ECET 211 Electric Machines and Controls Lecture Feb. 12, 2013  AC Circuits (steady state)  Complex numbers  Voltage source o v(t) = Vm sin (ωt + θ); ω = 2πf o V = Vrms∟θ – Polar form o Vrms = Vm / √2 = Vm * 0.707  The Sinusoidal Forcing Function o v(t) = Vm*sin(2πft + θ), ω = 2πf = 2π/T o sin(ωt + 90º) = sin(ωt + π/2) = cos ωt o cos(ωt ‐ 90º) = cos(ωt – π/2) = sin ωt o sin(‐θ) = ‐sin θ o cos(‐θ) = cos(θ)  Frequencies: 60 Hz, 50 Hz  The Phasor o v(t) = Vm sin(ωt + θ1) → V = Vm/√2 ∟θ1 o i(t) = Im sin(ωt + θ2) → I = Im/√2 ∟θ2  Phasor Relationships for R, L, and C  Impedance o Z = R + j(XL – XC) – ractangular form (Impedance triangle)  θ = tan-1 (XL‐XC)/R o Z = |Z| ∟θ = |Z| [cosθ + j sin θ] o R = Z cos θ o X = Z sin θ o ZR = R∟0º o ZL = XL∟90º; XL = j2πf L o ZC = XC∟‐90º; XC = ‐j1/2πf C o Zt = Z1 + Z2 + ... + Zn  I = E/Z  Complex power o Apparent power = P + jQ = EI cosθ + j EI sinθ (Unit: VA = watts + vars; KVA, KW, KVAR) o Power triangle o Power factor – cosθ; power factor correction Homework #3 Assigned date: 2013/2/5 MATLAB Solution Problem 2‐1. The voltage and current in a given system are given by: ⁄ ⁄ .
Write phasor expressions for the voltage and current and draw a phasor diagram showing the voltage and current for the following cases: (a). Using sin(377t) as the reference vector (b). Using the voltage as the reference phasor (c). Using the current as the reference phasor %ecet211_prob2_1.m
% Paul Lin
% Chapter 2
f = 60;
T = 1/f;
% 16.67 ms
dt = T/100;
t = 0: dt: 2*T;
w = 2*pi*f; % 377.7 rad/sec
Vm = 169;
Vrms = Vm /sqrt(2) % 119.5 Volts
Vt = Vm*sin(2*w*t + pi/6);
VrectForm = Vrms*exp(j*pi/6) % 103.5 + j59.8 Volts
Im = 282.8;
Irms = Im/sqrt(2); %199.97 Amperes
It = Im*sin(w*t - pi/6);
IrectForm = Irms*exp(-j*pi/6)
% 173.2 -j99.9 Amperes
figure(1), plot(t, Vt,'red', t, It,'black'), grid on
title('V(t) 169 sin(wt+pi/6), I(t) =282.8sin(wt -pi/6)')
xlabel('time - second')
% theta = pi/6;
% compass (z) => compass(REAL(z), IMAG(z)
% rose()
% feather()
% qyuiver()
theta = 0: pi/6 : 2*pi;
VmP = Vm*sin(theta + pi/6)
figure(2), subplot(2,1,1),compass(VrectForm, '--r');
figure(2), subplot(2,1,2),compass(IrectForm,'--r');
%
A = [VrectForm, IrectForm];
figure(3), compass(A)
V(t) 169 sin(wt+pi/6), I(t) =282.8sin(wt -pi/6)
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Problem 2‐2. Find the complex power in both rectangular form and polar form for the system of Problem 2‐1. Indicate the values of P, Q, and |S|. Ans: Use equation on page 68 ∗
∗
Rectangular form: Complex Power Apparent Power = Real Power + j*Reactive Power S = P +jQ Vrms = 169/sqrt(2) = 119.5 Volts Irms = 282.8/sqrt(2) = 199.97 Rectangular Form: Rectangular form: Apparent Power = Real Power + j*Reactive Power ∗ ∗
∗
unit => VA, KVA, MVA Vreal = Vrms*cos(0) = 119.5 Vimag = Vrms*sin(0) = j*0 Ireal = Irms*cos(‐pi/3) = 99.99 Iimag = Irms*sin(‐pi/3) = ‐j*173.18 I = 99.99 –j*173.18 I* = Complex Conjugate of I = 99.99 +j*173.18 (Change the sign of the imaginary part) So ∗ ∗
∗
= (119.5) * (99.99 +j*173.18) = 1.1948E4 + j*2.0693E4 = 11.948 + j20.695 KVA S = P + JQ P = 11.948 KW, Q = 20.695 KVAR Polar Form |S| = abs(S) = 23.896 KVA Angle = angle(S) = 1.0472 radians (60 degree) %ecet211_prob2_2.m
% Paul Lin
Vm = 169;
Vrms = Vm /sqrt(2); % 119.5 Volts
% Choosing voltage as a reference
VrectForm = Vrms * exp(j*0) % 119.5 + j0 Volts
Im = 282.8;
Irms = Im/sqrt(2); %199.97 Amperes
% V = V/_theta = V*exp(j*theta)
IrectForm = Irms * exp(-j*pi/3)
% 99.9 -j173.2 Amperes
% Irms*cos(-pi/3) + j*Irms(sin(-pi/3)
IrectForm2 = Irms*cos(-pi/3) -j*Irms*(sin(-pi/3))
%
Comp[lex Conjugate: 99.9 +j173.2 Ampere
% Equation 2-22
% S = VI*
or V times Complex Conjugate of I
PrectForm = VrectForm * IrectForm2
% 1.1948e+04 + 2.0695e+04i
P = real(PrectForm)
% 11.948 KW
Q = imag(PrectForm)
% 20.695 KVAR
S_abs = abs(PrectForm) % 2.3897E4 = 23.897 kVA
S_angle = angle(PrectForm) %1.0472 rad, 60 degree
Problem 2‐3. Figure 2‐29 shows a sinusoidal voltage and current. (a) Write the voltage and current as function of time. (b) Write the voltage and current as phasors using sin(ωt) as the reference phasor. (c) Write the voltage and current as the reference phasor. (d) Calculate the impedance of the load that this current is being delivered to. %ecet211_prob2_3.m
% Paul Lin
Vm = 170;
THETA = 2*pi;
dTheta = THETA/100;
theta = 0: dTheta : THETA;
Vrms = Vm /sqrt(2) % 120.2 Volts
Vt = Vm*sin(theta + 0);
% Choosing voltage as a reference
Im = 20;
Irms = Im/sqrt(2); %14.14 Amperes
% I leads V by 60 degree
It = Im*sin(theta + pi/3);
plot(theta, It, theta, Vt), grid on
V = Vrms*exp(j*0)
I = Irms*exp(j*pi/3)
% 7.07 + j12.25
Zrect = V/I
%Z = R + jXC = 4.25 - j 7.36 Ohms
Zpolar = abs(Zrect)
% 8.50 ohms
Zpolar_angle = angle(Zrect) % -1.0472 Radians, -60 degrees
V = 120.2082 I = 7.0711 +12.2474i Z = 4.2500 ‐ 7.3612i = 8.5/_‐60˚ Ohms 200
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Problem 2‐4. Consider a parallel R‐C circuit with R = 20 Ω and ‐JXc = ‐j13Ω, with an applied voltage of 240/0˚ V. (a) Calculate the real, reactive, and apparent power. (b) Calculate the total current from the source and the resistor and capacitor (inductor) currents. Solution: Given Z = R –jXC = 20 –j*13 Ohms and V = 240/_0˚ Volts Find I = V/Z = I_real + j*I_imaginary Compute complex power or apparent power (in KVA) S = V times (Conjugate of I) = VI* = P + JQ P = real power, Q = reactive power %ecet211_prob2_4.m
% Paul Lin
% Resistor and capacitor are in parallel
R = 20;
XC = -j*13;
Z = R + XC;
Vrms = 240;
Vrect= Vrms*exp(j*0)
% 240 + j0
% P = V^2/R
P = (V*V)/R
% 2.88KW
Q = j*V^2/XC
% -4.331 KVAR
S = P + j*Q
% S = P + jQ
KVA
S_abs = abs(S) % 5.285 KVA
IR = Vrms/R
% 12 amperes
IC = Vrms/XC
% +j18.46 Amperes
It = IR + IC
% 12 + j18.46 Amperes
It_mag = abs(It) % 22.02 Amp current measured by an Ammeter
It_angle = angle(It) % 0.9944 radians; 56.976 degree
Problem 2‐7. A single‐phase load has a voltage of 120/_26˚ V and a current of 45/_‐8˚. (a) Calculate the impedance of the load. (b) Calculate the apparent power, real power, reactive power, and power factor. Solution: Z = V/I = 120/45 /_26+8 = 2.67/_34˚ = 2.67 cos(34˚) + j*2.67sin(34˚) 180 PI ‐‐‐ = ‐‐‐‐ => X = (PI*34)/180 = 0.5934 radians 34 X %ecet211_prob2_7.m
% Paul Lin
% V = 120/_26 degree
% I = 45 /_-8 degree
Vtheta = (pi*26)/180
%0.4538 radians
Itheta = -(pi*8)/180
% -0.1396 radians
V = 120*exp(j*Vtheta) % 107.86 + j52.60
I = 45*exp(j*Itheta)
% 44.56 -j6.26
% Impedance of the Load
Z = V/I
% 2.21 +j1.49 Ohms
Z_abs = abs(Z)
% 2.667 Ohms
Z_angle = angle(Z)
% 34 defree (0.5934 radians)
% Appraent Power, Real Power, Reactive Power, Power Factor
I_conjugate = real(I) -j* imag(I)
% 44.56 + j6.26
S = V*I_conjugate
% Apparent Power: 4.48 +j3.02 KVA
P = real(S)
% Real Power: 4.48 KW
Q = imag(S)
% Reactive Power 3.02 KVAR
PF = cos(angle(S))
% Power Factor 0.829
Problem 2‐8. A single‐phase motor draws 5kW at a power factor of 0.88 when 230V is applied. (a) Calculate the apparent and reactive power drawn by the motor and the power factor angle. (b) Draw the power triangle. (c) Calculate the current drawn by the motor. Solution: %ecet211_prob2_8.m
% Paul Lin
P = 5000; % 5KW
PF = 0.88 % cos(theta)
V = 230;
theta = acos(PF)
% 0.4949 radians
% S = P + jQ = S*cos(theta) + j*S*sin(theta)
% from
P = S*cos(theta) = S*PF
% OR from Power TRiangle
%
Q = SQRT(S^2 - P^2)
% Part (a):
S = P/PF
% 5.68 KVA
Q = S*sin(theta)
% 2.70 KVAR
% Part (b) Draw Power Traingle
%
% Part (c) Current drawn by the motor
Imotor = S/V
% 24.7 amperes