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End Region Heating Limit 3

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charlierod

Electrical
Mar 16, 2004
71
I have two questions about the same topic:

Can anyone explain with some detail the phenomenon of end region heating limit in synchronous generators?

How can i obtain the equation for the curve associated with this limit in the synchronous machine capability curve?
 
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There are three distinct generator capability limits in the field, armature and end region.
1. Field Capability: The field current heating limit determines the reactive power that can be generated at low load. This is normally represented by the upper circle that start at the Q-axe equal at Qo =-V2/Xd, (often V = 1 pu). Then the locus of P and Q is what would be obtained by swinging a vector of length Vf/Xd over an angle ?.

2. Armature heating limit: Max. permissible heating in armature winding due to copper loss (RI2) of generator stator. The current limit for a given terminal voltage is a circle with the centre at the origin of the plane P-Q.

3. End Region Heating Limit: Occur in the underexcited region with notorious increase in the armature end leakage flux. This flux enters and leaves in a direction perpendicular to the stator laminations causing eddy currents in the laminations, and hence heating. This can limit the capability, particularly for round rotor machines.

4. Other limits:
a. Cooling States: multiples curves represent different capability limits for example in cooling hydrogen pressure of large machines.
b. Stator heating: large synchronous generators typically have a problem with heating of the stator iron when they attempt to operate in highly underexcited conditions (q strongly negative).
c. Stability limit.
d. Prime mover limit.

The enclose sites may have additional info in this topic.
 
CharlieRod,

There is an old CEGB document which gives equations and construction details for generator capability curves. It is quite heavy going, but very informative.

The specific question of end-winding heating is, according to this document, derived from practical tests. There are equations listed for most of the dominant parameters affecting the capability curve.



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I found the following expression for the small signal stability limit on the P-Q plane:

Q = [0.5*V^2*((1/Xsys)-(1/Xd))^2]^0.5 + 0.5*V^2*((1/Xsys)-(1/Xd)) (1)

How can i obtain this expression from those corresponding to delivered active and reactive powers?

P = (E*V/Xd)*sin(delta) + 0.5*U^2*((1/Xq)-(1/Xd))*sin(2*delta) (2)

Q = (E*V/Xd)*cos(delta) - U^2*((cos^2(delta)/Xd) + sin^2(delta)/Xq)) (3)

I assume that the procedure is something like this:

1. Assume fixed values for E, V and machine-system parameters.

2. Find a limiting value for delta corresponding to a given P from (2).

3. Use the value of delta computed above to find Q from (3).

Doesn`t the limiting value of delta result in unique values for P and Q?
 
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