Load Sharing by Two Alternators in Parallel Operation

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Load Sharing by Two Alternators in Parallel Operation

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

Load Sharing by Two Alternators in Parallel Operation

Let

  • f1(nl) = No load frequency of alternator 1
  • f2(nl) = No load frequency of alternator 2
  • f1(fl) = Full load frequency of alternator 1
  • f2(fl) = Full load frequency of alternator 2
  • f = Common operating frequency when the two alternators are operating in parallel
  • W1 = Full load power rating of alternator 1
  • W2 = Full load power rating of alternator 2
  • P1 = Power shared by alternator 1
  • P2 = Power shared by alternator 2
  • P = Total power delivered by the two alternators

For Alternator 1

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

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

$$\mathrm{\therefore \: \text{ Drop in frequency for a load of power }\:P_{1} \:=\: \frac{f_{1(nl)} \:-\: f_{1(fl)}}{W_{1}} \: \times\: P_{1} \:\: \dotso\: (1)}$$

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

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

$$\mathrm{\Rightarrow \: f_{1} \:=\: f_{1(nl)} \:-\: \left(\frac{f_{1(nl)} \:-\: f_{1(fl)}}{W_{1}}\:\times\: P_{1}\right) \:\: \dotso\: (2)}$$

For Alternator 2

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

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

$$\mathrm{\therefore \: \text{ Drop in frequency for a load of power }\:P_{2} \:=\:\frac{f_{2(nl)} \:-\: f_{2(fl)}}{W_{2}} \: \times \: P_{2} \:\: \dotso\: (3)}$$

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

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

$$\mathrm{\Rightarrow\:f_{2} \:=\: f_{2(nl)} \:-\: \left(\frac{f_{2(nl)} \:-\: f_{2(fl)}}{W_{2}}\:\times\: P_{2} \right) \:\: \dotso\: (4)}$$

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

$$\mathrm{f_{1} \:=\: f_{2} \:=\: f}$$

$$\mathrm{\therefore\:f \:=\: f_{1(nl)} \:-\: \left(\frac{f_{1(nl)} \:-\: f_{1(fl)}}{W_{1}}\:\times \:P_{1} \right)}$$

$$\mathrm{=\:f_{2(nl)} \:-\:\left(\frac{f_{2(nl)} \:-\: f_{2(fl)}}{W_{2}}\:\times\: P_{2} \right) \:\: \dotso\: (5)}$$

Also, the total power delivered by the two alternators is

$$\mathrm{P \:=\: P_{1}\:+\:P_{2} \:\: \dotso\: (6)}$$

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

  • PA = load shared by alternator A in MW
  • PB = load shared by alternator B in MW

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

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

Alternator A −

For a load of 30 MW,

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

For a load of 1 MW,

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

∴ For a load of PA MW,

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

Hence, the operating frequency of alternator A is,

$$\mathrm{f_{A} \:=\: f_{nl} \:-\: \text{ (Drop in Frequency)}}$$

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

Alternator B −

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

$$\mathrm{f_{B} \:=\: f_{nl} \:-\: \text{ (Drop in Frequency)}}$$

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

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

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

$$\mathrm{f_{A} \:=\: f_{B}}$$

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

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

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

And,

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

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

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

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

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

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