1 Subject :- Electrical Power System-2 Topic :- Power Flow Throuh Transmission Lines Topics Covered 3 Introduction. Power flow through transmission line. Single - line diagram of three phase transmission. Derivation. Circle diagram Analytical method. Graphical method. Summary. Introduction 4 The electric power generated in the generating station is transmitted using transmission lines. Transmission lines are conductors designed to carry electricity over a long distances with minimum losses and distortion. The parameters associated with these transmission lines are inductance, capacitance, resistance and conductance. 5 Power Flow Through Transmission Lines SS =PS + jQS G Generating station VS ∠ δ VR ∠0 Transmission SR =PR + jQR line ABCD Bus-1 LOA D Bus-2 Fig:- Single line diagram of three phase transmission Assuming, VR = Receiving End voltage = |VR| ∠0°(VR is reference phasor) VS = |VS| ∠δ°= Sending End voltage(δ is the phase angle between sending and receiving end voltage) 6 Power Flow Through Transmission Lines Generalised line constants are : A = |A|∠ α ; B = |B| ∠β; C = |C| ∠γ; D = |D| ∠Δ; Complex power at receiving end SR =PR + jQR = VR IR* …eq (1) Here, IR* is conjugate of Receiving end current IR We know that VS = AVR + BIR 7 Power Flow Through Transmission Lines From the above equation ⸫I R = VS −AVR B = |VS| ∠δ −|A|∠ α|VR| ∠0° |B| ∠ β = |VS| |A||VR| ∠(δ – β ) ∠(α – β) |B| |B| |V | |A||V | i.e. = IR* = |B|S ∠(β − δ ) - |B| R ∠(β − α) …eq (2) Power Flow Through Transmission Lines 8 Now we put the value IR* in equation of …eq (1), we get SR = VR IR* 2 = |Vs||VR| ∠(β − δ ) - |A||VR | ∠(β − α) |B| |B| …eq (3) Now, we separate real and imaginary parts, then we get the values of PR and QR So, Receiving end True power, 2 PR = |Vs||VR| cos (β − δ ) - |A||VR | cos(β − |B| |B| α) Receiving end Reactive power, 2 QR = |Vs||VR| sin (β − δ ) - |A||VR | sin(β − |B| |B| α) …eq (4) …eq (5) Power Flow Through Transmission Lines 9 For fixed values of Vs and VR, Power Received will be maximum when cos(β − δ) =1 or when δ= β, So |Vs||V R| |A||V R2| PR(max) = - |B| cos(β − |B| α) and QR(max) = - |A||V R2| sin(β − |B| α) In transmission line A=D=1∠0° B=Z ∠θ …eq (7) …eq (8) Power Flow Through Transmission Lines 10 Now we substitute above values in eq (4)&eq (5), We get ⸫P = Vs VR cos(θ − δ) - VR 2 cos θ R Z Z Q = Vs VR sin(θ − δ) - VR 2 sin θ β = 0 Resistance of transmission line is usually very small as compared to R ⸫α = 0 Z Z reactance. Hence Z = X and θ = 90 ° VsItVis R sin δ ⸫ Pangle. = R Z QR = small δ=1) V V s X R 2 - VR X (⸫ δ is the power angle. It is usually very small ⸫ cos Methods Of Finding The Performance Of Transmission Line. Basically two methods 11 Analytical method. Graphical method. Analytical methods are found to be laborious, while graphical method is convenient. Graphical method or circle diagram are helpful for determination of active power P, Reactive power Q, power angle δ and power factor for given load condition. Relations between the sending end and receiving end voltage and currents are given below. VS = AVR + BIR A, B, C, D are generalised constants of transmission. IS = CVR + DIR VS = sending endvoltage, Methods Of Finding The Performance Of Transmission Line. 12 VR = Receiving end voltage IS = Sending end current, IR =Receiving end current. By taking either VS, VR, IS or IR as a reference these characteristics can be plotted. These characteristics are nothing but representing circles, hence such diagrams are called circle diagrams. Circle diagram is drawn by taking active power P on X- axis and reactive power on Y- axis. Receiving End Power Circle Diagram : 13 Receiving end true power – Horizontal coordinates Reactive power component – Vertical coordinates From the equation, VS = AVR +BIR ⸫IR = = VS −AVR B VS ∠δ A ∠δ V ∠0 B ∠ β B ∠β R = Vs ∠(δ − β) - AVR ∠(α−β) B B Receiving End Power Circle Diagram : 14 |A||VR| ∠(β − α) |B| |Vs| IR* = |B| ∠(β − δ ) - Volt- ampere at the receiving end will be SR = PR + jQR = VR IR* 2 |Vs||V | |A||V | R R = ∠(β − δ ) ∠(β − α) |B| |B| R| [cos (β − δ ) + j sin (β − δ )] = |Vs||V 2 |A||V|B| R| |B| [ cos(β − α)+j sin (β − δ )] 2 |Vs||V | |A||V | R R SR = cos (β − δ ) cos(β − α) + |B| |B| |Vs||VR| j sin (β − δ ) - |A||VR|2 j sin (β − δ ) |B| |B| Receiving End Power Circle Diagram : 15 By separating real and imaginary parts, we have |Vs||VR| |A||VR|2 PR = cos (β − δ ) cos(β − |B| |B| α) |Vs||VR| |A||VR|2 QR = sin (β − δ ) sin(β − |B| |B| α) The power component can be expressed as |A||VR|2 PR + cos(β − α) |B| = |Vs||VR| cos (β − δ |B| ) |A||VR|2 QR + sin(β − α) |B| = |Vs||VR| sin (β − δ |B| ) Receiving End Power Circle Diagram : 16 Squaring and adding these equations will give {PR + |A||V |2 |A||VR|2 cos( β − α)}2 + {QR + |B|R sin(β − α) |B| } 2 2 = |Vs|2|VR| { cos2( β − δ )+ sin2 (β − |B| α)} 2 = |Vs|2|VR| |B| It is an equation of a circle. The coordinates of centre of a circle are: |A||VR|2 X-coordinate of the circle = - |B| cos(β − α) |A||V |2 Y- coordinate of the circle = - |B|R sin(β − α) |Vs||V | Radius of the circle = |B| R Construction of circle diagram: 17 Plot the centre of the circle N on a suitable scale. From N draw an arc of a circle with the calculated radius From the origin O draw the load line OP inclined at angle ϕ R with the VSVR . B horizontal. Let it cut the circle at P, then the receiving end true power and reactive power will be represented by OP and PQ respectively. If the voltages VS and VR are taken phase voltage in volts then the powers indicated on X-axis and Y-axis will be in watts and VARs per phase respectively. Construction of circle diagram: 18 If the voltages VS and VR are taken line voltage in volts then the powers indicated on X-axis and Y-axis will be in watts and VARs for all three phases respectively. If the VS and VR are taken from line to line and in kV then the power indicated will be in MW and MVAR and for all the three phases. To determine the maximum power a horizontal line is drawn from the centre of the circle intersecting vertical axis at the point L and the circle at the point M. Distances LM represents the maximum power for the receiving end. Receiving End Power Circle Diagram : 19 Sending End Power Circle Diagram : 20 We have seen… 21 Receiving end True power, 2 PR = |Vs||VR| cos (β − δ ) - |A||VR | cos(β − |B| |B| α) Receiving end Reactive power, Q = |Vs||VR| sin (β − δ ) - |A||VR | sin(β − α) 2 R |B| |B| |Vs||VR| |A||VR2| cos(β − PR(max) = |B| |B| α) |A||VR2| QR(max) = sin(β − α) |B| Construction of circle diagram Receiving End Power Circle Diagram Sending End Power Circle Diagram :