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(d) If the swing generator trips off the line, the other three generators would have to supply all 290 MW of the load Therefore, the system frequency will become
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PLOAD = s P1 ( f nl1 f sys ) + s P 2 ( f nl2 f sys ) + s P 3 ( f nl3 f sys ) 290 MW = (34 )(6221 - f sys ) + (34 )(6221 - f sys ) + (34 )(6221 - f sys ) 8529 = 18663 3 f sys f sys = 5937 Hz
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Each generator will supply 967 MW to the loads 6-8 Suppose that you were an engineer planning a new electric co-generation facility for a plant with excess process steam You have a choice of either two 10 MW turbine-generators or a single 20 MW turbine generator What would be the advantages and disadvantages of each choice SOLUTION A single 20 MW generator will probably be cheaper and more efficient than two 10 MW generators, but if the 20 MW generator goes down all 20 MW of generation would be lost at once If two 10 MW generators are chosen, one of them could go down for maintenance and some power could still be generated
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7: Induction Motors
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7-1 A dc test is performed on a 460-V -connected 100-hp induction motor If VDC = 21 V and I DC = 72 A, what is the stator resistance R1 Why is this so SOLUTION If this motor s armature is connected in delta, then there will be two phases in parallel with one phase between the lines tested
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VDC R1 R1
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Therefore, the stator resistance R1 will be
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VDC R (R + R1 ) 2 = 1 1 = R1 I DC R1 + (R1 + R1 ) 3
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R1 =
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7-2
3 VDC 3 21 V = = 0438 2 I DC 2 72 A
A 220-V three-phase six-pole 50-Hz induction motor is running at a slip of 35 percent Find: (a) The speed of the magnetic fields in revolutions per minute (b) The speed of the rotor in revolutions per minute (c) The slip speed of the rotor (d) The rotor frequency in hertz SOLUTION (a) The speed of the magnetic fields is
nsync =
120 f e 120(50 Hz ) = = 1000 r/min P 6
The speed of the rotor is
nm = (1 s ) nsync = (1 0035)(1000 r/min ) = 965 r/min
(c) The slip speed of the rotor is
nslip = snsync = (0035)(1000 r/min ) = 35 r/min
(d) The rotor frequency is
fr =
7-3
nslip P 120
( 35 r/min )( 6) = 175 Hz
Answer the questions in Problem 7-2 for a 480-V three-phase four-pole 60-Hz induction motor running at a slip of 0025 SOLUTION (a) The speed of the magnetic fields is 114
nsync =
120 f e 120 ( 60 Hz ) = = 1800 r/min P 4
The speed of the rotor is
nm = (1 s ) nsync = (1 0025)(1800 r/min ) = 1755 r/min
(c) The slip speed of the rotor is
nslip = snsync = ( 0025)(1800 r/min ) = 45 r/min
(d) The rotor frequency is
fr =
7-4
nslip P 120
( 45 r/min )( 4) = 15 Hz
A three-phase 60-Hz induction motor runs at 715 r/min at no load and at 670 r/min at full load (a) How many poles does this motor have (b) What is the slip at rated load (c) What is the speed at one-quarter of the rated load (d) What is the rotor s electrical frequency at one-quarter of the rated load
SOLUTION (a) This machine has 10 poles, which produces a synchronous speed of
nsync =
120 f e 120 ( 60 Hz ) = = 720 r/min P 10 720 670 100% = 694% 720
The slip at rated load is
nsync nm nsync
100% =
The motor is operating in the linear region of its torque-speed curve, so the slip at load will be
s = 025(00694) = 00174
The resulting speed is
nm = (1 s ) nsync = (1 00174 )( 720 r/min ) = 707 r/min
(d) The electrical frequency at load is
f r = sf e = ( 00174 )( 60 Hz ) = 104 Hz
7-5 A 50-kW 440-V 50-Hz two-pole induction motor has a slip of 6 percent when operating at full-load conditions At full-load conditions, the friction and windage losses are 520 W, and the core losses are 500 W Find the following values for full-load conditions: (a) The shaft speed nm (b) The output power in watts (c) The load torque load in newton-meters (d) The induced torque ind in newton-meters 115
(e) The rotor frequency in hertz SOLUTION (a) The synchronous speed of this machine is
nsync =
120 f e 120 ( 50 Hz ) = = 3000 r/min P 2
Therefore, the shaft speed is
nm = (1 s ) nsync = (1 006)( 3000 r/min ) = 2820 r/min
(b) (c) The output power in watts is 50 kW (stated in the problem) The load torque is
load =
POUT
50 kW = 1693 N m 2 rad 1 min (2820 r/min ) 1 r 60 s
The induced torque can be found as follows:
Pconv = POUT + PF&W + Pcore + Pmisc = 50 kW + 520 W + 500 W = 512 kW
ind =
Pconv
512 kW = 1734 N m 2 rad 1 min (2820 r/min ) 1 r 60 s
The rotor frequency is
f r = sf e = ( 006)( 50 Hz ) = 300 Hz
7-6 A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load speed of 3440 r/min Calculate the slip and the electrical frequency of the rotor at no-load and full-load conditions What is the speed regulation of this motor [Equation (4-57)] SOLUTION The synchronous speed of this machine is 3600 r/min The slip and electrical frequency at noload conditions is
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