Module 5 - SNAMeLearning



2005 Professional Engineer Review Course

Unit V – Electrical; Instructor: J. Hennings

Introduction and Assignment

Module 5.3: Electrical Energy Conversion

Single Phase A/C: Sinusoids and Phasors

Consider the sinusoidal voltage:

[pic]

[pic]

Figure 5.3.1

Where:

Vm is the wave amplitude

( is the angular frequency in radians/seconds

( is the phase angle

The frequency in hertz is:

f = [pic]

and the wave period is:

T = [pic]

Although the average value of the voltage over a full period is zero

Vavg = 0

the more important voltage is the root mean square (RMS) or effective value. It is this RMS value that actually delivers power to a circuit.

For sinusoids:

VRMS = .707 ( Vm = [pic]

A/C currents can also be expressed as sinusoids:

I (t) = Im cos (( t + ()

It is the RMS value of current that actually delivers power to a circuit.

IRMS = .707 ( Im

The RMS or the effective value of A/C current is the value of an equivalent value of D/C current that would deliver the same average power to a resistor. See Reference 4, Chapter 10.

Example:

Given v(t) = 169.7 cos (377 t + 30() volts

Find VRMS , ( , f and T

Solution: VRMS = .707 ( 169.7V = 120V

◆ = 377 [pic]

f = [pic]

T = [pic]

A/C circuit calculations are greatly simplified by the use of phasor math. A phasor is a complex number that represents A/C voltages and currents by their amplitudes and phase angle. They look like vectors and the math using them is quite similar. The amplitude that is used for power calculations and other A/C circuit analysis is ALWAYS the RMS value of the voltage, and the phase is the angle ( where:

V(t) = Vm cos (( t + () this sinusoidal expression has an

equivalent RMS phasor of

V = VRMS [pic] [pic]in polar form

Example:

Express v (t) = 169.7 cos (377 t + 30() volts as a phasor

Solution:

[pic]

Note that frequency content is “lost” in the phasor expression, but so long as frequency remains constant, it will not effect calculations. Also note that the original phasor must be a cosine wave. If the v(t) is a sine wave then a phase shift of 90( must be included.

Thus:

V(t) = Vm Sin (( t + () has a phasor representation of

[pic]

Phasor relationships for circuit elements

Resistors oppose the flow of current through them as given by Ohm’s Law

VR = IR ( R

Where: VR and IR are phasors

Because resistors instantaneously change electrical energy into heat, the voltage across resistors is always in phase with the current through them. In A/C circuits resistors are given complex or phasor notations just as are voltages and currents. As a complex number:

R = R [pic]

Example:

A 10( resistor has a current of I = 5 [pic]amps flowing through it. Calculate the voltage (VR) across it.

Solution:

VR = I ( R = (5 [pic]) ( (10 [pic])

VR = 50 [pic] volts.

Note that when polar phasors are multiplied together, their magnitudes are multiplied as in simple arithmetic, but their angles are added.

Inductors (L) have units of Henries. Inductors oppose changes in current through them. This leads to the A/C current through an inductor being 90( out of phase with the voltage across it. The amount of opposition in ohms that an inductor has to A/C is directly proportional to its value of L and the frequency of the changing A/C. The inductors opposition to A/C is called inductive reactance (XL).

Where XL = (L = 2(fL

Inductive reactance in polar form has an angle of + 90( because of the fact that its current lags its voltage by 90(.

Example:

An inductor L = .0265 Henries = 26.5mH

(m = 10-3) is operated in a 60 hz system

Calculate XL as a polar phasor.

XL = 2(fL [pic] = 6.28 ( 60 ( 26.5 ( 10-3 [pic]

XL = 10 [pic] (

Capacitors have units of Farads. Capacitors oppose changes in voltage across them. This leads to the A/C voltage across a capacitor being 90( out of phase with the capacitor current. The amount of opposition in ohms that a capacitor has to A/C is inversely proportional to the product of its value and the frequency of the changing A/C. The capacitors opposition to A/C is called capacitive reactance XC.

Where: XC = [pic]

Capacitance reactance in polar form has an angle of -90(.

Example:

A capacitor C = 5.3 ( 10-4 F = 530( F

( ( = 10-6 ) is operated in a 60hz system.

Calculate: XC as a polar phasor.

XC = [pic] [pic]

XC = 5 [pic] (

Like inductors, capacitors also follow Ohm’s Law in A/C circuits.

Example:

If XC = 10 [pic] ( and VC = 25 [pic] V

Find IC

Solution:

IC = [pic]

IC = 2.5 [pic] AMPS.

Note that when polar phasors are divided, their magnitudes are divided as in simple arithmetic, but the angle of the denominator is subtracted from the angle of the numerator.

Example:

If XC = 5 [pic] ( and IC = 10 [pic] Amps

Find VC

Solution: VC = 10 [pic] ( 5 [pic] = 50 [pic] Volts

A/C Series Circuits

The solution of A/C series circuits follows all the rules for D/C series circuits but using complex or phasor math (Reference 4, Chapter 11)

In a series A/C circuit the total opposition is called impedance (Z) and has units of ohms. Impedance is also a complex value and can be expressed in either polar or rectangular form.

[pic]

Figure 5.3.2

As is always the case, the total opposition in any series circuit is the sum of the oppositions. But R cannot simply be added to XL and XC because of the angular relationships. In complex form R at zero degrees is all real, but XL at +90( is purely imaginary. In rectangular form the +90( for the inductor is replaced by +j, and the -90( of the capacitor is replaced by –j. (Note: do not confuse –j in phasor calculations with the unit vector j)

For the above R – L – C circuit where R = 4(, XL = 6(, and XC = 3( the impedance is:

Z = R + j XL - jXC or Z = 4 + j6 - j3 = 4 + j3 (

Using the Pythagorean Theorem, Z can also be expressed in polar form.

Z = [pic]

And [pic]

Where:

( is the impedance angle

Z = [pic]

← = tan-1 (¾) = 36.9º

so: Z = 4 + j3 = 5[pic]

Had XC = -j6( and XL = +j3( then the impedance would have been

Z = 4 - j3 = 5[pic]

As is also the case in all series circuits the current is the same through all oppositions connected in series (same magnitude and same angle).

Example:

For the single phase series A/C circuit shown, find Z, I, VR, VL, and VC

[pic]

Figure 5.3.3

Solution: (using the discussed methods)

Z = 4 +j10 –j5 = 4 + j5 = 6.4 [pic]

from Ohm’s Law: I = [pic]

VR = [pic]

VL = [pic]

VC = [pic]

It should be noted that the phasor sum of VR + VL + VC must equal VS. This is called Kirchhoff’s Voltage Law (KVL).

[pic]

A/C Parallel Circuits

The solution of A/C parallel circuits follows all the rules for D/C parallel circuits but using complex or phasor math (Reference 4, Chapter 12). As is always the case in all parallel circuits, the voltage across each parallel branch is the same as the source voltage (same amplitude and same phase angle). The phasor sum of IR, IL, and IC will equal IT. (Kirchhoff’s Current Law, KCL). Then the total impedance (ZT) will be the ratio of VS/IT .

[pic]

Figure 5.3.4

For the above parallel circuit find: VR , VL , VC , IR , IL , IC and ZT .

VS = VR = VL = VC = 120 [pic]

IR = [pic]

IL = [pic]

IC = [pic]

IT = [pic]

ZT = [pic]

It should be noted that the phasor sum of [pic]must equal [pic]. This is called Kirchhoff’s Current Law (KCL).

[pic]

Three Phase Loads

Most of the loads aboard ship are 3 phase loads. See Reference 4, Chapter 21.

Three phase loads can be either delta or wye connected. We will consider them to be “balanced” loads. That is to say that each side of the delta has the same magnitude and angle of impedance. Each leg of the wye also has the same magnitude and angle of impedance. When a set of 3 phase voltages is applied to the loads, currents flow in the lines to the loads. The phasor sum of all balanced, 3 phase voltages and currents is always zero. Notice that the lines are lettered A, B and C.

[pic]

Figure 5.3.5 A

[pic]

Figure 5.3.5 B

For both delta and wye, the three voltages between each pair of lines are called Line voltage (VLine). VLine is always the value of rated or nameplate voltage for the load. The three line voltages are:

VAB, VBC and VCA.

They are all equal in magnitude but mutually 120( out of phase with each other.

For a typical 440V, 3 phase load:

VAB = 440 [pic], VBC = 440 [pic], and VCA = 440 [pic],

Note [pic]

For both delta and wye, the balanced impedances will draw 3 Line currents when connected to the 3 Line voltages. The 3 Line currents are:

IA, IB, and IC

They are all equal in magnitude but mutually 120( out of phase with each other for balanced loads.

For a typical 3 phase load drawing 50A:

IA = 50 [pic], IB = 50 [pic], and IC = 50 [pic],

Note [pic]

For both the delta and wye, the voltage across and the current through one side of the delta or one leg of the wye are called phase voltage and phase current. Phase voltage and phase current are functions of Line voltage. There are three 3 phase voltages and three 3 phase currents for each load. The relationships between Line and Phase values for each type of load are the following:

For delta:

VLine = Vphase

I Line = [pic] Iphase

Iphase = [pic]

[pic] [pic]

For wye:

VLine = [pic] Vphase

ILine = Iphase

Iphase = [pic]

[pic] [pic]

Example:

A balanced 3 phase delta load is rated at [pic]

Is connected to a 440V, 3 phase system.

Calculate VLine , Vphase , ILine , Iphase

VLine = 440V always the system voltage

Vphase = VLine = 440V

Iphase = [pic]

ILine = [pic] ( 44 = 76.1 A

The nameplate voltage and nameplate current for this load would be 440v and 76.1A respectively.

Example:

A balanced 3 phase wye load is rated at [pic]

Is connected to a 440v, 3 phase system.

Calculate VLine , Vphase , ILine , Iphase

VLine = 440v always the system voltage

Vphase = [pic]

Iphase = [pic]

ILine = Iphase = 25.4 A

The nameplate voltage and nameplate current for this load would be 440v and 25.4A respectively.

A/C Power

As discussed in Module 5.2, there are three types of powers associated with A/C systems:

Real or Average power, P in KW

Apparent power, S in KVA

Reactive power, Q in KVAR

And power factor pf in KW/KVA

For single phase circuits:

Real power (KW) is the power to resistors and can be found with:

P = I2 R or by P = VS ( I ( pf

where pf can also be calculated by pf = cos of the impedance angle.

Apparent power (KVA) is the power to the circuit impedance and can be found with: S = I2 Z = VS ( I or S = P/pf

Reactive power (KVAR) is the power to XL and XC and can be found with:

Q = I2 X

where: X = XL - XC or Q = [pic]

The circuit power factor is said to be “lagging” if the impedance angle is positive and “leading” if the impedance angle is negative. See Reference 4, Chapter 13.

Example:

For the series single phase circuit solved previously, find P, S, Q and pf.

[pic]

Figure 5.3.6

Pf = cos (= 51.3() = 0.625 lagging

Lagging because of the positive impedance angle

P = 120 ( 18.75 ( .625 = 1406 W = 1.4 KW

or

P = (18.75)2 ( 4 = 1.4 KW

S = 120 ( 18.75 = 2250 VA = 2.25KVA

or

S = P/pf = [pic]

Q = I2 X where X = +10 –5 = 5 (

Q = (18.75)2 ( 5 = 1758 VAR approx. 1.76KVAR

or Q = [pic]

Note that no angles are used with voltage or current for calculating powers.

For three phase circuits

P = [pic] ( VLine ( ILine ( pf (KW)

S = [pic] ( VLine ( ILine (KVA)

Q = [pic] (KVAR)

And pf = P/S

The circuit power factor is said to be “lagging” if the impedance angle is positive and “leading” if the impedance angle is negative. P, Q and S are the total values to the entire load.

Example:

For the balanced delta circuit previously solved, find P, Q, S and pf.

[pic]

Figure 5.3.7

Solution:

pf = cos (+37() = .8 lagging

The pf is lagging because of the positive impedance angle.

P = [pic] ( 440 ( 76.1 ( .8 = 46.3 KW

S = [pic] ( 440 ( 76.1 = 57.9 KVA

Q = [pic]

Example:

For the balanced wye circuit previously solved, find P, Q, S and pf.

[pic]

Figure 5.3.8

Solution:

pf = cos (+37() = .8 lagging

the pf is lagging because of the positive impedance angle.

P = [pic] ( 440 ( 25.4 ( .8 = 15.5 KW

S = [pic] ( 440 ( 25.4 = 19.3 KVA

Q = [pic]

Again, note that no angles for voltage and current are used in power calculations.

Three phase power calculations can also be done for loads when only nameplate values are given. Consider the induction motor example of Module 5.2.

Example:

A 3 phase squirrel-cage induction motor has the following nameplate data:

440V, 50hp, ( = 90%, pf = .83, 60hz

find P, S, Q, ILine and Z

P = [pic]

S = [pic]

Q = [pic]

ILine = [pic]

If the motor is delta connected:

VLine = 440V , Vphase = 440V

ILine = 65.6A, Iphase = [pic]

Z( = [pic]

The impedance angle is cos-1 (pf)

If the motor is wye connected

VLine = 440V , Vphase = [pic]

ILine = 65.6A, Iphase = 65.6A

Zy = [pic]

For both delta and wye calculations, the impedance angle is [pic] (pf).

Generators

Ships generators are typically three phase 450V, 60hz, synchronous machines. Their power ratings run from 500KVA to 1500KVA. They can be either driven by steam or diesel prime movers. There are always an even number of magnetic poles on the rotor carrying a nominally small amount of D/C current. This rotor circuit is called the field or excitor circuit. The 3 phase winding in the stator is called the armature. Ship service generators are each capable of carrying the total at sea load. For peak loads, two generators are usually put on line at once.

For a synchronous machine shaft speed nm is the same as synchronous speed nS (the speed of the rotating magnetic field).

nm = nS = [pic] where f is frequency, usually 60hz and P is the number of poles, usually 2, 4 or 6.

Like the 3 phase loads they power, generators can be either delta or wye connected. In either case the nameplate voltage and current are always VLine and ILine .

The power can be given in either KW or KVA. While the generator voltage and frequency are kept relatively constant, the current and power outputs change according to load demand changes.

[pic]

Figure 5.3.9 A

[pic]

Figure 5.3.9 B

The time sequence in which the three Line Voltage (VAB, VBC and VCA) are induced as phasors is called the phase sequence. The only two possible phase sequences are:

A - B - C

and

C - B - A

In order for generators to be paralleled they must have the same phase sequence as well as frequency and voltage. The direction of rotation of 3 phase motors depends on the phase sequence of the applied voltages. If a 3 phase motor is installed and is turning in the wrong direction, reverse either of the two input leads to correct the problem.

Example:

A 3 phase, 4 pole, 60hz, 450V synchronous generator is rated at 1200A and .8 lagging power factor with ( = 92%.

Calculate the rated power out (P,S and Q) and the required primer mover speed and horse power.

Solution:

The given voltage and current on the nameplate are always Line values

S = [pic]

P = [pic]

Q = [pic]

Shaft speed:

nm = nS = [pic]

The calculated P = 747.4KW is the power output of the generator. Therefore the power required of the prime mover is

Pin = [pic]

Pin = [pic]

If the generator is wye connected:

VLine = 450V , Vphase = [pic]

ILine = 1200A, Iphase = 1200A

If the generator is delta connected:

VLine = 450V , Vphase = [pic]450V

ILine = 1200A, Iphase = [pic]

Transformers:

Transformers are input/output devices that allow A/C voltages to be transformed from one voltage level to another. They are highly efficient (usually ( > 90% and do not cause any change in frequency and very minimal change from Power-in to Power-out. For the purpose of these discussions, we will assume ( = 100%.

[pic]

Figure 5.3.10

Transformers are constructed by wrapping insulated turns of copper wire on a laminated steel core. See Reference 2, Chapter 7. The number of turns is designated N. Primary quantities are: N1 # primary turns

V1 rated primary voltage

I1 primary current

S1 primary KVA

Secondary quantities are: N2 # secondary turns

V2 rated secondary voltage

I2 secondary current

S2 secondary KVA

In both the primary and secondary, the voltage levels are fixed by the turns ratio (a). Turns ratio determines the ratio of V2 to V1 as follows:

a = [pic]

If N1 > N2 then V1 > V2 and the transformer is a step-down device.

If N1 < N2 then V1 < V2 and the transformer is a step-up device. For both step-up and step-down transformers

S1 = S2 (KVAin = KVAout)

So, V1 I1 = V2 I2

Therefore, P1 = P2

So, V1 ( I1 ( pf = V2 ( I2 ( pf

The power factor is the same on both primary and secondary and depends on the load impedance connected to the secondary. Where:

Z2 = Zload = [pic]

Transformers, as is the case with most other electrical equipment do not necessarily run at full-rated nameplate load. Although V1 and V2 are always fixed at rated values, I2 depends on the power demand of the load and I1 depends on I2.

For step-down transformers:

I1 < I2

For step-up transformers:

I1 > I2

I2 can be zero (when the secondary is open circuited providing no power out). If I2 = 0 then I1 = 0 but both V1 and V2 are rated so long as the transformer is energized.

Note: S1 = S2

Example:

A single phase 2.5KVA, 60hz, step-down transformer is rated at 440V/120V. If it is providing rated KVA at .8 lagging pf, find:

V1 , V2 , the turns ration, I1 , I2 , P1 , P2 , and Zload

Solution: Because it is a step-down device

V1 > V2

So: V1 = 440V

V2 = 120V

Turns ratio = [pic]

If N1 = 370 turns then N2 = 100 turns.

I2 = [pic]

P2 = S2 ( pf = 2.5KVA ( .8 = 2.0KW

[pic]

P1 = S1 ( pf = 2.5KVA ( .8 = 2.0KW

Note from the current calcs

[pic]

note: turns ratio is [pic]

Z2 = Zload = [pic] = [pic]

Z2 = 5.77 [pic]

Where: + 37( is cos-1 (pf)

Three phase transformers

Marine 3 phase transformers are made of 3 identical single phase transformers connected together to form one 3 phase device. Typically the primaries are connected in delta and the secondaries in either delta or wye.

[pic]

Figure 5.3.11 A

[pic]

Figure 5.3.11 B

Both the primary and secondary nameplate voltage are both always the line values, as are the nameplate primary and secondary currents. The rated KVA on the nameplate is the addition of the separate KVA values of each single phase device. The three phase turns ratio is always:

Turns ratio = [pic]

Where VLine 1 is the rated 3 phase primary Line voltage

And VLine 2 is the rated 3 phase secondary Line voltage.

Calculations for 3 phase transformers are similar to single phase devices.

Example: A 3 phase, 60hz, 440v/120v, step-down transformer is connected delta-wye. The three phase transformer is providing rated KVA of 15KVA to a .8 lagging power factor load. Find VLine1 , Vphase1 , VLine2, Vphase2 , ILine1 , ILine2. Solution:

VLine1 = 440V and VLine 2 = 120V given on nameplate.

Because the primary is delta: Vphase 1 = VLine 1 = 440V

Because the secondary is wye: Vphase 2 = [pic]

Turns ratio = [pic]

S2 = [pic]

So: ILine 2 = [pic]

Similarly: ILine 1 = [pic]

Note again that S1 = S2 and that [pic]is the turns ratio.

Problems:

1) Given the two sinusoids:

v1(t) = 155.6 cos (377t + 30() v

v2(t) = 325.3 sin (2512 t) v

find the RMS phasor equivalent of both

find the phase angle (() for both

V1: a) 0 V b) 110 V c) 120 V d) 155.6 V

V2: a) 0 V b) 208 V c) 220 V d) 230 V

f1: a) 50 hz b) 60 hz c) 377 hz d) 400 hz

f2: a) 50 hz b) 60 hz c) 400 hz d) 2512 hz

Φ1: a) 0o b) 377o c) 180o d) 30o

Φ2: a) 0o b) 2512o c) 180o d) –90o

2) An inductor of L = 5.3mH and a capacitor of C = 100(F are operated at 60hz.

find XL and XC. If the system frequency somehow increases what would happen to XL and XC?

XL: a) 5.3Ω │90o b) 5.3 Ω │-90o c) 2Ω │90o d) 2Ω │-90o

XC: a) 100Ω │-90o b) 100Ω │90o c ) 26.5Ω │90o d) 26.5Ω │-90o

3) For a single phase series A/C circuit

R = 4Ω , [pic] , [pic]

find: Z, I, VR, VL, P, Q, S, and the pf.

Z: a) 6Ω │37o b) 6Ω │-37o c) 4.47Ω │26.6o d) 4.47Ω │-26.6o

I: a) 16.7A │37o b) 16.7A │-37o c) 22.4A │26.6o d) 22.4A │-26.6o

VR: a) 66.8V │37o b) 66.8V │-37o c) 89.6V │26.6o d) 89.6V│-26.6o

VL: a) 33.4V │53o b) 33.4V│-53o c) 44.8V 63.4o d) 44.8V │-63.4o

P: a) 1 KW b) 2 KW c) 3 KW d) 4 KW

Q: a) 1 KVAR b) 2 KVAR c) 3 KVAR d) 4 KVAR

S: a) 1.1 KVA b) 2.24 KVA c) 3.36 KVA d) 4.47 KVA

4) For a single phase parallel A/C circuit

R = 10Ω , [pic], [pic]

Find: VR, VC, IR, IC, IT, Z, P, Q, S, and the pf.

VR: a) 50 V b) 100 V c) 150 V d) 200 V

VC: a) 50 V b) 100 V c) 150 V d) 200 V

IR: a) 5A │0o b) 10A│0o c) 5A │26.6o d) 10A │26.6o

IC: a) 5A │90o b) 5A │-90o c) 10A │26.6o d) 10A│-26.6o

IT: a) 5A │26.6o b) 11.2A -26.6o c) 11.2A │26.6o d) 5A │-26.6o

Z: a) 15Ω │26.6o b) 15Ω │-26.6o c) 8.9Ω │26.6o d) 8.9Ω │-26.6o

P: a) .5 KW b) 1.0 KW c) 1.12 KW d) 1.5 KW

Q: a) .5 KVAR b) 1.0 KVAR c) 1.12 KVAR d) 1.5 KVAR

S: a) .5 KVA b) 1.0 KVA c) 1.12 KVA d) 1.5 KVA

5) A balanced 3 phase delta load has phase current of Iphase = 30A at .8 lagging power factor, and is connected to a 450V, 3 phase system.

Calculate or find: VLine , Vphase , ILine, Z( , P, Q and S.

VLine: a) 150 V b) 260.1 V c) 450 V d) 778.5 V

VPhase: a) 150 V b) 260.1 V c) 450 V d) 778.5 V

ZΔ: a) 5Ω │36.9o b) 8.7Ω │36.9o c) 15Ω │-36.9o d) 15Ω │36.9o

P: a) 24.3 KW b) 32.3 KW c) 40.4 KW d) 55.6 KW

Q: a) 24.3 KVAR b) 32.3 KVAR c) 40.4 KVAR d) 55.6 KVAR

S: a) 24.3 KVA b) 32.3 KVA c) 40.4 KVA d) 55.6 KVA

6) A balanced 3 phase, wye load is connected to a 208V, 3 phase system. The load draws 1.5KW and 0.8 KVAR. Calculate or find VLine , Vphase , ILine, Iphase , ZY , P, Q , S and the pf.

VLine: a) 69.3 V b) 120 V c) 208 V d) 360 V

VPhase: a) 69.3 V b) 120 V c) 208 V d) 360 V

ILine: a) 2.73 A b) 4.72 A c) 8.16 A d) 14.2 A

IPhase: a) 2.73 A b) 4.72 A c) 8.16 A d) 14.2 A

P: a) 1.5 KW b) .8 KW c) 1.7 KW d) 2.3 KW

Q: a) 1.5 KVAR b) .8 KVAR c) 1.7 KVAR d) 2.3 KVAR

S: a) 1.5 KVA b) .8 KVA c) 1.7 KVA d) 2.3 KVA

7) A 3 phase, 450V, wye connected synchronous generator is supplying rated 1500KVA to a .8 lagging power factor load. The prime mover is at rated speed of 900RPM and the generator is at rated frequency of 60hz.

Calculate or find:

VLine , Vphase , ILine, Iphase , P, Q , and the number of poles.

VLine: a) 150 V b) 260.1 V c) 450 V d) 778 V

VPhase: a) 150 V b) 260.1 V c) 450 V d) 778 V

ILine: a) 633 A b) 1098 A c) 1927 A d) 3295 A

IPhase: a) 633 A b) 1098 A c) 1927 A d) 3295 A

P: a) 900 KW b) 1200 KW c) 1500 KW d) 2700 KW

Q: a) 900 KVAR b) 1200 KVAR c) 1500 KVAR d) 2700 KVAR

Poles: a) 2 b) 4 c) 6 d) 8

8) Repeat problem No. 7, but with a delta generator. Same 450V, 1500KVA, .8 lagging pf, 60hz and 900RPM.

VLine: a) 150 V b) 260.1 V c) 450 V d) 778 V

VPhase: a) 150 V b) 260.1 V c) 450 V d) 778 V

ILine: a) 633 A b) 1098 A c) 1927 A d) 3295 A

IPhase: a) 633 A b) 1098 A c) 1927 A d) 3295 A

P: a) 900 KW b) 1200 KW c) 1500 KW d) 2700 KW

Q: a) 900 KVAR b) 1200 KVAR c) 1500 KVAR d) 2700 KVAR

Poles: a) 2 b) 4 c) 6 d) 8

9) A single phase, 60hz step-down transformer operates at rated load and voltage of

230V/115V, 10KVA

and .8 lagging power factor

Calculate or find

V1 , V2 , I1 , I2 , Zload , S2 , S1 , P2 , and P1 .

If this transformer has N1 = 200 turns, Calculate N2.

V1: a) 115 V b) 230 V

V2: a) 115 V b) 230 V

I1: a) 43.5 A b) 87 A

I2: a) 43.5 A b) 87 A

ZLoad: a) 2.64Ω │37o b) 1.32Ω │37o c) 2.64Ω │-37o d) 1.32Ω │-37o

S2: a) 5 KVA b) 10 KVA c) 20 KVA

S1: a) 5 KVA b) 10 KVA c) 20 KVA

P2: a) 4 KW b) 8 KW c) 10 KW

P1: a) 4 KW b) 8 KW c) 10 KW

N2: a) 100 b) 200 c) 300

10) A 3 phase delta to wye transformer is a step-down device. It operates at its rated nameplate values of:

208V/120V and 12KVA at power factor .8 lagging.

Calculate or find for both the primary and secondary:

VLine , Vphase , ILine, Iphase , P, S, and Q.

If the three phase transformer consists of 3 identical single phase devices, what is the KVA rating of each single phase transformer?

For the primary:

VLine: a) 69.3 V b) 120 V c) 208 V d) 360 V

VPhase: a) 69.3 V b) 120 V c) 208 V d) 360 V

ILine: a) 11.1 A b) 19.2 A c) 33.3 A d) 57.8 A

S1: a) 7.2 KVA b) 9.6 KVA c) 12 KVA d) 16.8 KVA

P1: a) 7.2 KW b) 9.6 KW c) 12 KW d) 16.8 KW

Q1: a) 7.2 KVAR b) 9.6 KVAR c) 12 KVAR d) 16.8 KVAR

For the secondary

VLine: a) 69.3 V b) 120 V c) 208 V d) 360 V

VPhase: a) 69.3 V b) 120 V c) 208 V d) 360 V

ILine: a) 11.1 A b) 19.2 A c) 33.3 A d) 57.8 A

S2: a) 7.2 KVA b) 9.6 KVA c) 12 KVA d) 16.8 KVA

P2: a) 7.2 KW b) 9.6 KW c) 12 KW d) 16.8 KW

Q2: a) 7.2 KVAR b) 9.6 KVAR c) 12 KVAR d) 16.8 KVAR

KVA Per Each Single Phase: a) 4 KVA b) 6.94 KVA c) 12 KVA d) 20.8 KVA

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