01․ The essential condition for parallel operation of two D.C generators is that they have
The following conditions must be satisfied for the parallel operation of DC generator. They are,
1. The terminal voltage of all generators must be same.
2. The polarities of all generators must be same.
3. The prime movers must have the same and stable characteristics from the rotation point of view.
4. The percent voltage drop with respect to the load variation must be same for all generators at all time.
02․ Which of the following load normally needs starting torque more than rated torque?
The conveyor belts are used to transfer a set of things from one place another place. During running condition, we can’t place any of the materials. So, before starting the conveyor belt, we need to place everything on the conveyor belt. Hence, we need more starting torque to run that load. Basically, in the conveyor belts, three phase slip ring induction motors (or) synchronous motors are used to achieve the high starting torque.
03․ The working principle of transformer depends on
The transformer is working under the mutual inductance principle. This principle is derived from the Faradays law of electromagnetic induction. Hence, it depends on that principle.
04․ The power developed by salient pole synchronous machine is/are
The power developed P = EV/Xd * sinδ + V²/2 * (Xd-Xq)/(Xd*Xq) *sin2δ
Where E = Excitation voltage
V = Terminal voltage
Xd = Direct axis reactance, Xq = Quadrature axis reactance
δ = load angle
Power due to excitation = EV/Xd * sinδ
Power due to saliency or reluctance = V²/2 * (Xd-Xq)/(Xd*Xq) *sin2δ
Therefore, the power developed by salient pole synchronous machine contains two components.
05․ In salient pole synchronous machine, if direct axis reactance and quadrature axis reactance is equal, then power developed due to reluctance is
The power developed in salient pole synchronous machine
P = EV/Xd * sinδ + V²/2 * (Xd - Xq)/(Xd*Xq) *sin2δ
Where E = Excitation voltage
V = Terminal voltage
Xd = Direct axis reactance, Xq = Quadrature axis reactance
δ = load angle
Power developed due to reluctance = V²/2 * (Xd - Xq)/(Xd*Xq) *sin2δ
Therefore, when direct axis reactance Xd is equal to quadrature axis reactance Xq, the power developed due to reluctance is zero.
06․ In salient pole synchronous machine, the reluctance power varies as
The power developed in salient pole synchronous machine
P = EV/Xd * sinδ + V²/2 * (Xd - Xq)/(Xd*Xq) *sin2δ
Where E = Excitation value
V = Terminal voltage
Xd = Direct axis reactance, Xq = Quadrature axis reactance
δ = load angle
Power due to reluctance = V²/2 * (Xd - Xq)/(Xd*Xq) *sin2δ
Therefore, in salient pole synchronous machine, the reluctance power varies as sin2δ.
07․ In salient pole synchronous machine, maximum power occurs at a load angle of
The power developed in salient pole synchronous machine
P = EV/Xd * sinδ + V²/2 * (Xd-Xq)/(Xd*Xq) *sin2δ
Excitation power varies as sinδ and power due to reluctance varies as sin2δ.
Maximum excitation power occurs at 90° and maximum power due to reluctance occurs at 45°.
Therefore, the maximum power developed in salient pole synchronous machine occurs at load angle less than 90°.
08․ The maximum power developed in cylindrical synchronous machine is occurs at a load angle of
The power developed in cylindrical synchronous machine is
Maximum power developed Pmax = EV/Xs
Where E = Excitation voltage
V = Terminal voltage
Xs = Synchronous reactance
δ = load angle
Therefore, maximum power developed in cylindrical synchronous machine is at a load angle of 90°.
09․ Maximum power developed in a cylindrical synchronous machine depends on
The power developed in cylindrical synchronous machine is
P = EV/Xs * sin δ
Where E = Excitation voltage
V = Terminal voltage
Xs = Synchronous reactance
δ = load angle
Maximum power developed Pmax = EV/Xs
Maximum power developed when sinδ=1, i.e. when δ = 90°, maximum power developed in cylindrical synchronous machine.
Therefore, maximum power developed only depends on excitation and synchronous reactance.
10․ Two generators operating in parallel with rating 50MW and 100MW respectively. Both their respective governor settings are identical as 4%, consider same no load frequency. How will the machine share a common load of 100 MW
System frequency f = f0 - (f0 - f1)/ Prated *P1
Where, f0 is rated power frequency.
System frequency f = 50 - 0.04*50/50* P1 -------------- equation number 1
and also
System frequency f = 50 - 0.04*50/100* P2 -------------- equation number 2
Given that P1 + P2 = 100 ----------------equation number 3
By solving the above equations
P1 = 33.33 MW
P2 = 66.67 MW
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