(a) A DC motor has an armature resistance of 0·5 Ω and Kφ of 3 Vs. The motor is driven by a single-phase thyristorized full converter. The input to the converter is an AC source of 230 V, 50 Hz. The motor is used as a prime mover of a forklift. In th

Question

(a) A DC motor has an armature resistance of 0·5 Ω and Kφ of 3 Vs. The motor is driven by a single-phase thyristorized full converter. The input to the converter is an AC source of 230 V, 50 Hz. The motor is used as a prime mover of a forklift. In the upward direction, the mechanical load is 69 Nm and the triggering angle is α = 15°. In the downward direction, the load torque is 180 Nm. Calculate the triggering angle required to keep the downward speed equal in magnitude to upward speed. Assume continuous motor current for all operation. Also calculate the triggering angle to keep the motor at holding position while it was moving upward. 12 marks

(b) The primary side of an ideal transformer (having 400 turns in primary winding and 720 turns in secondary winding) is excited by a 1000 V, 50 Hz AC source. The secondary of the transformer is connected to a resistive load of 80 kW. There is one tapping in secondary winding at 480 turns and this tapping is supplying a pure inductive load of 100 kVA. Determine the primary current and its power factor. 12 marks

(c) (i) Obtain an expression for the total average power of a sinusoidal AM wave
v_c = V_c sin ω_c t
v_m = V_m sin ω_m t

(ii) An AM transmitter broadcasts a carrier power of 100 kW. Determine the radiated power at the amplitude modulation index of 0·8. 12 marks

(d) Given a unity feedback system with G(s) = K/s(s+a) as shown in the figure :

(i) Find the values of K and a, when the closed-loop system has K_v = 100 and admits 20% peak overshoot.

(ii) Find the values of K and a, when the closed-loop system has settling time (2% tolerance band) of 2 seconds and admits 10% peak overshoot. 12 marks

(e) Two relays R_1 and R_2 are connected in two sections of a feeder as shown in the following figure. CTs are of ratio 1000/5. The plug setting of relay R_1 is 100% and of R_2 is 125%. The operating time characteristics of the relay is given in the following table :

Operating time characteristics for TMS = 1
PSM | 2 | 4 | 5 | 8 | 10 | 20
Operating time (seconds) | 10 | 5 | 4 | 3 | 2·8 | 2·4

The time multiplier setting of the relay R_1 is 0·3. The time grading scheme has a discriminative margin of 0·5 s between the relays. A three-phase short circuit at F results in a fault current of 5000 A. Find the actual operating time of R_1 and R_2. What is the time multiplier setting (TMS) of R_2? 12 marks

UPSC Answer Check · Accepted Answer

(a)
For a single-phase fully controlled converter with continuous current, the average output voltage is
Vdc = (2Vm/π) cos α, where Vm = √2 × 230 = 325.27 V.
Thus Vmax = 2Vm/π = 2 × 325.27/π = 207.07 V.

Upward motion:
Vdc = 207.07 cos 15° = 207.07 × 0.96593 = 200.0 V.
Motor torque T = Kφ Ia, so Ia = 69/3 = 23 A.
Back emf: Eb = Vdc − IaRa = 200 − 23 × 0.5 = 188.5 V.
Speed: ωup = Eb/Kφ = 188.5/3 = 62.83 rad/s.

Downward motion at same speed magnitude:
The back emf reverses, so Eb,down = −188.5 V. The downward load torque assists motion; the motor must brake, so it develops +180 Nm torque.
Ia,down = 180/3 = 60 A.
Required converter voltage: Vdc,down = Eb,down + IaRa = −188.5 + 60 × 0.5 = −158.5 V.
Hence
cos αd = −158.5/207.07 = −0.7654.
So αd = arccos(−0.7654) = 139.95° ≈ 140°.

Holding position while moving upward:
At holding, speed = 0, so Eb = 0.
Load torque is 69 Nm, hence Ia = 23 A.
Vdc = IaRa = 23 × 0.5 = 11.5 V.
cos αh = 11.5/207.07 = 0.05554.
So αh = arccos(0.05554) = 86.82°.

Final answers: downward α = 139.95°, holding α = 86.82°.

(b)
Primary turns N1 = 400, full secondary N2 = 720, tapping N3 = 480.
V1 = 1000 V.
Full secondary voltage: V2 = 1000 × 720/400 = 1800 V.
Tapping voltage: V3 = 1000 × 480/400 = 1200 V.

Resistive load current:
I2 = 80,000/1800 = 44.44 A, in phase with V2.

Inductive load current:
I3 = 100,000/1200 = 83.33 A, lagging by 90°, so phasor I3 = −j83.33 A.

Reflecting secondary currents to primary:
I1 = (N2 I2 + N3 I3)/N1
= [720 × 44.44 + 480 × (−j83.33)]/400
= [32000 − j40000]/400
= 80 − j100 A.

Magnitude:
|I1| = √(80² + 100²) = √16400 = 128.06 A.
Power factor = cos φ = 80/128.06 = 0.6247 lagging.

Final answers: primary current = 128.06 A, power factor = 0.6247 lagging.

(c)(i)
AM wave:
v = (Vc + Vm sin ωm t) sin ωc t
= Vc(1 + m sin ωm t) sin ωc t, where m = Vm/Vc.

For resistive load R, instantaneous power p = v²/R.
Averaging over an RF cycle and over the modulating cycle:
average sin²ωc t = 1/2,
average sin ωm t = 0,
average sin²ωm t = 1/2.

Thus
Ptotal = Vc²/(2R) × (1 + m²/2)
= Pc(1 + m²/2), where Pc = Vc²/(2R).

Equivalently,
Ptotal = Vc²/(2R) + Vm²/(4R).
The total sideband power is m²Pc/2.

(c)(ii)
Pc = 100 kW, m = 0.8.
Ptotal = 100 × (1 + 0.8²/2)
= 100 × (1 + 0.32) = 132 kW.

Final answer: radiated power = 132 kW.

(d)(i)
For unity feedback, closed-loop characteristic equation:
1 + G(s) = 0
1 + K/[s(s+a)] = 0
s² + a s + K = 0.

Compare with s² + 2ζωn s + ωn² = 0:
a = 2ζωn, K = ωn².
Velocity error constant:
Kv = lim s→0 sG(s) = K/a.

Given Kv = 100,
K/a = 100 ⇒ K = 100a.

Peak overshoot 20%:
Mp = e^(−πζ/√(1−ζ²)) = 0.20.
Solving gives ζ ≈ 0.456.

Using K = 100a and a = 2ζωn, K = ωn²:
a = 400ζ², K = 40000ζ².
ζ² ≈ 0.2079.
Therefore a ≈ 83.16 s⁻¹, K ≈ 8316 s⁻².

Final answers: K ≈ 8.32 × 10³, a ≈ 83.16 s⁻¹.

(d)(ii)
For 2% settling time:
ts = 4/(ζωn) = 2 s ⇒ ζωn = 2.

Peak overshoot 10%:
Mp = e^(−πζ/√(1−ζ²)) = 0.10.
Solving gives ζ ≈ 0.591.

Then a = 2ζωn = 2 × 2 = 4 s⁻¹.
ωn = 2/ζ = 2/0.591 ≈ 3.383 rad/s.
K = ωn² ≈ 11.45 s⁻².

Final answers: K ≈ 11.45, a = 4 s⁻¹.

(e)
CT ratio = 1000/5, so secondary current for 5000 A fault:
Ict = 5000/(1000/5) = 5000/200 = 25 A.

Assume relay rated current = 5 A.

For R1:
Plug setting = 100% ⇒ pickup = 1.00 × 5 = 5 A.
PSM = 25/5 = 5.
From table, for PSM = 5, operating time at TMS = 1 is 4 s.
TMS1 = 0.3.
Actual time of R1:
t1 = 4 × 0.3 = 1.2 s.

For R2:
Plug setting = 125% ⇒ pickup = 1.25 × 5 = 6.25 A.
PSM = 25/6.25 = 4.
From table, for PSM = 4, operating time at TMS = 1 is 5 s.

Grading margin = 0.5 s. Since R2 is upstream backup,
t2 = t1 + 0.5 = 1.2 + 0.5 = 1.7 s.

Therefore TMS of R2:
TMS2 = t2/5 = 1.7/5 = 0.34.

Final answers: actual operating time of R1 = 1.2 s, actual operating time of R2 = 1.7 s, TMS of R2 = 0.34.

(a) A DC motor has an armature resistance of 0·5 Ω and Kφ of 3 Vs. The motor is driven by a single-phase thyristorized full…

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