Lecture (20) DC Machine Examples Start of Synchronous Machines Energy Systems Research Laboratory, FIU All rights reserved. 20-1
Energy Systems Research Laboratory, FIU All rights reserved. 20-2
Ra R f + I m V t T e, ω m Series DC Motor - Ra + I m V t Series DC Generator - Energy Systems Research Laboratory, FIU All rights reserved. 20-3
Ra + I a I a R f I t V t T e, ω m Shunt DC Motor Ra + I a I a R f I t V t Shunt Generator Energy Systems Research Laboratory, FIU All rights reserved. 20-4
+ S 2 I t R S S 1 F 1 I a Ra V t - R f A 1 F 2 A 2 Short Shunt Compound DC Motor + S 2 I t R S S 1 F 1 I a Ra V t - R f A 1 F 2 A 2 Short Shunt Compound DC Generator Energy Systems Research Laboratory, FIU All rights reserved. 20-5
+ S 2 R S S 1 I t I f F 2 I a Ra V t R f A 1 _ F 1 A 2 Long Shunt Compound DC Motor + S 2 R S S 1 I t I f F 2 I a Ra V t R f A 1 _ F 1 A 2 Long Shunt Compound DC Generator Energy Systems Research Laboratory, FIU All rights reserved. 20-6
Example Shunt DC Machine V t = 440 volts R f = 110 Ω R a = 0.15 Ω Connected to 22 kw load Calculate the developed power as a motor and as a generator Energy Systems Research Laboratory, FIU All rights reserved. 20-7
Motor P Generator Ra = 0.15 I a I f R I t f V t R f =110 Ω L o a d 22 kw T e, ω m Energy Systems Research Laboratory, FIU All rights reserved. 20-8
Motor P develop = I a I a = I t - I f = Load V t V R t f I a = 22, 000 440 440 110 = 46 A = V t - I a R a = 440 (46) (0.15) = 433.1 V. P develop = 433.1 46 = 19,922.6 W Energy Systems Research Laboratory, FIU All rights reserved. 20-9
For a Generator Ra = 0.15 Load I a I f R f I t V t 22 kw = V t + I a R a I t = 22kW = 50 A 440 Energy Systems Research Laboratory, FIU All rights reserved. 20-10
440 I f = = 4 A 110 I a = 50 + 4 = 54 A = 440 + ( 54 0.15 ) = 448.1 V. P developed = 448.1 54 = 24.197 kw Energy Systems Research Laboratory, FIU All rights reserved. 20-11
Problem #2. 5 hp, shunt DC machine, 1800 rpm I t = 36 A, R a = 0.3 Ω, R f = 120 Ω. Ra I a I f R f I t = 36 A s o V t u r c e Energy Systems Research Laboratory, FIU All rights reserved. 20-12
Find the rotational losses. Find η =?, Assume V t = 120 V. P out = 5 ( 746 ) = 3730W P in = 120 ( 36 ) = 4320W 3730 η = 100 % = 86. 3 % 4320 Energy Systems Research Laboratory, FIU All rights reserved. 20-13
P losses = P in - P out = 4320 3730 = 590 W. 120 I f = = 1 A 120 I a = 36-1 = 35 A Energy Systems Research Laboratory, FIU All rights reserved. 20-14
P copper loss = P f loss + P a loss = I f 2 R f + I a 2 R a = (1) 2 (120) + (35) 2 (0.3) = 487.5 W P rotational = 590 487.5 = 102.5 W Energy Systems Research Laboratory, FIU All rights reserved. 20-15
Problem #3. Series Motor, R a + S = 0.2 Ω, I a rated = 50 A, V t = 240 V. Find the Starting Current. Ra + R s = 0.2 Ω + I a - Energy Systems Research Laboratory, FIU All rights reserved. 20-16
I a starting = V t E R t a = 240 0 0. 2 = 1200 A = 1200 100% = 2400% 50 Energy Systems Research Laboratory, FIU All rights reserved. 20-17
Calculate the Resistance Now limit the starting current to 150% (of full load rated current) by inserting an external resistance. 240 0 1.5 50 = 0. 2 + R ext. 75 ( 0.2 + R ext. ) = 240 R ext. = 3 Ω Energy Systems Research Laboratory, FIU All rights reserved. 20-18
Example in class A DC Motor Connected to, 600-V, 18 kw Source. A. Calculate the Starting Current drawn from the source B. Limit the Starting Current to equal the rated current obtained by 18 kw, 600 Voltage source. (Hint: a Starting resistance is added to the armature) Ra=0.1 I a V t V t =600V I f R f =150 Energy Systems Research Laboratory, FIU All rights reserved. 20-19
Synchronous Machines e = d λ = dt N d φ dt E = 4.44 f N ph φ p k w k w = k d k p k s Distribution pitch skewing Factor factor factor n s = 120 f 1 n s α, n s α f P P Energy Systems Research Laboratory, FIU All rights reserved. 20-20
Example P = 6, f = 60 Hz, find n s at 50 Hz operation. n s = 120 f P ( ) 120 60 n s (60HZ) = 6 = 1200 rpm Energy Systems Research Laboratory, FIU All rights reserved. 20-21
( 120 n s (50HZ) = 6 50) = 1000 rpm E c 120 120 = 0 120 E b = - 120 E b E c = 120 Energy Systems Research Laboratory, FIU All rights reserved. 20-22
Example 600 V, n s = 375 rpm, P = 16 poles n s = 120 f P E = 4.44 f φ p N ph k w 375 = 120 f 16 Variables constant f = 50 Hz Energy Systems Research Laboratory, FIU All rights reserved. 20-23
Example 3-φ, synchronous generator, has = 620 V. at 60 Hz. If we 1. Reduce φ p by 15%, and 2. Increase n s by 10%, then 3. Find nd f E E = 2 E 2 1 620 3 Energy Systems Research Laboratory, FIU All rights reserved. 20-24
( )( φ ) E 4.44 1.1f 0.85 2 = 620 4.44 f φ p N 3 ph p K N w ph K w E 2 620 1.1 0.85 = 3 1 ( ) ( ) = 334.7 V Energy Systems Research Laboratory, FIU All rights reserved. 20-25
1. 1 n n s n s = s = 120 f P 120 P 120 P f new f 1.1 = f new f f new = 1.1 60 Hz = 66 Hz Energy Systems Research Laboratory, FIU All rights reserved. 20-26