Power System Design of Super Absorbent Plant at Al-Jubail, KSA · 2020. 2. 15. · Figure 1 – Super Absorbent Polymer Plant 3D model snapshot 4 Figure 2 – Super Absorbent Polymer

i

POWER SYSTEM DESIGN OF SUPER ABSORBENT POLYMER

PLANT AT AL-JUBAIL, KSA

(REVISED)

By:

Ryan M. Magalang

ii

Letter of Transmittal and Approval

iii

May 31, 2011

Eng’r. Fortunato C. Leynes

Chairman

The Board of Electrical Engineering

Professional Regulation Commission

P. Paredes St., Sampaloc, Manila

Philippines

Subject: PROPOSED TECHNICAL REPORT TITLE AND OUTLINE

Dear Sir,

In accordance with Article IV, Rule 14, Item #8, of the

Implementing Rules and Regulation of Republic Act No. 7920, known as the “New Electrical Engineering Law”; I hereby submit

my proposed technical report entitled “ POWER SYSTEMS DESIGN OF SUPER ABSORBENT POLYMER PLANT AT AL-

JUBAIL, KSA ” including report outline for your evaluation.

I hope you will find this proposed technical report in order and satisfactory to your requirement.

Thank you.

Respectfully Yours,

RYAN M. MAGALANG

REE No. 0038154

PEE Examinee

iv

v

Preface

This Technical Engineering Report is submitted as a requirement for

the Professional Electrical Engineer board examination. It contains

work done from June to July 2011. This report has been made

solely by me; preliminary data however were based on the project

which is the subject of this independent study and I have done my

best to mention this in the report.

I would like to express my sincere gratitude to the three

Professional Electrical Engineers who served as my consultants and

mentors; Engr. Cesar R. Buensuceso Jr., Engr. Charles P. Pante and

Engr. Anthony G. Quiogue.

Finally, I wish to express my greatest thanks to my family, friends

and colleagues who have supported and helped me. To my dearest

Marina I give my especial thanks for her continued support,

understanding and for inspiring me always to achieve greater

heights. Above all, to God Almighty is the glory.

vi

Table of Contents Page

Title Page i

Letter of Transmittal and Approval ii

Preface v

List of Figures vii

Report Executive Summary ix

Chapter 1 – Introduction 1

Chapter 2 – Design Criteria 6

Chapter 3 – Electrical Calculations 7

Part 1 - Short Circuit Calculations 7

Part 2 – Overcurrent Protection Device (OCPD), 16

Primary and Secondary Feeder, Neutral Conductor and

Equipment Grounding Conductor

Part 3 – Time – Current Curve (TCC) Coordination 30

Part 4 – Load Flow Calculations 38

Section 1 – Motor starting Analysis 38

Section 2 – Power Factor Capacitor Sizing 40

Calculations

Section 3 – System Harmonic Analysis and 41

Calculations

Chapter 4 – Auxiliary Systems 46

Part 1 – Fire Pump and Fire Alarm System Diagram 46

Part 2 – Paging System Diagram 50

Part 3 – Telephone and I.T. System Diagram 50

Part 4 – Security System (CCTV) Diagram 50

Chapter 5 – Conclusions / Recommendations 55

References 56

Affidavits 57

Curriculum Vitae 65

vii

List of Figures

Figures

Page

Figure 1 – Super Absorbent Polymer Plant 3D model snapshot 4

Figure 2 – Super Absorbent Polymer Plant Key Single Line

Diagram

5

Figure 3 – System diagram of SAP plant

8

Figure 4 –Equivalent impedance diagram with 1BUS-101A/B as reference faulted bus

10

Figure 5 –Equivalent impedance diagram with 2BUS-101A/B as

reference faulted bus

12

Figure 6 –Equivalent impedance diagram with 2BUS-202 as reference

14

Figure 7 - Schematic Diagram of a Solidly Grounded System

29

Figure 8 - Schematic Diagram of a Resistance Grounded

System

29

Figure 9 – System diagram showing LV, MV and HV bus for the TCC coordination

30

Figure 10 – Equivalent system impedance diagram with utility as sole source and 2BUS-202 as faulted bus

31

Figure 11 – Time-Current Curve Coordination analysis

37

Figure 12 – Motor starting analysis single line diagram

39

Figure 13 – Harmonic analysis single line diagram

42

Figure 14 – The 2.7MVAr capacitor bank

44

Figure 15 – Harmonic filter diagram

45

Figure 16 – Fire Pump Motor Starting diagram

48

viii

Figure 17 – Fire Alarm System Diagram 51

Figure 18 – Paging System Diagram 52

Figure 19 – Telephone System and I.T. Diagram 53

Figure 20 – Security System (CCTV) Diagram

54

ix

Report Executive Summary

The subject of this report is the Super Absorbent Polymer Plant, a

project currently in the design stage being executed by Fluor

Philippines. The plant will be constructed in Al-Jubail, KSA. This is a

manufacturing plant intended to produce super absorbent polymer

resins which are the major components of baby diapers.

Preliminary data were used by the author as the basis of design. This

data includes the key single line diagram of the electrical system

where transformer sizes were already defined during the project’s

front-end engineering design stage. From there, the author made the

calculations necessary in order to size the protective devices, feeders

and buses sizes using the Philippines Electrical Code (PEC) and other

supplementary standards and assumptions.

Part 1 of Chapter 3 is the Short Circuit calculations. Using the key

single line diagram, data and assumptions the result of the calculations

were able to determine that the existing bus 1BUS-101A/B rated 40kA

from Substation 1, where Substation 2 will tap for its power source,

can still handle the additional short circuit current contributions from

the new Substation 2 loads. The bus short circuit ratings for Substation

2 were determined to be 50kA for the MV (4.16kV) bus and 70kA for

the LV (400V) bus.

Part 2 of Chapter 3 is the sizing of Overcurrent Protection Devices,

Primary and Secondary Feeders, Neutral Conductors and Equipment

Grounding Conductors. Using the applicable PEC provisions, the sizes

of the primary and secondary breakers of the two 20MVA, 34.5/4.37kV

transformers were determined to be 1250A, 40kA and 3150A, 50kA

x

respectively. The sizes of the primary and secondary breakers of the

six 2.5MVA, 4.16/0.42kV transformers were determined to be 1250A,

50kA and 4000A, 70kA respectively. The MV and LV bus tie-breakers

ratings were determined to be 1250A, 50kA and 4000A, 70kA

respectively.

The 20MVA transformers primary feeder size were determined to be

two single core conductors per phase of 250mm2 copper and the

secondary feeder to be four single core conductors per phase of

500mm2 copper. For the 2.5MVA transformer primary feeder size were

determined to be two single core conductors per phase of 250mm2

copper and the secondary feeder to be six single core conductors per

phase of 500mm2 copper. The neutral conductor of the 2.5MVA

transformers is the same size as the secondary feeder since the

transformers are supplying non-linear loads. These conductor sizes

were calculated taking into considerations the continuous loads, the

short circuit capability of the cables and the voltage drop.

Also, in this part of the report the Neutral Grounding Resistor (NGR) of

the 20MVA transformers were sized to be 400A, 10sec taking into the

consideration the system capacitive charging. Based on this NGR size

the System Grounding Conductor size was determined to be a single

core 125mm2 copper.

The determined sizes of primary and secondary Equipment Grounding

Conductors (EGC) for both the 20MVA and 2.5MVA transformers are

100mm2 copper and 80mm2 copper respectively. These sizes were

based on the applicable provisions of the PEC.

xi

Part 3 of Chapter 3 is the Time-Current Curve Coordination. Here the

author presented a possible scheme of coordination of protective

devices. Four breakers from LV to HV bus were coordinated using the

IEC Extremely Inverse Curve of ABB RXIDK 4 relays. Only the through

fault current or short circuit contributions from the utility/source was

considered in the relay coordination calculations. The result of the

calculations was able to set the Time Overcurrent and Instantaneous

settings of the relays such that they would operate correctly. The relay

curves, transformer damage curves and inrush currents were plotted

(see Figure 11, page 37) to show the results of the calculations.

Part 4 of Chapter 3 is the Load Flow Calculations. This part of the

chapter covers the Motor starting Analysis, Power Factor Capacitor

Sizing Calculations and the System Harmonic Analysis and

Calculations.

The subject of the Motor Starting Analysis is the largest motor rated

1239kW connected to the MV bus. Scenario and assumptions were

made to determine if the motor will start in the worst possible system

condition. The result of the calculations showed that there will be no

problem starting the motor.

Power Factor Capacitor sizing calculations were conducted to

determine the size of the capacitor bank needed to improve the

system power factor from an assumed value of 80% to 96%. The

result of the calculations suggested installing two banks of 2.7MVAr,

3-Phase capacitors.

xii

In the System Harmonic Analysis, calculations were made to

determine if a problem brought about by the VSD driven motors would

exist. The calculations showed that the probability of harmonic

problem is low since the calculated Short Circuit Ratio (SCR) is greater

than 20. However a possible parallel resonance might occur in the

order of 7th harmonic therefore a harmonic filter tuned to this

frequency might be needed to be installed. A harmonic filter for the 5th

order harmonic, which has larger harmonic current contributions

compared with the 7th order harmonic, was designed in this part of the

report. The filter is a series combination of an inductor and a capacitor.

The filter was designed in such a way that the Power Factor Capacitor

was used as part of the filter making the capacitor dual purpose

equipment.

Chapter 4 is the Auxiliary Systems. This chapter shows Fire Alarm

System Diagram, Paging System Diagram, Telephone and I.T. System

Diagram, Security System (CCTV) Diagram. These diagrams were

based on the design philosophy developed by Fluor Philippines Control

Systems Department.

In Section 1 of Chapter 4 the author presented a guideline for the

calculations of feeder and breaker size, power supply and other

considerations in the installation of fire pump based on the provisions

of the PEC. An estimated pump motor size of 120kW, 400V, 3Phase

was used as an example in sizing the breaker and conductors. Voltage

drop during motor starting and running at 115% of full load conditions

were also evaluated and the results showed that there will be no

problem installing a fire pump in the LV bus.

1

Chapter 1 – Introduction

Power system study is necessary for every project in order to make

a design which is reliable, efficient and cost effective. Every design

is expected to meet the requirements mandated by the governing

codes and standards that the specific project follows.

Project Background and Description

Super Absorbent Polymer (SAP) plant is located in Al-Jubail,

Kingdom of Saudi Arabia. As the name implies, its products are

super absorbent polymer resins which are the major components in

making baby diapers.

The plant is composed of main production building which is a seven

storey building, a warehouse, a Central Control Room and a

substation. Refer to Figure 1, page 4 for the 3D model view of the

plant.

The electrical system of SAP plant is based on the following

concept:

1. One new dedicated substation (Substation 2),

34.5/4.16/0.4kV for the SAP facilities.

2. Substation 2 will be powered from two 34.5kV spare circuit

breakers available from existing Substation 1.

3. For Substation 2 the following setup will apply:

a. Medium Voltage (MV) distribution switchgear is

2BUS-101A/B radial, 100% redundant transformers.

b. Low Voltage (LV) switchgear/MCC, 2BUS-201A/B is

radial, 100% redundant transformers.

c. Four (4) LV switchgear /MCCs for process units, each

corresponding to one process unit. Each of the

process LV switchgear/MCCs is fed from single

transformer and has one incoming feeder from the

2

stand-by transformer 2TR-206. The purpose of the

stand-by transformer 2TR-206 is to take over the

load from any single transformer when this is out of

operation. The load transfer will be done manually.

The parallel conditions are insured at the MV

switchgear which in normal operation has the bus-

coupler normally closed.

4. All bus-couplers will be operated manually.

5. Typical General Feeders include at least the following

panels supply:

a. Lighting Panel

b. Convenience Outlet Panel

c. Heat Tracing Panel

d. Main Distribution Panels

6. An Interlock between ACB1/B, ACB2/B, ACB3/B and

ACB4/B shall be provided so that the stand-by transformer

2TR-206 shall be connected only to one process LV

SWITCHGEAR / MCC at a time.

Refer to Figure 2, page 5, for the single line diagram.

Project Objectives

This technical report will present a power system design of the

Super Absorbent Polymer Plant based on engineering principles and

applicable codes. For this report, the primary code that will be used

as the basis of the design is the Philippine Electrical Code (PEC).

The intent of this report is to present a power system design for the

project in such a way that the plant is assumed to be constructed in

the Philippines and because of that this independent study is

conducted using the PEC as the new basis in sizing the buses,

3

feeders, breakers and all the components of the electrical system of

the SAP plant.

Scope and Delimitation

This technical report covers power system design of Super

Absorbent Polymer Plant. The design covers the following:

1. Main bus and feeder size calculations.

2. Breaker sizing using short circuit calculations.

3. Neutral Grounding Resistor (NGR) sizing

4. Load flow analysis and calculations.

5. Motor starting analysis.

6. Power factor capacitor sizing.

7. System harmonics analysis

8. Time-Current Curve coordination.

9. Fire pump motor circuit design.

All of the calculations pertaining to the scope of the design will be

based on the preliminary data from the project. These data includes

the transformer sizes, system frequency, voltages and single line

diagram showing the transformers’ connections as per the client’s

requirements. These parameters form part of the design

delimitations.

Auxiliary systems, namely; paging system, fire alarm system,

telephone system and security systems will be presented in this

report as adherence to the National Building code that mandates

the inclusion of such auxiliary systems plan drawings along with the

submission of electrical plan drawings.

This report does not cover and discuss maintenance of the electrical

equipments.

4

Figure 1 – Super Absorbent Polymer Plant 3D model snapshot.

5

Figure 2 – Super Absorbent Polymer Plant Key Single Line Diagram

6

Chapter 2 – Design Criteria

Basic Design Codes

Although the actual project is located in Al-Jubail, KSA, a territory

outside Philippines, for the purpose of this report all calculations

pertaining to the design will adhere to the PEC. Other supporting

standards that the PEC recognizes shall be used such as IEC, NEC,

IEEE, NEMA and ANSI standards will be considered as

supplementary codes.

System Frequency and Voltage

The system frequency is 60Hz. There are three utilization voltage

levels, namely 34.5kV for the high voltage (HV), 4.16kV for the

medium voltage (MV) and 400V for the low voltage (LV) system.

Design Assumptions

Assumptions will be made in some parts of the calculations. These

assumptions will be mentioned in the specific parts of this report

where they are needed.

7

Chapter 3 – Electrical Calculations

This chapter is divided into three major parts namely;

Part 1 - Short Circuit Calculations

Part 2 – Overcurrent Protection Device (OCPD), Primary and

Secondary Feeder, Neutral Conductor and Equipment Grounding

Conductor

Part 3 - Time-Current Curve Coordination

Part 4 - Load Flow Calculations

Part 1 - Short Circuit Calculations

The results of the short circuit calculations will determine the sizes

of the buses, feeders, breakers and the Neutral Grounding Resistor.

Case 1: For 1BUS-101A and 1BUS-101B

Verify if the existing bus can still handle the short circuit currents

when the new substation is to be served by the spare breakers.

Assumptions:

1. As per project design basis document assume that about 80%

or 16MVA of the total plant load consists of motor loads.

2. Assume that the two 20MVA transformers 2TR-101A&B are

running in parallel during the fault occurrence.

3. Assume that 80% or 2MVA of the load of each of the six

transformers 2TR-201A, 201B, 202, 203, 204 and 205 are

motor loads with 0.25 per unit reactance based on the

combined (2MVA) rating as per IEEE Std 141.

4. As per project design basis document neglect the cable

impedance and treat all transformer and motor impedance to

be purely reactance. This will provide more conservative

results.

8

5. Transformer 2TR-206 is excluded in the computation since its

purpose is a back-up power for the four LV transformers.

Calculations:

Based on the assumptions, compute for the induction motor loads

that are not served by any of the six transformers connected to bus

2BUS-101A&B. These motors are directly connected to the MV bus.

Let IM-1 be the composite induction motors directly connected to

the MV bus,

MVAIM-1 = 16 – 2(6) = 4MVA

Use 0.25 per unit reactance for IM-1.

Let IM-201A, IM-201B, IM-202, IM-203, IM-204 and IM-205 be the

composite induction motors connected to 2TR-201A, 2TR-201B,

2TR-202, 2TR-203, 2TR-204 and 2TR-205 respectively.

Using the Per-Unit Method, calculate for the fault current at the bus.

1. Draw system diagram. Refer to Figure 3 below for the

simplified system diagram based on Figure 2 and the

assumptions.

Figure 3 – System diagram of SAP plant.

9

2. Select MVA base. Use the 20MVA transformer rating as the

MVA base and the following base voltages:

For HV Bus: 34.5kV

For MV Bus: 4.37kV

For LV Bus: 0.441kV4.16

0.424.37 =

3. Compute for the PU impedance of Utility/Source.

0.01063pu1882

20ZU =

=

4. Compute for the PU impedance of the transformers.

1 101B-2TR101A-2TR 0.14puZZ ==

2201A-2TR 0.435pu

4.37

4.16

2.5

20

100

6Z =

=2

5. Compute for the PU impedance of the motors.

1.1327pu4.37

4.16

4

200.25Z 1-IM =

=2

3201A-IM 2.056pu

0.441

0.4

2

200.25Z =

=2

1 Value is typical to the two HV transformers. 2 Value is typical to the six LV transformers. 3 Value is typical to the six LV motor loads.

Equation 1

Equation 2

Equation 4

Equation 3

Equation 5

10

6. Compute for the short circuit currents.

a. Draw impedance diagram.

b. Draw simplified diagram with values. Compute for the

total impedance (ZT) at the fault point.

c. Compute short circuit currents, ISC, (symmetrical)

d. If Asymmetrical values of short circuit currents are

needed, multiply the symmetrical values by 1.25 for LV

and 1.6 for MV/HV systems.

a. For impedance diagram refer to Figure 4 below.

Figure 4 –Equivalent impedance diagram with

1BUS-101A/B as reference faulted bus.

b. Compute for ZT

HGAT Z

1

Z

1

Z

6

Z

1 +

++

+=

−−− 111

14.0

1

14.0

1

ZA = ZB = ZC = ZD = ZE = ZF

2.4917pu0.4352.056ZA =+=

Equation 6

Equation 7

11

Solving for ZT,

0.01063

1

0.14

1

0.14

1

1.1327

1

2.4917

6

Z

1

T

+

++

+=−− 11

0.01034puZT =

c. Compute for the fault current,

voltage base3

10 MVA base

Z pu

voltage puI

6

T

SC ×××=

1pu34.5

34.5voltage pu ==

RMS 32,369A345003

10 20

0.01034

1I

6

SC =×

××=

Results:

Based on the short circuit calculations, the bus can still handle the

additional load since it is rated at 40kA.

Equation 8

12

Case 2: For 2BUS-101A and 2BUS-101B

Assumptions:

For the purpose of this calculation, all assumptions made for Case 1

will apply.

a. For impedance diagram refer to Figure 5 below.

Figure 5 –Equivalent impedance diagram with

2BUS-101A/B as reference faulted bus.

b. Compute for ZT

HGFEDCBAT Z

1

Z

1

Z

1

Z

1

Z

1

Z

1

Z

1

Z

1

Z

1 +++++++=

From Equation 7 page 10, 2.4917puZA =

0.08063pu0.01063Z

-

H =

++=1

14.0

1

14.0

1

Equation 9

Equation 10

13

Solving for ZT,

0.08063

1

1.1327

1

2.4917

1

Z

1

T

++

= 6

0.06327puZT =

c. Compute for the fault current,

voltage base3

10 MVA base

Z pu

voltage puI

6

T

SC ×××=

0.9519pu4.37

4.16voltage pu ==

RMS 40,149A43703

10 20

0.06327

0.9519I

6

SC =×××=

d. Compute for the bus current rating. The bus must be

able to carry the full load current of the transformer.

2775.7A41603

1020I

6

B =××=

Say, 3150A.

Results:

Based on the short circuit calculations, the bus rating for 2BUS-

101A and 2BUS-101B should be greater than or equal to the ISC,

40,149A RMS. Say, 50,000A or 50kA RMS Symmetrical. For the

withstand time, use 1sec.

Case 3: For 2BUS-201A, 2BUS-201B, 2BUS-202, 2BUS-203,

2BUS-204, 2BUS-205 and 2BUS-206

Assumptions:

For the purpose of this calculation, all assumptions made for Case 1

will apply.

Equation 11

Equation 12

14

Calculations:

Since the seven buses are served by two MV buses connected in

parallel via a Normally Closed (N.C.) tie breaker and having

identical impedances, we can assume that they have the same

magnitude of fault currents. We only need to compute for the fault

at one bus. Let’s take 2BUS-202 as the reference faulted bus.

Using the same MVA base, base voltages and the per unit

impedances of utility/source, transformers and motors from Case 1,

compute for the short circuit current.

a. Draw the system impedance diagram using 2BUS-202 as

the reference bus. Refer to Figure 6 below.

Figure 6 –Equivalent impedance diagram

with 2BUS-202 as reference.

15

b. Compute for ZT,

RIT Z

1

Z

1

Z

1 +=

M

1

QPJ

R ZZ

1

Z

1

Z

15Z +

++

=

−

From Equation 7 page 10 and Equation 10 page 12,

ZJ = ZA = 2.4917pu

ZQ = ZH = 0.08063pu

0.5pu0.4350.08063

1

1.1327

1

2.4917

15Z

1

R =+

++

=−

0.402pu0.5

1

2.056

1Z

1

T =

+=−

c. Compute for the short circuit current,

voltage base3

10 MVA base

Z pu

voltage puI

6

T

SC ×××=

0.9070pu0.441

0.4voltage pu ==

RMS 59,076A4413

10 20

0.402

0.9070I

6

SC =×××=

d. Compute for the bus current rating. The bus must be able

to carry the full load current of the transformer.

3608.4A4003

102.5I

6

B =××=

Say, 4000A.

Results:

Based on the short circuit calculations, the bus rating for 2BUS-

201A, 2BUS-201B, 2BUS-202, 2BUS-203, 2BUS-204, 2BUS-205 and

2BUS-206 should be greater than or equal to the ISC, 59,076A RMS.

Equation 13

Equation 14

Equation 15

16

Say, 70,000A or 70kA RMS Symmetrical. For the withstand time,

use 1sec.

Part 2 – Overcurrent Protection Device (OCPD), Primary and

Secondary Feeder, Neutral Conductor and Equipment

Grounding Conductor

In general, calculations for the OCPD, feeder conductors, neutral

conductors and equipment grounding conductors shall comply with

the following:

1. Calculations for the size of primary and secondary overcurrent

protection devices must comply with PEC Art 4.50.1.3.

2. Calculations for the size of primary and secondary feeder

conductors must comply with “PEC Art 2.40.1.4, conductors

shall be protected against overcurrent in accordance with their

ampacities specified in 3.10.1.15”.

Feeder conductors must be sized such that the voltage drop is

limited to 3% as per Art 2.15.1.2(1)(3) FPN No. 2 and 3.

The estimated length of primary feeders for 2TR-101A and

2TR-101B is 70m while the length of secondary feeders of the

same transformers including the length of primary and

secondary feeders of the other seven MV transformers (i.e.

2TR-202) are estimated to be 15m.

3. Calculations for the size of neutral conductors of MV

transformers must comply with “PEC Art 2.20.3.22(c)(2),

where the neutral conductor of a 3-phase, 4-wire system is

supplying nonlinear loads no reduction in ampacity is

required”.

4. Calculations for the size of system grounding conductor must

comply with PEC Art 2.50.2.17 for the impedance grounded

system applicable to the HV transformers 2TR-101A and 2TR-

17

101B. For the solidly grounded system of MV transformers

(i.e. 2TR-202), calculations must comply with PEC Art

2.50.2.1(b)1.

5. Calculations for the size of main equipment grounding

(bonding) conductors must comply with PEC Art 2.50.6.13

and Art 2.50.2.9(d) where applicable.

Section 1 – Primary and Secondary Overcurrent Protection

Device

The determining factor in sizing the interrupting rating of all the

circuit breakers is the short circuit rating of the bus where they are

connected. In general,

a. All breakers connected to the MV bus must be rated 50kA

RMS Symmetrical.

b. All breakers and disconnect switches connected to the LV bus

must be rated 70kA RMS Symmetrical.

Case 1: For transformers 2TR-101A and 2TR-101B.

a. Primary Overcurrent Protection Device.

Determine primary full load current (FLA).

334.7A345003

1020I

6

P =×

×=

The spare breaker used as primary OCPD which is 1250A is

300% of the primary FLA. Based on Table 4.50.1.3(a) and

note 1 of the same table, this is permitted.

b. Secondary protection.

Based on Table 4.50.1.3(a), where the primary OCPD is 300%

of primary current the secondary conductor is considered

protected and does not require secondary OCPD.

Equation 16

18

Although secondary OCPD is not required in this situation, we

are required to install OCPD for the protection of panelboards

located downstream of secondary conductor.

Where a feeder supplies continuous loads, the rating of the

secondary OCPD may not be less than 125% of the

continuous load as per PEC Art 2.15.1.3.

Determine secondary full load current.

2775.7A41603

1020I

6

s =××=

125% x 2775.7 = 3469.63A

Say, 3150A.

The choice of 3150A is logical since all loads were assumed to

be continuous, which is not the case in the actual operation of

the plant.

Case 2: For transformers 2TR-201A, 2TR-201B, 2TR-202,

2TR-203, 2TR-204, 2TR-205 and 2TR-206.

a. Primary Overcurrent Protection Device.

Determine primary full load current (FLA).

346.97A41603

102.5I

6

P =×

×=

From Table 4.50.1.3(a), we can use 300% as the multiplying

factor for the primary current.

300% x primary FLA

300% x 346.97 = 1040.91A

Say, 1250A.

Equation 18

Equation 17

19

b. Secondary protection.

The same rules applies here as with the case of MV

transformers. Secondary OCPD is not required since primary

OCPD is size at 300% of the primary current.

However OCPD is required for the panelboards connected to

the transformer secondary.

Determine secondary full load current.

3608.4A4003

1020I

6

s =××=

125% x secondary FLA

125% x 3608.4 = 4510.5A

Say, 4000A.

Case 3: For Tie Breakers.

The tie breakers must be rated to be the same as the bus rating.

a. The bus tie breaker of 2BUS-101A and 2BUS-101B should

have a rating of 3150A, 50kA.

b. The bus tie breaker of 2BUS-201A and 2BUS-201B should

have a rating of 4000A, 70kA.

Case 4: For Breakers of Typical Motors and the Spare

Breakers at 2BUS-101A and 2BUS-101B.

For the simplicity and flexibility of the design it is advisable to size

all the other breakers to be the same as the transformer breaker.

Usually switchgear are manufactured by vendors in standard sizes

and buying one with customize sizes oftentimes more costly that’s

why it is advisable to maximize the sizes of the needed equipment.

Equation 19

20

Section 2 – Primary and Secondary Feeder

Sizing the feeder conductors does not only involve determining its

continuous current carrying capacity but also taking into

consideration its capacity to withstand the high magnitude of

currents during fault or short circuit.

Calculations for the size of primary and secondary feeder

conductors must comply with “PEC Art 2.40.1.4, conductors shall be

protected against overcurrent in accordance with their ampacities

specified in Art 3.10.1.15”. Feeder conductors should be protected

by an overcurrent device. The conductor ampacity shall not be less

than the overcurrent device setting or rating.

Derating factors for cables rated 2001V or over shall comply with

PEC Art 3.92.1.13. For feeder cables rated 2000V or less derating

factors shall comply with PEC Art 3.92.1.11.

To insure proper functioning of utilization equipment, it is

recommended that feeder conductors be sized such that the voltage

drop is limited to 3% as per Art 2.15.1.2(1)(3) FPN No. 2 and 3.

The estimated length of primary feeders for 2TR-101A and 2TR-

101B is 70m while the length of secondary feeders of the same

transformers including the length of primary and secondary feeders

of the other seven MV transformers (i.e. 2TR-202) are estimated to

be 15m.

21

Case 1: Feeder Size for Transformers 2TR-101A and 2TR-

101B

a. Primary Feeder (34.5kV).

1. Size according to OCPD.

Primary feeder ampacity rating shall not be less than the

rating of its OCPD which is 1250A.

From PEC Table 3.10.1.69, a single insulated copper

conductor size 250mm2, XLPE (Type MV-90), 15001-

35000V has an ampacity of 680A.

Use two (2) conductors per phase. Install in a single layer

in an uncovered cable tray maintaining a space not less

than one cable diameter between individual conductors in

accordance with PEC Art 3.92.1.13(b)(2).

2. Size according to short circuit current.

Using the thermal equation of a copper conductor

determine the minimum size of feeder that can handle the

short circuit current passing through the transformer

primary side.

++

×=

i

f10

5

2

SC

T234.5

T234.5log101.18t

A

I

Where: ISC = Short circuit current in amperes

A = Size of cable in mm2

t = Tripping time of the protective device in

seconds

Tf = Short circuit current temperature rating of

the cable in OC (250 OC for XLPE)

Ti = Continuous temperature rating of the

cable in OC (90 OC for XLPE)

Equation 20

22

The maximum amount of short circuit current that the

primary feeder can be subjected to is the same as the

fault current at the bus where it is connected. Therefore,

let us use the short circuit current rating of 1BUS-101A.

++×=

90234.5

250234.5log101.181

A

4000010

5

2

A=279.09mm2

Since the value determined from the first calculation is

larger than this, then the requirement is satisfied.

b. Secondary Feeder (4.16kV).


Secondary feeder ampacity rating shall not be less than

the rating of its OCPD which is 3150A.


conductor size 500mm2, XLPE (Type MV-90), 2001-5000V

has an ampacity of 870A.

870 x 4 = 3180A

Use four (4) conductors per phase. Install in a square

formation maintaining free air space not less than 2.15

times the diameter (2.15 x O.D.) of the cable in


2. Size according to short circuit current.

Using the available fault current at the 2BUS-101A

calculate for the minimum size of feeder that can handle

23

the short circuit current passing through the transformer

secondary side.

++×=

90234.5

250234.5log101.181

A

5000010

5

2

A=348.86mm2

The requirement is already satisfied from the first

calculation.

3. Voltage drop calculation.

Per PEC Table 9.1.1.8, the DC resistance of a single

500mm2 copper conductor at 75OC is 0.0132 ohm/305m.

3lengthresistance4

FLA secondaryVD ×××=

0.78V315305

0.0132

4

2775.7VD =×××=

%VD = 0.78/4160 x 100 = 0.019%

Cable reactance is not considered in the calculation but

results shows that even if it is included the voltage drop is

likely to be below 3%.

Equation 21

24

Case 2: Feeder size for transformers 2TR-201A, 2TR-201B,

2TR-202, 2TR-203, 2TR-204, 2TR-205 and 2TR-206

a. Primary feeder (4.16kV).


Primary feeder ampacity rating shall not be less than the

rating of its OCPD which is 1250A.




Use two (2) conductors per phase. Install in a single layer

in an uncovered cable tray maintaining a space not less

than one cable diameter between individual conductors in


2. Size according to short circuit current rating.

The minimum size of primary feeder conductor that can

handle the short circuit current passing through the

transformer primary side is the same size with the

secondary conductor of 2TR-101A which is 348.86mm2 as

determined from Equation 21 page 23. And since this is

smaller than the size from the first calculation then this

requirement is satisfied.

b. Secondary Feeder (400V).


Secondary feeder ampacity rating shall not be less than

the rating of its OCPD which is 4000A.

25


conductor size 500mm2, XLPE (Type XHHW-2), 0-2000V


Applying adjustment factor from Art 3.92.1.11(b)(1).

6 x 995 x 0.75 = 4477.5A

Use six (6) conductors per phase. Install in accordance

with the provisions of Art 3.92.1.11(b)(1) and

requirements of Art 3.92.1.10.

2. Size according to short circuit current rating.

Using the available fault current at the 2BUS-202 (similar

to all seven MV transformer) calculate for the minimum

size of feeder that can handle the short circuit current

passing through the transformer secondary side.

++×=

90234.5

250234.5log101.181

A

7000010

5

2

A=488.41mm2

The requirement is already satisfied from the first

calculation.

3. Voltage drop calculation.

Per PEC Table 9.1.1.8, the DC resistance of a single

500mm2 copper conductor at 75OC is 0.0132 ohm/305m.

3lengthresistance6

FLA secondaryVD ×××=

26

0.68V315305

0.0132

6

3608.4VD =×××=

%VD = 0.68/400 x 100 = 0.17%

Cable reactance is not considered in the calculation but

results shows that even if it is included the voltage drop is

likely to be below 3%.

Section 3 -Neutral Grounding Resistor, Neutral Grounding

Conductor, System Grounding Conductor and Equipment

Grounding Conductor

Case 1: For transformers 2TR-101A and 2TR-101B

a. Neutral Grounding Resistor

Sizing the neutral grounding resistor (NGR) is dependent on

the decision of the design engineer. The common practice is

to use a 200A or 400A NGR for a medium voltage system.

Other consideration is to size the NGR higher than the

capacitive charging current of the system to avoid transient

overvoltage

“Every system has a capacitance value, mostly due to the

system's cables and surge arresters/capacitors. The general

rule of thumb for estimating this charging current is to use 1A

per 1,000kVA (Excerpt from www.ecmweb.com)”.

Let us calculate for the estimated capacitive charging current.

20A1000kVA

20000kVAIC ==

27

Since the recommended NGR rating is much higher than the

capacitive charging current then there would be no problem.

Let’s choose 400A, 10sec.

b. System Grounding Conductor

The system (neutral) grounding conductor must be able to

carry the ground fault current which is 400A.



with an ampacity of 435A can be used. The conductor must

also have an insulation level as the phase conductors as per

PEC Art 2.50.10.7(b).

c. Equipment Grounding Conductor

1. Primary. Size the equipment grounding (bonding)

conductor for the transformer primary based on the size of

primary OCPD, per PEC Table 2.50.6.13.

For a 1250A OCPD, the recommended equipment

grounding size conductor is a 100 mm2 copper.

2. Secondary. Size the equipment grounding (bonding)

conductor in accordance with Table 2.50.3.17. The total

area of secondary conductor is 2000mm2 (4 x 500mm2)

copper.

Based on the table, the recommended system grounding

and bonding conductor size is an 80 mm2 copper.

28

Case 2: For transformers 2TR-201A, 2TR-201B, 2TR-202,

2TR-203, 2TR-204, 2TR-205 and 2TR-206

a. System Grounding and Bonding Conductor

The system grounding and bonding conductor must be size in

accordance with Table 2.50.3.17. The total area of secondary

conductor is 3000mm2 (6 x 500mm2) copper.

Based on the table, the recommended system grounding and

bonding conductor size is an 80 mm2 copper.

b. Neutral Conductor

As per PEC Art 2.20.3.22(c)(2), where the neutral conductor

of a 3-phase, 4-wire system is supplying nonlinear loads no

reduction in ampacity is required.

We have to size the neutral conductor to have an ampacity

equal to the ampacity of phase conductors. Therefore the

neutral conductor is also six 500mm2 copper.

c. Equipment Grounding Conductor

1. Primary. Size the equipment grounding (bonding)

conductor for the transformer primary based on the size of

primary OCPD, per PEC Table 2.50.6.13.

For a 1250A OCPD, the recommended equipment

grounding conductor size is a 100 mm2 copper.

2. Secondary. The secondary side equipment grounding

conductor is size in the same manner as the system

grounding (bonding) conductor. Therefore the size is also

an 80 mm2 copper.

29

Section 4 –Miscellaneous Electrical Details

Grounded conductors (neutral) should be identified as white or gray

(PEC Art 2.0.1.6) and equipment grounding/bonding conductors

shall be identified as green or green with yellow stripes (PEC Art

2.50.6.10). Phase or ungrounded conductors on the other hand

must be readily identifiable with other colors.

The color identification scheme below for the phase conductors is a

common practice in the USA, using Black, Red and Blue for Phase A,

Phase B and Phase C respectively.

a. Figure 7 - Schematic Diagram of a Solidly Grounded

System

b. Figure 8 - Schematic Diagram of a Resistance Grounded

System

30

Part 3 – Time – Current Curve (TCC) Coordination

For the purpose of our analysis and calculations, let us consider the

possible scheme of coordination and settings of the protective

devices from the 34.5kV HV bus down to the 400V LV bus. Refer to

Figure 9 below for the illustration.

The four circuit breakers CB1, CB2, CB3 and CB4 are to be

coordinated to achieve selective coordination. These breakers

should operate progressively from LV bus up to the HV bus should a

fault occurs at the LV bus. Another objective is that whenever a

fault occurs, the immediate breaker located to the upstream of the

fault should be the one to operate first in order to clear the

downstream fault.

Figure 9 – System diagram showing LV, MV

and HV bus for the TCC coordination.

31

Short Circuit Calculations:

For a simpler analysis let us consider the utility to be the sole

source of fault current, neglect contributions from motor loads.

Refer to Figure 10 below for the system impedance diagram.

Figure 10 – Equivalent system impedance diagram

with utility as sole source and 2BUS-202 as faulted bus

The fault currents per bus are:

1. For 2BUS-202.

0.58563pu0.4350.140.01063ZT =++=

RMS 40.5kA4413

10 20

0.58563

0.9070I

6

SC =×××=

2. For 2BUS-101A/B

0.15063pu0.140.01063ZT =+=

RMS 16.7kA43703

10 20

0.15063

0.9519I

6

SC =×××=

The Extremely Inverse Time-Current Curve

The Extremely Inverse TCC is best selected when the application is

transformer protection because it matches the transformer damage

curve profile. Below is the equation of the Extremely Inverse TCC.

=1-I

80TDt

2

32

Where: t = tripping time of relay in sec

I = multiple of Setting Current

TD = Time Dial

Setting for CB1 Relay

a. Time Overcurrent (TOC) Setting

1. Tap Setting.

Assume setting = 125% IS

TAP = 1.25 x 3608.44 (1/4000)

= 1.13A

Use: TAP = 1.2A

2. Time Dial (TD).

Use: TD = 0.1

3. Compute for relay tripping time (t) for a fault at 2BUS-202.

( )[ ] 0.114sec1-40001.21040.5

800.1t

23=

×÷×=

The calculated time seems a good setting. This will give

enough time for the protective devices (MCCBs) of the loads

connected to 2BUS-202. The estimated clearing time of

MCCBs is 0.025sec (1.5cycles).

b. Instantaneous Setting

This function should be disabled in order to achieve selectivity

for faults on outgoing lines from 2BUS-202.

Refer to Figure 11 page 37 for the TCC of CB1 Relay using

ABB RXIDK 4 relay set to Extremely Inverse curve

characteristic.

33



1. Tap Setting.

Assume setting = 200% IP

TAP = 2 x 346.96 (1/400)

= 1.73A

Use: TAP = 1.7A

2. Time Dial (TD).

The Time Dial must be coordinated with transformer 2TR-

202 magnetizing inrush and its damage curve. It must also

have at least 0.3sec margin above CB1 Relay allowing the

later to trip first before it can operate. Using the TCC

equation, calculate for TD.

( )

×÷×=+

1-4001.7)4160

420(1040.5

80TD0.3)(0.114

2

3

TD = 0.1819

Use: TD = 0.2


This setting should be above the calculated fault at 2BUS-202

so that it will allow CB1 Relay to operate first for a fault at the

mentioned bus.

Setting = 175% IF(2BUS-202) (1/TR ratio)

= 1.75 x 40,500 (1/400)(0.42/4.16)

= 17.89A

34

For the Multiple (M) of TOC setting:

M = 17.89/1.7

= 10.52

Use: 11



characteristic and the Instantaneous Pick-up set to 20.



1. Tap Setting.

Assume setting = 125% IS

TAP = 1.25 x 2775.7 (1/3000)

= 1.16A

Use: TAP = 1.2A

2. Time Dial (TD).

A Time Dial must be selected to allow at least 0.3sec delay

for a fault at 2BUS-101A/B since large motors are

connected here and uses relays as tripping unit for their

OCPDs.

( )[ ]

×÷×=

1-30001.21016.7

80TD(0.3)

23

TD = 0.0769

Use: TD = 0.1

3. Compute for relay tripping time (t) for a fault at 2BUS-

101A/B.

( )[ ] 0.39sec1-30001.21016.7

800.1t

23=

×÷×=

35


This function should be disabled in order to achieve selectivity

for faults on outgoing lines from 2BUS-101A/B.



characteristic.



1. Tap Setting.

Assume setting = 180% IP

TAP = 1.8 x 334.7 (1/400)

= 1.51A

Use: TAP = 1.5A

2. Time Dial (TD).

The Time Dial must be coordinated with transformer 2TR-

101A magnetizing inrush and its damage curve. It must

also have at least 0.3sec margin above CB3 Relay allowing

the later to trip first before it can operate.

( )

×÷

×

=+

1-4001.51016.7

80TD0.3)(0.39

2

3

5.34

37.4

TD = 0.0985

Use: TD = 0.1

36


This setting should be above the calculated fault at 2BUS-

101A so that it will allow CB3 Relay to operate first for a fault

at the mentioned bus.

Setting = 175% IF(2BUS-101A) (1/TR ratio)

= 1.75 x 16,700 (1/400) (4.37/34.5)

= 9.25A

For the Multiple (M) of TOC setting:

M = 9.25/1.5

= 6.17

Use: 7



characteristic and the Instantaneous Pick-up set to 20.

37

Figure 11 – Time-Current Curve Coordination analysis.

38

Part 4 – Load Flow Calculations

Load flow calculations are divided into the following sections:

Section 1 – Motor starting Analysis

Section 2 – Power Factor Capacitor Sizing Calculations

Section 3 – System Harmonic Analysis and Calculations

Section 1 – Motor starting Analysis

The subject of our analysis is the 1239kW motor connected to the

MV bus since it is the largest motor.

The required voltage across the motor terminal should be at least

80% of the motor rated voltage during starting.

Assumptions/conditions:

1. The motor is to be started when the 20MVA transformer is

loaded to about 80% of its rating, excluding the motor itself.

2. Only one (1) – 20MVA transformer is operating at the time

when the motor is started. This will provided the greatest

voltage drop since transformer impedance is greatest in this

condition.

3. The motor and loading profile is as follows:

a. Motor – 1239kW, 4160V, 90% efficiency, 80% power

factor and 600% LRC.

b. Other load (composite) – 16MVA, 4160V and 80% power

factor.

39

Solution:

Refer to Figure 12 below for the equivalent single line diagram.

Figure 12 – Motor starting analysis single line diagram.

1. At 1882MVA power base; 34.5kV and 4.37kV voltage bases;

ZU = j1pu

j13.174pu20

188214ZTR =

= .0

pu j63.9685.274.37

4.16

36.8716

1882Z

2

L +=

°−∠=

( ) pu j99.11132.146

1

4.37

4.16

36.870.90.81.239

1882Z

2

M +=

°−∠÷÷=

Since Load and Motor impedances are in parallel, hence,

pu j38.8751.83ZZ

ZZ

ML

ML +=

+•

40

pu 45.9774.57

1

13.74j1j38.8751.83

1

puZ

pu VI

T ∠=

+++==

Then the starting voltage (VST) is,

pu 0.873 45.9774.57

9013.174 9011VST =

∠∠+∠−=

VST = 4370 (0.873) = 3815.01V

91.71% 4160

3815.01 =

Based on the result, there will be no problem starting the

motor.

Section 2 – Power Factor Capacitor Sizing Calculations

Given the following data calculate for the size of capacitor needed to

improve the average power factor from 80% (typical value if data is

not available) to 96% (Client requirements).

The following data were based on the plant load list:

a. Motors on continuous duty.

11,200kW (Duty Factor = 100%)

b. Motors on intermittent duty.

167kW (Duty Factor = 30%)

c. Motors on standby duty.

1979kW (Duty Factor = 10%)

The total active load is,

11,200(100%) + 167(30%) + 1979(10%) = 11,448kW

°== − 36.87 0.8cosPFθ 1OLD

41

°== − 16.26 0.96cosPFθ 1NEW

Calculate for the rating of the capacitor,

( ) ( )[ ]NEWOLDC PFθtanPFθtanPowerMVAr −=

( )°−°= tan16.26tan36.8711.448MVArC

MVArC = 5.247

Install two (2) banks of 2.7MVAr, 3-phase capacitor having 900kVAr

per phase each. Install one (1) for 2BUS-101A and one (1) for

2BUS-101B.

Section 3 – System Harmonic Analysis and Calculations

The possibility of harmonic problem is present due to the fact that

many of the motor loads are driven by Variable Speed Drives

(VSD). Variable Speed Drives are known contributors of 5th and 7th

order harmonics. Of the two, the 5th order harmonic has the largest

contribution of harmonic currents. This section will present an

analysis of the system harmonics and a design of single tuned

harmonic filter for the 5th order harmonic.

Data and assumptions:

1. Based on the load list about 3200kW of the motors used for

process areas are driven by variable speed drives (VSD).

2. In addition to the 3200kW harmonic power source coming

from VSD driven motors, let’s assume a 300kW harmonic

power source coming from lighting and computer loads.

42

Analysis and design:

1. Determine Short Circuit Ratio (SCR).

The SCR is a ratio of the short circuit MVA to the harmonic

power source at the Point of Common Coupling (PCC). “If the

SCR is less than 20, and there is a parallel resonance

condition near to a characteristic harmonic of the non-linear

load, there will probably a problem. A study should be made

to determine the possible addition of harmonic filters to

eliminate any problem (excerpt from Power System Design by

Benigno S. Jimenez, PEE)”.

Refer to Figure 13 below for the equivalent single line diagram

showing harmonic power source.

Figure 13 – Harmonic analysis single line diagram.

43

H

SC

MW

MVASCR =

From Equation 12 page 13, ISC = 40,149A.

82.65103.5

4014941603

MW

MVASCR

6H

SC =×

××==

Since the SCR is greater than 20, the probability of harmonic

problem is low. However, try to evaluate for possible

resonance problem.

2. Evaluate for the possibility of parallel resonance.

Parallel resonance occurs when the system inductive

reactance and capacitive reactance are equal at some

frequency. If the combination of capacitor banks and the

system inductance result in parallel resonance near one of the

characteristic harmonics generated by the non-linear loads,

that harmonic current will excite the system, thereby causing

an amplified current to oscillate between the energy at the

inductance and the energy in the capacitance. This high

oscillating current can cause excessive voltage distortion.

To determine whether the possibility of harmonic problem

exists, the formula can be used:

C

SCP

MVAr

MVAH =

7.325.4

289.29HP == (Near 7th order harmonic)

Result of the calculation shows that a possible parallel

resonance might occur in the 7th order harmonic frequency.

44

Earlier it was mentioned that VSDs are contributor of 7th order

harmonic. Though majority of its contribution is in the 5th

order harmonics but analysis shows that additional filter tuned

to the 7th order harmonics might be needed if investigation

during the plant operation shows problems caused by this

order of harmonic.

3. Details of the 2.7MVAr capacitor.

Figure 14 – The 2.7MVAr capacitor bank.

The current per phase is,

347.72A41603

102.7I

6

C =×

×=

The reactance per phase is,

j6.91Ω347.72

34160/XC −==

45

4. Size the reactor and design the filter for 5th harmonic.

Figure 15 – Harmonic filter diagram.

j0.276Ω5

6.91

H

XX

22

CL ===

362.04Aj6.91-j0.276

34160/

XX

EI

CL

T ==+

=

The 5th harmonic current is,

97.15A34160

103.5

5

1

3V

Power

H

1I

6

L

5 =

×=

=

The voltage across the capacitor is,

+=H

XIXIV C5CTC

( ) 2635.96V5

6.9197.156.91362.04VC =

+=

Allowing a factor of 1.2 for the filter,

2

5

2

TFILTER RMS II1.2 I +=

449.82A97.15362.041.2 I 22FILTER RMS =+=

Use: 450A, 2.7kV Reactor, in series with the 900kVAr

capacitor in each phase.

46

Chapter 4 – Auxiliary Systems

This chapter is divided into four (4) parts namely;

Part 1 – Fire Pump and Fire Alarm System Diagram

Part 2 – Paging System Diagram

Part 3 – Telephone and I.T. System Diagram

Part 4 – Security System (CCTV) Diagram

Part 1 – Fire Pump and Fire Alarm System

Section 1- Fire Pump Motor

The fire pump motor circuit protection, conductor size, power

source and installation must comply with PEC Art 6.95. The fire

pump motor size that will be used in this section is an estimated

size based on the common practice in the industry. Should the final

size of the pump as determined by the responsible discipline differ

from what is presented here, the calculation method presented here

shall in turn serve as a guide. A motor used to drive a 1000gpm at

150psi fire pump has the following specifications:

120kW, 400V, 3Phase, 60Hz, 0.8Power Factor, 90% Efficiency

Power Source (Art 6.95.1.4). The power source for the fire pump

motor shall provide continuity of power. For a 120kW, 400V motor,

the possible power source can be any of the six LV MCC bus since

all of them provides redundancy of power source.

Disconnecting Means (Art 6.95.1.4). The disconnecting means

must:

1. Be identified as suitable for use as service equipment.

2. Be lockable in the closed position.

47

3. Not be located within equipment that feeds loads other than

the fire pump.

4. Be located sufficiently remote from other building or other

fire pump source disconnecting means.

5. Be marked “Fire Pump Disconnecting Means.” The letters

must be at least 25mm in height and be visible without

opening enclosure doors or covers.

Power Wiring (Art 6.95.1.6). Route the supply conductors outside

buildings, where possible. When this is not possible, install in

accordance with Art 6.95.1.4(b). In either case, install these

conductors as service entrance conductors in accordance with Art

2.30.

Overcurrent Protection (Art 6.95.1.5). Overcurrent protection

devices (OCPDs) must be sized to carry the sum of the locked-rotor

current of the fire pump and pressure maintenance pump motor(s)

indefinitely, and 100% of the ampere rating of the fire pump's

accessory equipment. For a 120kW, 400V, 0.8pf and 0.9Eff.,

240.6A0.90.84003

120,000I =

×××=

6 x 240.6 = 1443.6A

Use: 1600A Breaker

Voltage Drop. The voltage at the line terminals of the controller,

when the motor starts (locked-rotor current), must not drop more

than 15% below the controller's rated voltage. The voltage at the

motor terminals must not drop more than 5% below the voltage

rating of the motor when the motor operates at 115% of the fire

pump full-load current rating.

48

a. Evaluate the voltage drop during starting.

1. The conductor size is 125% x full load current.

1.25 x 240.6 = 301A

Use: 250mm2 (Table 3.10.1.16)

2. The estimated conductor length is 150m. The resistance

from Table 9.1.1.8 is 0.0265ohm/305m.

3. The fault current at LV bus is 59076A at 400V from

Equation 15 page 15.

4. Refer to Figure 16 below for the equivalent impedance

diagram.

Figure 16 – Fire Pump Motor Starting diagram.

Calculate for the starting current (Is) using per unit method:

The short circuit MVA at the bus is,

40.93590764003

MVABUS =××=610

At 40.93MVA base; 400V base voltage;

pu j1 ZBUS =

pu j20.4652

40.93 ZL ==

49

pu j24.5632.746

1

36.87-0.9)0.8(0.12

40.93 ZM +=

°∠÷÷=

3.33pu59076)3(400/

1500.0265/305 ZCABLE =

÷×=

Since Load and Motor impedances are in parallel, hence,

pu j14.804.54ZZZ

)Z(ZZ

MCABLEL

MCABLEL +=

+++•

pu j15.804.54

1

j1 j14.804.54

1

puZ

pu VI

T +=

++==

Then the starting voltage (VST) is,

pu 0.9416 73.9716.44

9011VST =

∠∠−=

VST = 400 (0.9416) = 377V

94.25% 400

377 =

Based on the result, there will be no problem starting the

motor.

b. Evaluate the voltage drop (VD) at 115% FLA.

( )R1.15I 3VD FL=

( )( ) 6.25V1500.0265/305240.61.153VD =××=

98.44% 400

6.25-400 =

50

Section 2 - Fire Alarm System Diagram

The Fire Alarm System Diagram in Figure 17 page 51 is based on

the design philosophy developed by Fluor Philippines Control

Systems Department. Some modifications were made in order to

suit the just needed requirement of this report.

Part 2 – Paging System Diagram

The Paging System Diagram in Figure 18 page 52 is based on the

design philosophy developed by Fluor Philippines Control Systems

Department. Some modifications were made in order to suit the just

needed requirement of this report.

Part 3 – Telephone and I.T. System Diagram

The Telephone System and I.T. Diagram shown in Figure 19 page

53 is based on the design philosophy developed by Fluor Philippines

Control Systems Department. Some modifications were made in

order to suit the just needed requirement of this report.

Part 4 – Security System (CCTV) Diagram

The Security System (CCTV) Diagram shown in Figure 20 page 54 is

based on the design philosophy developed by Fluor Philippines

Control Systems Department. Some modifications were made in

order to suit the just needed requirement of this report.

51

Figure 17 – Fire Alarm System Diagram

52

Figure 18 – Paging System Diagram

53

Figure 19 – Telephone System and I.T. Diagram

54

Figure 20 – Security System (CCTV) Diagram

55

Chapter 5 – Conclusions / Recommendations

After conducting all the calculations, analysis and design, the author

have the following observations and recommendations:

1. An additional harmonic filter for the 7th order harmonic must

be installed after verifying that a parallel resonance occur in

this frequency and may cause problems.

2. Providing an emergency generator can be a good idea in order

to provide continuity of power to the fire pump.

3. The production area is of considerable distance from the

substation. A better way of providing power to the loads

located is this area is to install distribution boards and MCCs

inside the building. This will result in a huge reduction of cable

trays needed to be routed since only panelboards and MCC

feeders are coming directly from the substation. This will also

minimize the available fault in the load side of the motors and

reduce voltage drop.

56

References

The following are the references used by the author:

1. Philippine Electrical Code (PEC).

2. National Electrical Code (NEC).

3. International Electrotechnical Commission (IEC)

4. National Electrical Manufacturers Association Standard

(NEMA).

5. Institute of Electrical and Electronics Engineers Standards

(IEEE).

6. Other references, books, engineering articles in the World

Wide Web.

57

Affidavits

1. Affidavit of Sole Authorship.

2. Affidavits of the three vouching PEEs.

58

59

60

61

62

63

64

65

Curriculum Vitae

66

67

68

69

70

71

Documents

Power System Design of Super Absorbent Plant at Al-Jubail, KSA · 2020. 2. 15. · Figure 1 – Super Absorbent Polymer Plant 3D model snapshot 4 Figure 2 – Super Absorbent Polymer