- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
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.
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,
$$\mathrm{\frac{75}{134}𝑃_{𝐵} + 𝑃_{𝐵} = 60\:MW}$$
$$\mathrm{\Rightarrow\:𝑃_{𝐵 }= 38.71\:MW}$$
$$\mathrm{𝑃_{𝐴} = 60 − 38.71 = 21.29\:MW}$$
- Related Articles
- Parallel Operation of Alternators
- Synchronization of Alternators by Synchroscope
- How to Setup HAproxy Load Balance Server for Sharing Web Traffic
- Synchronization of Alternators by Synchronizing Lamps Method
- Parallel Operation of Transformers – Reasons and Conditions
- Plot two horizontal bar charts sharing the same Y-axis in Python Matplotlib
- Time-Sharing Operating system
- Manipulate two selects on page load with jQuery
- Prove that two lines are respectively perpendicular to two parallel lines , they are parallel to each other.
- Connected Load, Average Load, and Maximum Demand Load
- Potier Triangle Method – Determining the Voltage Regulation of Alternators
- C program to perform union operation on two arrays
- C program to perform intersection operation on two arrays
- What is the name of the chemical bond formed:(a) By the sharing of electrons between two atoms?(b) By the transfer of electrons from one atom to another?
- What is sharing, blogging and publishing?
