Question 8.4: Consider a symmetrical two-phase 4/2 VRM whose λ-i character......
Consider a symmetrical two-phase 4/2 VRM whose λ-i characteristic can be represented by the following λ-i expression (for phase 1) as a function of θ_m over the range 0 ≤ θ_m ≤ 90°
λ_1=\left(0.005+0.09\left(\frac{90°- θ_m}{90°}\right) \left(\frac{8.0} {8.0+i_1} \right) \right) i_1
Phase 2 of this motor is identical to that of phase 1, and there is no significant mutual inductance between the phases. Assume that the winding resistance is negligible.
a. Using MATLAB, plot a family of λ_1-i_1 curves for this motor as θ_m varies from 0 to 90° in 10° increments and as i_1 is varied from 0 to 30 A.
b. Again using MATLAB, use Eq. 8.19
\frac{\text{Inverter volt-ampere rating}}{\text{Net output area}}= \frac{\text{area}(W_{rec} + W_{net})}{\text{area}(W_{net})} \quad \quad \quad (8.19)
and Fig. 8.14 to calculate the ratio of the inverter volt-ampere rating to the VRM net power output for the following idealized operating cycle:
(i) The current is instantaneously raised to 25 A when θ_m = -90°.
(ii) The current is then held constant as the rotor rotates to θ_m = 0°.
(iii) At θ_m = 0°, the current is reduced to zero.
c. Assuming the VRM to be operating as a motor using the cycle described in part (b) and rotating at a constant speed of 2500 r/min, calculate the net electromechanical power supplied to the rotor.
a. The λ_1-i_1 curves are shown in Fig. 8.15a.
b. Figure 8.15b shows the areas W_{net} \text{and} W_{rec}. Note that, as pointed out in the text, the λ-i curves are symmetrical around θ_m = 0° and thus the curves for negative values of θ_m are identical to those for the corresponding positive values. The area W_{net} is bounded by the λ_1-i_1 curves corresponding to θ_m = 0° \text{and} θ_m = 90° \text{and the line} i_1 = 25 ~A. The area W_{rec} is bounded by the line λ_1= λ_{max} \text{and the} λ_1-i_1 curve corresponding to θ_m = 0°, \text{where} λ_{max}= λ_1 (25 ~A, 0°).
Using MATLAB to integrate the areas, the desired ratio can be calculated from Eq. 8.19 as
\frac{\text{Inverter volt-ampere rating}}{\text{Net output area}}= \frac{\text{area}(W_{rec} + W_{net})}{\text{area}(W_{net})}=1.55
c. Energy equal to area(W_{net}) is supplied by each phase to the rotor twice during each revolution of the rotor. If area(W_{net}) is measured in joules, the power in watts supplied per phase is thus equal to
P_{phase}= 2 \left(\frac{area(W_{net})}{T} \right) W
where T is the time for one revolution (in seconds).
From MATLAB, area(W_{net}) = 9.91 joules and for 2500 r/min, T = 60/2500 = 0.024 sec,
P_{phase}= 2 \left(\frac{9.91}{0.024} \right) =825 ~W
and thus
P_{mech} = 2 P_{phase} = 1650 ~W
Here is the MATLAB scribt :
Script File
clc
clear
%(a) First plot the lambda i characteristics
for m = 1 : 10
theta (m) = 10 *(m-1) ;
for n=1:101
i (n) = 30*(n-1)/100;
Lambda (n) = i (n)*(0.005 + 0.09* ( (90-theta (m) ) /90) *(8/ (i (n) +8))) ;
end
plot (i, Lambda)
if m==1
hold
end
end
hold
xlabel (' Current [A] ')
ylabel (' Lambda [Wb] ')
title('Family of lambda-i curves as theta_m varies from 0 to 90 degrees ')
text(17,.7 , 'theta_m = 0 degrees ')
text (20,.06, 'theta_m = 90 degrees ')
% (b) Now integrate to find the areas.
%Peak lambda at 0 degrees, 25 Amps
lambdamax = 25*(0.005+0.09*(8/ (25+8) ) ) ;
AreaWnet = 0 ;
AreaWrec 0 ;
% 100 integration step
deli = 25/100 ;
for n=1 : 101
i (n) = 25* (n-1)/100 ;
AreaWnet = AreaWnet + deli*i (n)*(0.09)*(8/(i (n)+8) ) ;
AreaWrec = AreaWrec + deli* (lambdamax - i(n)*(0.005+0.09* (8/ (i(n)+8)))) ;
end
Ratio = (AreaWrec + AreaWnet) /Areawnet ;
fprintf ( '\nPart (b) Ratio =%g',Ratio)
%(c) Calculate the power
rpm = 2500 ;
rps = 2500/60 ;
T = 1/rps ;
Pphase = 2*AreaWnet/T ;
Ptot = 2*Pphase ;
fprintf ('\n\nPart (c) AreaWnet = %g [Joules]',AreaWnet)
fprintf ('\n Pphase = %g [W] and Ptot = %g [W]\n' , Pphase, Ptot)
