Terminal voltage above emf and armature reaction

Yes. The terminal voltage $V$ of an alternator can be greater than the internally generated emf $E$ when it supplies a leading power factor (capacitive) load.

Why

The voltage equation is

$$\bar E = \bar V + \bar I(R_a + jX_s) \quad\Rightarrow\quad E = \sqrt{(V\cos\phi + IR_a)^2 + (V\sin\phi \pm IX_s)^2}$$

with $+$ for lagging and $-$ for leading current. With leading current the term $V\sin\phi - IX_s$ is small, so $E < V$. Physically, the armature reaction of a leading current is magnetising: it strengthens the main field flux, so the voltage rises with load.

 Leading pf load (E < V)
             I      E
              \    /'.
               \  /   '.  j I Xs
                \/      '.
                O----------+ V

Effect of armature reaction: resistive (unity pf) load

• Current is in phase with emf; armature mmf is at 90° to the field mmf.
• Effect is cross-magnetising: flux is distorted (stronger at one pole tip, weaker at the other) with little change in its total value.
• Terminal voltage falls slightly with load; $E$ is a little more than $V$.

 Unity pf
                    E
                  / |
                /   |  j I Xs
              /     |
            O-------+  V
            O------->  I

Effect of armature reaction: inductive (lagging pf) load

• For zero pf lagging, current lags emf by 90°, the armature mmf directly opposes the field mmf.
• Effect is demagnetising; air-gap flux falls and the terminal voltage drops sharply with load.
• For practical lagging pf (e.g. 0.8), it is partly demagnetising and partly cross-magnetising. More field current is needed to hold rated voltage; regulation is large and positive.

 Lagging pf (E > V)
                      E
                     /|
                   /  |  j I Xs
                 /    |
               O------+ V
                `-.
                   `-. I