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EM-Problems

DC- Machines

First Semester 2018-2019

Problem 1 An 8-pole, wave-connected armature has 600 conductors and is driven at 625rev/min. If the flux per pole is 20mWb. - Determine the generated e.m.f. Solution: 𝑍 = 600, 𝑐 = 2 (𝑓𝑜𝑟 𝑎 𝑤𝑎𝑣𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔), 𝑝 = 4 𝑝𝑎𝑖𝑟𝑠 𝑛=

625 𝑟𝑒𝑣 [ ], 60 𝑠

𝜙 = 20 ∗ 10−3 𝑊𝑏

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒. 𝑚. 𝑓. , 𝐸 =

2𝑃𝜙𝑛𝑍 𝑐

625 2 (4) (20 ∗ 10−3 ) ( ) (600) 60 𝐸= = 500 𝑉 2 Problem 2 A 4-pole generator has a lap-wound armature with 50 slots with 16 conductors per slot. The useful flux per pole is 30mWb. - Determine the speed at which the machine must be driven to generate an e.m.f. of 240V. Solution: 𝐸 = 240 𝑉, 𝑐 = 2𝑝 ( 𝑓𝑜𝑟 𝑎 𝑙𝑎𝑝 𝑤𝑖𝑛𝑑𝑖𝑛𝑔), 𝑍 = 50 ∗ 16 = 800, 𝜙 = 30 ∗ 10−3 𝑊𝑏 𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒆𝒅 𝒆. 𝒎. 𝒇. , 𝑬 = 𝑹𝒆𝒂𝒓𝒓𝒂𝒏𝒈𝒊𝒏𝒈 𝒈𝒊𝒗𝒆𝒔, 𝒔𝒑𝒆𝒆𝒅, 𝒏 =

𝟐𝒑𝝓𝒏𝒁 𝟐𝒑𝝓𝒏𝒁 = = 𝝓𝒏𝒁 𝒄 𝟐𝒑

𝑬 𝟐𝟒𝟎 𝒓𝒆𝒗 𝒓𝒆𝒗 = = 𝟏𝟎 𝒐𝒓 𝟔𝟎𝟎 (𝟑𝟎 ∗ 𝟏𝟎−𝟑 ) (𝟖𝟎𝟎) 𝝓𝒁 𝒔 𝒎𝒊𝒏

Problem 3 An 8-pole, lap-wound armature has 1200 conductors and a flux per pole of 0.03Wb. - Determine the e.m.f. generated when running at 500rev/min. Solution: 𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒆𝒅 𝒆. 𝒎. 𝒇. , 𝑬 =

𝟐𝑷𝝓𝒏𝒁 𝟐𝑷𝝓𝒏𝒁 = = 𝝓𝒏𝒁 𝒄 𝟐𝑷 𝑪 = 𝒑 𝒇𝒐𝒓 𝒍𝒂𝒑 − 𝒘𝒐𝒖𝒏𝒅 𝒎𝒂𝒄𝒉𝒊𝒏𝒆

𝑬 = 𝝓 𝒏 𝒁 = (𝟎. 𝟎𝟑) (

𝟓𝟎𝟎 ) (𝟏𝟐𝟎𝟎) = 𝟑𝟎𝟎 𝑽 𝟔𝟎

1

Problem 4 - Determine the generated e.m.f. in Problem 3 if the armature is wave-wound. Solution:

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒. 𝑚. 𝑓. , 𝐸 =

2𝑃𝜙𝑛𝑍 2𝑃𝜙𝑛𝑍 = 𝑐 2 "𝒄 = 𝟐 𝒇𝒐𝒓 𝒘𝒂𝒗𝒆 − 𝒘𝒐𝒖𝒏𝒅 𝒎𝒂𝒄𝒉𝒊𝒏𝒆"

𝑬 = 𝑷 𝝓 𝒏 𝒁 = (𝟒) (𝝓 𝒏 𝒁) = 𝟑𝟎𝟎 𝑽 𝑬 = (𝟒) (𝟑𝟎𝟎) = 𝟏𝟐𝟎𝟎 𝒗𝒐𝒍𝒕𝒔 Problem 5 A d.c. shunt-wound generator running at constant speed generates a voltage of 150 V at a certain value of field current. - Determine the change in the generated voltage when the field current is reduced by 20%, assuming the flux is proportional to the field current. Solution: The generated e.m.f. E of a generator is proportional to Ф ω, i.e. is proportional to Ф n, where Ф is the flux and n is the speed of rotation. It follows that E = k Ф n, where k is a constant. At speed n1 and flux Ф1, E1 = k Ф1 n1 At speed n2 and flux Ф2, E2 = k Ф2 n2 Thus, by division: 𝐸1 𝑘 𝜙1 𝑛1 𝜙1 𝑛1 = = 𝐸2 𝑘 𝜙2 𝑛2 𝜙2 𝑛2 The initial conditions are E1 =150V, Ф = Ф1 and n = n1. When the flux is reduced by 20%, the new value of flux is 80/100 or 0.8 of the initial value, i.e. Ф2 = 0.8 Ф1. Since the generator is running at constant speed, n2 = n1. 𝑇ℎ𝑢𝑠 𝑡ℎ𝑎𝑡 𝑖𝑠,

𝐸1 𝜙1 𝑛1 𝜙1 𝑛1 1 = = = 𝐸2 𝜙2 𝑛2 0.8 𝜙1 𝑛1 0.8 𝐸2 = 150 ∗ 0.8 = 120 𝑉

Thus, a reduction of 20% in the value of the flux reduces the generated voltage to 120 V at constant speed. 2

Problem 6 A d.c. generator running at 30 rev/s generates an e.m.f. of 200V. - Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s. Solution: E ∝ Ф ω and since ω =2πn, E ∝ n.

Let E1 = 200 V, n1 = 30 rev/s and flux per pole at this speed be Ф1 Let E2 = 250 V, n2 = 20 rev/s and flux per pole at this speed be Ф2 𝑆𝑖𝑛𝑐𝑒 𝐸 ∝ 𝜙 𝑛 𝑡ℎ𝑒𝑛 𝐻𝑒𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝜙2 =

𝐸1 𝜙1 𝑛1 = 𝐸2 𝜙2 𝑛2

200 𝜙1 (30) = 250 𝜙2 (20) 𝜙1 (30) (250) = 𝟏. 𝟖𝟕𝟓 𝝓𝟏 (20) (200)

Hence the increase in flux per pole needs to be 87.5 %

Problem 7 - Determine the terminal voltage of a generator which develops an e.m.f. of 200 V and has an armature current of 30 A on load. Assume the armature resistance is 0.30 Ω Solution: With reference to the Figure, terminal voltage, VT = E – (Ia * Ra) = 200 – (30) (0.30) = 200 – 9 = 191 V Problem 8 A generator is connected to a 60 Ω load and a current of 8 A flows. If the armature resistance is 1Ω - Determine (a) the terminal voltage, and (b) the generated e.m.f. Solution: a- Terminal voltage, V = Ia * RL = (8) (60) = 480 V b- Generated e.m.f., E = V + (Ia* Ra) = 480 + (8 *1) = 480 + 8 = 488 V 3

Problem 9 A separately-excited generator develops a no-load e.m.f. of 150V at an armature speed of 20 rev/s and a flux per pole of 0.10Wb. Determine the generated e.m.f. when: (a) the speed increases to 25 rev/s and the pole flux remains unchanged (b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08Wb. (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07Wb. Solution:

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ,

(0.10)(20) 150 = (0.10)(25) 𝐸2

Hence,

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ,

𝐸1 ∅1 𝑛1 = 𝐸2 ∅2 𝑛2

𝐸2 =

(150)(0.10)(25) = 187.5 𝑉 (0.10)(20)

(0.10)(20) 150 = (0.08)(20) 𝐸3 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑒. 𝑚. 𝑓. ,

𝐸3 =

(150)(0.08)(20) = 120 𝑉 (0.10)(20)

(0.10) (20) 150 = (0.07) (24) 𝐸4 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑒. 𝑚. 𝑓. ,

𝐸4 =

4

(150)(0.07)(24) = 126 𝑉 (0.10)(20)

Problem 10 A shunt generator supplies a 20 kW load at 200 V through cables of resistance, R= 100 mΩ. If the field winding resistance, Rf =50 Ω and the armature resistance, Ra =40 mΩ, Determine: (a) the terminal voltage (b) the e.m.f. generated in the armature.

Solution: (a) The circuit is as shown in the Figure.

𝐿𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼 =

20000 𝑊 = 100 𝐴 200 𝑉

𝑉𝑜𝑙𝑡 𝑑𝑟𝑜𝑝 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑏𝑙𝑒𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 = 𝐼 ∗ 𝑅 = (100 𝐴) ∗ (100 ∗ 10−3 𝛺) = 10 𝑉

𝐻𝑒𝑛𝑐𝑒 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒, 𝑉 = 200 𝑉 + 10 𝑉 = 210 𝑉 𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼𝑎 = 𝐼𝑓 + 𝐼

𝐹𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑓 =

𝑉 210 𝑉 = = 4.2 𝐴 𝑅𝑓 50 Ω

𝐻𝑒𝑛𝑐𝑒 𝐼𝑎 = 𝐼𝑓 + 𝐼 = 4.2 𝐴 + 100 𝐴 = 104.2 𝐴 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒. 𝑚. 𝑓.

𝐸 = 𝑉 + 𝐼𝑎 𝑅𝑎

𝐸 = 210 𝑉 + (104.2 𝐴) (40 ∗ 10−3 Ω) = 210 𝑉 + 4.168 𝑉 = 214.17 𝑉

5

Problem 11 A short-shunt compound generator supplies 80 A at 200 V. If the field resistance, Rf = 40 Ω, the series resistance, Rse = 0.02 Ω and the armature resistance, Ra = 0.04 Ω, determine the e.m.f. generated. Solution: The circuit is shown in the Figure. 𝑉𝑜𝑙𝑡 𝑑𝑟𝑜𝑝 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝐼 𝑅𝑆𝑒 = (80 𝐴) (0.02 Ω) = 1.6 𝑉 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑓𝑖𝑒𝑙𝑑 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑉1 = 200 𝑉 + 1.6 𝑉 = 201.6 𝑉 𝐹𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼𝑓 =

𝑉1 𝑅𝑓

=

201.6 𝑉 40 Ω

= 5.04 𝐴

𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑎 = 𝐼 + 𝐼𝑓 = 80 𝐴 + 5.04 𝐴 = 85.04 𝐴 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒. 𝑚. 𝑓. , 𝐸 = 𝑉1 + (𝐼𝑎 ∗ 𝑅𝑎 ) 𝐸 = 201.6 𝑉 + (85.04 𝐴 ∗ 0.04 Ω) = 201.6 𝑉 + 3.4 𝑉 = 205 𝑉

Problem 12 A 10kW shunt generator having an armature circuit resistance of 0.75 Ω and a field resistance of 125 Ω, generates a terminal voltage of 250V at full load. - Determine the efficiency of the generator at full load, assuming the iron, friction and windage losses amount to 600W. Solution: The circuit is shown in the Figure. 𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 = 10000 𝑊 = 𝑉 ∗ 𝐼 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑙𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐹𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑓 =

𝑉 𝑅𝑓

=

250 𝑉 125 Ω

𝐼=

10000 𝑉

=

10000 𝑊 250 𝑉

= 40 𝐴

=2𝐴

𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑎 = 𝐼𝑓 + 𝐼 = 2 𝐴 + 40 𝐴 = 42 𝐴 Efficiency, 𝑉∗𝐼 𝜂= ( ) ∗ 100 % 𝑉 ∗ 𝐼 + 𝐼𝑎2 𝑅𝑎 + 𝐼𝑓 ∗ 𝑉 + 𝐶 𝜂= (

10000 10000 ∗ 100 % = ∗ 100 % = 80.5 % ) 10000 + (422 )(0.75) + (2)(250) + 600 12423 6

Problem 13 A d.c. motor operates from a 240 V supply. The armature resistance is 0.2 Ω. - Determine the back e.m.f. when the armature current is 50 A. Solution: 𝐹𝑜𝑟 𝑎 𝑚𝑜𝑡𝑜𝑟, 𝑉 = 𝐸 + 𝐼𝑎 ∗ 𝑅𝑎 ℎ𝑒𝑛𝑐𝑒 𝑏𝑎𝑐𝑘 𝑒. 𝑚. 𝑓. ,

𝐸 = 𝑉 − (𝐼𝑎 ∗ 𝑅𝑎 )

𝐸 = 240 𝑉𝑜𝑙𝑡𝑠 − (50 𝐴 ∗ 0.2 Ω) = 240 𝑉𝑜𝑙𝑡𝑠 − 10 𝑉𝑜𝑙𝑡𝑠 = 230 𝑉𝑜𝑙𝑡𝑠 Problem 14 The armature of a d.c. machine has a resistance of 0.25 Ω and is connected to a 300 V supply. - Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A. (b) as a motor taking 80 A. Solution: (a) As a generator, generated e.m.f., 𝐸 = 𝑉 + (𝐼𝑎 ∗ 𝑅𝑎 ) 𝐸 = 300 𝑉 + (100𝐴 ∗ 0.25 Ω) = 300 𝑉 + 25 𝑉 = 325 𝑉

(b) As a motor, generated e.m.f. (or back e.m.f.), 𝐸 = 𝑉 − (𝐼𝑎 ∗ 𝑅𝑎 ) 𝐸 = 300 𝑉 − (80 𝐴 ∗ 0.25 Ω) = 280 𝑉 Problem 15. An 8-pole d.c. motor has a wave-wound armature with 900 conductors. The useful flux per pole is 25 mWb. Determine the torque exerted when a current of 30A flows in each armature conductor. Solution: P = 4, c = 2 for wave winding, Ф = 25 * 10-3 Wb, Z= 900, Ia = 30 A (4)(25 ∗ 10−3 )(900)(30) 𝑝 𝜙 𝑍 𝐼𝑎 𝑡𝑜𝑟𝑞𝑢𝑒 𝑇 = = = 429.7 𝑁𝑚 𝜋𝑐 𝜋(2) 7

Problem 16. - Determine the torque developed by a 350V d.c. motor having an armature resistance of 0.5 Ω and running at 15 rev/s. The armature current is 60 A. Solution: 𝑉 = 350 𝑉,

𝑅𝑎 = 0.5 Ω,

𝑛 = 15

𝑟𝑒𝑣 , 𝑠

𝐼𝑎 = 60 𝐴

𝐵𝑎𝑐𝑘 𝑒. 𝑚. 𝑓. 𝐸 = 𝑉 − (𝐼𝑎 ∗ 𝑅𝑎 ) = 350 𝑉𝑜𝑙𝑡𝑠 − (60 𝐴 ∗ 0.5 Ω) = 320 𝑉𝑜𝑙𝑡𝑠

𝑡𝑜𝑟𝑞𝑢𝑒 𝑇 =

(320 𝑉𝑜𝑙𝑡𝑠) (60 𝐴) 𝐸 𝐼𝑎 = = 203.7 𝑁𝑚 𝑟𝑒𝑣 2𝜋𝑛 2 𝜋 (15 𝑠 )

Problem 17 A six-pole lap-wound motor is connected to a 250V d.c. supply. The armature has 500 conductors and a resistance of 1Ω. The flux per pole is 20mWb. 2p = one pair of poles

Calculate: -

This means

(a) the speed

C = number of poles

(b) the torque developed when the armature current is 40A. Solution: 𝑉 = 250 𝑉, 𝑍 = 500, 𝑅𝑎 = 1 Ω, Φ = 20 ∗ 10−3 Wb, 𝐼𝑎 = 40 A, c = 2P for lap winding 𝐵𝑎𝑐𝑘 𝑒. 𝑚. 𝑓. 𝐸 = 𝑉 − (𝐼𝑎 ∗ 𝑅𝑎 ) = 250 𝑉 − (40 𝐴 ∗ 1 Ω) = 210 𝑉

𝐸. 𝑚. 𝑓.

2𝑝𝜙𝑛𝑍 2𝑝 (20 ∗ 10−3 𝑊𝑏) 𝑛 (500) 𝐸= = 𝑖. 𝑒. 210 𝑉 = 𝑐 2𝑝

𝐻𝑒𝑛𝑐𝑒 𝑠𝑝𝑒𝑒𝑑 𝑛 =

210 𝑉 𝑟𝑒𝑣 𝑟𝑒𝑣 (21 = 21 𝑜𝑟 ∗ 60) = 1260 (20 ∗ 10−3 𝑊𝑏)(500) 𝑠 𝑚𝑖𝑛

𝑡𝑜𝑟𝑞𝑢𝑒 𝑇 =

(210 𝑉)(40 𝐴) 𝐸 𝐼𝑎 = 𝑟𝑒𝑣 = 63.66 𝑁𝑚 2𝜋𝑛 2 𝜋 (21 𝑠 ) 8

Problem 18 The shaft torque of a diesel motor driving a 100V d.c. shunt-wound generator is 25Nm. The armature current of the generator is 16 A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15%, the torque increases to 35Nm. - Determine the armature current at this new value of torque. Solution: The shaft torque T of a generator is proportional to Ia, where is the flux and Ia is the armature current. Thus, T = k Ф Ia, where k is a constant. The torque at flux Ф1 and armature current Ia1 is T1 = k Ф1 Ia1. Similarly, T2 = k Ф2 Ia2 𝑇1 𝑘 ∅1 𝐼𝑎1 ∅1 𝐼𝑎1 = = 𝑇2 𝑘 ∅2 𝐼𝑎2 ∅2 𝐼𝑎2

𝐵𝑦 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛

𝐻𝑒𝑛𝑐𝑒

𝑖. 𝑒.

25 ∅1 ∗ 16 = 35 0.85 ∅1 ∗ 𝐼𝑎2

𝐼𝑎2 =

16 ∗ 35 = 26.35 𝐴 0.85 ∗ 25

That is, the armature current at the new value of torque is 26.35 A.

9

Problem 19 A 100V d.c. generator supplies a current of 15A when running at 1500 rev/min. If the torque on the shaft driving the generator is 12 Nm. Determine: (a) the efficiency of the generator. (b) the power loss in the generator. Solution:

The efficiency of the generator =

𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 ∗ 100 % 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

The output power is the electrical output, i.e. V I [Watts]. The input power to a generator is the mechanical power in the shaft driving the generator, i.e. T ω or T (2πn) [Watts], where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator

𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦,

𝑖. 𝑒.

𝜂=

𝜂=

𝑉∗𝐼 ∗ 100 % 𝑇 (2𝜋𝑛)

(100) (15) ∗ 100 % = 79.6 % 1500 (12) (2𝜋) ( 60 )

𝑇ℎ𝑒 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 + 𝑙𝑜𝑠𝑠𝑒𝑠 𝐻𝑒𝑛𝑐𝑒, 𝑇 (2𝜋𝑛) = (𝑉 ∗ 𝐼) + 𝑙𝑜𝑠𝑠𝑒𝑠 𝑖. 𝑒.

𝑙𝑜𝑠𝑠𝑒𝑠 = 𝑇 (2𝜋𝑛) − (𝑉 ∗ 𝐼)

𝐿𝑜𝑠𝑠𝑒𝑠 = 12 (2 𝜋

1500 ) − (100 ∗ 15) 60

𝑖. 𝑒. 𝑝𝑜𝑤𝑒𝑟 𝑙𝑜𝑠𝑠𝑒𝑠 = 1885 𝑊 − 1500 𝑊 = 385 𝑊

10

Problem 20 A 240V shunt motor takes a total current of 30 A. If the field winding resistance Rf =150 Ω and the armature resistance Ra = 0.4 Ω. Determine: (a) the current in the armature. (b) the back e.m.f. Solution: 𝐹𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑓 =

𝑉 𝑅𝑓

=

240 𝑉 150 Ω

= 1.6 𝐴

𝑆𝑢𝑝𝑝𝑙𝑦 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼 = 𝐼𝑓 + 𝐼𝑎 𝐻𝑒𝑛𝑐𝑒 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑎 = 𝐼 − 𝐼𝑓 = 30 𝐴 − 1.6 𝐴 = 28.4 𝐴 𝐵𝑎𝑐𝑘 𝑒. 𝑚. 𝑓.

𝐸 = 𝑉 − 𝐼𝑎 𝑅𝑎

𝐸 = 240 𝑉 − (28.4 𝐴) (0.4 Ω) = 228.64 𝑉

Problem 21 A 200V, d.c. shunt-wound motor has an armature resistance of 0.4 Ω and at a certain load has an armature current of 30A and runs at 1350 rev/min. If the load on the shaft of the motor is increased so that the armature current increases to 45A, determine the speed of the motor, assuming the flux remains constant. Solution: 𝑇ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝 𝐸 ∝ 𝜙 𝑛 𝑎𝑝𝑝𝑙𝑖𝑒𝑠 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟𝑠 𝑎𝑛𝑑 𝑚𝑜𝑡𝑜𝑟𝑠. 𝐹𝑜𝑟 𝑎 𝑚𝑜𝑡𝑜𝑟𝑠, 𝐸 = 𝑉 − 𝐼𝑎 𝑅𝑎 𝐸1 = 200 𝑉 − (30 𝐴) (0.4 Ω) = 188 𝑉 𝐸2 = 200 𝑉 − (45 𝐴) (0.4 Ω) = 182 𝑉 𝑇ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝,

𝐸1 𝜙1 𝑛1 = 𝐸2 𝜙 2 𝑛2

𝑎𝑝𝑝𝑙𝑖𝑒𝑠 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟𝑠 𝑎𝑛𝑑 𝑚𝑜𝑡𝑜𝑟𝑠. 𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑓𝑙𝑢𝑥 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝜙1 = 𝜙2 1350 𝜙1 ∗ ( ) 188 60 𝐻𝑒𝑛𝑐𝑒 = , 182 𝜙1 ∗ 𝑛2

𝑖. 𝑒. 𝑛2 =

22.5 ∗ 182 𝑟𝑒𝑣 = 21.78 188 𝑠

𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 45𝐴 𝑖𝑠 21.78 ∗ 60 11

𝑟𝑒𝑣

, 𝑖. 𝑒. 1307 𝑚𝑖𝑛

𝑟𝑒𝑣 𝑚𝑖𝑛

Problem 22 A 220V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 Ω. Determine: (a) the maximum value of armature current if the flux is suddenly reduced by 10 %. (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant. Solution: (a) For a d.c. shunt-wound motor, E = V – Ia Ra. Hence initial generated e.m.f., E1 = 220 – 30 × 0.4 = 208V. The generated e.m.f. is also such that E ∝ n, so at the instant the flux is reduced, the speed has not had time to change, and E = 208 × 90/100 = 187.2V. Hence, the voltage drop due to the armature resistance is 220 −187.2, i.e. 32.8V. The instantaneous value of the current is 32.8/0.4, i.e. 82A. This increase in current is about three times the initial value and causes an increase in torque, (T ∝ Ia). The motor accelerates because of the larger torque value until steady state conditions are reached. (b) T ∝ Ia and since the torque is constant, Ф1 Ia1 = Ф2 Ia2. The flux is reduced by 10 %, hence Ф2 = 0.9 Ф1 Thus, Ф1 × 30 = 0.9 Ф1 × Ia2 i.e. the steady state value of armature current,

𝐼𝑎2 =

30 1 = 33 𝐴 0.9 3

12

Problem 23 A series motor has an armature resistance of 0.2 Ω and a series field resistance of 0.3 Ω. It is connected to a 240V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux. Solution: With reference to the Figure, generated e.m.f., E, at initial load, is given by

𝐸1 = 𝑉 − 𝐼𝑎 (𝑅𝑎 + 𝑅𝑓 ) = 240 𝑉 − [(15𝐴) ∗ (0.2 Ω + 0.3Ω)] = 232.5 𝑉 𝑊ℎ𝑒𝑛 𝑡ℎ𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑡𝑜 30 𝐴, 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑒. 𝑚. 𝑓. 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦: 𝐸2 = 𝑉 − 𝐼𝑎 (𝑅𝑎 + 𝑅𝑓 ) = 240 𝑉𝑜𝑙𝑡𝑠 − [(30𝐴) ∗ (0.2 Ω + 0.3Ω)] = 225 𝑉𝑜𝑙𝑡𝑠 𝑁𝑜𝑤 𝑒. 𝑚. 𝑓. 𝐸 ∝ Φ𝑛 𝐸1 𝜙1 𝑛1 = 𝐸2 𝜙 2 𝑛2

𝑇ℎ𝑢𝑠

𝑖. 𝑒.

232.5 𝜙1 ∗ (24) = (2𝜙1 ) ∗ (𝑛2 ) 225

𝐻𝑒𝑛𝑐𝑒 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑚𝑜𝑡𝑜𝑟,

𝑛2 =

𝑠𝑖𝑛𝑐𝑒 𝜙2 = 2𝜙1

(24) (225) 𝑟𝑒𝑣 = 11.6 (232.5) (2) 𝑠

As the current has been increased from 15A to 30A, the speed has decreased from 24 rev/s to 11.6rev/s. Its speed/current characteristic is similar to the Figure.

13

Problem 24 A 320V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron, friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 Ω and the armature resistance is 0.2 Ω. - Determine the overall efficiency of the motor. Solution: The circuit is shown in the Figure.

𝐹𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼𝑓 =

𝑉 𝑅𝑓

=

320 𝑉 40 Ω

=8𝐴

𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼𝑎 = 𝐼 − 𝐼𝑓 = 80𝐴 − 8𝐴 = 72𝐴

C = iron, friction and windage losses = 1500 W Efficiency, 𝑉 ∗ 𝐼 − 𝐼𝑎2 𝑅𝑎 − 𝐼𝑓 ∗ 𝑉 − 𝐶 𝜂= ( ) ∗ 100 % 𝑉∗𝐼

(320)(80)− (722 )(0.2)− (8)(320)−1500

𝜂= (

(320)(80)

) ∗ 100 % =

14

20503 25600

∗ 100 % = 80.1 %

Problem 25 A 250V series motor draws a current of 40 A. The armature resistance is 0.15 Ω and the field resistance is 0.05Ω. - Determine the maximum efficiency of the motor. Solution: The circuit is as shown in the Figure. The efficiency,

𝑉 𝐼 − 𝐼𝑎2 𝑅𝑎 − 𝐼𝑓 𝑉 − 𝐶 𝜂= ( ) ∗ 100 % 𝑉𝐼

𝐻𝑜𝑤𝑒𝑣𝑒𝑟 𝑓𝑜𝑟 𝑎 𝑠𝑒𝑟𝑖𝑒𝑠 𝑚𝑜𝑡𝑜𝑟, 𝐼𝑓 = 0 𝑎𝑛𝑑 𝑡ℎ𝑒 𝐼𝑎2 𝑅𝑎 𝑙𝑜𝑠𝑠 𝑛𝑒𝑒𝑑𝑠 𝑡𝑜 𝑏𝑒 𝐼 2 (𝑅𝑎 + 𝑅𝑓 ) 𝑉 𝐼 − 𝐼 2 (𝑅𝑎 + 𝑅𝑓 ) − 𝐶 𝐻𝑒𝑛𝑐𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 = ( ) ∗ 100 % 𝑉𝐼

𝐹𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝐼 2 (𝑅𝑎 + 𝑅𝑓 ) = 𝐶

𝑉 𝐼 − 2𝐼 2 (𝑅𝑎 + 𝑅𝑓 ) 𝐻𝑒𝑛𝑐𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 = ( ) ∗ 100 % 𝑉𝐼 (250)(40) − 2(40)2 (0.15 + 0.05) 𝜂= ( ) ∗ 100 % (250)(40)

𝜂= (

𝜂= (

10000 − 640 ) ∗ 100 % 10000

9360 ) ∗ 100 % = 93.6 % 10000

15

Problem 26 A 200V d.c. motor develops a shaft torque of 15 Nm at 1200 rev/min. If the efficiency is 80%. - determine the current supplied to the motor. Solution: The efficiency of a motor =

𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 ∗ 100 % 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

The output power of a motor is the power available to do work at its shaft and is given by T ω or T (2πn) [watts], where T is the torque in Nm and n is the speed of rotation in [rev/s]. The input power is the electrical power in watts supplied to the motor, i.e. VI [watts]. Thus for a motor, efficiency, 𝜂=

i.e.

𝑇 (2 𝜋 𝑛) 𝑉𝐼

∗ 100 %

80 = [

(15) (2 𝜋) (

1200 ) 60

(200) (𝐼)

] (100)

𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑, 𝐼 =

(15) (2𝜋) (20) (100) (200) (80)

= 11.8 𝐴

Problem 27 A d.c. series motor drives a load at 30 rev/s and takes a current of 10 A when the supply voltage is 400V. If the total resistance of the motor is 2 Ω and the iron, friction and windage losses amount to 300W, determine the efficiency of the motor.

Solution: 𝑉 𝐼 − 𝐼 2 𝑅−𝐶

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 = (

𝑉𝐼

) ∗ 100 %

(400)(10)− (10)2 (2)−300

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 = (

(400) (10)

4000−200−300

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦, 𝜂 = (

4000

) ∗ 100 %

) ∗ 100 % = 87.5 % 16

Problem 28 A 500V shunt motor runs at its normal speed of 10 rev/s when the armature current is 120A. The armature resistance is 0.2 Ω. (a) Determine the speed when the current is 60A and a resistance of 0.5 Ω is connected in series with the armature, the shunt field remaining constant. (b) Determine the speed when the current is 60A and the shunt field is reduced to 80 % of its normal value by increasing resistance in the field circuit. Solution: With reference to the Figure, back e.m.f. at 120A, 𝐸1 = 𝑉 − (𝐼𝑎 ∗ 𝑅𝑎 ) = 500 𝑉 − [(120𝐴) ∗ (0.2Ω)] = 476 𝑉

𝑊ℎ𝑒𝑛 𝐼𝑎 = 60𝐴, 𝐸2 = 500 𝑉 − [(60𝐴) ∗ (0.2Ω + 0.5Ω)] = 458 𝑉 𝑁𝑜𝑤 𝑖. 𝑒.

476 458

=

𝜙1 (10)

𝐸1 𝜙1 ∗ 𝑛1 = 𝐸2 𝜙2 ∗ 𝑛2

𝑠𝑖𝑛𝑐𝑒 𝜙1 = 𝜙2

𝜙1 (𝑛2 )

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑠𝑝𝑒𝑒𝑑 𝑛2 =

(10) (458) (476)

= 9.62

𝑟𝑒𝑣 𝑠

𝐵𝑎𝑐𝑘 𝑒. 𝑚. 𝑓. 𝑤ℎ𝑒𝑛 𝐼𝑎 = 60𝐴 𝐸2 = 500𝑉 − [(60𝐴) ∗ (0.2Ω)] = 488 𝑉 𝑁𝑜𝑤

𝑖. 𝑒.

𝐸1 𝜙1 ∗ 𝑛1 = 𝐸2 𝜙2 ∗ 𝑛2

(𝜙1 ) (10) 476 = , (0.8 𝜙1 ) (𝑛3 ) 488

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑠𝑝𝑒𝑒𝑑 𝑛3 =

𝑠𝑖𝑛𝑐𝑒 𝜙2 = 0.8 𝜙1

(10) (488) 𝑟𝑒𝑣 = 12.82 (0.8) (476) 𝑠

17

Problem 29 On full-load a 300 V series motor takes 90 A and runs at 15 rev/s. The armature resistance is 0.1Ω and the series winding resistance is 50 mΩ. - Determine the speed when developing full load torque but with a 0.2 Ω diverter in parallel with the field winding. (Assume that the flux is proportional to the field current.) Solution: 𝐴𝑡 300 𝑉, 𝑒. 𝑚. 𝑓.,

𝐸1 = 𝑉 − 𝐼(𝑅𝑎 + 𝑅𝑠𝑒 ) = 300𝑉 − (90𝐴)(0.1Ω + 0.05Ω) = 286.5 𝑉

𝑊𝑖𝑡ℎ 𝑡ℎ𝑒 0.2 Ω 𝑑𝑟𝑖𝑣𝑒𝑟 𝑖𝑛 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑤𝑖𝑡ℎ 𝑅𝑠𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑟𝑒𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑅 =

(0.2)(0.05)

Electric Circuits (0.2)(0.05)

= = 0.04 Ω (0.2)+(0.05) (0.25) 0.2 𝐵𝑦 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒊𝒗𝒊𝒔𝒊𝒐𝒏, 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼1 = ( ) 𝐼 = 0.8 𝐼 0.2 + 0.05 𝑇𝑜𝑟𝑞𝑢𝑒, 𝑇 𝛼 𝐼𝑎 ∅ 𝑎𝑛𝑑 𝑓𝑜𝑟 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑡𝑜𝑟𝑞𝑢𝑒, 𝐼𝑎1 ∅1 = 𝐼𝑎2 ∅2 𝑠𝑖𝑛𝑐𝑒 𝑓𝑙𝑢𝑥 𝑖𝑠 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑡𝑜 𝑓𝑖𝑒𝑙𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 ∅1 ∝ 𝐼𝑎1 𝑎𝑛𝑑 ∅2 ∝ 0.8 𝐼𝑎2 𝑇ℎ𝑒𝑛 (90)(90) = (𝐼𝑎2 )(0.8 𝐼𝑎2 ) 𝑓𝑟𝑜𝑚

2 𝑤ℎ𝑖𝑐ℎ, 𝐼𝑎2

(90)2 90 = 𝑎𝑛𝑑 𝐼𝑎2 = = 100.62 𝐴 0.8 √(0.8)

𝐻𝑒𝑛𝑐𝑒 𝑒. 𝑚. 𝑓. 𝐸2 = 𝑉 − 𝐼𝑎2 (𝑅𝑎 + 𝑅) = 300𝑉 − (100.62𝐴)(0.1Ω + 0.04Ω) = 285.9 𝑉 𝑁𝑜𝑤 𝑒. 𝑚. 𝑓. 𝐸 ∝ ∅𝑛 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝐻𝑒𝑛𝑐𝑒

𝐸1 ∅1 𝑛1 𝐼𝑎1 𝑛1 = = 𝐸2 ∅2 𝑛2 0.8 𝐼𝑎2 𝑛2

(286.5) (90)(15) = 285.9 (0.8)(100.62) 𝑛2

(285.9)(90)(15) = 16.74 𝑟𝑒𝑣/𝑠 (286.5)(0.8)(100.62) 𝑟𝑒𝑣 𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟 ℎ𝑎𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑓𝑟𝑜𝑚 15 𝑠𝑒𝑐 𝑎𝑛𝑑 𝑛𝑒𝑤 𝑠𝑝𝑒𝑒𝑑, 𝑛2 =

𝑖. 𝑒. 900 𝑟𝑒𝑣/𝑚𝑖𝑛

𝑡𝑜 16.74 𝑟𝑒𝑣/𝑠𝑒𝑐 i.e., 1004 rev/min by inserting a 0.2 Ω diverter resistance in parallel with the series winding. 18

Problem 30 A series motor runs at 800 rev/min when the voltage is 400 V and the current is 25 A. The armature resistance is 0.4 Ω and the series field resistance is 0.2 Ω. - Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current. Solution: With reference to the Figure, at 800 rev/min,

𝑒. 𝑚. 𝑓., 𝐸1 = 𝑉 − [𝐼 ∗ (𝑅𝑎 + 𝑅𝑠𝑒 )] = 400 𝑉 − [(25𝐴) ∗ (0.6Ω)] = 385 𝑉 At 600 rev/min, since the current is unchanged, the flux is unchanged. 𝑻𝒉𝒖𝒔 𝑬 ∝ 𝝓𝒏, 𝑬 ∝ 𝒏, 𝒂𝒏𝒅 𝐻𝑒𝑛𝑐𝑒

385 𝐸2

=

𝑬𝟏 𝒏𝟏 = 𝑬𝟐 𝒏𝟐

800 600

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝐸2 =

(385) (600) (800)

= 288.75 𝑉

𝐸2 = 𝑉 − [𝐼 ∗ (𝑅𝑎 + 𝑅𝑠𝑒 + 𝑅)]

𝑎𝑛𝑑

288.75 = 400 − [(25) ∗ (0.4 + 0.2 + 𝑅)]

𝐻𝑒𝑛𝑐𝑒

𝑅𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑖𝑛𝑔 𝑔𝑖𝑣𝑒𝑛;

0.6 + 𝑅 =

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑒𝑥𝑡𝑟𝑎 𝑠𝑒𝑟𝑖𝑒𝑠 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒,

400 − 288.75 = 4.45 25 𝑅 = 4.45 − 0.6 = 3.85 Ω

Thus, the addition of a series resistance of 3.85 Ω has reduced the speed from 800 rev/min to 600 rev/min.

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Appendix

20

[𝑉]

𝐸 = 𝑘𝑎 ∅ 𝜔𝑚 𝑇=

𝑃 𝐸 𝐼𝑎 = = 𝑘𝑎 ∅ 𝐼𝑎 𝜔𝑚 𝜔𝑚

[𝑁. 𝑚]

The principal losses of machines are: (i) Copper loss, due to I 2R heat losses in the armature and field windings. (ii) Iron (or core) loss, due to hysteresis and eddy current losses in the armature. This loss can be reduced by constructing the armature of silicon steel laminations having a high resistivity and low hysteresis loss. At constant speed, the iron loss is assumed constant. (iii) Friction and windage losses, due to bearing and brush contact friction and Losses due to air resistance against moving parts (called windage). At constant speed, these losses are assumed to be constant. (iv) Brush contact loss between the brushes and commutator. This loss is approximately proportional to the load current.

Efficiency of a d.c. generator The efficiency of an electrical machine is the ratio of the output power to the input power and is usually expressed as a percentage. The Greek letter ‘η’ (eta) is used to signify efficiency and since the units are power/power, then efficiency has no units. Thus 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚, 𝜂 = (

𝒐𝒖𝒕𝒑𝒖𝒕 𝒑𝒐𝒘𝒆𝒓 ) ∗ 100% 𝒊𝒏𝒑𝒖𝒕 𝒑𝒐𝒘𝒆𝒓

If the total resistance of the armature circuit (including brush contact resistance) is Ra, then the total loss in the armature circuit is (𝑰𝟐𝒂 ∗ 𝑹𝒂 ) If the terminal voltage is V and the current in the shunt circuit is If, then the loss in the shunt circuit is (If *V).

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If the sum of the iron, friction and windage losses is C then the total losses is given by:

𝐼𝑎2 𝑅𝑎 + 𝐼𝑓 𝑉 + 𝐶 𝐼𝑎2 𝑅𝑎 + 𝐼𝑓 𝑉 𝑖𝑠, 𝑖𝑛 𝑓𝑎𝑐𝑡, 𝑡ℎ𝑒 "𝑐𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠𝑒𝑠" Various power stages in the case of a d. c. generator are shown below: A

C

B

Mechanical Power Input = Output of Driving Engine

Iron & Friction Losses

Electric Power Developed in Armature = Eg * Ia Watt

Electric Power output CU Losses

= V * I Watt

Following are the three generator efficiencies: 1- Mechanical Efficiency

𝜂𝑚 =

𝐵 𝑡𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐸𝑔 ∗ 𝐼𝑎 = = 𝐴 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝐸𝑛𝑔𝑖𝑛𝑒

2- Electrical Efficiency 𝜂𝑒 =

𝐶 𝑤𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑉∗𝐼 = = 𝐵 𝑡𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝐸𝑔 ∗ 𝐼𝑎

3- Overall or Commercial Efficiency 𝜂𝑐 =

𝐶 𝑤𝑎𝑡𝑡𝑠 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑙𝑜𝑎𝑑 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑉∗𝐼 = = 𝐴 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝐸𝑔 ∗ 𝐼𝑎

It is obvious that overall efficiency ηc = ηm * ηe. For good generator, its value by as high as 90%. Note: Unless specified otherwise, commercial efficiency is always to be understood.

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Condition for Maximum efficiency Generator output = V * I Generator input = output + losses 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑛𝑝𝑢𝑡 = 𝑉 ∗ 𝐼 + 𝐼𝑎2 𝑅𝑎 + 𝑊𝑐 = [𝑉 ∗ 𝐼] + [(𝐼 + 𝐼𝑠ℎ )2 𝑅𝑎 ] + [𝑊𝑐 ] ∵ 𝐼𝑎 = 𝐼 + 𝐼𝑠ℎ However, if Ish is negligible as compared to load current, then Ia = I (approx.) ∴

𝜂=

𝑜𝑢𝑡𝑝𝑢𝑡 𝑉∗𝐼 𝑉∗𝐼 = = (𝑉 ∗ 𝐼) + 𝐼𝑎2 𝑅𝑎 + 𝑊𝑐 (𝑉 ∗ 𝐼) + (𝐼 ∗ 𝑅𝑎 ) + 𝑊𝑐 𝑖𝑛𝑝𝑢𝑡 𝜼=

𝟏 𝑰 ∗ 𝑹𝒂 𝑾𝒄 𝟏+[ + ] 𝑽 𝑽∗𝑰

Now, efficiency is maximum when denominator is maximum i.e. when 𝒅 𝑰 ∗ 𝑹𝒂 𝑾𝒄 + [ ]=𝟎 𝒅𝑰 𝑽 𝑽∗𝑰

𝒐𝒓

𝑹𝒂 𝑾𝒄 − =𝟎 𝑽 𝑽 ∗ 𝑰𝟐

𝒐𝒓

𝑰𝟐 𝑹𝒂 = 𝑾𝒄

Hence, generator efficiency is maximum when

Variable losses = Constant losses The load current corresponding to maximum efficiency is given by the relation. 𝐼 2 𝑅𝑎 = 𝑊𝑐

𝑜𝑟

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𝑊𝑐 𝐼= √ 𝑅𝑎

Maximum efficiency for Generator & Motor If the output current is I, then the output power is VI. Total input power = 𝑉 𝐼 + 𝐼𝑎2 𝑅𝑎 + 𝐼𝑓 𝑉 + 𝐶. Hence

The efficiency of a generator is a maximum when the load is such that: 𝐼𝑎2 𝑅𝑎 = (𝑉 ∗ 𝐼𝑓 ) + 𝐶 i.e. when the variable loss = the constant loss The efficiency of a d.c. motor The efficiency of a d.c. machine is given by: Also, the total losses = 𝐼𝑎2 𝑅𝑎 + 𝐼𝑓 𝑉 + 𝐶 (for a shunt motor) where C is the sum of the iron, friction and windage losses. For a motor, the input power = VI and the output power = VI – losses Hence

The efficiency of a motor is a maximum when the load is such that:

Also, the total losses = 𝐼𝑎2 (𝑅𝑎 + 𝑅𝑓 ) + 𝐶 (for a series motor) where C is the sum of the iron, friction and windage losses.

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