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v D = ω e λ Q size 12{v rSub { size 8{D} } = - ω rSub { size 8{e} } λ rSub { size 8{Q} } } {} (10.25)

v Q = ω e λ D size 12{v rSub { size 8{Q} } =ω rSub { size 8{e} } λ rSub { size 8{D} } } {} (10.26)

v F = R f i F size 12{v rSub { size 8{F} } =R rSub { size 8{f} } i rSub { size 8{F} } } {} (10.27)

Finally, we can write Eq. 10.21 as

T mech = 3 2 poles 2 ( λ D i Q λ Q i D ) size 12{T rSub { size 8{ ital "mech"} } = { {3} over {2} } left [ { { ital "poles"} over {2} } right ] \( λ rSub { size 8{D} } i rSub { size 8{Q} } - λ rSub { size 8{Q} } i rSub { size 8{D} } \) } {} (10.28)

From this point on, we will focus our attention on machines in which the effects of saliency can be neglected. In this case, the direct- and quadrature-axis synchronous inductances are equal and we can write

L d = L q = L s size 12{L rSub { size 8{d} } =L rSub { size 8{q} } =L rSub { size 8{s} } } {} (10.29)

where L s size 12{L rSub { size 8{s} } } {} is the synchronous inductance. Substitution into Eqs. 10.22 and 10.23 and then into Eq. 10.28 gives

T mech = 3 2 poles 2 [ ( L s i D + L af i F ) i Q L s i Q i D ] size 12{T rSub { size 8{ ital "mech"} } = { {3} over {2} } left [ { { ital "poles"} over {2} } right ] \[ \( L rSub { size 8{s} } i rSub { size 8{D} } +L rSub { size 8{ ital "af"} } i rSub { size 8{F} } \) i rSub { size 8{Q} } - L rSub { size 8{s} } i rSub { size 8{Q} } i rSub { size 8{D} } \]} {}

= 3 2 poles 2 L af i F i Q size 12{ {}= { {3} over {2} } left [ { { ital "poles"} over {2} } right ]L rSub { size 8{ ital "af"} } i rSub { size 8{F} } i rSub { size 8{Q} } } {} (10.30)

Equation 10.30 shows that torque is produced by the interaction of the field flux (proportional to the field current) and the quadrature-axis component of the armature current, in other words the component of armature current that is orthogonal to the field flux. By analogy, we see that the direct-axis component of armature current, which is aligned with the field flux, produces no torque.

This result is fully consistent with the generalized torque expressions which are derived in Chapter 4. Consider for example the equation which expresses the torque in terms of the product of the stator and rotor mmfs ( F s size 12{F rSub { size 8{s} } } {} and F r size 12{F rSub { size 8{r} } } {} respectively) and the sine of the angle between them.

T = poles 2 μ 0 π Dl 2g F s F r sin δ sr size 12{T= - left [ { { ital "poles"} over {2} } right ] left [ { {μ rSub { size 8{0} } π ital "Dl"} over {2g} } right ]F rSub { size 8{s} } F rSub { size 8{r} } "sin"δ rSub { size 8{ ital "sr"} } } {} (10.31)

where δ r size 12{δ rSub { size 8{r} } } {} is the electrical space angle between the stator and rotor mmfs. This shows clearly that no torque will be produced by the direct-axis component of the armature mmf which, by definition, is that component of the stator mmf which is aligned with that of the field winding on the rotor.

Equation 10.31 shows the torque in a nonsalient synchronous motor is proportional to the product of the field current and the quadrature-axis component of the armature current. This is directly analogous to torque production in a dc machine for which the equations can be combined to show that the torque is proportional to the product of the field current and the armature current.

The analogy between a nonsalient synchronous machine and dc machine can be further reinforced. Consider the equation, which expresses the rms value of the line-toneutral generated voltage of a synchronous generator as

E af = ω e L af i F 2 size 12{E rSub { size 8{ ital "af"} } = { {ω rSub { size 8{e} } L rSub { size 8{ ital "af"} } i rSub { size 8{F} } } over { sqrt {2} } } } {} (10.32)

Substitution into Eq. 10.30 gives

T mech = 3 2 poles 2 E af i Q ω e size 12{T rSub { size 8{ ital "mech"} } = { {3} over {2} } left [ { { ital "poles"} over { sqrt {2} } } right ] { {E rSub { size 8{ ital "af"} } i rSub { size 8{Q} } } over {ω rSub { size 8{e} } } } } {} (10.33)

This is directly analogous to Eq. T mech = E a I a / ω m size 12{T rSub { size 8{ ital "mech"} } =E rSub { size 8{a} } I rSub { size 8{a} } /ω rSub { size 8{m} } } {} for a dc machine in which the torque is proportional to the product of the generated voltage and the armature current.

The brushes and commutator of a dc machine force the commutated armature current and armature flux along the quadrature axis such that I d size 12{I rSub { size 8{d} } } {} = 0 and it is the interaction of this quadrature-axis current with the direct-axis field flux that produces the torque. A field-oriented controller which senses the position of the rotor and controls the quadrature-axis component of armature current produces the same effect in a synchronous machine.

Although the direct-axis component of armature current does not play a role in torque production, it does play a role in determining the resultant stator flux and hence the machine terminal voltage, as can be readily shown. Specifically, from the transformation equations of Appendix C,

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Source:  OpenStax, Electrical machines. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10767/1.1
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