Figure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given a follows: G : 90MVA 22kV x = 0 .18 per unit T 1 : 50MVA 22/220kV x = 0 .10 per unit T 2 : 40MVA 220/11kV x = 0 .06 per unit T 3 : 40MVA 22/110kV x = 0 .064 per unit T 4 : 40MVA 110/11kV x = 0 .08 per unit M : 66 .5MVA 10 .45kV x = 0 .185 per unit Lines I and 2 have series reactance’s of 48.4 and 65.43 Ω , respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.
Figure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given a follows: G : 90MVA 22kV x = 0 .18 per unit T 1 : 50MVA 22/220kV x = 0 .10 per unit T 2 : 40MVA 220/11kV x = 0 .06 per unit T 3 : 40MVA 22/110kV x = 0 .064 per unit T 4 : 40MVA 110/11kV x = 0 .08 per unit M : 66 .5MVA 10 .45kV x = 0 .185 per unit Lines I and 2 have series reactance’s of 48.4 and 65.43 Ω , respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.
Figure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given a follows:
G
:
90MVA
22kV
x
=
0
.18
per
unit
T
1
:
50MVA
22/220kV
x
=
0
.10
per
unit
T
2
:
40MVA
220/11kV
x
=
0
.06
per
unit
T
3
:
40MVA
22/110kV
x
=
0
.064
per
unit
T
4
:
40MVA
110/11kV
x
=
0
.08
per
unit
M
:
66
.5MVA
10
.45kV
x
=
0
.185
per
unit
Lines I and 2 have series reactance’s of 48.4 and
65.43
Ω
, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.
I would like to get help to resolve the following case
Last Chance Securities
The IT director opened the department staff meeting today by saying, "I've got some good news and
some bad news. The good news is that management approved the payroll system project this morning.
The new system will reduce clerical time and errors, improve morale in the payroll department, and avoid
possible fines and penalties for noncompliance. The bad news is that the system must be installed by
January 1st in order to meet new federal reporting rules, all expenses from now on must be approved in
advance, the system should have a modular design if possible, and the vice president of finance would
like to announce the new system in a year-end report if it is ready by mid-December."
Tasks
1. Why is it important to define the project scope? How would you define the scope of the payroll
project in this case?
2. Review each constraint and identify its characteristics: present versus future, internal versus exter-
nal, and mandatory versus desirable.
3. What…
2. Signed Integers
Unsigned binary numbers work for natural numbers, but many calculations use negative
numbers as well. To deal with this, a number of different methods have been used to represent
signed numbers, but we will focus on two's complement, as it is the standard solution for
representing signed integers.
2.1 Two's complement
• Most significant bit has a negative value, all others are positive. So, the value of an n-digit
-2
two's complement number can be written as: Σ2 2¹ di 2n-1 dn
• Otherwise exactly the same as unsigned integers.
i=0
-
• A neat trick for flipping the sign of a two's complement number: flip all the bits (0 becomes 1,
or 1 becomes 0) and then add 1 to the least significant bit.
• Addition is exactly the same as with an unsigned number.
2.2 Exercises
For questions 1-3, answer each one for the case of a two's complement number and an
unsigned number, indicating if it cannot be answered with a specific representation.
1. (15 pts) What is the largest integer…
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