single and three phase transformers and power flow and three... · single and three phase...

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Syed A. Rizvi Summer’12-Revised Spring’16 ENS 441/ELT 437

Single and Three Phase Transformers and Power Flow

Caution: In these experiments high voltages are involved, which could be harmful or fatal

if one is exposed to them. Adhere to electrical safety rule at all times. Make sure that all of

your connections are correct before turning on the power.

A transformer is an electrical device that is used to raise or lower the voltage (or current) level. A

single-phase transformer has two electrically isolated windings. The winding that is connected to

the electrical power source is called the “primary” winding and the winding that is used to draw

electrical power is called the “secondary” winding. Fig. 1 shows a schematic diagram of a single-

phase transformer.

Figure 1: Schematic diagram of a single phase transformer.

Let us examine the ideal transformer shown in Fig. 1 where,

EP = induced voltage on the primary side (RMS)

ES = induced voltage on the secondary side (RMS)

NP = Number of turns on the primary windings

NS = Number of turns on the secondary windings

Φm

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Φm = magnetic flux created by the source voltage E

When the primary winding is connected to an AC power source it produces an instantaneous

magnetic flux “Φm (t)” linking the primary as well as secondary windings of the transformer. The

flux linkage induces an instantaneous voltage “Ep (t)” in the primary winding in accordance with

the Faraday’s law of electromagnetic induction. That is,

𝐸𝑝(𝑡) = 𝑁𝑝

𝑑∅𝑚(𝑡)

𝑑𝑡 (1)

Since the voltage applied at the primary is sinusoidal, therefore, the magnetic flux Φm (t) is also

sinusoidal and mathematically it can be expressed as

∅𝑚(𝑡) = ∅𝑚𝑎𝑥 sin 𝜔𝑡 (2)

therefore,

𝐸𝑝(𝑡) = 𝑁𝑝

𝑑∅𝑚(𝑡)

𝑑𝑡= 𝑁𝑝 × ∅𝑚𝑎𝑥

𝑑(sin 𝜔𝑡)

𝑑𝑡 (3)

or

𝐸𝑝(𝑡) = 𝑁𝑝 × ∅𝑚𝑎𝑥 × 𝜔 cos 𝜔𝑡. (4)

Substituting cos 𝜔𝑡 = sin(𝜔𝑡 + 90𝑜) and 𝜔 = 2𝜋𝑓 in Eq. 3 we get

𝐸𝑝(𝑡) = 2𝜋𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥 sin(𝜔𝑡 + 90𝑜). (5)

Eq. 4 shows that the induced voltage Ep (t) leads the magnetic flux Φm (t) by 90o. The RMS value

of Ep (t), denoted by Ep is given by

𝐸𝑝 = 𝑅𝑀𝑆[𝐸𝑝(𝑡)] =

|𝐸𝑝(𝑡)|𝑚𝑎𝑥

√2=

2𝜋𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥

√2 (6)

or

𝐸𝑝 = 4.44 𝑓 × 𝑁𝑝 × ∅𝑚𝑎𝑥. (7)

Since the magnetic flux Φm (t) also links the secondary winding that has Ns turns, the RMS value

of the induced voltage at the secondary side of the transformer, denoted by Es is given by

𝐸𝑠 = 4.44 𝑓 × 𝑁𝑠 × ∅𝑚𝑎𝑥. (8)

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Comparing Eqs. (7) and (8) yields

𝐸𝑝

4.44 𝑓 × 𝑁𝑝=

𝐸𝑠

4.44 𝑓 × 𝑁𝑠 (9)

or

𝐸𝑝

𝑁𝑝=

𝐸𝑠

𝑁𝑠 (10)

𝐸𝑝

𝐸𝑠=

𝑁𝑝

𝑁𝑠= 𝑎 (11)

or

𝐸𝑝 = 𝑎 × 𝐸𝑠 (12)

where, “a” is called the turn ratio of the transformer.

Figure 2: An ideal transformer with a load on the secondary side.

When secondary side is open (no load condition) under ideal conditions no current flows through

the primary winding and the primary current “IP” is zero. When a load is connected to the

secondary side (see Fig. 2), a current, Is, flows out of the secondary windings. The secondary

current, Is, creates a magnetic flux “Φs” in the transformer’s core. Note that the source voltage, E,

is unchanged, which mean the flux, Φm, must remain the same as it was before the load was

connected. Therefore, the new flux (Φs) due to the secondary the current, Is, must be countered

by an equal and opposite flux. The counter flux “Φp” is produced by causing a current, Ip, to flow

into the primary windings. Under ideal conditions, the fluxes Φs and Φp cancel each other and the

total flux through the core remains Φm. Therefore, the primary and secondary induced voltages

(Ep and Es) remain unchanged.

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The flow of the primary current, Ip, represents the power transfer from primary side to the

secondary side of the transformer to support the load Zs. Under ideal conditions there is no loss

of power during the power transfer from the primary side to the secondary side of the

transformer. Therefore, the total power, Sp, on the primary side of the transformer must be equal

to the total power, Ss, on the secondary side of the transformer, where

𝑆𝑝 = 𝐸𝑝 × 𝐼𝑝 (13)

and

𝑆𝑠 = 𝐸𝑠 × 𝐼𝑠 (14)

but Sp = Ss, therefore,

𝐸𝑝 × 𝐼𝑝 = 𝐸𝑠 × 𝐼𝑠. (15)

Combining Eqs. 12 and 14 yeilds

𝑎 × 𝐸𝑠 × 𝐼𝑝 = 𝐸𝑠 × 𝐼𝑠 (16)

or

𝐼𝑝 =

1

𝑎× 𝐼𝑠. (17)

The load impedance as seen by the primary side, denoted by Zp, can be given by

𝑍𝑝 =

𝐸𝑝

𝐼𝑝=

𝑎 × 𝐸𝑠

1𝑎 × 𝐼𝑠

= 𝑎2 ×𝐸𝑠

𝐼𝑠 (18)

but

𝑍𝑠 =

𝐸𝑠

𝐼𝑠 (19)

therefore,

𝑍𝑝 = 𝑎2 × 𝑍𝑠. (20)

Equation 20 shows an important property of the transformer; that is, it can be used to change the

effective impedance of any load and thus can also act as an impedance transformer. Equation 20

is used to express impedances on the secondary side as the equivalent impedances on the primary

side to greatly simplify the circuit analysis involving transformers (see Fig. 3).

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Figure 3: An ideal transformer with the load shifted to the primary side.

So far we have considered an ideal transformer with no primary current when there was no load

attached to the secondary side. That is, the ideal transformer is lossless. In a real transformer,

however, the primary current is not zero under no-load condition due to the “iron losses” and

“magnetizing reactance” of the transformer. The iron losses are due to eddy currents and

hysteresis losses (described below).

Eddy Current Losses:

When time-varying magnetic flux passes through a solid metal core, the solid metallic core acts

like a bundle of concentric conductors and a voltage is induced in each of these sections of the

metallic core. Note that in a solid metal plate each of these sections would behave like a short-

circuited conductor and, therefore, the induce voltages would cause a strong currents which swirl

back and forth in the core. These currents are called eddy currents. Fig. 4 shows a cross-section

of a solid metal core with circulating eddy currents. Eddy currents are a major source of

generating heat in transformers. In a solid metal core these currents can be so strong that they

render the core red hot. To minimize eddy current, cores are designed with thin laminated stripes

instead of using a solid piece of metal. The lamination insulates the strips from each other, thus

greatly reducing the length of the conducting path, which in turn, significantly reduces the

magnitudes of the induced voltages in the core thereby diminishing eddy current losses. In

general, eddy current losses are reduced proportional to the square of the number of strips used

in the core.

Figure 4: Eddy currents in a solid metallic core.

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Hysteresis Losses

The magnetic flux is caused by magnetic field strength, H. The resulting flux density, B, in any

material is related to the magnetic field strength (H) by the following equation:

𝐵 = 𝜇𝑜 × 𝜇𝑟 × 𝐻 (21)

where,

B = magnetic flux density (weber/m2 or tesla)

H = magnetic field strength (A-turns/m)

µo = permeability of free space = 4𝜋 × 10−7

µr = relative permeability of the material

Note that the value of µr changes with flux and, therefore, the relationship between B and H is

expressed in terms of a B-H curve. Under an AC current the iron core is continuously

magnetized in one direction, demagnetized, and then magnetized again in the opposite direction.

However, the relationship between B and H is not linear. Fig. 5 shows the change in flux density

with the magnetic field intensity for one complete cycle of AC current. Initially, B reaches its

peak (let’s assume the positive peak) as H reaches the maximum field strength (at maximum

current in the positive direction). After that H decrease with the decreasing current and so does

B; however, B decreases at a slower rate than H following the path a-b. The result is that as the

current reduces to zero bringing H to zero, there still remains some residual flux density in the

core (“+Br” in Fig. 5). As the current starts increasing in the negative direction, producing a

negative magnetic field strength, B continue to decrease following the path b-c and eventually

reduces to zero at a field strength of “-Hc.” As the current continue to decrease in negative

direction B starts to decrease in negative direction following the path c-d and eventually reach its

maximum value in the negative direction (-Bmax) as H reaches its negative peak. As the current

starts decreasing again reducing H, B decreases as well, but with a slower rate following the path

d-e. Consequently, when the current and H reduce to zero, B still maintains a negative flux of –

Br in the core. B eventually reduces to zero at when H = +Hc following the path e-f. B then

increases to its peak value following the path f-a, as H continues to increase and eventually peaks

(point “a” in Fig. 5).

The closed B-H curve shown in Fig. 5 is called a hysteresis loop and the area under this curve

represents the energy that changes into heat during each cycle (joules per cubic meters). To

minimize the hysteresis losses materials with relatively narrow hysteresis loop are used to build

transformer cores. Since the iron losses (eddy current and hysteresis losses) only generate heat

and do not affect the induced voltage they can be represented as resistance (Rm) in parallel to the

primary terminals of the transformer.

Furthermore, a current must flow through the primary windings to generate the magnetic flux

Φm. This current is called magnetizing current and depends on the permeability of the material

used to build the core. Higher permeability means lower magnetic current and vice versa.

Permeability can be represented by a parallel inductance (Xm) to the primary terminal because

magnetizing current does not affect the amount of induced voltage. This inductance is called

magnetizing inductance.

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Figure 5: Hysteresis loop.

In a real transformer, the winding of the transformer has a finite resistance that must be taken

into account while analyzing or modeling a practical transformer. Since these winding

resistances result in a potential drop when current passes through the windings, they are

represented as the series resistors (Rp and RS), both at the primary as well as at the secondary

side of the transformer.

In the analysis of the ideal transformer we assumed the flux due to the load current (Φs) and the

counter flux (Φp) generated by the primary remain completely inside the core. However, in a real

transformer some of the flux Φs and Φp does leak in the air. This flux leakage induces voltages

both in the primary winding and secondary windings, which is counter to the voltages induced by

the flux Φm. Consequently, it results in a voltage drop both at the primary as well as the

secondary side. Accordingly, the effect of the leakage flux is represented as the series

impedances (Xlp and Xls) at the primary as well as the secondary side of the transformer of the

model of a practical transformer. Fig. 6 shows the complete model of practical transformer that

incorporate the iron losses, magnetizing current, primary and secondary winding resistances, and

the effect of the leakage flux under load conditions.

Figure 6: Model of a practical transformer.

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The model of the practical transformer shown in Fig. 6 becomes a lot easier to analyze if the

elements on the secondary side are shifted to the primary side using equation (20). Figure 7

shows the model of the practical transformer when everything is viewed from the primary side.

We will use the model in Fig. 7 in our analysis of practical transformers.

Figure 7: Model of a practical transformer when the circuit is viewed from the primary side.

Let’s now examine the model for the practical transformer shown in Fig. 7 for two special load

conditions: (1) No-load condition and (2) full-load condition.

No-load Condition:

Under no-load condition the secondary is open and, therefore, no current flows through the

secondary terminals. Consequently, I/p = 0, which means that Rs and Xls can be eliminated from

the circuit. Also, Ip = Io, that is the branch containing Rp and Xlp now appear in series with the

parallel combination of Rm and Xm. Note that both Rp and Xlp are significantly smaller than Rm

and Xm that they can safely be ignored when compared to Rm and Xm. This simplifies the model

of the practical transformer under no-load condition to the one given in Fig. 8.

Full-load Condition:

Note under the full-load condition 𝐼𝑝 ≫ 𝐼𝑜 and, therefore, we can ignore Rm and Xm and the

model for the practical transformer reduces to the one shown in Fig. 9. The model of the

practical transformer can further be simplified by adding the resistances and inductances as

follows:

𝑅𝑒𝑝 = 𝑅𝑝 + 𝑎2 × 𝑅𝑠 (22)

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Figure 8: Model for the practical transformer under no-load condition.

Figure 9: Model of a practical transformer under full-load condition.

and

𝑋𝑒𝑝 = 𝑋𝑙𝑝 + 𝑎2 × 𝑋𝑙𝑠. (23)

Figure 10 shows the simplified circuit for the practical transformer with full load. Furthermore,

Rlp and Xlp can be represented by the complex impedance Zp as follows:

𝑍𝑝 = 𝑅𝑒𝑝 + 𝑗𝑋𝑒𝑝 (24)

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Figure 10: Simplified circuit of the practical transformer under full-load condition.

Zp is referred to as transformer impedance and is an important parameter that is used to compute

the internal voltage drop of a transformer for varying load conditions (as long as Ip ≥ 10 Io).

Example: Specifications of a power transformer are as follows:

Sn (KVA)

Enp (V)

Ens (V)

Inp (A)

Ins

(A) Rp

(Ω) Rs

(Ω) Xlp

(Ω) Xls

(Ω) Xm (Ω) Rm (Ω) Io (A)

1000 69000 6900 14.5 145 27.2 0.25 151 1.5 505000 432000 0.210

Compute the full-load Vs, Is, as well as heat dissipation in the transformer using the simplified

model of Fig. 10. Also, compute Iron losses using the model under no-load condition of Fig. 8.

Solution: Let’s assume it’s a pure resistive load and full-load secondary voltage is the same as

Vns and full-load secondary current is the same as Ins. Now we can calculate the load resistance,

RL, as follows:

𝑅𝐿 = 𝐸𝑛𝑠

𝐼𝑛𝑠=

6900

14.5= 47.6 Ω

The turn ratio is given by

𝑎 =𝐸𝑛𝑝

𝐸𝑛𝑠=

69000

6900= 10

Now we can compute Rep and Xep

𝑅𝑒𝑝 = 𝑅𝑝 + 𝑎2 × 𝑅𝑠 = 27.2 + 102 × 0.25 = 52.2 Ω

𝑋𝑒𝑝 = 𝑋𝑝 + 𝑎2 × 𝑋𝑠 = 151 + 102 × 1.5 = 301 Ω

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The resistance shifted to primary, RLp, is given by

𝑅𝐿𝑝 = 𝑎2 × 𝑅𝐿 = 102 × 47.6 = 4760 Ω.

Now we can compute the total impedance of the model, ZTp, as follows

𝑍𝑇𝑝 = 𝑅𝑒𝑝 + 𝑅𝐿𝑝 + 𝑗𝑋𝑒𝑝 = 52.2 + 4760 + 𝑗301 = 4812.2 + 𝑗301 = 4821.6 ⟨0.0186𝑜 Ω

Now we can compute primary current, Ip, primary voltage, Ep, secondary current,

Is, and secondary voltage, Es as follows

𝐼𝑝 =𝐸𝑛𝑝

𝑍𝑇𝑝=

69000

4821.6= 14.31 𝐴

𝐸𝑝 = 𝐼𝑝 × 𝑍𝑇𝑝 = 14.31 × 4821.6 = 68118.4 𝑉

𝐸𝑠 =𝐸𝑝

𝑎=

68118.4

10= 6811.84 𝑉

𝐼𝑠 =𝐸𝑠

𝑅𝐿=

6811.84

47.6= 143.1 𝐴

Now we can compute the heat loss due to the load as follows

𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 = 𝐼𝑝2 × 𝑅𝑒𝑝 = 14.312 × 52.2 = 10689.3 𝑊 = 10.69 𝐾𝑊

Finally, we’ll compute iron losses at no load and at full load as follows

𝐼𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑡 𝑛𝑜 𝑙𝑜𝑎𝑑 =𝐸𝑛𝑝

2

𝑅𝑚=

690002

432000= 11020.83 𝑊 = 11.02 𝐾𝑊

𝐼𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 =𝐸𝑝

2

𝑅𝑚=

68118.42

432000= 10741 𝑊 = 10.74 𝐾𝑊

Total power loss at full load = 10.69 + 10.74 = 21.43 KW.

This power loss produces a significant amount of heat that must be removed from the

transformer to avoid any damage to the transformer. There are a number of techniques that are

used in the transformers to remove heat, such as through the natural circulation of air (type AA),

through forced air circulation (type AFA), oil-immersed self-cooled (OA), and oil immersed self-

cooled/forced air cooled (type OA/FA) to name a few. Figure 11 shows the front view of an

OA/FA type power transformer and Fig. 12 shows the rear view of the same transformer where

cooling fans can be seen, which remove heat from the transformer through forced air circulation.

A transformer is called a “step-down” transformer if the secondary voltage, ES, is lower than the

primary voltage, EP. On the hand, if the secondary voltage is higher than the primary voltage, it

is called a “step-up” transformer.

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Figure 11: Front view of an OA/FA type power transformer.

Figure 12: Rear view of an OA/FA type power transformer. The cooling fans can be in, which

are used for forced air circulation to remove heat from the power transformer.

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Terminal Marking on Power Transformers:

In power transformers, the high voltage (HV) terminals are marked by symbols H1 and H2; the

low voltage (LV) terminals are marked by symbols X1 and X2. These terminals are mounted on

the transformer tank in a way that they either have additive polarity or they have subtractive

polarity. The transformer is said to have additive polarity when the terminal H1 is mounted

diagonally opposite to the terminal X1. On the other hand, if the terminal H1 is mounted directly

opposite to the terminal X1, it is said to have subtractive polarity (see Fig. 13). When you face

the voltage side of a power transformer, the terminal at the extreme right is H1, and H2, and so

on. The location of the low voltage side terminal would depend on whether the transformer has

an additive or subtractive polarity. Subtractive polarity is standard on transformer above 200

KVA with HV rating of above 8660 V. Otherwise the transformer would have additive polarity.

Note that it is imperative to know the correct polarity of the transformers if you are connecting

them in parallel or configuring a 3-phase transformer bank using single phase transformers.

Figure 13: Terminal marking and polarity of the power transformers.

Voltage Regulation:

Voltage regulation is a measure of variation in the secondary voltage from no-load to full-load

condition when the primary voltage is kept constant. Voltage regulation is expressed in percent

(%) and is defined by the following equation

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =

𝐸𝑠 (𝑛𝑜−𝑙𝑜𝑎𝑑) − 𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)

𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)× 100 (25)

Power in AC Systems:

In AC systems the power is divided into three different categories:

Apparent Power:

It is the product of the voltage and the complex conjugate of the current and is expressed in

“volt-amperes” or VA. If the current I = IR + j IX then the conjugate of I, denoted by I*, is given

by I*=IR – j IX. The apparent power, PA, is given by

𝑃𝐴 = 𝑉 × 𝐼∗

(26)

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Real Power:

The real power (PR) is the power utilized by a user and is expressed in watts (W). When the

apparent power is expressed in rectangular form, the real part of the apparent power is the real

power. The real power always flows from the source to the user.

Reactive Power:

The reactive power (PX) is needed to establish a magnetic field in inductors. An electric field in a

capacitor acts as a source of reactive power. Reactive power can be supplied as well as absorbed

by an electric power source. It means that the reactive power can flow from the source to the

circuit or from the circuit to the source. When the reactive power flows from the source to the

circuit it is considered “positive” reactive power. However, it is considered “negative” reactive

power when it flows from the circuit to the source.

Mathematically, the apparent power (PA), the real power (PR), and the reactive power (PX) are

related as follows:

𝑃𝐴 = 𝑃𝑅 ± 𝑗𝑃𝑋

Assignment 1.1: Compute Iron losses and magnetizing reactance of a single-phase transformer.

Also, find the polarity of a single phase transformer and measure its voltage regulation.

Components needed:

Single phase transformer (N-6U): 1

Single phase transformer (VPS230-190): 1

Resistor 10 Ω: 1

Resistor 5 Ω: 1

Resistor 2 Ω: 1

1. Connect the transformer (N-6U) and meters as shown in Fig. 14. Make sure that all of

your connections are correct before turning on the power.

2. Turn the power on and measure P through the power meter. Also measure the current Ip

and Vp. Since there is no load connected to the secondary, P is solely due to the iron

losses.

3. Compute apparent power as follows:

𝑆 = 𝑉𝑝𝐼𝑝

4. Compute the reactive power, Q, as follows:

𝑄 = √𝑆2 − 𝑃2

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5. Compute magnetizing reactance “Xm” as follows:

𝑋𝑚 = 𝑉𝑝

2

𝑄

Figure 14: Circuit diagram for measuring iron losses and magnetizing reactance

Voltage Regulation:

1. Build the circuit shown in Fig. 15 using the transformer VPS230-190. Use 17 Ω for the

load. Note at full load Is must be less than 1.6 A. If this is not the case, add more

resistance to the load until Is becomes 1.6 A or less.

2. Measure the voltage across the load. This is Es (full-load).

3. Now disconnect the load and measure the voltage across the secondary terminals. This is

Es (no-load).

Figure 15: Circuit diagram for measuring voltage regulation of a transformer.

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4. Compute the voltage regulation as follows

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 =𝐸𝑠 (𝑛𝑜−𝑙𝑜𝑎𝑑) − 𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)

𝐸𝑠 (𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑)× 100.

Polarity of a transformer:

1. Connect the HV side of the transformer VPS230-190 to 50 V AC supply.

2. Connect any two adjacent HV and LV terminals using a jumper.

3. Measure the voltages Ep and Ex (see Fig. 16).

4. If Ex > Ep, the polarity of the transformer is additive.

5. However, if Ex < Ep, the polarity of the transformer is subtractive.

Figure 16: Circuit for polarity test of a transformer.

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