zxing barcode reader java download Y = j 6 10 6 /mile (100 miles) = j 00006 S in Software

Making PDF 417 in Software Y = j 6 10 6 /mile (100 miles) = j 00006 S

Y = j 6 10 6 /mile (100 miles) = j 00006 S
Generate PDF417 In None
Using Barcode maker for Software Control to generate, create PDF-417 2d barcode image in Software applications.
PDF417 Reader In None
Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications.
The ABCD constants for this line are:
Painting PDF417 In C#.NET
Using Barcode drawer for Visual Studio .NET Control to generate, create PDF-417 2d barcode image in Visual Studio .NET applications.
Drawing PDF417 In .NET
Using Barcode creation for ASP.NET Control to generate, create PDF 417 image in ASP.NET applications.
Z = ( 020 + j 085 /mile )(100 miles ) = 20 + j85
PDF-417 2d Barcode Generator In Visual Studio .NET
Using Barcode creator for VS .NET Control to generate, create PDF417 image in Visual Studio .NET applications.
PDF417 Generation In VB.NET
Using Barcode creation for Visual Studio .NET Control to generate, create PDF 417 image in .NET framework applications.
ZY ( 20 + j85 )( j00006 S) + 1 = 09745 035 +1 = 2 2
Encoding Code-128 In None
Using Barcode creation for Software Control to generate, create Code 128 Code Set B image in Software applications.
Encode Data Matrix In None
Using Barcode creator for Software Control to generate, create Data Matrix 2d barcode image in Software applications.
( 20 + j85 )( j 00006 S) ZY + 1 = ( j 00006 S) + 1 = 000059 902 S C =Y 4 4
UPC A Generation In None
Using Barcode maker for Software Control to generate, create UPC Code image in Software applications.
Creating Barcode In None
Using Barcode generation for Software Control to generate, create bar code image in Software applications.
B = Z = 20 + j85 = 873 768
UPC - 13 Generation In None
Using Barcode generator for Software Control to generate, create EAN-13 image in Software applications.
EAN / UCC - 13 Maker In None
Using Barcode maker for Software Control to generate, create EAN 128 image in Software applications.
ZY ( 20 + j85 )( j00006 S) + 1 = 09745 035 +1 = 2 2
Print ISSN In None
Using Barcode drawer for Software Control to generate, create International Standard Serial Number image in Software applications.
EAN / UCC - 13 Printer In Java
Using Barcode printer for Java Control to generate, create GS1-128 image in Java applications.
Assuming that the receiving end voltage is at 0 , the receiving end phase voltage and current are
Barcode Maker In Java
Using Barcode creation for Java Control to generate, create barcode image in Java applications.
Code128 Encoder In None
Using Barcode maker for Word Control to generate, create Code 128A image in Microsoft Word applications.
VR = 110 0 kV/ 3 = 635 0 kV P 60 MW IR = = = 370 A 3V cos 3 (110 kV )( 085)
Code 128 Code Set B Creation In Visual Basic .NET
Using Barcode encoder for VS .NET Control to generate, create Code 128C image in .NET applications.
ECC200 Creation In VB.NET
Using Barcode creation for Visual Studio .NET Control to generate, create ECC200 image in .NET framework applications.
I R = 370 317 A
Scanning USS Code 39 In Java
Using Barcode reader for Java Control to read, scan read, scan image in Java applications.
Drawing Code 3 Of 9 In Java
Using Barcode encoder for Eclipse BIRT Control to generate, create Code 39 Full ASCII image in Eclipse BIRT applications.
The receiving end power factor is 085 lagging The receiving end line voltage and current are 110 kV and 370 A, respectively (b) The sending end voltage and current are given by
VS = AVR + BI R = A ( 635 0 kV ) + B ( 370 317 A ) VS = 878 1535 kV I S = CVR + DI R = C ( 635 0 kV ) + D ( 370 317 A ) I S = 3424 26 A
The sending end power factor is cos 1535 ( 26 ) = cos ( 414 ) = 0751 lagging The sending end line voltage and current are 3 ( 878 kV ) = 152 kV and 342 A, respectively (c) The power at the sending end of the transmission line is
PS = 3V ,S I ,S cos = 3 ( 87,800)( 342 )( 0751) = 677 MW
The power at the receiving end of the transmission line is
PR = 3V ,R I ,R cos = 3 ( 63,500)( 370)( 085) = 599 MW
Therefore the losses in the transmission line are approximately 78 MW (d) The angle is 1535 It is about 1/4 of the way to the line s static stability limit
10: Power System Representation and Equations
10-1 Sketch the per-phase, per-unit equivalent circuit of the power system in Figure 10-2 (Treat each load on the systems as a resistance in series with a reactance) Note that you do not have enough information to actually calculate the values of components in the equivalent circuit
SOLUTION The per-phase, per-unit equivalent circuit would be:
Bus 1 T1 Line T2 Bus 2
Load A
Load B
10-2
A 20,000 kVA, 110/138 kV, Y- three phase transformer has a series impedance of 002 + j008 pu Find the per-unit impedance of this transformer in a power system with a base apparent power of 500 MVA and a base voltage on the high side of 120 kV SOLUTION The per-unit impedance to the new base would be:
per-unit Z new
V S = per-unit Z given given new Vnew Sgiven
(10-8)
110 kV 500,000 kVA per-unit Z new = ( 002 + j 008) = 042 + j168 pu 120 kV 20,000 kVA
10-3
Find the per-phase equivalent circuit of the power system shown in Figure P10-1
SOLUTION The per-phase equivalent circuit must be created on some system base voltage and apparent power Since this problem has not specified the system base values, we will use the ratings of generator G1 as the system base values at that point Therefore, the system base apparent power is Sbase = 30 MVA , and the system base voltages in each region are:
Vbase,1 = 138 kV 115 kV 115 kV Vbase,2 = (138 kV ) = 120 kV Vbase,1 = 132 kV 132 kV 125 kV 125 kV Vbase,3 = V = (120 kV ) = 125 kV 120 kV base,2 120 kV
The base impedance of Region 2 is:
Z base,2
(V =
LL , base,2
S3 , base
(120,000 V ) 2
30,000,000 VA
= 480
The per unit resistance and reactance of G1 are already on the proper base:
Z G1 = 01 + j10 pu
The per unit resistance and reactance of T1 are:
per-unit Z new
V S = per-unit Z given given new Vnew Sgiven
132 kV 30,000 kVA Z T 1 = ( 001 + j 010) = 000784 + j 00784 pu 138 kV 35,000 kVA
The per unit resistance and reactance of the transmission line are:
Z line =
Z 5 + j 20 = = 00104 + j 00417 pu Z base 480
The per unit resistance and reactance of T2 are already on the right base:
Z T 2 = 001 + j 008 pu
The per unit resistance and reactance of M1 are: 194
125 kV 30, 000 kVA Z M 1 = ( 01 + j11) = 015 + j165 pu 125 kV 20,000 kVA
The per unit resistance and reactance of M1 are:
125 kV 30,000 kVA Z M 2 = ( 01 + j11) = 030 + j 330 pu 125 kV 10,000 kVA
The resulting per-phase equivalent circuit is:
T1 j00784 01 j10 Line 00104 j008 T2 001 015 030
000784 j00417
j165
j330
10-4
Two 416 kV three-phase synchronous motors are connected to the same bus The motor ratings are: Motor 1: 5,000 hp, 08 PF lagging, 95% efficiency, R = 3%, X S = 90% Motor 2: 3,000 hp, 10 PF, 95% efficiency, R = 3%, X S = 90% Calculate the per-unit impedances of these motors to a base of 20 MVA, 416 kV (Note: To calculate these values, you will first have to determine the rated apparent power of each motor considering its rated output power, efficiency, and power factor) SOLUTION The rated input power of Motor 1 is
P= 1
Copyright © OnBarcode.com . All rights reserved.