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zxing barcode reader java download f nl1 = 6191 Hz in Software
f nl1 = 6191 Hz Generate PDF 417 In None Using Barcode drawer for Software Control to generate, create PDF 417 image in Software applications. Read PDF417 In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. (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 Paint PDF417 2d Barcode In C# Using Barcode printer for Visual Studio .NET Control to generate, create PDF417 2d barcode image in .NET framework applications. Generating PDF 417 In VS .NET Using Barcode generation for ASP.NET Control to generate, create PDF417 2d barcode image in ASP.NET applications. 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 Generating PDF417 In .NET Using Barcode encoder for VS .NET Control to generate, create PDF417 2d barcode image in .NET framework applications. Making PDF 417 In Visual Basic .NET Using Barcode creator for .NET framework Control to generate, create PDF417 image in VS .NET applications. Each generator will supply 967 MW to the loads 68 Suppose that you were an engineer planning a new electric cogeneration facility for a plant with excess process steam You have a choice of either two 10 MW turbinegenerators 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 UPCA Supplement 2 Generator In None Using Barcode creation for Software Control to generate, create UPCA Supplement 5 image in Software applications. Encoding DataMatrix In None Using Barcode drawer for Software Control to generate, create Data Matrix ECC200 image in Software applications. 7: Induction Motors
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A 220V threephase sixpole 50Hz 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 =
73 nslip P 120
( 35 r/min )( 6) = 175 Hz
Answer the questions in Problem 72 for a 480V threephase fourpole 60Hz 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 =
74 nslip P 120
( 45 r/min )( 4) = 15 Hz
A threephase 60Hz 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 onequarter of the rated load (d) What is the rotor s electrical frequency at onequarter 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 torquespeed 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
75 A 50kW 440V 50Hz twopole induction motor has a slip of 6 percent when operating at fullload conditions At fullload conditions, the friction and windage losses are 520 W, and the core losses are 500 W Find the following values for fullload conditions: (a) The shaft speed nm (b) The output power in watts (c) The load torque load in newtonmeters (d) The induced torque ind in newtonmeters 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
76 A threephase 60Hz twopole induction motor runs at a noload speed of 3580 r/min and a fullload speed of 3440 r/min Calculate the slip and the electrical frequency of the rotor at noload and fullload conditions What is the speed regulation of this motor [Equation (457)] SOLUTION The synchronous speed of this machine is 3600 r/min The slip and electrical frequency at noload conditions is

