U:51) A 50 kW dc shunt generator delivers an output voltage of 210V at no load when running at 1500rpm. The generator has an armature resistance of 40 m2, and a field resistance of 1582 whichincludes a rheostat for field current control.a) Sketch the electrical circuit of this generator showing all the relevant voltage and currentsincluding direction of currents and voltage polarities.b) If the generator is delivering 18 kW at a terminal voltage of 200V, determine the generatorspeed and torque.c) While maintaining the terminal voltage at 200V the rheostat was adjusted to increase the fieldcurrent by 2%. In addition the speed was maintained at the value determined from the previousstep (b), what will be the terminal power delivered by the generator under these conditions?

According to question: a) The electrical circuit with the direction of current and voltage…

Answered: A 50 kW de shunt generator delivers an…

Introductory Circuit Analysis (13th Edition)

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ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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U:51)

A 50 kW dc shunt generator delivers an output voltage of 210V at no load when running at 1500
rpm. The generator has an armature resistance of 40 m2, and a field resistance of 1582 which
includes a rheostat for field current control.
a) Sketch the electrical circuit of this generator showing all the relevant voltage and currents
including direction of currents and voltage polarities.
b) If the generator is delivering 18 kW at a terminal voltage of 200V, determine the generator
speed and torque.
c) While maintaining the terminal voltage at 200V the rheostat was adjusted to increase the field
current by 2%. In addition the speed was maintained at the value determined from the previous
step (b), what will be the terminal power delivered by the generator under these conditions?

Expert Solution

Step 1

According to question:

The electrical circuit with the direction of current and voltage polarities:

Step 2

$R_{a} = 40 mΩ R_{f} = 158 Ω P = 18 kW V = 200 V ∵ I_{L} = \frac{P}{V} \Rightarrow I_{L} = \frac{18000}{200} \Rightarrow I_{L} = 90 A ∵ I_{f} = \frac{V}{R_{f}} \Rightarrow I_{f} = \frac{200}{158} \Rightarrow I_{f} = 1.265 A ∵ I_{a} = I_{f} + I_{L} \Rightarrow I_{a} = 1.265 + 90 \Rightarrow I_{a} = 91.265 A ∵ E_{b} = V + I_{a} R_{a} \Rightarrow E_{b} = 200 + 91.265 \times 0.04 \Rightarrow E_{b} = 203.6506 V..........................(1) ∵ E_{b} = \frac{N P ϕ Z}{60 A} & (E$ _b)nl=210V∴210=1500PϕZ60A..............(2)Also203.6506=NPϕZ60A.................(3)

From (2) and (3)

$∴ \frac{210}{203.6506} = \frac{1500}{N} \Rightarrow N = \frac{1500 \times 203.6506}{210} \Rightarrow N = 1454.64 rpm ∵ ω T = E_{b} I_{a} \Rightarrow T = \frac{E_{b} I_{a}}{ω} \Rightarrow T = \frac{203.6506 \times 91.265}{\frac{2 π \times 1454.64}{60}} \Rightarrow T = 122 N-m$

Step 3

$V = 200 V N = 1454.64 rpm ∵ I_{f}^{'} = I_{f} + \frac{2}{100} \times I_{f} \Rightarrow I_{f}^{'} = 1.02 I_{f} \Rightarrow I_{f}^{'} = 1.02 \times 1.265 \Rightarrow I_{f}^{'} = 1.2903 A ∵ E_{b} = \frac{N P ϕ Z}{60 A} \Rightarrow E_{b} = K (N I_{f}) [∵ ϕ α I_{f}] \Rightarrow 203.6506 = K (1454.64 I_{f}) . . . . . . . . . . . . (4) & E_{b} = K (1.02 I_{f} \times 1454.64) . . . . . . . . . . . . . . . . . . (5)$

From (4) and (5)

$∴ \frac{203.6506}{E_{b}} = \frac{K (1454.64 I_{f})}{K (1.02 I_{f} \times 1454.64)} \Rightarrow E_{b} = 207.723 V ∵ I_{a} = \frac{E_{b} - V}{R_{a}} \Rightarrow I_{a} = \frac{207.723 - 200}{0.04} \Rightarrow I_{a} = 193.075 A ∵ I_{L} = I_{a} - I_{f}^{'} \Rightarrow I_{L} = 193.075 - 1.2903 \Rightarrow I_{L} = 191.78 A$

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