Dr Audih al-faoury
Philadelphia University
Electrical department
2015-2016
Transient Stability
Dr audih
0
Power system stability refers to the ability of synchronous
machines to move from one steady-state operating point following
a disturbance to another steady-state operating , without losing
synchronism. There are three types of power stability:
1-Steady-state: involves slow or gradual changes in operating due to
variation on loads.
2-Transient: Involves major disturbances such as( loss of
generation due to faults and sudden load changes).This case
deviations from synchronous frequency (50 Hz) ,and changing
the machine power angles( our subject) .In many cases,
transient stability is determined during the first swing of
machine power angles following a disturbance( typically 1s).
3-Dynamic: After a fault, an individual synchronous machine may
remain in synchronism during the first swing . After this, strong
electromechanical oscillations occur may cause the system to have
natural oscillations. The system is said to be dynamically
stable .
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Classification of stability
Classification is based on the following considerations:
physical nature of the resulting instability
size of the disturbance considered
processes, and the time span involved
Power
system
stability
2
Usefully Formula
o
o
 p
1o mechanical    electrical
 2
P  No. of poles
 2
o
1 electical   mechancal
 P
P o
'
(mechanical)  2 n (mech.rad / sec) One revolution 360  elect.deg.
2
'
n  speed in revolution / sec.(rps) or
2 .n

60
n  speed in revolution / min .(rpm)
(electrical)  Angular frequancy
(electrical)  2 f (elect.rad / sec)
o
(electrical)  360 f (elect.deg / sec)
o
 2
1 electical   360o (revolution)
 P
o
o
 2 1
1 elec. / sec   o (rev/ sec)
 P 360
o
o
o
 2  1 1   2 60
o
1 elec. / min   o   
 P 360 60  P 360
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Rotation Mechanics-vs-Linear Mechanics
Rotational mechanical is very important in stability studies. The following table show us comparison between the linear and rotational mechanical formula.
Rotational mechanics
  Angular deplacement ( rad .)
d
dt
  Angular accelaration ( rad 2 / sec 2 )
L inear m echanics
S = D epalcem ent(distance)(m )
ds
dt
a  accelaration ( m / sec)
  Angular velocity ( rad / sec) 
V  V elocity ( m / sec) 
d  d 2


dt
dt 2
J  Moment of inertia ( kg .m 2 )
dv
d 2
mv 

dt
dt
m  M ass ( kg )
F  F orce  m .a ( N )
T = Torque (N.m) = J.
M  Angular momentum
2
( kg .m .rad / sec)  J 
1
K . E  J  2 (Joule)
2
P  power ( watt )  T 
W  work 
 Pe d    T
d
 T
dt
M  M om entum ( kg .m / sec) = m .v
1
K . E .  .m .v 2 ( Joule )
2
d
P=
( F . s )  F .v ( N . m / s )
dt
W ork = F .s
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Transient stability limit: depends on the type of disturbance,
location and magnitude of disturbance, clearing time and the
method of clearing.
After every disturbance, the machines must adjust the relative
angles of their rotors to meet the condition of the power transfer
involved. The problem is mechanical as well as electrical.
The kinetic energy(KE) of an electric machine is given by:
1
J . 2 ,  J
2
1
KE 
M  m ech
2
w h e re :
KE =

,
w h ere
J 
M

J - Is t h e M o m e n t o f In e r t i a i n ( k g . m 2 )
M - Is t h e a n g u l a r M o m e n t u m i n J , a n d d e p e n d s o n t h e s i z e o f t h e m a c h i n e a n d i t s t y p e .  r a d m ech 

 sec . 
 r a d e le c t 
r o to r s p e e d in 

 sec . 
 -Is t h e r o t o r a n g u la r v e lo c i t y i n 
 P 
 elect  
  m e c h a n ic a l
2


o r (2 f ), H z
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The Inertia constant H is defined as the Mega Joules of stored energy
of the machine at synchronous speed per MVA of the machine.
H 
KE ( energy stored in rotor in MJ )
G ( Machine power rating in MVA )
The relation between the Angular Momentum M and the Inertia
constant H can be derived as follows.
1
1
1
2 f
 2 
 2 
M  m ech  M    el ectl  M    2  f   M .
2
2
2
P
P
P
W here P is n um b er o f m ach in e p oles , if w e assu m e P  2 
K E  G .H 
, J
K E  M . f  M . . f

or
M 
KE
G .H

 f  f
,
( Jo u le
)
ra d e lect sec
If th e p ow er is ex pressed in p er un it, th en G = 1 . 0 pu . an d
H
M=
o
180  f
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SWING EQUATION : The differential
equation that relates the angular momentum
(M) the acceleration power (Pa) and the rotor
angle (δ) is known as swing equation.
Consider the generator shown in Fig.(a). It receives mechanical
power Pm at the shaft torque Tm and the angular speed ω via.
shaft from the prime-mover. It delivers electrical power Pe to the
power system network.
The generator develops electromechanical torque Te in
opposition to the shaft torque Ts. At steady state, Tm = Te
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accelerating torque acting on the rotor is given by Ta = Tm – Te
Multiplying by ω on both sides, we get power acceleration (Pa) : Pa = Pm – Pe
In case of motor Ta = Te – Tm and Pa = Pe – Pm
Pa  Ta .  J . .  M .
d  d 2
 
= 2  angular acceleration
dt
dt
d
d 2
Pa  M .
 M . 2 ,(   angular displacement of rotor ( rad .)
dt
dt
d 2
 Pa  M . 2
dt
, (  it ' s m ore convenient to measure the angular position
of rotor with respect to synchrou ously rotation frame
of reference  t  0 
(      t ) th en
t
d 2 d 2


2
2
dt
dt
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d 2
G.H d 2
Pa  Pm  Pe  M . 2 
. 2 Known as swing equation or in perunit
o
dt
180 . f dt
d 2
H d 2
Pa  M . 2 
. 2
o
dt
180 . f dt
In case damping power (D) is to be included, then equation is
modified as
2
Pa  Pm  Pe  M .
d
d

D
dt 2
dt
Swing curve, plot of power angle δ vs time t, can be obtained by
solving the swing equation. Fig. (a) stable system and (b) unstable
system.
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Swing curves are used to determine the stability of the system. If
the rotor angle δ reaches a maximum and then decreases, then it
shows that the system has transient stability. On the other hand if
the rotor angle δ increases indefinitely, then it shows that the
system is unstable.
We are going to study the stability of a generator connected to
infinite bus and a synchronous motor drawing power from
infinite bus. We know that the complex power is given by:
S=P + j Q = V. I* or P – j Q = V*.I
Thus real power P = Re {V*I}
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Consider a generator connected to infinite bus.
Where:
o
V - is the voltage at infinite bus. V = V  0
o
E

E


E - is internal voltage of generator. X - is the total reactance
The internal voltage E leads V by angle δ. Thus
V + j X I = E (for generator)
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o
o
E  V E   V 0
1
I


jX
jX
jX
  E .cos  j E .sin    V 
Electric output power Pe  ReV .I  
E .V
.sin   Pmax .sin 
X
Consider a synchronous motor drawing power from infinite bus
V – j X I = E (for motor)
o
E

E


Internal Voltage E lags the voltage V by angle δ. Thus
o
o
V  E V 0  E   
1
I


V   E .cos   j E .sin  
o
jX
X 90
jX


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As result of swing equation for generator is
d 2
Pa  Pm  Pe  Pm  Pmax .sin   M . 2
dt
d 2
M . 2  Pm  Pmax .sin 
dt
an d
d 2
M . 2  Pmax .sin   Pm
dt
swing equation for generator
for m o t o r
Notice that the swing equation is second order nonlinear
differential equation since the power acceleration have sine term.
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The graphical plot of power angle equation in Pe  Pmax .sin 
for generator and motor is shown Fig.
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EXAMPLE 1: Generator per-unit swing equation and power
angle during a short circuit.
A three-phase, 60-Hz, 500-MVA, 15-kV, 32-pole hydro-electric generating with H constant of 2.0 p.u. and D=0. (a) Determine ωe and ωm. (b) Give the per-unit swing equation for this unit. (c) The unit is initially operating at pm(pu)=pe(pu)=1 :ω=ωsyn,
and δ=10 when a three-phase-to-ground bolted short circuit at
the generator terminals causes pe(pu) to drop to zero for
t ≥ 0.Determine the power angle 3 cycles after the short circuit
commences. Assume pm(pu) remains constant at 1.0 per unit.
Also assume ωpu(t)=1 in the swing equation.
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S o lu tio n :
( a )  e  2 f  2  6 0  3 7 7 r a d / s
 P 
s in c e  e  
 . m th e n ;
 2 
 2 
 2 
 m    . e  
  3 7 7  2 3 .5 6 r a d / s
 P 
 32 
( b ) H = 2 p .u .
Pm ( pu )  Pe ( pu )
H
d 2
2
d 2

. 2 
. 2
o
o
180 . f dt
180  60 dt
(c ) Intial power angle is  (0)  10 o  0.1745 rad .
for t  0 
d  (0)
dt
0
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Using Pm(pu) =1 , Pe(pu) = 0
and (pu) (t)  1
H d 2
2
d 2 (t)
then the swing equation ( pu) is
. 2  o
. 2  Pm  Pe  1 , t  0
o
180 . f dt 180  60 dt
d 2 (t)
1

Integrating twice and using the above intial conditions,
2
2
dt


 o

 180  60 
we get from first integral for t :
And from second integral
d (t)  180o  60 

t  0 
dt
2 

 180o  60  t 2 
 (t)  
.    (0)
2  2

3 cycles
at t  3 cycle 
 0.05 second
60cycle / sec ond
deg. 
 180o  60 
2
o
 (0.05)  

0.05

0.1745

0.2923
rad

16.75

4


Dr Audih
rad.
180o
17
Example 2:
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA.
Find:
(a) The stored energy in the rotor at synchronous speed.
(b) If the mechanical input is suddenly raised to 80 MW for an
electrical load of 50 MW, find rotor acceleration, neglecting
mechanical and electrical losses.
(c) If the acceleration calculated in part (b) is maintained for 10
cycles, find the change in torque angle and rotor speed in
revolutions per minute(rpm) at the end of this period.
Solution:
(a) Stored energy = GH = 100 × 8 = 800 MJ
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d 2
b  P a  8 0  5 0  3 0 M W  M . 2
dt
G .H
100  8
4
M J .s
M 


,
(
)
1 8 0 o. f
180  50
45
e l e c .d e g r e e
d 2
4 d 2
 Pa  M .

.
 30 
dt2
45 dt2
d 2
30  45
e le c t d e g .


 3 3 7 .5 , (
)
2
dt
4
sec .
d 2
e le c t d e g .
A n d s in c e  



3
3
7
.
5
,
(
)
dt2
sec .
cycles 10

 0.2 sec.
freq. 50
And changing in   is :
1
   . .(t ) 2 where   is tourqe angule and  is acceleration
2
1
     337.5  (0.2) 2  6.75 elect .deg .
2
( c ) tsec. 
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rad
1
)
 rpm / s
sec 2 (rad/ rev.)
2
1
for electrical in min.  e  (rad / sec) 
 60  rpm / min
o
P
360 (rad / rev.)
2
1
60
for 4 pols  337.5  

60

337.5

 28.125rpm / min
o
o
4 360
2  360
  in rpm angular velocity is  Nm  m (
At the end of 10 cycle rotor speed is 

120 f
+ ns .ts 
P
120  50
 28.125  0.2  1505.6 rpm
4
**Note: We can also identify the mechanical speed of rotation in rpm according to
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3
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Further Considerations of the Swing Equation
The mega-voltampere (MVA) base Smach ,which is introduced by the definition of H, with many synchronous machines only one MVA base common to all parts of the system can be chosen. In this case:
 G mach 
H system  H mach . 

G
 system 
For a large system with many machines over a wide area it
is desirable to minimize the number of swing equations to
be solved. In such cases the machines within the plant can
be combined into a single equivalent machine , and only one
swing equation must be written for them.
Consider a power plant with two generators connected to the
same bus which is electrically remote from the network
disturbances.
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The swing equations on the common system base are
H1 d 21
. 2  Pm1  Pe1 ( pu )
o
180 . f dt
H 2 d 2 2
. 2  Pm 2  Pe 2 ( pu )
o
180 . f dt
 1   2   ( sincethe
rotors swing together ), then;
H equ . d 2
. 2  Pm  Pe ( pu )
o
180 . f dt
where : H equ . = H 1 + H 2 , Pm  Pm1  Pm 2 and
Pe  Pe1  Pe 2
Machines swing together called coherent machines
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Example: Tow synchronous machine the first one is 400MVA and H1=4.6MJ/MVA and
the second is1200MVA with H2=3MJ/MVA. The two machine operate in
parallel in power plant. Find Heq relative to a 100MVA base
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Equal area criterion
The accelerating power in swing equation will have sine term.
Pe  Pmax .sin 
Therefore the swing equation is non-linear differential equation .
For two machine system and one machine connected to infinite
bus bar, it is possible to say whether a system has transient
stability or not, without solving the swing equation. Using of
equal area in power angle diagram from sine wave which is
known as EQUAL AREA CRITERION.
A1  A2
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If the rotor angle δ oscillates, then the system is stable fig.(a). For
δ to oscillate, it should reach a maximum value and then should
decrease. At that point dδ/dt= 0. Because of damping present in
the system oscillations will be smaller and smaller. Thus while δ
changes, if at one instant of time, dδ/dt= 0, then the stability is
ensured.
d
0
dt
Stable system
Unstable system
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Let us find the condition which dδ/dt =0. The swing equation for 2
the generator connected to the infinite bus is: P  P  P  M . d 
a
m
e
dt 2
Multiplying both sides by dδ/dt , we get
d
d 2 d 
 M.
.
 m a th m a tic a lly
 Pm  Pe  .
2
dt
dt
dt
Pm
Pm
d
1
d
 P e .

M .
dt
2
dt
 Pe
M
2
.d 
2  Pm  Pe 
M
2  Pm  Pe 
M
d

dt
dt
d 
.

 dt 
d  d 

.

dt  dt 
d

d
2
d 
.
 
 dt 

2
2
 d 
.

 dt 
2
2
d  d   dt
d
d
 d 

.
.

.




dt
dt  dt  d 
 dt  d
2
2
 2  Pm  Pe   d 
 d 
 
.
  
M
 dt 

 d
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Before the disturbance occurs,δ0 was the torque angle (initial
power angle). At that time dδ/dt= 0.
When disturbance occurs, dδ/dt is no longer zero and δ starts
changing. Torque angle δ will change and the machine will again
be operating at synchronous speed after a disturbance, when
dδ/dt = 0 or

2(Pm - Pe )
 M .d   0 and
o

2(Pa )
 M .d   0
o
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The area of accelerating
power Pa versus δ must be
zero for some value of δ, the
positive (or accelerating) area
A1 under the graph must be
equal to the negative area A2
( decelerating) .
(a)corresponding to the initial steady-state operating point. At this point, Pm0 = Pe0 and δ =δo
(b) When a sudden fault accrue Pa becomes positive the rotor
accelerates and the power angle begins to increase δ1
At this point Pm = Pe and δ =δ1
(c) After the fault is cleared Pa is negative and δ reaches a
maximum value δ2, or point (c) and then swing back towards (b)
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At this point Pm = Pe and δ =δ2 which is ultimate steady-state stable operating point .In
accordance with eqn.



 P .sin.d    P  P .d    P  P
a
o
m
o
e
m
max
.sin .d  0
o
Note: We have assumed that the machine is connected to a large power system so that
| Vt | and xd does not change field current maintains | Eg | constant.
The equal-area criterion requires that, for stability from figure
before , A1=A2 or
1
2
  P
m
o
 Pmax .sin .d    Pmax .sin  Pm .d 
1
 Pm (1 o )  Pmax (cos1  coso )  Pm (1 2 )  Pmax (cos1  cos2 )
But during th fault Pe =0  Pm = Pmax sin1 then;
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 Pmax .sin1 ( 1 o )  Pmax (cos1  coso )  Pmax .sin1 ( 1 2 )  Pmax (cos1  cos2 ) 
sin1 (o )  coso  sin1 (2 )  cos2 ) or
( 2  o )sin1  cos( 2   0 )  0
This is stability equation for equals area
Example: A synchronous generator supply 500MW per phase
with power angle =8o.By how much can the input shaft power
(Pm) be increased suddenly without loss of stability. Assume
Pmax remain constant
Solution :
Initially at  0  8o ,the power
Pe (0)  Pmax sin  0  500sin 8o  69.6MW
Let δm is power angle to which the rotor can swing before losing
synchronism. If this angle is exceeded Pm will again become
greater than Pe, and the rotor will once again be accelerated and
synchronism will be lost as shown in Fig. , therefore, the equal
area criterion requires that equation to be satisfied with δm
replacing δ2.thus:
(2 o )sin1  cos(2 0 )  0 ,Where 2   1 replacing we get;
( 1 o )sin1  cos( 1)  cos0  0 Substituation o  8o  0.139 rad 
(  0.139 1)sin1  cos( 0.139)  cos1  0
(31)sin1  cos( 3)  cos1  0
(31)sin1  0.99  cos1  0
Solving eqn. iteratively, we get
1  50 .NowPefect  Pa
Pefect  Pmax.sin1 
 500sin(50o )  383.02MW
Dr Audih
34
The machine was 69.6 MW. Hence, without loss of stability, the system can accommodate a sudden increase of
Pef  Pe (0)  383.02  69.6  313.42 M W / phase or
 3  313.42  940.3M W of input shaft power
Fig. shows three different cases: case (a) is STABLE. Case (b)
indicates CRITICALLY STABLE while case (c) falls under
UNSTABLE. Note that the areas A1 and A2 are obtained by
finding the difference between INPUT and OUTPUT.(Pm=Pe)
Dr Audih
35
If a fault occurs in a system, δ begins to increase under the
influence of positive accelerating power, and the system will
become unstable.
if δ becomes very large. There is a critical angle within which the fault must be cleared. This angle is known as the critical clearing angle. Consider a system in figure operating with mechanical input Pm at
steady angle δ0 (Pm=Pe) at point 'a' on the power angle diagram
shown in the diagram in the next slide this is steady state.
Pm
If a three phase short circuit
occurs at the point F of the
outgoing radial line, the
terminal voltage goes to zero
and electrical power output
of the generator reduces to
zero(Pe = 0) and the state
point drops to 'b'.
The acceleration area A1 starts to increase while the state point
moves along bc. At time tc corresponding clearing angle δC, the
fault is cleared by the circuit breaker. tc is called clearing time
and δC is called clearing angle. After the fault is cleared, the
system becomes healthy and transmits power Pe = Pmax sinδ , the
state point shifts to "d" on the power angle curve. The rotor now
decelerates and the decelerating area A2 begins to increase while
the state point moves along de.
Dr Audih
37
For stability, the clearing angle, δC, must be such that area A1 =A2.
mathematically, we have
A1  A2 where;
1
A1  Pm ( c   o ) and
A2 
  P  P .d  
e
m
c
1

Pm ( c   o )    Pe  Pm .d  
c


 Pe  Pmax .sin 

1
Pm ( c   o ) 
  P
max
.sin   Pm .d 
c
Dr Audih
38
1
1
1
   Pmax .sin   Pm .d    Pmax .sin   Pm .d   Pm .d 
c
c
1
c
1
Pm ( c   o )    Pmax .sin   Pm .d   Pm .d
c
or
c
Pm c  Pm o  Pmax   cos 1  cos  c   Pm (1   c ) 
Pm c  Pm o  Pmax   cos 1  cos  c   Pm1  Pm c 
Pmax  cos  c  cos 1   Pm 1   o 
 At the intial Pinput = Poutput  Pm  Pe where :


 Pe  Pm = Pmax .sin o

Dr Audih
39
Pm ax  cos  c  cos  1   Pm ax .   1   o  sin o 
 cos  c  cos  1    1   o  sin o or
cos  c  cos  1   1   o  sin o 
 c  co s  1  cos  1   1   o  s i n o 
W here :
 c  is clearing angle
 o  is intial pow er angle
 1  is pow er angle to w hich the rotor
overshoots be y o nd  c .
Dr Audih
40
In order to determine the clearing time, we rewrite the swing equation, H
d 2
Pa  Pm  Pe 
. 2
o
180 . f dt
H d 2
 Pm 
. 2
 . f dt
or
( Pe  0, since we have a three phase fault )
 . f .Pm
d 2
H

P

m
.f
dt 2
H
d 2  . f .Pm
d   . f .Pm


integrating


.t
2
dt
H
dt
H
to find  the second integral   
 . f .Pm t 2
H
.
2
 o
If t  tc and    c 
c 
 . f .Pm
2H
2
c
.t   o
and
tc 
2 H  c   o 
 . f .Pm
Dr Audih
41
Note that δc can be solved also by the following equation
 c  cos 1  cos 1  1   o  sin o 
If the clearing time is larger than this value, the system would
be unstable.
The maximum allowable value for the system to remain stable
are known as critical clearing angle and critical clearing time,
from the figure
 m     o and note
 m  1 in equation above
Then;
Dr Audih
42
cr  cos  cos  o     o  o  sino  
1
 cos  cos  o     2o  sino  
1
 cos  coso    2o  sino  or
1
1
cr  cos
  2  sin
tcr  tc 
o
o
 coso 
and
2H c  o 
. f .Pm
Dr Audih
43
Example:50Hz,synchrounous generator capable to supply 400MW
connected to a large power system and delivering 80MW when
three phase fault occurs at its terminals, determine:
(a) The time in which the fault must be cleared ,if the maximum
power angle is to be 85o
(b) The critical clearing angle.
Solutio n :
o
85

o
When the fault occurs    1  85 
 1.4835 r ad .
o
180
we need to find the intial  o , thus ;
Pe  Pmax .sin  whe re Pmax  400 MW
and Pe  8 0 MW  Pm  steady sta te  
Dr Audih
44
Pe
80
   o  Pe  Pmax .sin  o  sin  o 

 0.2 
Pmax 400
11.54  
 o  sin (0.2)  11.54 
 0.201 rad . 
o
180
1
o
 o  11.5 4 o  0.201 r adian
Now to find  c , from equation
 c  cos  1  cos  1   1   o  sin o  
 cos  1  cos 1 .4835   1.4835  0.201 sin  0. 201  
 c os  1  0.09067   1.4835  0.201  0.198 67   
  c  cos 1 (0.345)  69.8 o  1.218 rad .
Dr Audih
45
The cleaning time t c is :
tc 
2 H  c   o 
 . f .Pm

2  7 1.218  0.201
  50  0.8
 0.336 sec .
(b ) For critical clearing angle  cr the equation is :
 cr  cos  1    2 o  sin o  cos  o  
 cos  1    2  0.2  sin (0.2)  cos(0.2)  
 cos  1 (  0.43)  115.46 o  2.01 rad .
The End
Dr Audih
46