ECE 5377 / 6377 Power Transmission and Distribution Project

7/25/2019 ECE 5377 / 6377 Power Transmission and Distribution Project

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What follows Refers to the Guide for the Project Task # 1:

1. (Almost identical with the first paragraph on the Guide, page4)

As it was specified on the Project Objective, power plants canbe built at sites A and C, with an upper limit on the active powersupplied from site C of 200 /MW/. How it is shown on the ProjectSolution, for the solved case it came out that the best version isthat one with a power plant at C (along with that one at A). But donot understand from here that also you must have a power plant atC. Your data are not the same with data given on the ProjectSolution. It means that it is possible that in your case, byconsidering technical and economical factors, to result that the bestversion is without a power plant at C.

2. (Almost identical with the second paragraph on the Guide,page 4)

How it is stated on the Project Tasks, page 4, item 1, the loadmust be supplied 100% just if one component of the new powersystem section is out of service. Such, do not forget to check that

just when a generating unit, or a transformer unit, or a lines circuitis out of service, what it is left on can carry the load, including activeand reactive power losses.

3. (Identical with the third paragraph on the Guide, page 4)

For the network components you have to check if eachcomponent that is in operation can carry the maximum apparentpower that it must be carried.

4. (A little different from the fourth paragraph of the Guide, page4, but read both, they are referred to the same topics)

In defining the number, rated power and voltage of different systemcomponents, it must be estimated what value has the apparentpower that must be carried by each component of the system. For

that reason, starting from the load side, the components loading isestimated.On the Project Solution the active and reactive powers lost havebeen defined in a more exact way. Here it is suggested a verysimple way to estimate the powers lost, and then the apparentpower at the sending end of an equivalent network component (hereby the equivalent network component it is understood the


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equivalent of the same type components parallel connected).Lets assume that the loads total active power consumed is 1,000/MW/ and, since the total active power lost is estimated to be 5%, itresults that the total active power lost is 50 /MW/. Lets assumethat for the configuration in discussion there are three powersubstations and four transmission lines that is a total of sevenequivalent network components. Then, as a simple estimation, itwill stated that each equivalent network component has an activepower lost of (50/7) = 7.14 /MW/.Similarly, starting with the loads total reactive power consumed andwith the percent of total reactive power lost (estimated as being10%), it is estimated the reactive power lost per equivalent networkcomponent.

Having both active and reactive powers lost for each equivalentnetwork component, and also the active and reactive power at thereceiving end of the equivalent component, now it can be estimated

the active and reactive, and then the apparent powers at the sendingend of each equivalent network component.

(Take care, on the Guide, page 4, fourth row, the correct apparentpower should be 1068.27 /MW/, not 1187.94 /MW/)

5. (Identical with the first distinct paragraph on the Guide, page4)

Again, as it was already stated, for units that are to operate inparallel (as generating units in a power plant, transformer units in apower substation, lines circuits for a line), it is very probable thatthe investment cost is lower as the number of these units is lowerand the corresponding units size and their rated/carrying capabilitypower is higher. But at the same time the supplementaryinvestment cost for the back-up units is higher. Such it must beseen how this cost is compared with the cost of more units inparallel, each of a lower rated power.

6. (Almost identical with the second distinct paragraph on theGuide, page 4)

On the above decision also it must be counted the circuit flexibility

and responsiveness to the actual loading.

Lets assume that for a power substation it is known that theannual peak load is 200 /MVA/, and the annual minimum load is 100/MVA/.

Version # 1: Lets assume that, offering a lower cost, it isdecided to install there only two transformer units, each of


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200 /MVA/ rated power. At the first view, for the peak load itlooks ok. But for the minimum loading the first temptationwill be to have in operation only one transformer, which canhandle that power. But that it is forbidden. If for somereasons that single transformer goes out of service theconnection through that power substation is interrupted.Always it must be at least two transformers in operation.That is, now at the minimum loading, each transformer isloaded with 50 /MVA/, that is 25% of its rated power, andthis is not an economical operation.

Also it must be counted that periodically units must go into amaintenance process. If on the above power substation thereare installed only two transformers, and since the operationcannot remain with only one transformer, it means that notransformer can go out for a maintenance schedule, which ofcourse cannot be accepted.

Version # 2: Now lets assume that for the same power

substation it is decided to have installed three transformers,each of a 100 /MVA/ rated power (if one unit is out, the othertwo can carry 200 /MVA/). For the minimum loading of 100/MVA/, it can remain in operation only two transformers,each being loaded with 50 /MVA/, which represents 50% ofits rated loading.

Now it is clear that there is a more economical loading thanbefore. Also transformers can be scheduled for maintenanceone by one with no problem. And just if one transformer ofthose two operating at minimum loading is going out ofservice, still it can be transferred 100 /MVA/ through thepower station. Such, just if this version results with highercost than the cost of Version # 1, this version must beselected.

7. (Identical with the last paragraph on the Guide, page 4)

The same judgment must be applied for generating units orlines circuits that are operating in parallel.

8. What follows is related to Fig. B7-1, page B-129:

Since the Line Loading in SIL curve was defined for an extra-highvoltage line, for a shorter lower voltage line the curve data must beadjusted.

Such the curve will be used for defining the Line Loading in SIL asfunction of the line length, but if it is coming out a multiple of SIL


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higher than 1.5, it will selected 1.5, not higher.


Page 5:---------

1. Take care, after you define the minimum needed cross section areafor conductor, from Table A.1. you have to select the correspondingconductor that it is available. If you do not find a conductor thatsatisfy exactly the calculated needed cross section, then you haveto selected the immediate next available conductor that has a crosssection higher that the minimum cross section area calculated.

Page 6, items 2.4 and 2.5:----------------------------------

2. For the parallel identical circuits installed on the same tower it mustbe defined parameters and build the corresponding sequenceequivalent circuits (for positive = negative and for zero sequences)for when only one circuit is in operation, and for when all circuitsinstalled on that tower are in operation.

3. If between terminal X and terminal Y there are parallel circuitsinstalled on more than one tower, then circuits installed on onetower define a line. Such, between terminal X and terminal Y will bemore parallel lines.For each of these lines it must be done what it is stated on the item2 above.Also, it must be calculated and built the corresponding sequenceequivalent circuits when all parallel lines between terminals X and Yare in operation.


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Page 2, item 3.2:---------------------

1. For identical transformers parallel connected between two terminalsX and Y it must be calculated parameters and build thecorresponding sequence equivalent circuits (for positive=negativeand for zero sequences) for when only one transformer is inoperation and for when all identical parallel transformers are inoperation.

2. If between two terminals X and Y there are connected transformersof different characteristics, then for the set of transformers ofidentical characteristics it is applied what it is stated on the aboveitem 1

Finally, it must be defined parameters and build the correspondingsequence equivalent circuits when all transformers parallelconnected between terminals X and Y are in operation.


Page 1, item 4.2:---------------------

On the equivalent circuits the series parameters to be defined andshownas impedances components (R and jX), and the shunt parameters tobe defined and shown as admittance components (G and jB)


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Page 2:----------

Here it is given another equation for defining the circuits current

Im, as(take care, on what follows all complex quantities, that are phasors,impedances, admittances, and the complex power are shown with

larger size letters),

Sn*Im = (1+ ZmnYm0) x --- + (Ym0+ Yn0+ Zmnx Ym0x Yn0)x Vn

Vn*


Page 1, item 7.1:---------------------

The generating units parameters are selected as the average valueof the range shown on Table A.4, Task # 3.

Page 1, item 7.2:---------------------

When the load parameters are calculated, to be considered thatnow by load it is understood the actual load and, where it the case,the shunt reactive compensation used for that load.

The reactive compensating devices are considered as passivecomponents Y connected and with the neutral grounded.


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Page 2, item 7.3:---------------------

Take care that, when the zero sequence impedance diagram is built itmust be counted the transformers winding connection, and that thisfact will decide which power system components will participate inthe zero sequence impedance diagram.

What follows Refers to the Project Solution

Page 41:-----------

The sentence to be read as: In the following it will be definedparameters for when one or for when both circuits are in operation.

Page 61:-----------

On the second paragraph it is stated that To satisfy this constraint,analytically it was found that .

Since you are not yet ready to solve this case analytically, you haveto solve the case by trying.

Page 101:------------

As it can be seen on Figure 8.4, the series parameters of each circuitof line AB are equal to two times (X = 0068 /pu/) parameters forboth circuits of line AB in operation (see Figure 4.1, page 57, wherefor line AB, X = 0.034 /pu/).

Since, just with the short circuit on to one of the circuits, bothcircuits are in operation, the shown procedure must be applied.


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O. Crisan, ECE 5377/6377

Project Updated Specifications-----------------------------------------

Spring 2012-----------------

To the given Project Guide, the following changes apply:

1. On the Section PROJECT OBJECTIVE, page 1, first paragraph, tobe read: to become operational on the year 2,016.

2. On page 2 of the Section Given Data:

- The data shown at item # 1 become:

d1= (30 + 2 n) /km/; d2= (200 - 2 n) /km/; d3= (100 + 2.5 n)/km/.

- The data shown at item # 2, become:

PA = (20 + n) /MW/, PB = (250 + 2 n) /MW/,

PC = (300 3 n) /MW/.

- Item # 4, first paragraph, becomes:

Load Curves for that day of the year 2016

- For the figure on page # 3 the title becomes:

Daily load curves, (2,016 2,025),

3. On the Section Project Tasks:

- Page # 4, Item # 1, second paragraph, to be read:

- Along a 10 years period, starting with year 2,016 andending with year 2,025, .


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O. Crisan, ECE 5377/6377

- 2 -

- Page 5, Item # 5, the shown table: on the table head, line Year,to

be read:

| 2,016 | 2,017 | |2,025|

- Page 6, Item # 8, second phrase, it would be read as:

For the section of the power system composed of the higher

voltage side bus of the step-up power substation at side A, and thetransmission line AB, design the primary and back-up protectionsystems.

4. General Comments: On the Sections Guide for the Project Task # 1, ,

8, and on the Section PROJECT SOLUTION, every time whenyears

are involved, it must be understood that now, along the designinterval the

first year is 2,016, and the last year is 2,025!

5. When defining the line parameters, for the return path resistance in the ground (see PROJECT SOLUTION, page 34), instead of 0.05

//km/, to

be considered Rgu= 0.06 //km/!!!! Also, for the equivalent distanceto the

ground, instead of 1000 /m/ (see the same section, page 37) to beconsidered

as Deq.cg

= 850 /m/!!!!!!

6. The Tasks Time-Schedule becomes:

Task #: Due Date (Along Year 2012):

1 Wednesday, February 15


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2 Wednesday, February 293 Monday, March 194 Monday, March 195 Monday, April 26 Monday, April 27 Monday, April 168 Monday, April 30


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Documents

ECE 5377 / 6377 Power Transmission and Distribution Project