mathrm{𝑓_{2} = No\:load\:frequency − Drop\:in\: frequency}$$

$$\mathrm{\Rightarrow\:𝑓_{2} = 𝑓_{2(𝑛𝑙)} − \left(\frac{𝑓_{2(𝑛𝑙)} − 𝑓_{2(𝑓𝑙)}}{𝑊_{2}}× 𝑃_{2} \right)… (4)}$$

For parallel operation, both the alternators must operate at the same frequency.Thus, from Eqns. (2) and (4), we have,

$$\mathrm{𝑓_{1} = 𝑓_{2} = 𝑓}$$

$$\mathrm{∴\:𝑓 = 𝑓_{1(𝑛𝑙)} −\left( \frac{𝑓_{1(𝑛𝑙)} − 𝑓_{1(𝑓𝑙)}}{𝑊_{1}}× 𝑃_{1} \right)}$$

$$\mathrm{= 𝑓_{2(𝑛𝑙)} −\left(\frac{𝑓_{2(𝑛𝑙)} − 𝑓_{2(𝑓𝑙)}}{𝑊_{2}}× 𝑃_{2} \right)… (5)}$$

Also, the total power delivered by the two alternators is

$$\mathrm{𝑃 = 𝑃_{1 }+ 𝑃_{2} … (6)}$$

Hence, the power shared by the two alternator 𝑃1, 𝑃2 and the common operating frequency (f) of the system can be determined using the eqns. (5) and (6).

Numerical Example

Two station alternators A and B operate in parallel. The Station capacity of A is 30 MW and that of B is 60 MW. The full-load speed regulation of station A is 4% and full-load speed regulation of B is 4.5%. Calculate the load sharing if the connected load is 60 MW. No-load frequency is 50 Hz.

Solution

Let

  • $𝑃_{𝐴}$ = load shared by alternator 𝐴 in MW

  • $𝑃_{𝐵}$ = load shared by alternator 𝐵 in MW

$$\mathrm{∴\:𝑃_{𝐴} + 𝑃_{𝐵} = 60\:MW … (1)}$$

Original frequency of the system at no-load is 𝑓𝑛𝑙 = 50 Hz

Alternator A:

For a load of 30 MW,

$$\mathrm{the\:drop\:in\:frequency = 4\%\:of\:𝑓_{𝑛𝑙 }=\frac{4}{100}\times 50 = 2\:Hz}$$

For a load of 1 MW,

$$\mathrm{the\:drop\:in\:frequency =\frac{2}{30}= 0.067\:Hz}$$

∴ For a load of 𝑃𝐴 MW,

$$\mathrm{the\:drop\:in\:frequency =\frac{2}{30}\times 𝑃_{𝐴}=0.067𝑃_{𝐴}\:Hz}$$

Hence, the operating frequency of alternator A is,

$$\mathrm{𝑓_{𝐴} = 𝑓_{𝑛𝑙} − (drop\:in\:frequency)}$$

$$\mathrm{\Rightarrow\:𝑓_{𝐴} = 50 − 0.067𝑃_{𝐴} … (2)}$$

Alternator B:

Similarly, the operating frequency of alternator B is given by,

$$\mathrm{𝑓_{𝐵} = 𝑓_{𝑛𝑙} − (drop\:in\:frequency)}$$

$$\mathrm{\Rightarrow\:𝑓_{𝐵} = 50 − \left( \frac{4.5}{100} \times 50\right)\frac{𝑃_{𝐵}}{60}}$$

$$\mathrm{\Rightarrow\:𝑓_{𝐵} = 50 − 0.0375\:𝑃_{𝐵} … (3)}$$

Since for parallel operation both the alternators must operate at the same frequency, we have,

$$\mathrm{𝑓_{𝐴} = 𝑓_{𝐵}}$$

$$\mathrm{\Rightarrow\:50 − 0.067𝑃_{𝐴} = 50 − 0.0375𝑃_{𝐵}}$$

$$\mathrm{\Rightarrow\:134𝑃_{𝐴} = 75𝑃_{𝐵}}$$

$$\mathrm{\Rightarrow\:𝑃_{𝐴} =\frac{75}{134}𝑃_{𝐵} … (4)}$$

And,

$$\mathrm{𝑃_{𝐵} =\frac{134}{75}𝑃_{𝐴} … (5)}$$

From Eqns. (1), (4) and (5), we get,

$$\mathrm{\frac{75}{134}𝑃_{𝐵} + 𝑃_{𝐵} = 60\:MW}$$

$$\mathrm{\Rightarrow\:𝑃_{𝐵 }= 38.71\:MW}$$

$$\mathrm{𝑃_{𝐴} = 60 − 38.71 = 21.29\:MW}$$

raja
Published on 14-Oct-2021 11:35:27
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