Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Engineering
Mechanical Engineering
Electrical Engineering
Aerospace Engineering
Nuclear Engineering
Materials Engineering
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Engineering
Mechanical Engineering
Electrical Engineering
Aerospace Engineering
Nuclear Engineering
Materials Engineering
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Engineering
Electrical Engineering
Power System, synchronous generator
Reply to thread
Message
[QUOTE="LagCompensator, post: 5438542, member: 588734"] Sorry for bad title(little informative regarding my question), however I came across the following question given as an exam question at some university. Question: [I]"A synchronous generator is running overexcited with excitation voltage Ef = 1.40 p.u and connected to the network at voltage of 1 p.u. This generator has synchronous reactance of 1.20 p.u., and delivering active power of 0.50 p.u. to the network. In the network, there is 1% increase in real power due to power drop from uncontrolled network connected renewable power source. Therefore, the prime mover (i.e. turbine) input is increased by 1%. The excitation or voltage controller of synchronous generator is not responding to this change. [B]Explain how the reactive power delivery from the synchronous generator will change under this condition[/B]" [/I] So, the Q delivered by the generator per phase is: [itex]Q = \frac{E_fV_t}{X_d}cos(\delta)-\frac{V_{t}^{2}}{X_d}-V_{t}^{2}(\frac{1}{X_d}-\frac{1}{X_q})sin^2(\delta)[/itex] (1) So according to this equation the Q will not change due to increase in turbine input. However if we take a look at this equation: [itex]P = \frac{E_fV_t}{X_d}sin(\delta)-\frac{V_{t}^{2}}{X_d}-V_{t}^{2}(\frac{1}{X_d}-\frac{1}{X_q})sin(2\delta)[/itex] (2) An increase of P will cause an increase in [itex]V_t[/itex], and an increase of [itex]V_t[/itex] will [B]increase the reactive power delivered by the sync. generator[/B] according to equation (1) Can someone tell me if this is the right way of thinking or if I am missing something please tell me. Best Regards [/QUOTE]
Insert quotes…
Post reply
Forums
Engineering
Electrical Engineering
Power System, synchronous generator
Back
Top