ETO COC NUMERICAL QUESTIONS AND ANSWERS - SOLVED II

SOLVED  NUMERICAL QUESTIONS IN ASKED IN ETO COC WRITTEN EXAM - PART II

CHAPTER 3: AC ALTERNATOR

3.1) Question: Two alternators are working in parallel supply a lighting load of 3000 KW and motor load aggregating to 5000 KW at 0.72 pf. One machine loaded up to 5000 KW at 0.8 pf lagging. What is the load and power factor of the other machine? -2015

Ans:

Lighting load = 3000kW (pf unity)

Cos ø₁ = 1

ø₁ = cos ^-1 1

= 0°

tan ø₁ =tan 0° = 0

Motor load = 5000kW at 0.72 pf lag Cos ø₂ = 0.72

Ø₂ = cos ^-1 0.72

       = 43.94°

tan ø₂ = tan 43.94° = 0.964

KVAr load = 5000 x 0.964 = 4820 KVAr

Total KVAr = Motor Load + Lighting Load

                 = 4820 + 0 = 4820 KVAr

Total load = 8000kW

Load on Gen 1 = 5000kW at 0.8 pf (lag)

Cos øm1 = 0.8

Øm1 = cos ^-1 0.8

        = 36.87°

tan Øm1 = tan 36.87°

             = 0.75

KVAr load = 5000 x 0.75

               = 3750 KVAr

Load on Gen 2 = 8000 – 5000

                             = 3000kW

KVAr load on machine 2 = 4820 -3750 = 1070

Phase angle Øm2 = tan ^-1 KVAr/kW

= tan -1 1070/3000

= 19.63

Power Factor pf = cos Øm2

= cos 19.63

= 0.942 (lag)

Answer:

Load of Gen B = 3000kW

Power Factor Gen B= 0.942 (lag)

3.2) Question: The following loads are supplied by two alternators running in parallel

(i) 1400 KW @ 0.86 pf lagging (ii) 900 KW @ 0.8 pf lagging (iii) 800 KW @ unity pf (iv) 500 KW @ 0.8 pf leading. If the load on one machine is adjusted to 2100KW @ pf of 0.92 find the load and power factor of the other alternator? (SEP 2015)

Solution:

	LOAD-1	LOAD-2	LOAD-3	LOAD-4
Kw	1400	900	800	500
Cos Ø =Pf	0.86 lag	0.8 lag	1	0.8 leading
Ø = Cos-1 Pf	30.68	36.86	0	-36.86
tan Ø	0.59	0.75	0	-0.75
KVAR = Kw tan Ø	826	675	0	-375

Now total Kw = 1400+900+800+500

=3600Kw

Total KVAR = 826+675+0+ (-375)

=1126 KVAR

Load on generator A = 2100 Kw at 0.92 lag

Cos Ø = 0.92

Ø = Cos^-1 0.92

= 23.07

tan Ø = 0.425 lag

KVAR = Kw tan Ø

= 2100 x 0.425

KVAR of generator A = 892.5 KVAR

So, KVAR of generator B = Total KVAR – Gen A KVAR

=1126 – 892.5

233.5KVAR

Load on alternator B = Total Kw – Gen A Load

= 3600-2100

= 1500 Kw

tan Ø_B = KVAR/Kw

= 233.5/1500

= 0.156

Ø_{B =} tan^-1 0.156

=8.867

Power factor of Gen B = Cos Ø_B

= cos 8.867

0.968 lag

Answer:

Load on alternator B = 1500 Kw

Power factor of Gen B = 0.968 lag

3.3) Question: Two 3ɸ alternators operate in parallel. The rating of A is 1000kW and B is 800kW. The droop setting of each generator is 4%. If the load to be shared by both the generators is 1000kW, Calculate the load sharing by generators A and B, if the original frequency at no-load is 62Hz. (8 marks) (Nov 2018, Jan, Apr 2019)

Solution:

The frequency/load characteristics (assumed straight) for alternators A & B as shown in the figure. Out of the combined load AB = 1000kW, A share is BM, and B share is AM. Hence AM + BM = 1000kW. PQ is the horizontal line drawn through point C, which is the point of intersection.

Total Load shared by Alternator

A & B = 1000kW

Or AB = 1000kW

Let PD = QE = Y

And Load taken by

Alternator A = BM = PC = X kW

So, Load taken by Alternator B

= AM = QC = (1000-X) kW

∆ BAD ~ ∆ PCD

So, BA/PC = BD/PD

1000/X = 0.04/Y

X/Y =1000/0.04

= 25000

Y = X/25000---------------------- (I)

∆ AFE ~ ∆ QCE

So, AF/QC = AE/QE

800 / (1000 – X) = 0.04/Y

20000Y = 1000 – X-------------- (II)

Substituting (I) & (II)

20000 * X/25000 = 1000 – X

X = 5000/9 = 555.56 kW

So, Load shared by Alternator A is 555.56kW

And Load shared by Alternator B is = 1000 – 555.56 = 444.44Kw

Answer:

Load shared by Gen A = 555.56kW

Load shared by Gen B = 444.44Kw

Conclusion: When governors’ drooping characteristics are the same, generators share the active power load in proportion to their capacities.

3.4) Question: The alternator is rated for 750 KVA at 0.85PF. What is the maximum load (KW) that can be put on it? Would you be overloading the alternator if the kilowatt reading was now 620KW and the power factor 0.80? (8 Marks) (Dec 2018, Feb, Aug, Oct, Nov, Dec 2019, Jan 2020)

Solution:

Case 1

Given That Apparent Power = 750KVA

Power factor = 0.85

True Power = Apparent Power x Power Factor

= 750 x 0.85

= 637.5 KW

So, Maximum load that can be put on the alternator is 637.5 KW

Case 2

True Power = 620 KW Power factor = 0.8

S0, Apparent power = True Power/ Power Factor

= 620/0.8  

= 775KVA

Yes, Alternator will be overloaded in the second case.

Answer:

The maximum load that can be put on the alternator is 637.5 KW

Yes, Alternator will be overloaded in the second case.

CHAPTER 4: TRANSFORMER

4.1) Question: A 100 KVA, 2400/240 V, 50 Hz, 1-phase transformer has no-load current of 0.64A and a core loss of 700 W, when its high voltage side is energized at rated voltage and frequency.

Calculate

(i) The two components of no-load current.

(ii) If this transformer supplies a load current of 40 amp at 0.8 lagging power factor at its low voltage side, determine the primary current and its power factor. Ignore the leakage impedance drop. (JUL 2016, OCT 2017, SEP 2018, MAR, JUN, SEP 2019, and MAR 2020).

Given Data:

100KVA, 2400/240 V, 50Hz

No load current Io = 0.64A

Secondary load current I2 = 40 A @ 0.8 lagging Pf

Solution:

Core loss W_o = V₁ I_o cos Ø_o

Cos Ø_o = W_o / V₁ I_o

= 700 / (2400*0.64)

No load power factor cos Ø_o = 0.455

Ø_o = cos^-1 0.455

Ø_o = 62.935

Sin Ø_o = 0.89

(i) The two components of no-load current:

I_w = I_o cos Ø_o =0.64 * 0.455 = 0.2912 A

I_µ = I_o sin Ø_o =0.64 * 0.89 = 0.5696 A

(ii) On load:

I₂’ cos Ø₂ = 4 * 0.8 = 3.2 A

I₂’ sin Ø₂ = 4 * 0.6 = 2.4 A

I_x = I_o sin Ø_o + I₂’ sin Ø₂

= 0.5696 A + 2.4

= 2.9696

I_y = I_o cos Ø_o + I₂’ cos Ø₂

0.2912 + 3.2

3.4912

I1 = √ (Ix²+Iy²)

= √ (2.96962+ 3.49122)

= 4.583 A

Cos Ø₁ = I_y / I₁

= 3.4912 / 4.583

= 0.76

Answer:

₍i₎ The two components of no-load current:

I_w = 0.2912 A (Active Component)

I_{µ =} 0.5696 A (Magnetizing Component)

(ii) On load:

Primary Current = 4.583 A

Power factor = 0.76 (lag)

4.2) Question: In a container ship, a 3-φ, delta/ delta connected 6600/ 440V, 60Hz transformer is feeding AMP supply from shore to 440 Volt switchboards. The transformer primary current takes a line current of 100 amp, when secondary Load of 0.8 lagging pf is connected. Determine each coil current and KW output of the transformer. (8 Marks) (Jun, Jul 2018, Feb 2020)

Solution:

Phase voltage on primary = Line voltage on primary = 6600V

Phase voltage on secondary = Line voltage on secondary = 440V

So, K = VL of secondary / VL of primary = 440/6600 = 1/15

Line current on primary = 100 Amp

So, Phase current on primary = 100/√𝟑 Amp

Thus, Phase current on secondary = Phase current on primary / K

= (100/√3) / (1/15)

= 1500/√3  Amp

So, Line current on secondary IL = 1500 Amp

Power output = √3 VL IL cosφ

= √3 x 440x 1500x 0.8

= 914.50 kW

Answer:

Line current on secondary IL = 1500 Amp

Power output = 914.50 kW

4.3) Question: A 440/110V single phase transformer supplies a load of 5KW at 0.8 power factor load. Calculate the primary and secondary currents. (Ignoring transformer power losses)? (6 marks) (AUG, SEP, OCT 2018, MAR 2019)

Given Data:

Primary Voltage Vp = 440V

Secondary Voltage Vs = 110V

Power P = 5000W

Pf = 0.8

Solution:

Power P = Vs Is cos ɸ

5000 = 110 x 0.8 x Is

Is = 56.81 Amp

Secondary Current = 56.81 Amp

K = Vs/Vp = Ip/Is

= 110/440 = Ip/56.81

Ip = 14.20 Amp

Primary Current = 14.20 Amp

Answer:

Primary Current = 14.20 Amp

Secondary Current = 56.81 Amp

4.4) Question: A Single-phase power transformer supplied a load of 20 KVA at a P.F. of 0.81 (lagging). The iron loss of the transformer is 200W and the copper loss at this load is 180W. Calculate (a) the efficiency (b) if the load is now changed to 30 KVA at a p.f. of 0.91 (lagging), calculate the new efficiency. (8 marks) (Nov 2018, Jan, Apr 2019)

Solution:

(a) Output (kW)

= 20KVA x 0.81 = 16.2 kW

Fe loss (PFe) + Cu Loss (PCu) = 200 + 180 = 0.380kW

Efficiency = Output (kW) / [Output (kW) + Losses (kW)

Or     

ƞ = kVA Cos Ø / [kVA Cos Ø + PFe + PCu]

= 16.2 / [(16.2) + (0.38)]

= 16.2/16.58

= 0.977

Full load Efficiency = 97.7 %

(b) Since the kVA rating is now 30kVA and it can be assumed that the voltage remains constant, therefore iron loss remains constant.

The new current is 3/2 or 1.5 times the original current.

Cu Loss is proportional to Current Square,

Thus new copper loss = New Cu Loss = (3/2)²

= 9/4 x 180

= 405W = 0.405kW

ƞ = kVA Cos Ø / [kVA Cos Ø + PFe + PCu]

= 30 x 0.91 / [(30 x 0.91) + 0.2 + 0.405]

= 27.3 / (27.3 + 0.605)

= 27.3 / 27.905

= 0.9783

 New Efficiency = 97.83 %

Answers:

Full load Efficiency = 97.7 %

New Efficiency = 97.83 %

4.5) Question: In a 25kW, 3300/233V, 1-phase transformer the iron and full load copper losses are 350 watts and 400 watts respectively, calculate the efficiency at half load, 0.8 p.f. (6 Marks) (Aug, Oct 2019)

Solution:

Let Full Load Output = 25kW at 0.8 p.f.

Cu loss at Half Load = 400 x (1/2)2 = 100W

Iron loss will remain constant = 350W

Total Loss = 100 + 350 = 450W = 0.45kW

Half Load output at 0.8 p.f. = 12.5kW

So, Efficiency % = [Output / Output + Total Loss] x 100

Ƞ = [12.5 / (12.5 + 0.45)] x 100

= [12.5 / 12.95] x 100 = 96.5%

Answer:

The efficiency at half load = 96.5%

CHAPTER 5: POWER SYSTEMS

5.1) Question: (i) which has the greater equivalent resistance; two equal capacitors in series or in parallel? Explain with reasons.

(ii) A circuit has a resistance of 3Ω and an inductance of 0.01H. The voltage across its ends is 60V and the frequency is 50Hz. Calculate

a) The impedance, b) The power factor, c) The power absorbed

(16 Marks) (AUG, SEP, OCT, Nov, Dec 2018, Feb, Mar, Jun, Sep, Nov 2019, Jan, Mar 2020)

(i) Capacitance in series:

When the capacitor is connected in series; it increases the distance of plates, so the capacitance decreases. Suppose two equal capacitors of C Henry connected in series,

Then, 1/CT = 1/C1 + 1/C2 or CT = C/2

Where CT is the total capacitance

So, the capacitive reactance Xc = 1/2πfCT = 2/2πfC = 1/πfC ohm

Capacitance in parallel:

When the capacitor is connected in parallel; it increases the cross-section of plates, so the capacitance increases. Suppose two equal capacitors of C Henry connected in parallel,

Then, CT = C1 + C2 = 2C, where CT is total capacitance

So, the Capacitive reactance Xc = 1/2πfCT = 1/4πfC ohm

Therefore, series-connected capacitors are having 4 times more equivalent resistance (capacitive reactance) than parallel connected capacitors.

(ii) Data Given:

Voltage V = 60V

Resistance R = 3Ω

Inductance L = 0.01 H

Frequency f = 50 Hz

Solution:

a) The impedance 

XL = 2πfL

= 2π x 50 x 0.01

= 3.14 ohm

Impedance Z = √32 + 3.142

= 4.34 Ω ∠ 46.4°

(tan ɸ = XL/R = 3.14/3 = 1.05 ; ɸ = tan-1 1.05 = 46.4°)

b) The power factor

Cos ∅ = R/Z = 3/4.34 = 0.691(lag)

c) The power absorbed

Current I = V/Z = 60/4.34 = 13.82 A

P = VI Cos ∅

= 60 x 13.82 x 0.691

= 573 Watt

Answer:

(i) Therefore, series connected capacitors is having 4 times more equivalent resistance (capacitive reactance) than parallel connected capacitors.

(ii) a) The impedance = 4.34 Ω

b) The power factor = 0.691(lag)

c) The power absorbed = 573 Watt

5.2) Question: A 440V, 10KW, 0.8 p.f, 3 phase load is supplied as shown. Calculate short circuit fault current at the load and at the main switch board. (8 Marks) (Mar, Apr 2018)

Solution:

Short circuit fault location is load terminal

So, the total impedance = Zf = 0.025 + 0.01 + 0.015 = 0.05 Ω

Short circuit fault current = If = V/Zf

= 440/0.05

=8800A

So, prospective fault current level at the load is 8800 A

If, short circuit at the main switch board

The fault current = If = V/Zf

= 440/0.025

=17600A

So, prospective fault current level at the load is 17600 A

Answer:

Prospective fault current level at the load is 8800 A

Prospective fault current level at the load is 17600 A

5.3) Question: What would be the ohmic value of the NER to limit the earth fault to the full load rating of a 2 MW, 0.8pf, 3.3KV, 3 phase AC generator? (4 marks) (AUG, SEP, OCT 2018, JAN, Apr 2019)

Solution:

VL = 3.3KV, P = 2 MW, cos ɸ = 0.8

P = √3 VL IL cos ɸ

So, the generator full load current is

IL = 2*1000 / √3 * 3.3 * 0.8

= 437A

Under E/F condition a phase voltage of the generator winding will be

V_PH = 3300/√3 = 1905 V drives the fault current through the NER. So, its ohmic value has to be 1905V/437A = 4.4 ohm.

Answer:

The ohmic value of the NER has to be = 4.4 ohm

5.4) Question: A permanent magnet moving coil instrument has a coil of dimension 15mm x 12mm. the flux density in the air gap is 1.8 x 10 -3 wb/m2 and the spring constant is 0.4 x 10 -6 Nm/rad. Determine the number of turns required to produce an angular deflection of 90° when a current of 5 mA is flowing through the coil. (8 marks) (Oct 2018)

Solution:

Area A = 15 x 12 mm2

= 180 x 10^-6 m2

Flux Density B = 1.8 x 10^-3 wb/m2

Spring Constant K = 0.4 x 10-6 Nm/rad

Angular Deflection θ = 90° = 1.5708 radian

Current I = 5mA = 5 x 10^-3 Amp

Deflection Torque Td = NBA I

= N x 1.8 x 10^-3 x 180 x 10^-6 x 5 x 10^-3 N-m

= N x 1.62 x 10^-9

Controlling Torque Tc = K θ

= 0.4 x 10-6 x 1.5708

= 0.62832 x 10^-6

= 628.32 x 10^-9 N-m for the final steady state position,

Td = Tc

N x 1.62 x 10^-9 = 628.32 x 10^-9

N = (628.32 x 10^-9) / (1.62 x 10^-9)

So, N = 387.85 ≅ 388 turn

Answer: the number turns N ≅ 𝟑𝟖𝟖 turn

5.5) Question: A coil having a resistance of 10 ohm and an inductance of 0.15 H is connected in series with a capacitor across a 100V, 50Hz supply. If the current and the voltage are in phase what will be the value of the current in the circuit and the voltage drop across the coil? (10 Marks) (Jul 2019)

Data Given:

Input Voltage V = 100V Frequency f = 50Hz

Coil Resistance R = 10 Ω Coil Inductance L = 0.15H

Solution:

XL = 2πfL

= (2x22x50x0.15)/7

= 47.14 Ω

Coil Impedance Zc = √R2 + XL2

= √102 + 47.142

= 48.19 Ω

Therefore Current and Voltage Phase is same

So, Impedance of Total Circuit Z = R = 10 Ω (∵ Z =10 + j 47.14 – j 47.14)

Supply Current I = V/Z

= 100/10 = 10 A

Voltage across the coil = I x Zc

= 10 x 48.19

= 481.9

Answer:

Supply Current I = 10 A

Voltage across the coil = 481.9 V

CHAPTER 6: ELECTRONICS

6.1) Question: Diode half-wave rectifier supply a resistive load of 100Ω from a 100Vac R.M.S. voltage source. The diode is a resistance of 5Ω during conduction state. Calculate (i) The DC output voltage (ii) DC average load current. (8 Marks) (Jan, Apr, Jul 2019)

Given Data:

Vs = 100VAC (rms)

RD = 5Ω &

RL = 100Ω

Solution:

So, Vsm = √2 x Vs = 141.4V

Assume that PN Diode is Silicon, so, VB = 0.7V

Let Maximum Load Current = ILM

 So, ILM = Vsm –VB / RD + RL

= 141.4 – 0.7 / 5 + 100

= 1.34 Amp

Now, VLM = ILM x RL = 134V peak

(i) Let DC output voltage = VLdc

So, VLdc = VLM / π

= 0.318 x VLM

= 0.318 x 134

= 42.6V average

(ii) Let DC average Load Current = ILdc

So, ILdc = ILM / π

= 0.318 x ILM

= 0.318 x 1.34

= 0.426 Amp average

Alternate Method

ILdc = VLdc / RL

= 42.6/100

= 0.426 Amp average

Answer:

(i) The DC output voltage = 42.6 V average

(ii) DC average load current = 0.426 A average

Solved by The senior Electro Technical Officer Mr. Mobin ETO

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