A'C' Circuit Analysis

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A.C. Circuit Analysis

• Bigger generator produce a.c.
• Hence needless to change from d.c. to a.c.
• Equivalent a.c. as compared to d.c. was sought.
• While the average of an a.c. signal is zero
  (verify), it is practically true that an a.c.
  current causes heat dissipation in a resistor.

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Average Power
a.c. circuit
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Average Power

• Instantaneous power
• Average power

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Average Power

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Average Power
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Average Power
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d.c. equivalent circuit power
d.c. circuit
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d.c. equivalent power

• Average power

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Root-Mean-Square value
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Root-Mean-Square value

• r,m,s. value of
• Or

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Root-Mean-Square value
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Root-Mean-Square value
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r.m.s value of a.c. signal

• r.m.s value
• Example
• Domestic supply
• 240V, 50Hz

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r.m.s value of a.c. signal
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Power Factor

• The load is NOT purely resistive

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Power Factor
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Power Factor

• Most industrial loads are inductive

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Power Factor
i

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Power Factor

• Let

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Power Factor

• Exercise
• Clearly, i is a complex number.
• 1. On the same Argand diagram sketch V i.
• 2. Consider the phase angle between V i
  analyse its value as L and/or w change.

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Power factor

• Results
• 1. i lags V (or V leads i) by an angle called a
  phase angle
• 2. The phase angle is a function of the product
  wL.
• 3. Cosine of the angle is called the power factor

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Power factor

• The power factor is a tool used in power
  analysis Its value indicated how inductive a
  circuit is.
• Exercise
• On the same phasor diagram draw the real power,
  apparent power reactive power.

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Real, Apparent Reactive Power
Im
V
P1
Re
?
?
i
P Apparent power
P1 Real power
P2
P2 Reactive power
P
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Real Apparent Power

• It is clear that the Apparent Power is bigger
  than the real reactive power.
• In reality the reactive power is stored
  released from the L every half cycle. It has no
  practical usage,
• The Real Power is the one doing the intended work
  we wish it to be as high as possible.

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Power Factor Correction

• In order to decrease the Reactive Power the
  phase angle is decreased.
• This is easily checked from the drawing below. It
  is clear that Real Power is not changed but the
  Apparent Power is also reduced in the exercise.
• The decrease of the phase angle is called Power
  Factor Correction

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Power Factor Correction

• If we add a current I such that it is purely
  imaginary. Hence I leads the ref phasor (V) by 90
  deg. The vector sum will be
• Interestingly note that the magnitude of the new
  current is smaller than the old. This provides
  useful practical results.

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Power Factor Correction
Im
I
II
P1
Re
V
O
i
P22
Pn
P Apparent power
P1 Real power
P2 Reactive power
P2
P
P22 New Reactive Power
Pn New Apparent Power
I Current to reduce Reactive Power
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Power Factor Correction

• It is of engineering interest to correct the
  power factor the reasons are
• 1. The useless Reactive Power is reduced
• 2. The Apparent Power is reduced hence more
  power is available from source (for other users).
• 3. The Real Power is NOT changed load remains
  unaffected.

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Power Factor Correction

• Let us have a current lagging the voltage by an
  angle ? take the voltage as the reference
  phasor.
• i(t) I cos(wt?)
• The r.m.s. value can be obtained from the
  previous method of comparison with the power
  dissipated in R.

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Power Factor Correction

• Exercise
• Let the current in an inductive circuit be given
  by
• i(t) I cos(wt ?).
• Consider the power dissipated in R and equate it
  to an equivalent amount of power dissipated in R
  of a purely resistive load R. Find the r.m.s
  value.

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Power Dissipated in R

• Exercise Find the r.m.s. value of
• i(t) Icos(wt ?)
• Solution
• T 2p/w
• The instantaneous power dissipated in R is

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Power Dissipated in R
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Power Dissipated in R
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Power Dissipated in R
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Power Dissipated in R
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Power Dissipated in R

• The above finding shows that the r.m.s. value is
  not dependent on the angle (lag or lead).
• Hence we shall use the r.m.s. values without any
  further calculations.

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Average Power in r.m.s values

• We can therefore consider the average power
  absorbed as
• P ½VI cos?
• Where ? is the phase angle between v(t) i(t)
  and V I are the peak values of the sinusoidal
  signals. i.e.
• v(t) Vcos(wt)
• i(t) Icos(wt-?)

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Average Power in r.m.s values

• Clearly, the r.m.s. values of v(t) i(t) are
  (V/v2) (I/v2).
• Since i(t) lags v(t) the average power absorbed
  is
• P (V/v2) (I/v2)cos?
• ½VIcos?

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Average Power in r.m.s values

• But alternatively, and better still, if we
  considered the r.m.s. values
• P VIcos?
• Where V I are the r.m.s. values.
• Since the r.m.s. values are the ones normally
  stated (NOT the peak values), we shall use the
  r.m.s values only.

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Practical Example pf correction

• A 2kW motor with 0.8 power factor (lag) is
  connected to 415V, 3-phase, 50Hz power supply.
  Determine the current drawn by the motor. It is
  required to improve the pf to 0.9 (lag) by
  connecting a capacitor in parallel with the
  motor. Determine the size of capacitor, the new
  current drawn and the reduced volt-amperes.

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Practical Example pf correction

• Solution

I
415V
2 kW
50Hz
0.8 p.f
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Practical Example pf correction

• The solution requires the connection of a
  capacitor as shown below
• Let the currents be as shown
• I new current from supply
• I original current from supply (which still
  goes to the motor)
• i current drawn by the capacitor

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Practical Example pf correction

• Solution

I
I
i
415V
2 kW
C
50Hz
0.8 p.f
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Practical Example pf correction

• Let V supply voltage ref phasor.
• P VI cos? 2000
• cos? 0.8
• V 415 0
• w 2.p.50 100p

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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
V
C
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction

• Phasor diagram

V
I
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction

• N.B. The reduced Volt-amperes can be used to
  supply another load.
• Note The capacitor is used to reduce the
  current and yet the power ( and current) to the
  load remains unaltered.

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Practical Example 2 (Supply line losses)

• A large consumer of 30 kW is supplied power at
  o.5 p.f, and 240 V, 50Hz. The supply wires have
  resistance of 0.001 per km. If the source is
  located 200km away, determine (i) the source
  voltage
• (ii) The line losses.
• (b) In order to improve the system the p.f is to
  be increased to 0.85 by a capacitor bank
  connected in parallel with the load.

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Practical Example pf correction

• Determine the improvement i.e.
• reduction is VA
• reduction in line losses.

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Practical Example pf correction

• Let ref phosor be voltage at the load.

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Practical Example pf correction
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Practical Example pf correction

• (iii) The line losses

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Practical Example pf correction
R
30kW
C
0.5 pf.
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Practical Example pf correction

• (b)

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Practical Example pf correction
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Practical Example pf correction
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Practical Example pf correction
V
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Practical Example pf correction
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Practical Example pf correction

• Reduction in line losses

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Practical Example pf correction

• It is clear that the improvement in p.f. has two
  advantages
• (i) Reduces the supplied VA so that the balance
  is used to supply other consumers.

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Practical Example pf correction

• (iii) Reduces the line losses which
  is just a wastage to supplier.
• Hence most power supply company give a penality
  charge called Maximum Demand (MD) charge so that
  the consumer is persuaded to improve the p.f.

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Practical Example pf correction

• The present load shedding in Uganda could
  probably be reduced if industries improved their
  p.fs. However, the biggest load on supply
  network is domestic which is resistive.