LESSONS WE NEED TO LEARN
SIZING TRANSFORMERS
WITH LARGE MOTOR LOADS
By Doods A. Amora, PEE
December 1, 2006)
(First of Two Parts)
By Doods A. Amora, PEE
December 1, 2006)
(First of Two Parts)
1.0: INTRODUCTION
The purpose of this article is to help expand the engineer’s basic understanding on ‘short circuit capacity’ – how it affects voltage sags during starts-up of motors of significant sizes.
But why are Short Circuit & Motor Starting Calculations involved in transformer sizing?
One clue is:
"Starting up one thousand-1.0 hp motors is world apart from starting up one-1,000 hp motor"
SIDEBAR 1:
When I took my PEE board exams 22 years ago, there was one problem under the subject “ELECTRICAL DESIGN & CONSTRUCTION” which ran similar as follows:
A 250 hp motor driving a pump is served by 3 x 100 kVA 13.8kV- 480v transformer bank. Other data are as follows:
Transformer Impedance: 3.7% IZ
Short Circuit Capacity of Source Utility: 200 MVA
Motor Starting Current: 8 times of FLC
Length of Cable from Transformer to Motor Controller: 50 ft
Solve for:
a) Size of Cable & Conduit
b) Branch Circuit Breaker & Controller Size if Auto-Transformer type RVS
c) The voltage dip during motor start-up using full voltage starter
d) The voltage dip during motor start-up using Auto-Transformer Reduced Voltage Starter tapped at 65%.
To be honest, I wasn’t ready to solve the (c) and (d) parts of the problem. To my defense mechanism, I told myself, “the problem was irrelevant. Nandiyan na ‘yan, naka-install na kunyari; bakit kwentahin pa ang voltage dip?“
To make the story short, lumipas ang mahabang panahon… Then here comes now a real situation. What if one is asked to design a set-up for a 500 hp motor driving a pump somewhere? Would a 750 kVA transformer be sufficient? Maybe yes, maybe not! Try solving the problem and one will discover things that are assumed easy but are actually not quite simple!
SIDEBAR 2:
In one brewery I happened to work with, we had two – 450 hp 4 kV motors driving ammonia compressors controlled by Auto-Transformer RVS. Later, three – 700 hp motors were added but interestingly this time with across-the-line starters. The system was directly connected to the Power Plant’s 4,160 v bus through feeders. No problem was imagined because the Power Plant as a standard operating procedure, used to operate with significant spinning reserve capacity at any given time – and there were nine (9) generating units to support the loads.
However, when we started-up for the first time one of the three 700 hp compressors, a huge voltage dip occurred and the neighboring Glass Plant tripped off all its operating air compressors thus shutting down the entire plant. Pandemonium followed as the Power Plant struggled to recover the voltage by shedding off other major loads. Power Plant finally recovered but with all major manufacturing plants served by it at shutdown, including the brewery.
The event’s aftermath saw signs posted on all compressor control panels that reads: “NOTIFY POWER PLANT BEFORE STARTING”. Why? Because Power Plant had to add more generators on stream before any starting can be done. This was to provide the on-line generating capacity large enough to support the starting kVA needed, thus averting objectionable voltage dips. Interestingly, after the start-up the additional generators had to be shutdown. In the end, the full voltage starters were replaced with RVS units. In the succeeding breweries being built thereafter, MV Soft Starters were employed.
Again, there’s more to it than meets the eye. Recalling back my own PEE board exams, now I understand that after all, the problem on voltage dip was not irrelevant but a test of engineering competence why I should be given a license as PEE.
2.0: SUBSTATION TRANSFORMERS VS. LOADS
Generators and transformers react to large motor starting as manifested in voltage dips. Let us first discuss transformers.
In general terms, transformers in an electrical system are usually larger than the maximum demands they serve, in some instances even larger than the connected loads. In the industrial plant scenario, the obvious reason at first glance for this apparent oversizing is the anticipation for future load growth. Fine…
But more often than not, sizing the transformer with extra kVA capacity unwittingly addresses voltage sag problems, not for load growth for which it is intended originally. That’s why for newly constructed plants where load growth is not yet there, the problem of starting significantly large motors may not surface out. Why? Because the extra kVA capacity intended for load growth is taking care of it.
3.0: SYSTEM BEHAVIOR IN EVENTS OF FAULTS
Short circuit capacity calculation is used in many applications, some of which are the following:
a) selecting the interrupting capacity ratings of circuit breakers and fuses,
b) specifying the short-time withstand of switchgears,
c) determining if a line reactor is required,
d) and a host of many others.
But unknown to many, Short Circuit Calculations together with
Motor Starting Calculations are also used in Sizing Transformers.
Short Circuit condition brings down the voltage very dramatically. The amount of voltage during a full three phase fault at the terminals of the transformer is determined by the %IZ of the transformer, i.e., if the transformer is connected to an infinite source of short circuit currents.
Take for instance a 7,500 kVA, 8.0% IZ, 69-4.16 kV transformer. The 8.0% impedance rating would mean that full load currents of 1,041 amps will flow in the secondary if the secondary is short circuited and the primary voltage is raised from zero volts to a point at which 8.0% of 4,160 volts, or 332.8 volts, appears at the secondary terminals. Therefore, the impedance (Z) of the transformer secondary may now be calculated:
Z = V / I = 332.8 volts/1,041 amps = 0.3197 ohms
The exercise described above is what is known in textbooks as the “Short Circuit Test” which is actually performed in laboratories or electrical shops (drying up transformers). What if the short circuit at the secondary is real – meaning, the short circuit happened at full 4160v at the inception of the fault? If the transformer is connected directly to a source capable of supplying the transformer with an unlimited short circuit kVA capacity, the short circuit amperage capacity which the transformer can deliver from its secondary is:
4,160 volts /0.3197 = 13,012.5 amps
Another method of calculating short circuit capacity for the above transformer is:
The purpose of this article is to help expand the engineer’s basic understanding on ‘short circuit capacity’ – how it affects voltage sags during starts-up of motors of significant sizes.
But why are Short Circuit & Motor Starting Calculations involved in transformer sizing?
One clue is:
"Starting up one thousand-1.0 hp motors is world apart from starting up one-1,000 hp motor"
SIDEBAR 1:
When I took my PEE board exams 22 years ago, there was one problem under the subject “ELECTRICAL DESIGN & CONSTRUCTION” which ran similar as follows:
A 250 hp motor driving a pump is served by 3 x 100 kVA 13.8kV- 480v transformer bank. Other data are as follows:
Transformer Impedance: 3.7% IZ
Short Circuit Capacity of Source Utility: 200 MVA
Motor Starting Current: 8 times of FLC
Length of Cable from Transformer to Motor Controller: 50 ft
Solve for:
a) Size of Cable & Conduit
b) Branch Circuit Breaker & Controller Size if Auto-Transformer type RVS
c) The voltage dip during motor start-up using full voltage starter
d) The voltage dip during motor start-up using Auto-Transformer Reduced Voltage Starter tapped at 65%.
To be honest, I wasn’t ready to solve the (c) and (d) parts of the problem. To my defense mechanism, I told myself, “the problem was irrelevant. Nandiyan na ‘yan, naka-install na kunyari; bakit kwentahin pa ang voltage dip?“
To make the story short, lumipas ang mahabang panahon… Then here comes now a real situation. What if one is asked to design a set-up for a 500 hp motor driving a pump somewhere? Would a 750 kVA transformer be sufficient? Maybe yes, maybe not! Try solving the problem and one will discover things that are assumed easy but are actually not quite simple!
SIDEBAR 2:
In one brewery I happened to work with, we had two – 450 hp 4 kV motors driving ammonia compressors controlled by Auto-Transformer RVS. Later, three – 700 hp motors were added but interestingly this time with across-the-line starters. The system was directly connected to the Power Plant’s 4,160 v bus through feeders. No problem was imagined because the Power Plant as a standard operating procedure, used to operate with significant spinning reserve capacity at any given time – and there were nine (9) generating units to support the loads.
However, when we started-up for the first time one of the three 700 hp compressors, a huge voltage dip occurred and the neighboring Glass Plant tripped off all its operating air compressors thus shutting down the entire plant. Pandemonium followed as the Power Plant struggled to recover the voltage by shedding off other major loads. Power Plant finally recovered but with all major manufacturing plants served by it at shutdown, including the brewery.
The event’s aftermath saw signs posted on all compressor control panels that reads: “NOTIFY POWER PLANT BEFORE STARTING”. Why? Because Power Plant had to add more generators on stream before any starting can be done. This was to provide the on-line generating capacity large enough to support the starting kVA needed, thus averting objectionable voltage dips. Interestingly, after the start-up the additional generators had to be shutdown. In the end, the full voltage starters were replaced with RVS units. In the succeeding breweries being built thereafter, MV Soft Starters were employed.
Again, there’s more to it than meets the eye. Recalling back my own PEE board exams, now I understand that after all, the problem on voltage dip was not irrelevant but a test of engineering competence why I should be given a license as PEE.
2.0: SUBSTATION TRANSFORMERS VS. LOADS
Generators and transformers react to large motor starting as manifested in voltage dips. Let us first discuss transformers.
In general terms, transformers in an electrical system are usually larger than the maximum demands they serve, in some instances even larger than the connected loads. In the industrial plant scenario, the obvious reason at first glance for this apparent oversizing is the anticipation for future load growth. Fine…
But more often than not, sizing the transformer with extra kVA capacity unwittingly addresses voltage sag problems, not for load growth for which it is intended originally. That’s why for newly constructed plants where load growth is not yet there, the problem of starting significantly large motors may not surface out. Why? Because the extra kVA capacity intended for load growth is taking care of it.
3.0: SYSTEM BEHAVIOR IN EVENTS OF FAULTS
Short circuit capacity calculation is used in many applications, some of which are the following:
a) selecting the interrupting capacity ratings of circuit breakers and fuses,
b) specifying the short-time withstand of switchgears,
c) determining if a line reactor is required,
d) and a host of many others.
But unknown to many, Short Circuit Calculations together with
Motor Starting Calculations are also used in Sizing Transformers.
Short Circuit condition brings down the voltage very dramatically. The amount of voltage during a full three phase fault at the terminals of the transformer is determined by the %IZ of the transformer, i.e., if the transformer is connected to an infinite source of short circuit currents.
Take for instance a 7,500 kVA, 8.0% IZ, 69-4.16 kV transformer. The 8.0% impedance rating would mean that full load currents of 1,041 amps will flow in the secondary if the secondary is short circuited and the primary voltage is raised from zero volts to a point at which 8.0% of 4,160 volts, or 332.8 volts, appears at the secondary terminals. Therefore, the impedance (Z) of the transformer secondary may now be calculated:
Z = V / I = 332.8 volts/1,041 amps = 0.3197 ohms
The exercise described above is what is known in textbooks as the “Short Circuit Test” which is actually performed in laboratories or electrical shops (drying up transformers). What if the short circuit at the secondary is real – meaning, the short circuit happened at full 4160v at the inception of the fault? If the transformer is connected directly to a source capable of supplying the transformer with an unlimited short circuit kVA capacity, the short circuit amperage capacity which the transformer can deliver from its secondary is:
4,160 volts /0.3197 = 13,012.5 amps
Another method of calculating short circuit capacity for the above transformer is:
SC Capacity = Trafo FLC/%IZ
= 1,041 A / 0.08
= 13,012.5 amps,
or terms of kVA:
SC Capacity = 7,500 kVA/0.08 = 93,750 kVA
= 1,041 A / 0.08
= 13,012.5 amps,
or terms of kVA:
SC Capacity = 7,500 kVA/0.08 = 93,750 kVA
Such three-phase fault will bring down the voltage at the secondary terminals from 4,160v to approaching zero, while at the same time delivering 93.75 MVA of short circuit power to the point of fault. This condition however must of course be short-lived because some sort of protection upstream must operate to interrupt the circuit quickly, otherwise the transformer will fail. Transformer damage (known as the transformer short-time withstand) is dependent on the amount of fault current and the impedance of the transformer. Usually this transformer short-time withstand could only last for a few seconds.
4.0: SYSTEM BEHAVIOR DURING LARGE MOTOR START-UP
Why is motor starting related to short circuit condition? Partly similar to short circuit condition, motors have a high initial inrush current when energized and draw heavy load at a low power factor (0.30 lagging for 100 hp & about 0.16 for 1,000 hp motor) for a very short time. This sudden increase in the current flowing to the load causes a momentary increase in the voltage drop at the supply transformer terminals, the voltage drop along the distribution system, and a corresponding reduction in the voltage at the utilization equipment.
The magnitude of transient current involved in motor starting is however very much lower than the short circuit condition. But in effect, switching “on” to energize a large motor can be likened to a “soft short circuit”. Like short circuits, the effect of starting large motors results to voltage dips. The voltage drop at the transformer secondary terminals is proportional to motor starting kVA over the short-circuit capacity of the transformer. When motor starting kVA is drawn from a system, the voltage drop in percent of the initial voltage is approximately equal to the Motor Starting kVA divided by the Sum of this kVA and the Short Circuit kVA.
% Voltage Drop = (Motor Starting kVA) x 100 /(Motor Starting kVA + Short Circuit kVA)
5.0: MOTOR STARTING KVA
Thus, we need to know the starting kVA of the subject motor. In general, IEEE Red Book says, “the starting current or starting kVA of a standard motor draws several times its full-load running ratings. Without any specific information on a subject motor, a motor is always assumed to require about 1 kVA for each motor horsepower in normal operation, so the starting current of the average motor will be about 5 kVA for each motor horsepower. When the motor rating in horsepower approaches 5% of the secondary unit substation transformer capacity in kilovolt-amperes, the motor starting apparent power approaches 25% of the transformer capacity which, with a transformer impedance voltage of 6.07%, will result in a noticeable voltage sag on the order of 1%. This sudden increase in the current drawn from the power system may result in excessive drop in voltage unless it is considered in the design of the system. The motor-starting load in kilovolt-amperes, imposed on the power supply system, and the available motor torque are greatly affected by the method of starting used”.
The specific starting values for ac motors over hp are indicated in terms of Code Letters on the nameplates of North American-made motors. NEMA Code Letters indicate the motor starting characteristics as presented in the following table.
4.0: SYSTEM BEHAVIOR DURING LARGE MOTOR START-UP
Why is motor starting related to short circuit condition? Partly similar to short circuit condition, motors have a high initial inrush current when energized and draw heavy load at a low power factor (0.30 lagging for 100 hp & about 0.16 for 1,000 hp motor) for a very short time. This sudden increase in the current flowing to the load causes a momentary increase in the voltage drop at the supply transformer terminals, the voltage drop along the distribution system, and a corresponding reduction in the voltage at the utilization equipment.
The magnitude of transient current involved in motor starting is however very much lower than the short circuit condition. But in effect, switching “on” to energize a large motor can be likened to a “soft short circuit”. Like short circuits, the effect of starting large motors results to voltage dips. The voltage drop at the transformer secondary terminals is proportional to motor starting kVA over the short-circuit capacity of the transformer. When motor starting kVA is drawn from a system, the voltage drop in percent of the initial voltage is approximately equal to the Motor Starting kVA divided by the Sum of this kVA and the Short Circuit kVA.
% Voltage Drop = (Motor Starting kVA) x 100 /(Motor Starting kVA + Short Circuit kVA)
5.0: MOTOR STARTING KVA
Thus, we need to know the starting kVA of the subject motor. In general, IEEE Red Book says, “the starting current or starting kVA of a standard motor draws several times its full-load running ratings. Without any specific information on a subject motor, a motor is always assumed to require about 1 kVA for each motor horsepower in normal operation, so the starting current of the average motor will be about 5 kVA for each motor horsepower. When the motor rating in horsepower approaches 5% of the secondary unit substation transformer capacity in kilovolt-amperes, the motor starting apparent power approaches 25% of the transformer capacity which, with a transformer impedance voltage of 6.07%, will result in a noticeable voltage sag on the order of 1%. This sudden increase in the current drawn from the power system may result in excessive drop in voltage unless it is considered in the design of the system. The motor-starting load in kilovolt-amperes, imposed on the power supply system, and the available motor torque are greatly affected by the method of starting used”.
The specific starting values for ac motors over hp are indicated in terms of Code Letters on the nameplates of North American-made motors. NEMA Code Letters indicate the motor starting characteristics as presented in the following table.
TABLE 1: NEMA CODE LETTER DESIGNATIONS (STARTING KVA’s)
CODE LETTER KVA per HP
A: 3.15
B: 3.16 – 3.55
C: 3.56 – 4.0
D: 4.1 – 4.5
E: 4.6 – 5.0
F: 5.1 – 5.6
G: 5.7 – 6.3
H: 6.4 – 7.1
J: 7.2 – 8.0
K: 8.1 – 9.0
L: 9.1 - 10.0
M: 10.1 - 11.2
N: 11.3 - 12.5
P: 12.6 - 14.0
R: 14.1 - 16.0
S: 16.1 - 18.0
T: 18.1 - 20.0
U: 20.1 - 22.4
V: 22.5 - and up
The above table means that a 200 hp motor with a NEMA Code Letter ‘G’ requires a starting kVA of 1,140 to 1,260 kVA (1,200 kVA on the average) when using Full Voltage (Across-the-Line) motor controller. This starting kVA is also known as “Locked Rotor kVA” or sometimes “Locked Rotor Amperes”.
With the starting kVA of the motor such as described above, we must then determine the voltage dip caused by the motor inrush on start-up. The ensuing voltage during start-up must stay within the allowable operating voltage of the system. If such condition happens, then no oversizing of the transformer is required. But when the voltage dip exceeds the operating requirement of the system, then the transformer must provide extra kVA.
North American motors are rated for 230, 460, 2,300, 4,000, 6,600 or 13,200 volts for use with distribution sub-systems that are rated at 240, 480, 2400, 4,160, 6,900 or 13,800 volts respectively. Note the difference in motor nameplate voltages viz-a-viz the transformer terminal voltages. The apparent lower motor voltages than distribution nominal voltages are deliberately established by manufacturers to deal with the inherent voltage drops in the system such as: internal voltage drop of the transformer as dictated by its voltage regulation capability, voltage drop along the distribution cables and the impedance of the system. This could mean that the 4,160 v at the transformer at no-load condition may only be 3,800 v at the motor terminals when the system is already heavily loaded. In this condition, when a large motor is started-up somewhere in the system, the lower will the voltage be as felt by other loads in the system.
That’s why NEMA standard motors are designed to be capable of operating at plus or minus 10% of nameplate voltage. Therefore, the voltage drop on inrush should not be allowed to drop more than -10% of the rated voltage. This means 208v for 230v or 414v for 460 volt motors. Likewise, 2.07 kV or 3.6 kV for 2.3 or 4.0 kV motors, respectively. It means that a 4 kV motor can still operate satisfactorily at 3,600 v but any disturbance in the system that brings the system voltage lower, the affected motors may trip off as provided for by its protection – or if not, the motor burns.
(To be continued…)
Doods A. Amora
December, 2006
8 comments:
first i thank you for share ur idea. now i am also facing same ... i required motor detail on starting of large motor . pls give other referance.... and also i like to dicuss with u
Mr. Doods A. Amora, PEE,
I appreciate this artical on sizing transformers for large motor loads. I am an electrical engineer currently looking at a similar problem. I would like to ask you how the following equation for detemining voltage drop is derived:
Vd = kVA(motor)/kVA (transformer + motor)
I have seen this equation from other sources, and am merely curious on how it is derived using circuit analyis.
If you could respond, I would greatly appreciate it.
Thank you,
Edward P. Battle, P.E.
Pittsburgh, Pennsylvania
It is running well.I am one of user of this Transformer.
11KV Lightning Arresters
please let me know where i can find part 2 of this article sothat continuation can be possible.
Thank you very much sir. Further Can you please explain the basis of the expression
% voltage drop =100 * ( Motor starting kVA/(Motor starting kVA+ Short Circuit kVA)
Although you said the above equation calculates approximately the % Voltage Drop Value..
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