Effects of Changes in Shaft Load
Shaft load is doubled – Iacosθi and Efsinδ must double
Effects of Changes in Shaft Load
Since Field Excitation is constant, Ef traces a circular arc, δ increases
θi decreases, increasing the power factor – continues until δ=90 -- pullout
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
Apply step changes to Ef
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
Efsinδ must
be constant
Apply step changes to Ef
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
• Increasing the strength of the magnets
should cause a closer pole alignment and
a smaller power angle.
• Assuming a constant shaft load,
– apply a step increase to Ef
– Efsinδ increases, the rotor accelerates
– angle δ decreases until Efsinδ has the same
value as before (synchronous speed)
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
Iacosθ
constant
Efsinδ
constant
Apply step changes to Ef
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
• For constant shaft load,
– P is proportional to Efsinδ
– Ef1sinθ1 = Ef2sinθ2 = Ef3sinθ3 = Efsinθ
– Locus of the tip of the Ef phasor is parallel to
the VT phasor
– P is proportional to Iacosθ
– Ia1cosθ1 = Ia2cosθ2 = Ia3cosθ3 = Iacosθ
– Locus of tip of the Ia phasor is perpendicular
to the VT phasor
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
Iacosθ
constant
Efsinδ
constant
Apply step changes to Ef
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
• IMPORTANT!!!!
– Increasing the excitation from Ef1 to Ef3
caused the angle in the current phasor (and
hence the power factor) to go from lagging to
leading!
– Normal excitation when power factor = 1
– Excitation greater than normal is known as
overexcitation
– Excitation less than normal is known as
underexcitation
Effects of Changes in Field Excitation
Voltage on Synchronous-Motor Performance
Iacosθ
constant
Power factor changes from lagging
to leading
Efsinδ
constant
Apply step changes to Ef
V Curves
• Plot armature current as a function of field
current or armature current as a function of
excitation voltage.
V-Curves
Stability Limit is where angle
δ = -90° -- the rotor is still
synchronized
V-Curves continued
• Constant-Load V-Curves can be plotted from
laboratory data, phasor diagrams, or from the
following expression
0.5
1  2
Ia 
E f  VT2  2 E 2f VT2  X s2 Pin2,1 

Xs 
Example 8.2
• Referring to the V-curve for 100% load,
determine
– a) the minimum value of excitation that will maintain
synchronism
98V
100% rated load
Example 8.2 continued
• b) using Eq. (8 – 16)
Pin  3  
VT E f
Xs
 Pin X s
sin   E f 
3VT sin 
40  746  1.27
Ef 
 99V
220
3
 sin(90)
3
Example 8.2 continued
• Repeat (a) using Eq. (8-21)
0.5
1  2
Ia 
E f  VT2  2 E 2f VT2  X s2 Pin2,1 

Xs 
Make the quantity under the radical = 0 for
minimum excitation
Example 8.2 continued
E 2f VT2  X s2 Pin2,1  0
Ef 
X s Pin ,1
VT
E f  99V
746
1.27  40 
3

220
3
Example 8.2 continued
• d) the power angle if the field excitation voltage is
increased to 175% of the stability limit determined in (c).
E f 1 sin 1  E f 2 sin  2
1.75 E f 1 sin  2  E f 1 sin(90)
1
sin  2  
 0.571
1.75
 2  35