141. An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls below 48 Hz. If an additional load of 3.5 MW is connected, then frequency will settle down to

1. 1. 49.417 Hz

   
   
   

   2. 49.917 Hz

   
   
   

   3. 50.083 Hz

   
   
   

   4. 50.583 Hz

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   From the diagram we have,
   

   ⇒	0.05ƒ	=	15
   	ƒ - 50		P1

   
   

   ⇒	0.05ƒ	=	15
   	ƒ - 50		12

   
   ⇒ ƒ = 52.083
   Now, the new load is, P' = 11.5 MW. Then,
   

   ⇒	52.083 - ƒ'	=	11.5
   	25.083 - 50		12

   
   

   ⇒ 52.083 - ƒ' = 2.08 ×	11.05
   	12

   
   ⇒ ƒ' = 50.083 Hz

   Correct Option: C

   
   From the diagram we have,
   

   ⇒	0.05ƒ	=	15
   	ƒ - 50		P1

   
   

   ⇒	0.05ƒ	=	15
   	ƒ - 50		12

   
   ⇒ ƒ = 52.083
   Now, the new load is, P' = 11.5 MW. Then,
   

   ⇒	52.083 - ƒ'	=	11.5
   	25.083 - 50		12

   
   

   ⇒ 52.083 - ƒ' = 2.08 ×	11.05
   	12

   
   ⇒ ƒ' = 50.083 Hz

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142. A 230 V (phase), 50 Hz, three - phase, 4 – wire system has a phase sequence ABC. A unity power - factor load of 4 kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and a pure capacitor in the other two phase. The value of inductor and capacitor is

1. 1. 72.95 mH in phase C and 139.02 μF in phase B

   
   
   

   2. 72.95 mH in phase B and 139.02 μF in phase C

   
   
   

   3. 42.12 mH in phase C and 240.79 μF in phase B

   
   
   

   4. 42.12 mH in phase B and 240.79 μF in phase C

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Inductor is connected in phase-B and capacitor is connected in phase-C
   I_B lagging V_BN by 90°.
   I_C leading V_CN by 90°.
   V_AN = 230° ∠0° [V_AN is reference phasor)
   V_BN = 230° ∠– 120
   V_CN = 230° ∠– 240
   Given, I_A is drawn at unity pf.
   

   IA =	P	=	4 × 10³	= 17.4∠0°
   	VANcosφ		230

   
   The current into the central will be zero
   I_B = I_C
   as I_A = I_Bcos60 + I_C cos60
   = 2I_B sin 60_B = √3I_B
   

   ⇒ IB = IC =	IA	=	17.4	amp = 10.05 A
   	√3		√3

   
   

   as |IB| = |	VBN	| =	230
   	ωL		(314)L

   
   

   ⇒ L =	230	= 72.86 mH
   	314 × 10.05

   
   and |L_C| = |C_CN|. ωC = (230) × 314 × C
   

   ⇒ C =	10.05	= 139.08 uf.
   	230 × 314

   Correct Option: B

   
   Inductor is connected in phase-B and capacitor is connected in phase-C
   I_B lagging V_BN by 90°.
   I_C leading V_CN by 90°.
   V_AN = 230° ∠0° [V_AN is reference phasor)
   V_BN = 230° ∠– 120
   V_CN = 230° ∠– 240
   Given, I_A is drawn at unity pf.
   

   IA =	P	=	4 × 10³	= 17.4∠0°
   	VANcosφ		230

   
   The current into the central will be zero
   I_B = I_C
   as I_A = I_Bcos60 + I_C cos60
   = 2I_B sin 60_B = √3I_B
   

   ⇒ IB = IC =	IA	=	17.4	amp = 10.05 A
   	√3		√3

   
   

   as |IB| = |	VBN	| =	230
   	ωL		(314)L

   
   

   ⇒ L =	230	= 72.86 mH
   	314 × 10.05

   
   and |L_C| = |C_CN|. ωC = (230) × 314 × C
   

   ⇒ C =	10.05	= 139.08 uf.
   	230 × 314

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143. Consider the two power systems shown in Fig (a) below, which are initi ally not interconnected, and are operating in steady state at t he same frequency. Separate loadflow solutions are computed individually for the two systems, corresponding to this scenario. The bus voltage phasors so obtained are indicated on Fig (a). These two isolated systems are now interconnected by a short transmission line as shown in Fig (b), and it is found that P₁ = P₂ = Q₁ = Q₂ = 0
     
     The bus voltage phase angular difference between generator bus X and generator bus Y after the interconnection is

1. 1. 10°

   
   
   

   2. 25°

   
   
   

   3. – 30°

   
   
   

   4. 30°

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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144. Suppose we define a sequence transformation between “a-b-c” and “p-n-o” variables as follows:
     

1. 1. 

   
   
   

   2. 

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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145. Consider a synchronous generator connected to an infinite bus by two identical parallel transmission lines. The transient reactance x of the generator is 0.1 pu and the mechanical power input to it is constant at 1.0 pu. Due to some pr evious distur bance, the rotor angle (δ) is undergoing an undamped oscillation, with the maximum value of δ(t) equal to 130°. One of the parallel lines trips due to relay maloperation at an instant when δ(t) = 130° as shown in the figure given below. The maximum value of the per unit line reactance, x, such that the system does not lose synchronism subsequent to this tripping is
     

1. 1. 0.87
      

   
   
   

   2. 0.74

   
   
   

   3. 0.67

   
   
   

   4. 0.54

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   Pel =	EƒVt	sinδ, X	X	+ 0.1
   	X1		2

   
   

   PeII =	EƒVt	sinδ, when tripping occurs
   	X2

   
   At δ₁ = δ₂
   P_eII = P_m X₂ = X + 0.
   

   ⇒	EƒVt	sinδ2 = 1
   	X2

   
   ⇒ sin 130° = X₂
   ⇒ X₂ = 0.76
   ⇒ X = X₂ – 0.1 = 0.66

   Correct Option: C

   
   

   Pel =	EƒVt	sinδ, X	X	+ 0.1
   	X1		2

   
   

   PeII =	EƒVt	sinδ, when tripping occurs
   	X2

   
   At δ₁ = δ₂
   P_eII = P_m X₂ = X + 0.
   

   ⇒	EƒVt	sinδ2 = 1
   	X2

   
   ⇒ sin 130° = X₂
   ⇒ X₂ = 0.76
   ⇒ X = X₂ – 0.1 = 0.66

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