**TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS OF**

**SINGLE-MACHINE INFINITE BUS SYSTEM**

**AIM :**

To become familiar with various aspects of the transient and small signal stability analysis of Single-Machine Infinite Bus (SMIB) system.

**OBJECTIVES**

__:__ (i) To understand modeling and analysis of transient and small signal stability of a SMIB power system.

(ii) To examine the transient stability of a SMIB and determine the critical clearing time of the system through simulation by trial and error.

(iii) To determine transient stability margin (MW) for different fault conditions.

(iv) To obtain linearised swing equation and to determine the roots of characteristic equation , damped frequency of oscillation and undamped natural frequency.

**EXERCISE :**

(i) A power system comprising a thermal generating plant with four 555 MVA,

24kV, 60HZ units supplies power to an infinite bus through a transformer and two transmission lines. The data for the system in p.u on a base of 2220 MVA, 24 kV is given below. An equivalent generator representing the 4 units, characterized by classical model:

Xd’ = 0.3 p.u H= 3.5 MW-s/MVA Transformer : X = 0.15 p.u

Line 1 : X = 0.5 p.u Line 2 : X = 0.93 p.u

Plant operating condition: P = 0.9 p.u ; pf= 0.9(lag) ; E

_{t}= 1.0 p.u It is proposed to examine the transient stability of the system for a three-phase-to ground fault at the end of line 2 near H.T bus occurring at time t= 0 sec. The fault is cleared at 0.07 sec. by simultaneous opening of the two circuit breakers at both the ends of line 2.(case1)

(a) Calculate the initial conditions necessary for the classical model of the

machine for the above pre-fault operating condition, determine the

critical clearing angle and time for the fault using “Equal Area Criterion”

and hence

**comment on the stability of the system for this fault.**

*SOLUTION:**Computation of stator current*

*I*

_{t}= S^{* }/ E * T = (0.9-j0.436) / 1.0 = 0.9- j0.436# Computation of the terminal voltage

*E*

^{’}= Et + jXd^{’}It = 1.1305 + j0.27*Computation of infinite bus voltage*

*E*

_{B }= Et –Jx4(i4+Ji)*X4 = Xtr + X3 = 0.47 p.u.*

*Ir = P / Et = 0.9 p.u.*

*I = -Q / E = - 0.435p.u.*

*Eb = 0.7933-j0.4275*

*Computation of angle of separation between E*^{’}and Eb*δ = ∟E1*

^{’}- ∟Eb = 0.7282 rad*If the infinite bus is taken as reference then*

*E*

^{’}= 1.16 ∟41074*It = 0.99∟2.52 Et =1∟28.31*

*Critical clearing angle**:*

*Cosδ = {P*

_{M}(δ_{MAX}– δ_{0}) + P3_{MAX}cosδ_{MAX}– P2_{MAX}cosδ_{0}} / { P3_{MAX}– P2_{MAX}}*P3*

_{MAX}= Eb E^{’}/ (Xd^{’}+ Xtr + Xline-1) = 1.098*P2*

_{MAX}= 0

*To find δ*_{MAX}:*After the fault clearance*

*δ*

_{MAX }= 180 – sin^{-1 }(Pe / P3_{max}) = 124.94 degree = 2.18 rad##### Substituting in the above formula

*Cos δc*

_{ }= (0.9(2.18-0.728) + 1.09 * cos124.94)/1.09*δc = 51.88 deg = 0.905 rad*

*To find critical clearing time**:*

*T*

_{c}= *Ö(2H(*

*δ*

_{c}– δ_{0})) /*P*

*f*

_{0}P_{M}= 0.085s**b) Simulate the above sequence of fault occurrence and clearance using the**

Software available and plot the swing curve (rotor angle versus time) as

well as the curves showing angular velocity and real power delivered by

the plant versus time

(c) Determine the critical clearing angle and time for the above fault through

trial and error method by repeating the simulation in (b) for different fault

clearing times and compare the critical clearing angle and time obtained.(case2)

Case 3:

Three-phase-to-ground fault at the mid point of line 2 occurs at t=0 sec and is

cleared at t=0.07 sec by the simultaneous opening of two breakers in line 2.

Comment on the transient stability of the system under case 2 and case 3 and

compare the severity of the faults; cases 1,2 and 3 from the point of view of

maximum rotor swing and also by comparing the clearing time margin available.

3. Determine the steady-state stability margin (MW) available for the system

under the given operating condition in exercise 6.5.1. Also determine the

transient stability margin (MW) available for the operating condition given in

exercise 6.5.1. for the three cases of fault, case 1, case 2 and case 3. Can the

severity of the fault be measured using this margin?

Steady state stability margin = (P

_{max}– P_{0})* base .MW.Pmax = EV / X = E

^{’}Eb / X = 1.16 * 0.9 / (0.328 + 0.3 + 0.15 )

= 1.347 M.W.

Substituting the value in the above formula

Stability margin = (1.347 – 0.9) * 2220

= 992.34 M.W.

Transient stability margin = (P

_{max }(stable) – P_{0}) M.W.4. Is proposed to examine the small-signal stability characteristics of the

system given in exercise 1. about the steady-state operating condition

following the loss of line 2; Assume the damping coefficient KD = 1.5 p.u

torque / p.u speed deviation.

(a) Write the linearized swing equation of the system. Obtain the

characteristic equation, its roots, damped frequency of oscillation in Hz,

damping ratio and undamped natural frequency. Obtain also the force-free

time response Δδ (t) for an aintial condition perturbation Δδ = 5 degree and Δω(t) = 0.

5. Repeat the small-signal stability analysis carried out using the software

package in exercise 4 with the following parameters and comment on the

relative stability of each case:

(a) KD = 0 p.u and –1.5 p.u

(b) KD = 1.5 p.u but with P = 1.2, 1.5 and 2.0 p.u

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