Review of Chapter 2, Plus Matlab Examples

Transcription

1 Reiew of Chapter, Plus Matlab Examples. Power in single-phase circuits Let () t and () i t be defined as: () = cos ( ω + θ ) and () = cos ( ω + θ ) t V t i t I t m m i then the instantaneous power is gie by: () = ()() = cos( ω + θ ) cos( ω + θ ) p t t i t VmIm t t i VI m m = cos( t i) cos( i) ω + θ + θ + θ θ = V I t+ + { cos ( ω θ ) ( θ θ ) cos( θ θ )} ( ω θ ) ( θ θ ) ( ω θ ) ( θ θ ) ( θ θ ) rms rms i i { cos cos i sin sin i cos i } = V I t+ + t+ + () = cosθ 1 + cos ( ω + θ ) + sinθ sin ( ω + θ ) p t V I t V I t where θ = θ θi. The angle θ is called the power factor angle and is the angle between the oltage and current at a particular -port. Note: the power factor is gien by cosθ. Note that the cosine is an een function thus the sign of θ is lost! For an inductie circuit (memory helper: ELI) oltage leads the current, the carrent lags the oltage and we call this a "lagging" power factor because the current lags the oltage. For a capacitie circuit (memory helper: ICE) the current leads the oltage, and we call this a "leading" power factor. Thus, we always say clearly if the power factor is lagging (inductie) or leading (capacitie). The lag or lead is the angle by which the current lags or leads the oltage. V V It is also noted that θ is the impedance angle, thus: θ = = ( i) I I θ θ. The instantaneous power can be expressed in two parts: () = () + () p t p t p t R thus p () t = V I cosθ 1 + cos ( ωt+ θ ) R and p () t V I sinθsin ( ωt θ ) = +. It is common to define P and Q as follows: 1

2 ( ()) R () ( ) cos P = ag p t = ag p t = V I θ Q= V I sinθ Thus in terms of P and Q, the instantaneous power can be expressed as: ( ) = 1+ cos( ω + θ ) + sin( ω + θ ) p t P t Q t Thus P is the "aerage" alue of the cosine terms (counting DC as a zero frequency cosine term) while Q is the "magnitude" of the sine term. It is noted that besides the DC term, power has twice line frequency components. Example.1 A supply oltage t () = 100cosωt is applied across a load whose impedance is gien by = ! Ω. Determine i() t and p() t. Use Matlab to plot i() t, () t, p() t, pr () t and p () t oer an interal from 0 to π. The Matlab program is shown below: Vm = 100; theta = 0; % Voltage amplitude and phase angle = 1.5; gama = 60; % Impedance magnitude and phase angle thetai = theta - gama; % Current phase angle in degree theta = (theta - thetai)*pi/180; % Degree to radian Im = Vm/; % Current amplitude wt=0:.05:*pi; % wt from 0 to *pi =Vm*cos(wt); % Instantaneous oltage i=im*cos(wt + thetai*pi/180); % Instantaneous current p=.*i; % Instantaneous power V=Vm/sqrt(); I=Im/sqrt(); % rms oltage and current P = V*I*cos(theta); % Aerage power Q = V*I*sin(theta); % Reactie power S = P + j*q % Complex power pr = P*(1 + cos(*(wt + theta))); % Eq. (.6) px = Q*sin(*(wt + theta)); % Eq. (.8) PP=P*ones(1, length(wt)); % Aerage power with length w for plot xline = zeros(1, length(wt)); % generates a zero ector wt=180/pi*wt; % conerting radian to degree subplot(,,1), plot(wt,, wt, i,wt, xline), grid title(['(t)=vm coswt, i(t)=im cos(wt +', numstr(thetai), ')']) xlabel('wt, degree') subplot(,,), plot(wt, p, wt, xline), grid title('p(t)=(t) i(t)'), xlabel('wt, degree') subplot(,,3), plot(wt, pr, wt, PP, wt,xline), grid title('pr(t) Eq..6'), xlabel('wt, degree') subplot(,,4), plot(wt, px, wt, xline), grid title('px(t) Eq..8'), xlabel('wt, degree') subplot(111)

3 S =.0000e e+003i (t)=vm coswt, i(t)=im cos(wt +-60) p(t)=(t) i(t) wt, degree pr(t) Eq wt, degree wt, degree px(t) Eq wt, degree.3 Complex power * It is obsered that the term VI gies the result: (N.B. θ = θ θi) cos * VI = V I θ + j V I sinθ which is identical to * S = VI = P+ jq where S is the complex power. OTHER FORMS for S : V S = R I + j I = = I. * 3

4 For example, for an inductie load, the current lags the oltage and the phasor diagrams would be as shown below: V θ I θ S Q P Phasor diagrams (V, I) and the power triangle for an inductie load And for a capacitie load, the current would lead the oltage, and the phasor diagrams would be as shown below: I V θ P θ S Q Phasor diagrams (V, I) and the power triangle for a capacitie load, leading power factor angle.4 Complex power balance The total alue of S for a circuit is the SUM of S for the components of S. Example.: Three impedances in parallel are suppleid by a source V = 100 0! V where the impedances are gien by: 1 = 60 + j0, = 6+ j1 and 3 = 30 j30. Find the power absorbed by each load and the total complex power. As a check, compute the power supplied by the source. 4

5 Matlab program follows: V = 100; 1= 60; = 6 + j*1; 3 = 30 - j*30; I1 = V/1; I = V/; I3 = V/3; S1= V*conj(I1), S= V*conj(I), S3= V*conj(I3) S = S1 + S + S3 S = V*conj(I1+I+I3) S1 = 4000 S = e e+004i S3 =.4000e e+004i S = e e+004i S = e e+004i It is obsered that the complex power is consered. 5

6 .5 Power factor correction Adding a capacitor (usually in parallel) with an inductie load can improe the power factor. Example.3 Two loads 1 = j0 and = 10 + j0 are connected across a 00-V 60 Hz source. (a) Find the total real and reactie power, the power factor at the source, and the total current. (b) Find the capacitance of the capacitor connected across the loads to improe the oerall power factor to 0.8 lagging. S Q Q C θ d P Example.3 The Matlab program follows: V = 00; 1= 100; = 10 + j*0; I1 = V/1; I = V/; S1= V*conj(I1), S= V*conj(I) I = I1 + I S = S1 + S, P = real(s); Q = imag(s); PF = cos(angle(s)) thd = acos(0.8), Qd = P*tan(thd) Sc = -j*(q - Qd) c = V^/conj(Sc), C = 1/(*pi*60*abs(c)) Sd = P + j*qd; Id=conj(Sd)/conj(V) S1 = 400 S = e e+003i I = i S = 1.000e e+003i PF = 6

7 thd = Qd = 900 Sc = e+00i c = i C = 4.640e-005 Id = i Example.4 Three loads are connected in parallel across a 1400-V, 60 Hz single-phase supply. Gien that load 1 is inductie, 15 kva at 0.8 power factor, load is capacitie, 10 kw and 40 kar, and load 3 is resistie at 15 kw. (a) Find the total kw, kva and the supply power factor. (b) A capacitor is connected in parallel with the loads to improe the power factor to 0.8 lagging. Find the kar rating of the capacitor and its capacitance in µ F. The Matlab program follows: V = 1400; S1= j*10000; S = j*40000; S3 = 15000; S = S1 + S + S3, P = real(s); Q = imag(s); PF = cos(angle(s)) I = conj(s)/conj(v) thd = acos(0.8), Qd = P*tan(thd) Sc = -j*(q - Qd) c = V^/conj(Sc), C = 1E6/(*pi*60*abs(c)) Sd = P + j*qd; Id=conj(Sd)/conj(V) S = e e+004i PF = I = i thd = Qd = Sc = e+004i c = i 7

8 C = Id = i.6 Complex Power Flow This is a ery important part of chapter. Note the diagram below. It shows four power flows. They are: 1. S 1 is the complex power at node number 1 flowing in (i.e. from node 1 in the direction of node ).. S1 is the complex power at node number 1 flowing out (i.e. from node 1 in the direction away from node ). 3. S 1 is the complex power at node number flowing in (i.e. from node in the direction of node 1). 4. S1 is the complex power at node number flowing out (i.e. from node in the direction away from node 1). -S 1 S 1 S 1 -S 1 + I 1 + V 1 V - - Thus the power consumed by the impedance can be expressed as S = S1 + S1. Important Results from section.6: (will be shown in detail later in the lecture) S V1 V1 V 1 = γ γ + δ1 δ ( ) 8

9 V1 V1 V 1 = cosγ cos γ + δ1 δ P Q ( ) V1 V1 V 1 = sinγ sin γ + δ1 δ ( ) where V1 = V1 δ1 and V = V δ and the impedance is = γ. The angle δ defined as δ = δ1 δ is often called the "power angle". N.B. This is ery different from the "power factor angle" discussed earlier. If we assume that the circuit aboe represents two generators connected by a transmission line, then the equations aboe still apply. In particular the impedance of a transmission line may be assumed purely inductie, i.e. = 0 + j. In this case the angle γ = 90! and the equations aboe simplify to: V1 V P1 = sin ( δ1 δ ) V Q = V V cos ( δ δ ) Since the line is "lossless" then power consumed by the line is zero, i.e. P1 + P1 = 0. In terms of the power angle, these equations become: P = V V 1 1 sin δ V Q = V V 1 cos 1 1 δ Obserations (assuming R 0): 1. Usually δ is ery small (less than 10 degrees), thus P1 sinδ, i.e. small changes in power angle greatly change the real power flow and not the reactie power. If δ1 > δ then power flows from node one to node two. If δ1 < δ, then power flows in the opposite direction (from node to node 1). V. Maximum power transfer occurs when δ = 90! 1 V and is gien by: Pmax =. 9

10 3. Since δ 0, Q V1 V. Thus small changes in V1 V greatly affect Q but not P. Thus, to control real power, we need to change the power angle δ. This is done be increasing prime moer power (mechanical power driing the generator). To control reactie power, we need to change the difference in oltage magnitude. This is done by changing the DC excitation of one generator. Example.5 Assume V 1 = 10 5 and V = Let = 1+ j7 Ω. Determine the real and reactie power supplied or receied by each source and the power loss in the line. The Matlab code follows: R = 1; = 7; = R +j*; V1 = 10*(cos(-5*pi/180) + j*sin(-5*pi/180)); V = 100+j*0; I1 = (V1 - V)/, I1 = -I1; S1 = V1*conj(I1), S1 = V*conj(I1) SL = S1 + S1 PL = R*abs(I1)^, QL = *abs(i1)^ I1 = i S1 = e e+00i S1 = e e+00i SL = i PL = QL = Example.6: Repeat example.5 using a Matlab program such that the angle of source 1 is changed from -30 to 30 degrees in steps of 5 degrees each. We cannot execute "chpex6" from within the notebook because input alues are to be entered by the user. Hence we go to a Matlab window and execute this command. GO TO A MATLAB WINDOW AND EECUTE "chpex6". 10