My dear Students!: UNIT 2

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Sunday, February 14, 2016

U14EET402-UNIT 2- 11. FERRANTI EFFECT

Ferranti Effect in Power System

at light load or no load operation of transmission system, the receiving end voltage often increases beyond the sending end voltage, leading to a phenomena known as Ferranti effect in power system.

Why Ferranti Effect occurs in a Transmission Line?

A long transmission line can be considered to composed a considerably high amount of capacitance and inductor distributed across the entire length of the line. Ferranti Effect occurs when current drawn by the distributed capacitance of the line itself is greater than the current associated with the load at the receiving end of the line( during light or no load). This capacitor charging current leads to voltage drop across the line inductor of the transmission system which is in phase with the sending end voltages. This voltage drop keeps on increasing additively as we move towards the load end of the line and subsequently the receiving end voltage tends to get larger than applied voltage leading to the phenomena called Ferranti effect in power system. It is illustrated with the help of a phasor diagram below.

Thus both the capacitance and inductor effect of transmission line are equally responsible for this particular phenomena to occur, and hence Ferranti effect is negligible in case of a short transmission lines as the inductor of such a line is practically considered to be nearing zero. In general for a 300 Km line operating at a frequency of 50 Hz, the no load receiving end voltage has been found to be 5% higher than the sending end voltage.

Now for analysis of Ferranti effect let us consider the phasor diagrame shown above.

Here Vr is considered to be the reference phasor, represented by OA.

This is represented by the phasor OC.

Now in case of a long transmission line, it has been practically observed that the line electrical resistance is negligibly small compared to the line reactance, hence we can assume the length of the phasor Ic R = 0, we can consider the rise in the voltage is only due to OA - OC = reactive drop in the line.

Now if we consider c0 and L0 are the values of capacitance and inductor per km of the transmission line, where l is the length of the line.

Since, in case of a long transmission line, the capacitance is distributed throughout its length, the average current flowing is,

Thus the rise in voltage due to line inductor is given by,

From the above equation it is absolutely evident, that the rise in voltage at the receiving end is directly proportional to the square of the line length, and hence in case of a long transmission line it keeps increasing with length and even goes beyond the applied sending end voltage at times, leading to the phenomena called Ferranti effect in power system

Saturday, February 13, 2016

U14EET402- UNIT2-11. LOADABILITY LIMITS

Surge Impedance loading

Z_cand γ transmission line parameters Z and Y. The characteristics of the long line depends upon these two parameters.

As already said Z_cand γ are complex numbers.

Let us consider a case when the load impedance is just equal to the characteristics impedance Z_c then, the receiving end voltage V_r=I_rZ_c

substituting in above equations we get

$\large {\color{Cyan} V(x)=V_r e^{\gamma x}}$

and

$\large {\color{Cyan} I(x)=I_r e^{\gamma x}}$

let, γ = α+j β

Dividing V(x) by I(x) we get

$\large {\color{Cyan}\frac{V(x)}{I(x)}=\frac{V_r e^{\gamma x}}{I_r e^{\gamma x}}=\frac{V_r}{I_r}=Z_c}$

From the above equation it is easy to interpret that the impedance as seen at any point of the line is the same as the load impedance that is Z_c, the characteristics impedance. Moreover from the above equations of V(x) and I(x) it is clear that

$\large {\color{Cyan} V(x)=V_r e^{\gamma x}=V_r e^{(\alpha +j\beta )x}=V_r e^{\alpha x+j\beta x}}$

$\large {\color{Cyan} V(x)=V_re^{\alpha x}.e^{j\beta x}}$

The magnitude of voltage is $\large {\color{Cyan} V_r e^{\alpha x}}$

Clearly the magnitude of the voltage increases with x. But our x increases from receiving end to sending end. So the voltage increases exponentially from receiving end to sending end.

At the sending end the voltage is

$\large {\color{Cyan} V_r e^{\alpha l}}$

The other term $\large {\color{Cyan} e^{j\beta x}}$ is the phasor and only provides phase shift between the voltages at receiving end and at a point x distance from the receiving end. Similar argument can be made for equation of current I(x).

In the above formula of Zc and γ we put Z(series impedance) and Y(shunt admittance). (recall that Z and Y are the values per unit line length)

${\color{Cyan} Z_c =\sqrt{\frac{R+j \omega L}{G+jwC}}}$

${\color{Cyan} \gamma =\sqrt{(R+j \omega L)(G+jwC)}}$

Loss Less Line

What will happen for a loss less line. For loss less line R=0 and G=0.
So our above equations reduces to

${\color{Cyan} Z_c=\sqrt{\frac{L}{C}}}$

${\color{Cyan} \gamma =j\omega \sqrt{LC}}$

Observe that for loss less line, Zc becomes a pure resistance. We also know that γ = α+j β, for this loss less situation we get

${\color{Cyan} \alpha =0}$

${\color{Cyan} \beta =\omega \sqrt{LC}}$

Now our previous voltage and current equations for surge impedance loading reduces to

$\large {\color{Cyan} V(x)=V_r e^{j\beta x} }$

$\large {\color{Cyan} I(x)=I_r e^{j\beta x} }$

As in case of lossy line here also at any distance x from receiving end the ratio of voltage and current is always same that is Zc, the surge impedance of the line. Using the complex algebra you are sure that the magnitude of V(x) is Vr and I(x) is Ir. Which means that for loss less line the voltage and current at any distance x from the receiving end is same. It also implies voltage at sending end is same as voltage at receiving end which is same as voltage at any intermediate point. So Vs = V(x) = Vr. At Surge Impedance Loading the reactive power generated by the line capacitance is equal to the reactive power absorbed by the line inductance for every unit length of line. In Power industry it is said that the voltage profile is flat. So we conclude that for a lossless line the voltage magnitude is same throughout the length of line. As in case of lossy line the term ${\color{Cyan} e^{j\beta x}}$ is the phasor responsible for phase shift. It is simply giving phase shift to the voltage wave along the length of line. The phase angle between sending and receiving end voltage is ${\color{Cyan} e^{j\beta l}}$ . It is clear that if the distance between the sending and receiving end is more then the phase difference between the voltage phasors at both the ends of the line will be more.

In previous articles we already discussed that in case of transmission line how and when we can ignore the line resistance R and leakage conductance G. At least for rough estimate of the load carrying capability of transmission line we can presume it lossless. The surge impedance loading is the ideal loading of the line, which is desired keeping in view of the optimised (flat for lossless ideal case) voltage profile of the line.

For Loss less line the surge impedance loading (SIL) is

$\large {\color{Cyan} SIL=\frac{V_l^{2}}{Z_c}}$

Where

${\color{Cyan} V_l=\sqrt{3}.V_r}$

It should be recalled that Zc is pure resistance for lossless line.

Vr and V_l are the receiving end phase and line voltage respectively.

Approximate SIL for few nominal voltages

132/138 kV - 50 MW
230 kV - 150 MW
345 kV - 400 MW
400 kV - 500 MW
500 kV - 900 MW
765 kV - 2090 MW

Above values of SIL is true for both 50 Hz and 60 Hz systems.

Line Loadability

System planners usually use line loadabilty curve for deciding loading capability of the line. See Fig-D. The relationship between SIL and length in km shown in Fig-D is almost same for all voltage levels.

What is the capacity of the transmission line or how much power it can carry. The power that a transmission line can carry are based on three factors. These are

Thermal Limit
Voltage Drop Limit
Stability Limit

Due to the current flow heat is generated in the line and the line length changes which gives rise to more sag. Sometimes heating of the line is enough that, later cooling of the line due to less load or environment factors does not make the line regain its actual length. The sagging become permanent. Due to this the minimum clearance of the line to ground decreases which may violates the standard set by the local authority. Also if the load is very high the conductor may be damaged due to excessive heat. All transmission lines has thermal limits. But the thing is that only short lines can approach this limit. Voltage drop and stability limits situation usually do not arise here due to short length. Lines less than 80 km length falls in this category.

For medium length line the loading is mainly limited by allowable voltage drop (usually between 5 to 10 % as set in grid standard). For medium length line the steady state stability limit situation usually does not arises due to lesser length(discussed below). But the length is enough so that the medium length line can encounter the voltage drop limit before reaching thermal limit. By reactive compensation the voltage drop limit can be increased. Lines exceeding 80 km and less than 250 km long belong to this category.

For long line(above 250 km) we have shown that effort is made to operate the line with surge impedance loading. So for long lines the voltage profile may be made more or less flat with SIL loading. If the loading of line exceeds above SIL then the voltage at receiving end is less than sending end. If the loading of the line is less than SIL then the voltage at receiving end is more than sending end. This phenomenon is called Ferranti effect. For very lightly loaded or open long lines the voltage at receiving end may become very high. To avoid this situation Reactors are used at receiving end.

In both lossy and lossless lines it is clear that phase difference between sending and receiving end voltage arises. This angle is called power angle. The power that flows from sending to receiving end depends upon this angle and the magnitude of Vs and Vr. (The magnitude of a phasor V is represented as |V| )

The simplified formula for this power (P) transmitted is

$\large {\color{Cyan} P=\frac{ \left |V_s \right |.\left | V_r \right |}{X}sin\delta }$

X is the equivalent series reactance(ignoring resistance) of the line.

To increase the power to be transmitted, this power angle δ may increase up to 90 degrees. Increasing slightly further, the line becomes unstable and lose synchronism. It is a good practice to operate the lines with sending and receiving ends phase differnce angle (power angle) less than 30 degrees. Doing so, if in emergency load generation disbalance in adjacent areas occurse this line can take more load by increasing this power angle, so avoiding instability. So operating below 30 degrees we keep more than 60 degree angle margin. From the above formula you can say that if X is made smaller and smaller for any fixed small δ ( small sin δ imply small δ) then more power can be transmitted. Hence it is clear that small line reactance X is desired. This small reactance which cannot be made arbitrarily small by line design, definitely limit the power transmission in line. This loading limit for transmission line is quite less than thermal limit. So the long lines cannot approach thermal limit, before that other limits come to action. This small power angle corresponds to quite lesser load carrying capacity in comparison to thermal limit. Effort is made by power companies to push the limit towards thermal limit by employing reactive compensation so reducing effective series X further.

U14EET402-UNIT 2-11. SURGE IMPEDANCE LOADING

The surge impedance loading or SIL of a transmission line is the MW loading of a transmission line at which a natural reactive power balance occurs. The following brief article will explain the concept of SIL.

Transmission lines produce reactive power (Mvar) due to their natural capacitance. The amount of Mvar produced is dependent on the transmission line's capacitive reactance (XC) and the voltage (kV) at which the line is energized. In equation form the Mvar produced is:

Transmission lines also utilize reactive power to support their magnetic fields. The magnetic field strength is dependent on the magnitude of the current flow in the line and the line's natural inductive reactance (XL). It follows then that the amount of Mvar used by a transmission line is a function of the current flow and inductive reactance. In equation form the Mvar used by a transmission line is:

A transmission line's surge impedance loading or SIL is simply the MW loading (at a unity power factor) at which the line's Mvar usage is equal to the line's Mvar production. In equation form we can state that the SIL occurs when:

If we take the square root of both sides of the above equation and then substitute in the formulas for XL (=2pfL) and XC (=1/2pfC) we arrive at:

The term

in the above equation is by definition the "surge impedance. The theoretical significance of the surge impedance is that if a purely resistive load that is equal to the surge impedance were connected to the end of a transmission line with no resistance, a voltage surge introduced to the sending end of the line would be absorbed completely at the receiving end. The voltage at the receiving end would have the same magnitude as the sending end voltage and would have a phase angle that is lagging with respect to the sending end by an amount equal to the time required to travel across the line from sending to receiving end.

The concept of a surge impedance is more readily applied to telecommunication systems than to power systems. However, we can extend the concept to the power transferred across a transmission line. The surge impedance loading or SIL (in MW) is equal to the voltage squared (in kV) divided by the surge impedance (in ohms). In equation form:

Note in this formula that the SIL is dependent only on the kV the line is energized at and the line's surge impedance. The line length is not a factor in the SIL or surge impedance calculations. Therefore the SIL is not a measure of a transmission line's power transfer capability as it does not take into account the line's length nor does it consider the strength of the local power system.

The value of the SIL to a system operator is realizing that when a line is loaded above its SIL it acts like a shunt reactor - absorbing Mvar from the system - and when a line is loaded below its SIL it acts like a shunt capacitor - supplying Mvar to the system.

Figure 1 is a graphic illustration of the concept of SIL. This particular line has a SIL of 450 MW. Therefore is the line is loaded to 450 MW (with no Mvar) flow, the Mvar produced by the line will exactly balance the Mvar used by the line.

U14EET402- UNIT 2-11. POWER FLOW IN TRANSMISSION LINES

Consider the following model depicting the transfer of AC power between two buses across a line:

Figure 1. Simple AC power transmission model

Where $\boldsymbol{V_{s}} = V_{s} e^{-j\phi_{1}} \,$ is the voltage and phase angle at the sending end

$\boldsymbol{V_{r}} = V_{r} e^{-j\phi_{2}} \,$ is the voltage and phase angle at the receiving end

$\boldsymbol{Z} \,$ is the complex impedance of the line.

$\boldsymbol{I} = \frac{\boldsymbol{V_{s}} - \boldsymbol{V_{r}}}{\boldsymbol{Z}} \,$ is the current phasor

The complex AC power transmitted to the receiving end bus can be calculated as follows:

$\boldsymbol{S} = \boldsymbol{V_{r}}\boldsymbol{I}^{*}$

At this stage, the impedance is purposely undefined and in the following sections, two different line impedance models will be introduced to illustrate the following fundamental features of AC power transmission:

The power-angle relationship
PV curves and steady-state voltage stability

Power-Angle Relationship

In its simplest form, we neglect the line resistance and capacitance and represent the line as purely inductive, i.e. $\boldsymbol{Z} = j \omega L = jX \,$ . The power transfer across the line is therefore:

$\boldsymbol{S} = \boldsymbol{V_{r}} \left[ \frac{\boldsymbol{V_{s}} - \boldsymbol{V_{r}}}{jX} \right]^{*}$

$= \frac{V_{r} e^{-j\phi_{2}}(V_{s} e^{j\phi_{1}} - V_{r} e^{j\phi_{2}})}{-jX}$

$= j \frac{V_{s} V_{r}}{X} e^{-j (\phi_{2} - \phi_{1})} - j \frac{V_{r}^{2}}{X}$

$= \frac{V_{s} V_{r}}{X} \sin\delta + j \frac{V_{r}}{X} (V_{s} \cos\delta - V_{r})$

Where $\delta = \phi_{2} - \phi_{1} \,$ is called the power angle, which is the phase difference between the voltages on bus 1 and bus 2.

We can see that active and reactive power transfer can be characterised as follows:

$P = \frac{V_{s} V_{r}}{X} \sin\delta$

$Q = \frac{V_{r}}{X} (V_{s} \cos\delta - V_{r})$

Plotting the active power transfer for various values of $\delta \,$ , we get:

Figure 2. Active power transfer characteristic for a lossless line

The figure above is often used to articulate the Power-Angle Relationship. We can see that in this simple model, power will only flow when there is a phase difference between the voltages at the sending and receiving ends. Moreover, there is a theoretical limit to how much power can be transmitted through a line (shown here when the phase difference is 90^o). This limit will be a recurring theme in these line models, i.e. lines have natural capacity limits on how much power they can transmit.