TRANSFORMER VOLTAGE is the induced voltage of mutual induction in the secondary coil. TRANSFORMER ACTION is the action t
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TRANSFORMER VOLTAGE is the induced voltage of mutual induction in the secondary coil. TRANSFORMER ACTION is the action that creates the emf. PRIMARY COIL is the coil that’s connected to the primary source of supply or takes electrical energy from AC source of supply. SECONDARY COIL is the coil which the voltage of mutual induction is induced and which “feeds” energy to the load. It received energy by electromagnetic induction and deliver to loads. Mutual flux changes in magnitude and direction. There still be created an induced emf in secondary winding “Es” and induced
emf in primary winding “Ep”. Note: • both induced emf’s are created by the same mutual flux. •
coils are form wound and are of the cylindrical type. The general form of this coils maybe circular or oval. In small type size core type transformer, “a simple rectangular core is used with cylindrical coils which are either circular or rectangular in form. “but for large size core type transformer” round or circular cylindrical coils are used.
Where: Eav – average induced emf in coil.
ŋ–
number of turns in coils.
Ф – maximum flux t – time for flux to change by Ф m (mxwell) Note: electrical energy can be transformed by: a. INDUCTION – no electrical connection between source and load (Ex. transformer). b. CONDUCTION – when there is c conductor that links between source and load. (Ex. transmission and distribution). Basic information that is placed on the nameplate of large power transformer. a. b. c.
Maker’s Name (Brand) Voltage Rating KVA / MVA Rating
d. Percent Impedance Rating (%z) – To be able to calculate amount of fault current that due to transformer can withstand. e. B.I.L. Rating – (Basic Insulation Level) - KVA rating of insulating oil
the
Simplified Circuit Diagram
Ip = primary current (amps) Rp = primary resistance (ohms) Xp = reactance of primary winding (ohms) Rc = resistance representing of iron losses (ohms) Ic = current flowing (amperes) Xm = magnitizing reactance of primary winding (ohms) Im = current flowing in the magnetizing reactance (current) Rs = resistance in the secondary winding (ohms) Xs = reactance in the secondary winding (ohms) Np = primary turns Ns = secondary turns Ep = primary voltage Es = secondary voltage Is = secondary current
Divide:
Induced emf equation of a transformer applied equally to the primary and secondary voltages
EQUIVALENT CIRCUIT OF AN IDEAL TRANSFORMER:
I₂
I₁ Vg
Ep
E = 4.44 NФm x 10-8 (effective) E = 4.44 NpФm x 10-8 (primary) Es = 4.44 NsФm x 10-8 (secondary)
Es
VL
Note: ideal transformer if its core loss is less and has no leakage flux and has no cupper loss. (Ф m = ϐmA) Where: • • • • • • • • • •
E = RMS voltage induced or effective voltage induced. Ep = primary winding induced voltage. Es = secondary winding induced voltage = frequency (Hz) N = number of turns in coil Np = No. of turns in primary coil Ns = No. of turns in secondary Фm = mutual flux (maxwell) A = area in core (cm²) Βm = flux density
VOLTAGE, CURRENT AND TURNS RATIO OF TRANSFORMER
that if the primary ampere turn is equal to secondary ampere turn
Where: Np / Ns = turns ratio Ip / Is = current ratio a = transformer ratio KVA rating of transformer refers to its KVA output.
KVAp = KVAs
Where: Zp / Zs = impedence ratio Ep = voltage induced in primary winding Es = voltage induced in secondary winding Polarity Marking – to avoid mistakes in banking and paralleling of transformer. • •
Additive Polarity Subtractive Polarity
Example: The 2,300 volt primary winding of a 60-cycle transformer has 4,800 turns. Calculate: (a) the mutual flux; (b) the number of turns in the 230-volt secondary winding. Given: Ep = 2,300v Np = 4,800 Solution: a.) b.)
F = 60 hz
Example: the maximum flux in the core of a 60-cycle transformer that has 1,320 primary turns and 46 secondary turns is 3.76 x maxwells. Calculate the primary and secondary induced voltages. Given:
F = 60 Hz
Calculate Ep and Es Solution:
Np = 1,320turns
Ns = 46 turns
Example: the secondary winding of a 4,600/230-volt transformer has 36 turns. How many turns are there in the primary winding? Given: Vp = 4,600 v Solution:
Vs = 230 v
Ns = 36 turn
Example: The volts per turn in a 25-cycle 2,400/230-volts transformer is 8. calculate: (a) the primary and secondary turns; (b) the maximum flux in the core. Given: Vp = 2,400 v Solution: a.)
b.)
Vs = 230 v
VOLTAGE AND CURRENT RATIO: Primary Voltage (Vp) Secondary Voltage(Es) Primary current (Ip) Secondary Current (Is) Ep x Ip = Es x Is
Example: The secondary load current of a 2,300/115-volts/transformer is 46 amp. Calculate the primary current. Solution: Example: The primary and secondary currents of a transformer were measured and found to be 3.8 and 152 amp, respectively. If the secondary load voltage is 116 volts, what is the primary emf? Solution:
Transformation Ratio: The ratio of primary to secondary turns Np:Ns, which equals the ratio of primary to secondary induced voltages Ep:Es, indicates how much the primary voltage is lowered or raised. The turn ratio, or the induced-voltage ratio, is called the ratio of transformation, and is represented by the symbol a. thus
Where the no-load and full load voltages measured at the secondary terminals.
and
are those
Equivalent Resistance, Reactance and Impedance.
}
in secondary terms
}
in primary terms
SEATWORK Example 1 A 25-kva 2,300/230-volt distribution transformer has the following resistance and leakage-reactance values: Rp = 0.8; Xp = 3.2; Rs = 0.009; Xs = 0.03. calculate the equivalent values of resistance, reactance and impedance; (a) in secondary terms; (b) in primary terms. Example 2 A 25-kva 2,300/230-volt distribution transformer has the following resistance and leakage-reactance values: Rp = 0.8; Xp = 3.2; Rs = 0.009; Xs = 0.03. calculate the equivalent values of resistance, reactance voltage drops for a secondary load current of 109 amp: (a) in secondary terms: (b) in primary terms.
OIL – is primarily used for insulation and cooling of windings. ASKAREL OIL – is non-flammable insulating liquid w/c when decomposed by an electric are evolves non-explosive gases. OBJECTIVE QUESTIONS 1. The transformer is usually used to change the values of voltage. 2. Transformer operates at a power factor depending on the power factor of the load. 3. A good transformer oil should absolutely free from sulfur, alkalies & moisture. 4. The working principle of transformer is mutual induction. 5. The lamination of a core on a transformer are made of silicon steel sheet.
6.
The purpose of laminating the core of a transformer is to reduce eddy current losses.
TRANSFORMER CIRCUIT AND THEORY. Defined parameters Note: No load voltage – means the input voltage to the transformer when it is in operation or load PRIMARY SIDE Ip = rated primary current Vp = rated primary voltage (output) voltage Rp = resistance of the primary winding
Xp = reactance of the primary winding (because of the leakage flux in the primary). Ep = induced voltage in the primary winding (because of the mutual flux). Zp = impedance of the primary winding. Zp = Rp + j Xp Np = number of turns in the primary winding Vpnl = input voltage to the primary (real input voltage) SECONDARY SIDE Is = rated secondary current Vs = rated secondary voltage (output voltage) (real output) Rs = resistance of the secondary winding Xs = reactance of the secondary winding Es = induced voltage in the secondary (because of mutual flux)
Zs = impedance of the secondary winding Zs = Rs + j X s Ns = number of turns in the secondary winding Vsnl = input voltage to the secondary side. (no load secondary voltage) – fictious. GENERALLY Ф m = maximum mutual flux (common to primary and secondary). This flux travels around the core. = this is a useful flux = this is a working flux a
= transformation ratio = turns ratio = induced voltage ratio = rated voltage ratio
= no load voltage ratio = inverse current ratio REFER THE CIRCUIT TO THE PRIMARY SIDE (transfer all data from secondary to primary). Approximate: In = too small; temporary neglected
Ip
VPNL
Rp
Xp
a²Rs
a²Xs
Vp = aVs
Is/a
L O A D
ELECTRICAL LOSSES: (cupper losses)
;
Is = aIp
EL = ELP + ELS = Ip2Rp + Is2Rs = Ip2Rp + a2Ip2Rs = Ip2 (Rp + a2Rs) Rep – equivalent resistance of the transformer only referred to primary side. Rep = Rp + a²Rs
ELP = Ip2Rep Xep – equivalent reactance of the transformer only referred to the primary side. Zep = Rep + j Xep
SIMPLIFIED DIAGRAM:
Ip
Rep
Xep L O A D
VPNL
VPNL = Vp + IpZep
PERCENT VOLTAGE REGULATION
Vp
REFER THE CIRCUIT TO THE SECONDARY SIDE (transfer all data from primary to secondary)
Rp/a²
Xp/a²
Rs
Xs
Is
aIp
= VSNL
KVA LOAD = = KVAsec
LO A D
ELECTRICAL LOSSES;
EL = ELP + ELS = Ip2Rp + Is2Rs
Res - equivalent resistance of the transformer referred to secondary side.
EL = Is2Res
Xes – equivalent reactance of transformer referred to secondary side.
Zes – equivalent impedance of the transformer referred to the secondary side.
Zes = Res + j Xes Zes2 = R2es + X2es
SIMPLIFIED DIAGRAM:
Is
Res
VSNL
Xes
VS
VSNL = Vs + Is Zes PERCENT VOLTAGE REGULATION
L O A D
Additional Formulas: %Z = percent impedance rating %R = percent resistance rating %X = percent reactance rating
GENERALLY %Z = %R + j%X (%Z)2 = (%R)2 + (%X)2
Zpu = Rpu + jXpu (Zpu)2 = (Rpu)2 + (Xpu)2
EQUIVALENT FORMULAS USING PRIMARY VALVES
USING SECONDARY VALVES
PRIMARY SIDE
SECONDARY SIDE
STANDARD DESIGNATION WHEN TRANSFORMER KVA LOAD OR OUTPUT IS VARIED. (Transformer is operating at different loads). RULES: •
Use subscript “(1)” to designate transformer operating at full load or rated load.
•
Use subscript “(2)” to designate transformer operating at any KVA output (below rated or above rated).
•
Use subscript prime “(‘)” to designate transformer operating at maximum efficiency condition.
NOTE: when computing for efficiency and voltage regulation of a transformer regulation, the following should be clearly specified: 1. Power factor of the load 2. KVA output of the load TRANSFORMER LOSSES a) Electrical losses – is also called resistance or cupper losses. resistance
These of
losses
primary
are
primarily
(Rp)
windings. These losses vary with the KVA load”.
and
due
to
secondary
the (Rs)
the “Square of
•
•
•
1st condition; (@ full load or rated load)
EL1 = Ip12Rep
2nd condition; (below rated or above rated)
Divide ②/①
EL2 = Ip22Rep
b. CORE LOSS (iron losses) – this loss is a constant loss irregardless of any change in KVA. output or load power factor. NOTE: this is loss will only change or vary if there is change in the primary voltage (Vp) or supply frequency.
TYPES OF CORE LOSSES 1. HYSTERESIS LOSS (Ph) -
This loss can be minimized only
by using high quality of material to construct the core (such as high grade silicon steel) because it has high value permeability and low hysteresis. 1. EDDY CURRENT LOSS (Pe) – this loss can be minimized or
reduced only by laminating the core.
Coreloss = PH + Pe TYPES OF TRANSFORMER EFFICIENCY a) ORDIARY EFFICIENCY (ŋ) over all efficiency or conventional efficiency.
of
General Formulas:
Where:
TL = EL + COL Po = KVAOUTPUT x P.F. PIN = Po + TLOSSES
@ FULL LOAD:
Po1 = KVA1 x P.F. PIN = Po1 + TL1 TL1 = EL1 + CLOSS
@ ANY KVA LOAD OUTPUT (below rated or above rated)
Po2 = KVA2 x P.F. PIN = Po2 + TL2 TL2 = EL2 + COLOSS B. MAXIMUM EFFICIENCY
Po’ = KVA’ x P.F. PIN’ = Po’ + TL’ TL’ = EL’ + COLOSS
ŋ'mx = but EL’ = CL (max.) TL’ = 2EL’ = 2CL
EQUIVALENT FORMULAS:
Note: KVA’ Where:
KVAoutp
at maximum efficiency condition.
ut
KVA1 = Ip1 Vp / 1000
EL1 = Ip12 Rep
KVA2 = Ip2 Vp / 1000
EL’ = (Ip’)2 Rep = CL
EL2 = Ip22 Rep
OBJECTIVE QUESTION: when will maximum efficiency occur and at what KVA load? 1. It could happen at “full load”
KVA1 = KVA’ ŋ1 = ŋ’max EL1 = EL’ = CL 2. It could happen at either “overload or below rated load” Example problem: the full load cupper loss in a transformer is 400 watts. At half load, the cupper loss will be… Given:
@full load: KVA1 = ? EL1 = 400w @full load: KVA2 = ½ KVA1
Req’d:
EL2 = ? Solution:
EL2 = 100 watts Example problem: a 10 KVA, 2400/240 volts distribution transformer has a primary resistance of 1.2 ohms and a secondary resistance of 0.058 ohm. Determine the full load cupper loss.
Note: if there's only one condition in the problem that is a full load, you may not use anymore the subscript. Given: (nameplate data)
KVAoutput = KVAprimary = KVAsec = 10 KVA Rp = 1.2 ohm Rs = 0.058 ohm
Vp = 2400 v Vs = 240 v
Solution:
(using primary value) EL1 =? Rep = Rp1 + a2Rs
= (1.2) + (10)2 (0.058) = 7 ohms
EL = Ip2Rep
= 4.17 Amp.
= (4.17)2 (7) EL = 121.7 watts
Example problem: the full load cupper loss and core loss of a 20 KVA, 2500/250v transformer are 300 watts and 320 watts respectively. What is the efficiency at half load and unity power factor. Given: @ 1st condition: (full load)
@ 2nd Condition: (half load)
EL1
= 300 w
EL2
=?
CL
= 320 w
CL
= 320 w (constant)
KVA1 = 20 KVA
KVA2 = 20 KVA
Vp
= 2500 v
PF
= unity = 1
Vs
= 250 v
= cos-1 (1) = 0
Req’d: ŋ2 = ? (half load) Solution:
But KVA2 = ½ KVA₁
= 75 Watts
TL2 = EL2 + CL = 75 + 320 = 395 watts Subs. Values:
= 98.06%
REGULATION CALCULATIONS USING VOLTAGES VALUES: Transformer percent regulation (%Reg)
the no load and full load voltage
are those measure
at secondary terms. THREE FLUXES 1. Mutual flux (φ m) = links both the primary and secondary windings. 2. Primary leakage flux (φ₁) = links with primary winding only and is varrying I₁ 3. Secondary leakage flux (φ₂) = links with secondary winding only and is varrying I₂.
VOLTAGE DROPS IN TRANSFORMER (INTERNAL) a) Voltage drop due to leakage reactance = IsRs b) Voltage drop due to leakage reactance = IsXs Below are the phasor. Diagram showing how the resistance and leakage-reactance drop are subtracted from the induced secondary voltage to yield the secondary terminal voltage. Es
a) @ unity power factor
IsXs
Vs
Is IsRs
b.) @ lagging power factor Es
IsXs
V s
Is Rs
PF co
s
Vs sin
Is Rs
Is Is
c.) @ leading power factor
R Is
s V
c
s o
PF
s
IsXs
Es
Vs
si n
Vs
Below are phasor diagrams showing how the resistance and leakage – reactance drops are subtracted from the impressed primary voltage to yield primary induced voltage. Vp
a. @ unity power factor
IpXp Vp
Ep Ip IpRp
IpXp Ep
PF s
co
Ep
Ep
si n
b. @ lagging power factor
Ip Rp
Ip Rp
Ip
I
c. @ leading power factor p R Ip
Ep
c
os
PF
IpXp
Vp
Ep
si n
IpRp Ep
Equivalent circuit of a practical transformer
Rp
Iφ
Ig Eg
Ip
Xp
Rm
Xm
Ep
Es
V₂
Where: Rp = Resistance of primary winding Xp = Reactance of primary winding Rs
= Resistance of secondary
Xs
= Reactance of secondary winding
Eg = Supply voltage or primary voltage Ep = Induced emf of the primary winding Es
= Induced emf of the secondary winding
Ig
= Supply current
If
= Primary current
Is
= Secondary current
Rm = Resistance representing iron losses Pm = Iron losses Xm = Magnetizing reactance of primary winding Io
= Exciting current
Qm = Reactive power needed to set-up the mutual flux (VAR)
EQUIVALENT
CIRCUIT
DIAGRAM
OF
A
PRACTICAL
TRANSFORMER AT “NO LOAD”
Rp
Xp
Rs
Io
Ig = Io Eg
Ip = 0
Rm
Xs
Is = o Xm
Ep
Es
Vs = Es
@ no load, Is = 0 and so if, thus only the exciting current Io flows in Rp and Xp and these impedance are so small that the voltage drop across them are “negligible”
EQUIVALENT CIRCUIT OF A PRACTICAL TRANSFORMER AT “FULL LOAD”
Rp Eg
Xp Ep
Rs Es
Xs Vs
NOTE: @ full load, “Ig” is at least 20 times bigger than “Io”
EQUIVALENT RESISTANCE REACTANCE AND IMPEDANCE IN PRIMARY TERMS Rep = a²Rs + Rp
IN SECONDARY TERMS
Xep = a²Xs + Xp Zep = Rep + jXep Zes = Res + jXes
IN TERMS OF TURNS RATIO
Where: Res = equivalent resistance referred to the secondary. Rep = equivalent resistance referred to the primary. Xes = equivalent reactance referred to the secondary. Xep = equivalent reactance referred to the primary. Zes = equivalent impedance referred to the secondary. Zep = equivalent impedance referred to the primary. EQUIVALENT VOLTAGE DROP IN TRANSFORMER REFERRED TO PRIMARY AND SECONDARY SIDE •
RepIp = equivalent resistance voltage drop in the primary side.
•
XepIp = equivalent reactance voltage drop in the primary side.
But:
ResIs = equivalent resistance voltage drop in the secondary side. XesIs = equivalent drop in the secondary side. Notation: Rep = equivalent resistance in primary side. Xep = equivalent reactance in primary side. Ip = primary current. Res = equivalent resistance in secondary side. Xes = equivalent reactance in secondary side. Is = secondary load current.
PERCENT REGULATION OF TRANSFORMER AT DIFFERENT LOADING. 1) @ unity power factor load
2.)
@ lagging power factor load.
3.)
@ leading power factor load.
Note: if the given data of the transformer is in terms of Rep, Xep, and Zep. Use this formula:
If the given data of the transformer are in terms of percentage values like %Z, %R and %X. use the corresponding formulas: a) For unity power factor
b) For lagging power factor
c.)
For leading power factor
Thus:
Example problem: The 2300 volts primary winding of 60 cps transformer has 4800 turns. Calculate: a)
Mutual flux, Фm
b)
No. of turns in the 230V, sec.
c)
Transformation ratio of voltage and winding
Given:
Req’d: Ep = 2300 v Es = 230 v f = 60 cps Np = 4800 turns
Solution: a.)
Ep = 4.44 fNpФm x 10-8
Фm = 1.8 x 105 maxwells
Фm, Ns, a
b.)
NS = 480 turns c.)
or
Example Problem: a 25 KVA, 2300/230 volts distribution transformer has the following resistance and leakage reactance value of 0.8 and 3.2 respectively. For primary and 0.009Ω and 0.03Ω for secondary respectively. Calculate the equivalent values of resistance, reactance and impedance in secondary terms:
a) Secondary terms b) Primary terms c.) Equivalent resistance and reactance voltage drop for a load current of 109 amps in secondary and primary terms. d.) calculate the present voltage regulation for a unity power factor. Given: Pa = 25 KVA V = 2300/230v Rp = 0.8Ω Xp = 3.2Ω Rs = 0.009Ω Xs = 0.03Ω
KVAoutput =
KVApri = KVAsec
Solution: a) In secondary terms
Xes = Xs + Xp/a² = 0.03 + 3.2/102 = 0.062 Ω Zes = Res + j Xes = 0.017 + j 0.062 = 0.642 < 74.67 Ω
b.) In primary side. Rep = Rp + a2Rs = 0.8 + (10)2 (0.009) = 1.7 Ω Xep = Xp + a2Xs = 3.2 + (10)2 (0.03) = 6.2 Ω Zep = Rep + j Xep = 1.7 + j 6.2 = 6.43 < 74.67 Ω
c.)
VDPR, VDPX, and VOSR, VDSX (Ip = Is/a) PRIMARY VDPR = IpRep
SECONDARY VDSR = IsRes = (109)(0.017) = 1.853 volts VDSX = IsXes = (109)(0.062)
= 18.53 volts
= 6.75 volts
VDPX = IpXep = 67.5 volts
d.) Percent Regulation @ unity P.F. = 1
= cos-1 (1) = 00
= 10.87 amps For unity VPNL = VP + IpZep = 2300 00 + (10.87 00)(6.43 74.67) = 2319.46 1.660 volts
For 0.8 lagging ( = -36.87)
VPNL = Vp + IpZep = 2300 00 + (10.87 -36.87) x (6.43 74.67) = 2355.6 1.04 volts
For 0.866 leading (0) = cos-1 (0.866) = 300 VPNL = Vp + IpZep = 2300 00 + (10.87 300 + (6.43 74.67) = 2283.3 1.7 volts
Another solution: For Unity:
= 00
Cos = 1 Sin = 0
For 0.8 lagging: cos = 0.8 sin = 0.6
= 36.870
For 0.866 leading: = 300 cos = 0.866 sin = 0.5
EXAMPLE PROBLEM: a 10 KVA, 2400/240 volt distribution. Transformer has a primary resistance of 1.2 ohm and a secondary resistance of 0.058 ohm. What is the full load cupper loss (Electrical loss). Given:
Req’d:
Pa = 10 KVA E = 2400/240 V Rp = 1.2 ohm Xp = 0.058 ohm Solution:
( NOTE: STOP HERE)
Using primary values: Rep = Rp + a2Rs = 1.2 + (10)2 (0.058) =7 ELOSS = Ip2Rep = (4.17)2 (7) ELOSS = 121.7223 watts
EXAMPLE PROBLEM: The full load cupper loss and core loss of 20 KVA, 2500/250 volt transformer are 300 and 320 w respectively. What is the efficiency at haf-load and unity P.F.
Given: @ condition 1: (full load) KVA1 = 20 KVA
Ep = 2500 v
EL1 = 300 w
Es = 250 v
COL = 320 w
@ condition 2: (half load) KVA2 = ½ KVA1 = 10 KVA PO2 = KVA2 x P.F. = 10 x 1 = 10 KW
Req’d: ŋ₂ = ? Solution:
Sample Problem: In a 50 kva transformer, the full load cupper loss are exactly twice the iron losses, and that quarter load, the efficiency is 95%. Calculate the full load efficiency at unity P.F.
@ condition 1: (full load) KVA1 = 50 kva EL1 = 2 COL
---------------- ①
PO1 = KVA1 x P.F. = 50(1) = 50 KW @ condition 1: (quarter load) KVA2 = ¼ kva1 = ¼(50) = 12.5 kva ŋ₂ = 95% Req’d: ŋ₁ = ?
Solution:
PIN2 = TL2 + P02 TL2 = PIN2 – P02 = 13,157.89 – 12500 = 657.89 watts But: TL2 = EL2 + CL ------------ ②
Substitute ① to ② TL2 = EL2 + CL 657.89 = EL2 + 0.5 EL1
--------- ③
Subst. ④ in ③ 657.89 = EL2 + 0.5 EL1 = 0.0625 EL1 + 0.5 EL1 = (0.0625 + 0.5) EL1 = 0.5625 EL1
From eqn.① CL = 0.5 EL1 = 0.5 (1,169.6) = 584.79 watts
Example Problem The core loss of a 50 KVA single phase transformer with normal voltage applied to the primary is 75 watts. The max efficiency at 60% load of full load. What is the full load efficiency of the transformer at 0.8 P.F. Given: @ condition 1 : (full load) KVA₁ = 50 KVA CL = 75 watts (constant) Note: Core loss (CL) don’t vary at any time of change except when the applied voltage and frequency is change.
EL1 = ? PO1 = 50 KVA (0.8) = 4000 watts @ condition 2 : (maximum load efficiency) KVA’ = 60% KVA₁ = 0.6 (50) = 30 KVA Po’ = 30 (0.8) = 24,000 watts Req’d: ŋ₁ = ? Solution
@ max eff. (E ’ = C ) L L Pin1 = PO1 + TL1 but TL1 = EL1 + CL
Pin1 = PO1 + EL1 + CL = 4000 + 208.3 + 75 = 4283.3 watts
subs.values:
Example Problem: A 100 KVA transformer has a maximum efficiency of 98% at 10% underload at unity power factor. Evaluate the efficiency at quarter-load at 0.8 power factor. Given: @ max.load:
@ full load condition:
KVA’ = 0.1 KVA1
KVA1 = 100 KVA
ŋ'mx = 98%
EL1 = ?
PF = 1 @ quarter load: KVA2 = ¼ KVA1 = ¼ (100) = 25 KVA
PO2 = 25 KVA (0.8) = 20,000 watts PF = 0.8
Req’d:
ŋ₂ = ?
Solution: KVA’ = 0.1 KVA1 = 0.1 (100)
PO’ = 10 KVA (1) = 10,000 watts
= 10 KVA
T L’ = E L’ + C L But for maximum efficiency: (E ’ = C ) L L TL’ = 2 CL -------------- ① TL’ = Pin’ – Po’ --------- ②
Subs. ① in ① TL’ = Pin’ – Po’ 2CL = 10,204.08 – 10,000 2CL = 204.08 watts
@ quarter load:
TL2 = EL2 + CL = 637.75 + 102.04 = 739.79 watts
All day efficiency – also known as commercial efficiency. The all day efficiency of the transformer. The ration of electrical energy (kilowatt-hours) output delivered by the transformer in a 24hour period to the energy input in the same period of time (or a day operation.) Where: Wo – Electrical energy output - ЄP x time (kw-hr)
Win - Electrical input - ЄP x time (kw-hr) - Wo + WL
WL – Electrical Energy Losses = WCORELOSS + WCOPPERLOSS = WCL + WEL = (CL x Time) + (ЄEL x Time)
When transformer is loaded: - energy copper loss is present (but not constant) - energy coreloss is present (it is constant) - energy output is present. Note: 1. Full load values/rated values must always be given (reference data). 2. Solutions maybe in a tabulated form. 3. When transformer is loaded 24 hrs a day operation, energy coreloss is always constant to be multiplied by 24 hrs . Condition that may arise later in problem; a) No load condition (transformer is unloaded) – energy copper loss is zero. Energy coreloss is constant. Energy output is zero.
b.) when primary is connected from the source – energy copper loss is zero. Energy coreloss is zero. Energy output is zero. Example problem: calculate the all day efficiency of 100 kva transformer operating under the following conditions; – 6 hrs on load of 50 kw at 0.73 P.F. – 3 hrs on load of 30 kw at 0.82 P.F. – 15 hrs with no load on secondary, the iron losses is 1000 watts and the full load copper loss is 1060 watts.
Req’d: Note: for energy Solution:
TABULATION FORM KW LOAD
P.F.
Kw
50
.73
# OF
KVA LOAD
ENERGY
ENERGY
ENERGY
HRS
(Kw/P.F.)
OUTPUT
COL
CUL
Hr.
KVA
Kw-Hr
(CL - t)
(EL - t)
6
50/0.73
50 (6)
1 (6)
= 68.5
= 300
=6 = 2.98
30
.82
3
30/0.82
30 (3)
1 (3)
= 36.59
= 90
=3
-
-
1 (15)
0.43
N O
0
15
L
0
= 15
O A D
TOTAL:
390
24
3.41
Wo
WCL
WCUL
WO = 300 + 90 = 390 Kw-Hr WCL = 6 + 3 + 15 = 24 Kw-Hr WCU = 2.98 + 0.43 = 3.41 Kw-Hr Win = WO + WLOSSES = 390 + (24 + 3.41) = 417.41 Kw-Hr
Example problem: a 30 kva, 2400/240 volts, 60 hz distribution transformer has a full load P.F. of unity over a period of 24 hrs. the maximum efficiency is 95% and it occurs at full load. Calculate the all day efficiency if it is loaded as follows: 6 hrs at full load 6 hrs at quarter load 12 hrs at no-load Req’d: Solution: for maximum efficiency (EL = CL)
For energy copper loss
TABULATION FORM Kw Load
P.F.
# of Hours
Kva load (kw/PF) Kva
Energy Output Kw-hr
Energy CL Kw-hr
Energy CUL Kw-hr
Full Load
1
6
30
30 (6) = 180
0.789 (6) = 4.737
0.296
¼ full load
1
6
7.5
7.5 (6)
0.789 (6) = 4.737
0.296
No load
1
12
-
0
0.789 (12) = 9.474
0
WO = 180 + 45 = 225 kw-hr WCL = 4.737 + 4.737 + 9.474 = 18.948 kw-hr WEL = 4.737 + 0.296 = 5.033 kw-hr
Win = WO + WLOSSES = 225 + 5.033 + 18.948 = 230.033 kw-hr + 18.948 Win = 248.948 kw-hr
Example problem: a 100 kva, 6600/400 volts 60 hz, single phase, core type transformer has the following average daily load:
full load at 0.8 PF for 8 hrs
half-load at 0.7 PF for 10 hrs
no load for 6 hrs
Calculate the ratio of the full load cupper loss to the iron loss for the transformer to be “most economical” for the above loading. Req’d:
Solution: @ full load condition: KVA1 = 100 KVA CL1 = ?
(CL = EL1)
ЄL1 = ? Note: most economical at maximum eff.
TABULATED FORM Kva Load
P.F.
Full load (100)
0.8
Half load (50)
0.7
Kw load Kw
# of hrs Hr
Wo Kw-hr
100 (0.8) = 80
8
80 (8) = 640
100 (0.7) = 70/2 = 35
10
70 (10) = 700/2 =350
WCL
WEL
Kw-hr
kw-hr
CL (8)
EL1 (8)
= 8 CL
= 8 EL1
CL (10)
EL1 (10)
= 10 CL = 2.5 EL1
No load
-
0
6
-
CL (6)
0
= 6 CL 24 CL
10.5 EL1
WCL = 8CL + 10 CL + 6 CL = 24 CL WEL = 8 EL1 + 2.5 EL1 = 10.5 EL1
24 CL = 10.5 EL1
Example Problem: a 25 kva, single phase transformer operates for one hr @ 20% overload @ 0.8 PF lagging; three hrs @ full load @ 0.9 PF leading; 6 hrs @ half-load @ a unity power factor and 10% loaded @ unity power factor for the rest of the day, the cipper loss and core loss @ full load are 500 watts and 150 watts respectively. Find the all-day efficiency. Given: condition 1: (full load) t₁ = 3 hrs KVA₁ = 25 KVA •
@ condition 2: (overload) t₂ = 1 hr KVA2 = 1.2 (KVA1) = 1.2 (25) = 30 KVA
PO2 = 30 (0.8) = 24 kw
@ condition 3: (half-load) t₃ = 6 hrs. KVA₂ = 0.5 (KVA₁) = 0.5 (25) = 12.5 kva @ condition 4: (underload) t₄ = 14 hrs KVA₂ = 0.1 (KVA₁) = 0.1 (25) = 2.5 KVA EL1 = 500 watts (0.5 kw) Req’d:
ŋd = ?
CL1 = 150 watts (0.15 kw)
Solution:
Wo = ЄPo x time = W1 + W2 + W3 + W4 = (25)(0.9)(3) + (30)(0.8)(1) + (12.5)(1)(6) + (2.5)(1)(14) Wo = 67.5 + 24 + 75 + 35 = 201.5 kw-hr
WEL = ЄEL x time = EL1 + EL2 + EL3 + EL4 WEL = 3.04 kw-hr
WCL = ЄCL x time = CL1 + CL2 + CL3 + CL4 = (0.15)(3) + (0.15)(1) + (0.15)(6) + (0.15)(14) WCL = 3.6 kw-hr
Example problem: the all day efficiency of a 10 kva single phase transformer is 94.7%, when loading as follows: •
Full load @ unity P.F. for 4 hrs
•
No-load for the rest of the day
If the full load cupper loss and @ unity PF is 140 watts. Calculate the value of core loss. Given:
ŋd = 94.7%
CL1 = ?
KVA1 = 10 KVA
KVA2 = 0
PF1 = 1
PF2 = 1
t1 = 4 hrs.
T2 = 20 hrs
EL1 = 140 watts (0.14kw)
Req’d: CL1 = ? Solution: WO = ЄP x time = (10)(1)(4) + (0)(1)(20) = 40 kw-hr
Problems on percent voltage regulation (%VR) Case I: if given data are in terms on Zep, Rep and Xep.
Example Problem: A 10 kva, 2400/240 volt single phase transformer has the following resistances and leakage reactances. a) Find the voltage regulation @ full load and 0.8 lagging b) What is %VR if the transformer is operating @ 25% overload and 0.8 PF leading c) What is %VR of the transformer when it is operating @ maximum efficiency @ unity power factor. Assume coreloss of 60 watts. Given:
rp = 3 ohms rs = 0.03 ohms Xp = 15 ohms Xs = 0.15 ohm
Solution: KVAP = KVAS = KVAoutput (full or rated load)
Refer to primary side: Rep = Rp + a2Rs
Xep = Xp + a2Xs
= 3 + (10)2 (0.03)
= 15 + (10)2 (0.15)
= 6 ohms
= 30 ohms
Zep = Rep + jXep = 6 + j30 = 30.59 ∠ 78.7 ohms VPNL = Vp + IpZep = 2400 ∠ 0o + (4.17 ∠ -36.87) (30.6 ∠ 78.7) = 2496 ∠ 1.9500 volts a) %VR @ 0.8 lagging ( = -36.87)
b.) %VR @ 25% overload and 0.8 leading ( = 36.87)
c.) %VR @ maximum efficiency @ unity power factor
KVA’ = ? EL1 = IP2Rep = (4.17)2 (6) = 104.33 watts
VPNL = VP + IP’Zep = 2400 ∠ 00 + (3.16 ∠ 00)(30.6 ∠ 78.7) = 2420.79 ∠ 2.240 volts
Example Problem: a 200 kva transformer with impedance of 15% and the PF of 85% lagging. The primary voltage is 6000 volts while the full load cupper loss is 15 kw. Find the percentage regulation at full load. Given:
KVAP = KVAS = KVAout = 200 KVA %Z = 15% PF = 0.85 lag cos = 0.85 = 31.788 sin = 0.5268 VP = 6000 volts EL1 = 15,000 watts
Req’d:
%VR Solution:
By applying formula Case II id the given data of transformer are in terms of %Z, %R & %X.
DERIVATION OF FORMULAS: @ full load
@ any kva load
Where
Divide %R₂ / %R₁
It follows: @ full load
Divide %X₂ / %X₁
Voltage equation:
@ any kva load
Transformer Testing Assumed: (step-down transformer) Vp Vs Where: Vp = high voltage windage Vs = low voltage windage a) Short circuit test (sct) Purpose: To determine the rated full load cupper loss Procedure: a) The low voltage side is short circuited. b) All measuring AC instrument such as ammeter, voltmeter, wattmeter are placed on the high voltage side. c) Energize the high side by a small input voltage. d) Then solve for; Zep, Rep, Xep.
Figure: ammeter variable resistor
Isc
A wattmeter
V
shorted
+ Vin HVW
LVW
(primary)
(secondary)
Where: Ise = short circuit current on the low voltage side must be equal to rated secondary current Is. Observed readings in all instruments: A – ammeter reading (AMR) - Ip₁ - rated primary current V – voltmeter reading (VMR) - Vin – (2
5%) Vp
W – wattmeter reading (WMR) - EL₁ - rated cupper loss. Where:
Where:
@ full load:
b.)
OPEN CIRCUIT TEST (OCT) Purpose: To determine the total rated iron loss (CORE LOSS).
Procedure: a)
the high winding (HVW) is open circuited “Danger High Voltage.”
b) all AC measuring instruments are placed on low voltage side. c)
energized the low voltage side w/ only very small amount of current.
FIGURE:
A
W
V
Vin autotransformer
VARIATION OF CORE LOSS (separation of iron losses) Note: All fomulas are derived by “Steinsmitz law” thru theory and experiments.
VARIATION OF CORE LOSS (separation of iron losses) Note: All formulas are derived by “Steinsmitz Law” thru theory an d experiments. 1st topic: Variation of Copper Loss a) Voltage Regulation b) All Day efficiency c) Maximum and ordinary efficiency 2nd topic: Variation of Core Loss a)Steinsmitz Law b)KVA sizing
EFFECTS WHEN CORELOSS ARE VARIED a)
If there is a change in the KVA load output • Efficiency will change • %VR will change • Copper loss will change (very much affected) • Core loss is constant b.)
If there is change in load power factor
• Efficiency will change • %VR will change (very much affected) • Copper loss is affected slightly • Core loss is constant
c.) If there is c change in supply primary voltage • Slightly change in efficiency • %VR will change • Copper loss is slightly • Core loss will change (very much affected) d.) • Slight change in efficiency • No change of %VR • Copper loss is slightly affected • Core loss will change (very much affected)
Formulas Case 1: flux density is not constant
Ph = Khfβm1.6 Pe = Kefβm2
Where: f = frequency of the AC source
Kh = hysteresis constant of proportionality (depends on the Ke =
Eddy current constant of proportional (depends on the
quality and material of the core.) lamination of core)
Where:
B = maximum flux density (not constant)
m фm = maximum mutual flux
A = net cross-sectional area of the core
Ph = hysteresis loss; watts Pe = eddy current loss; watts From Ep: Ep = 4.44NpfФm x 10-8
Therefore:
CLOSS = Ph + Pe
Let K1 = design constant
Let K₂ = design constant
Note: in most cases approximate formula used:
Applying it to coreloss equivalent formula 1st Condition: (V₁ , f₁)
2nd Condition: (V₂ , f₂)
DIVIDE ②/① CASE 1: flux density (βm) is not assumed to be constant.
CASE 2: flux density (βm) is assumed to be constant. STEINMITZ LAW:
Ration:
Therefore:
Example Problem: a 4400v, 60 hz transformer has a core loss of 840 watts of which one-third is eddy current loss when the transformer is connected to a 4600v, 50 hz source. Determine the value of iron loss. Note: if the problem does not mentioned coreloss is constant therefore, do not assume constant.
Given: Vp₁ = 4400 v
Vp₂ = 4600 v
f₁ = 60 hz
f₂ = 50 hz
CL₁ = 840 w Pe₁ = ⅓ (CL₁) = ⅓ (840) = 280 w Req’d: CL₂ = ? Solution: (Case 1) CL₂ = Ph₂ + Pe₂ ---------- working formula.
Example Problem: a 25 cycles, 1000 kva transformer is applied to a 60 cps system. The full load efficiency of the transformer at 25 cps is 98%. Half of the core, the hysteresis and eddy current loss are equal at 25 cps. What would be the rating of this transformer at 60 cps if the transformer was operated at rated voltage. Given:
f1 = 25 hz
2nd Condition:
KVA1 = 1000 kva
f2 = 60 hz
ŋ₁ = 98%
KVA2 = ?
EL1 = ½ TL1
Vp2 = ?
CL = ½ TL1 (EL = CL) Ph1 = Pe1 = ½ CL1
Req’d:
KVA2 = ?
Solution: (Assume PF = 1) not mentioned.
Pe1 = Ph1 = 0.5 CL1 = 0.5 (10,205) = 5,102.5 watts
@ 2nd condition: (assume Vp₁ = Vp₂)
(assume TL1 = TL2) CL2 = Ph2 + Pe2 = 3017.56 + 5102.5 = 8116.08 watts
TL2 = CL2 + EL2 EL2 = TL2 – CL2 = 20410 – 8116.08 EL2 = 12,289.94 watts
Example Problem: In a 400 v, 50 cps transformer the total iron loss is 2500 watts. When the supplied potential voltage is 220 v at 25 cps, the corresponding loss is 850 watts. Calculate the eddy current loss at normal frequency and potential diff. Given: @ 1st month condition:
Req’d: Pe₁ = ?
@ 2nd condition:
VP1 = 400 v
VP2 = 220 v
f1 = 50 cps
f2 = 25 cps
CL1 = 2500 w
CL2 = 850 w
Solution: @ condition 1: CL1 = Pe1 + Ph1 = Ke Vp12 + Kh
2500 = Ke (400)2 + Kh 2500 = 160000 Ke + 1392.88 Kh Kh = 1.795 – 114.87 Ke ----- ①
@ condition 2: CL2 = Pe2 + Ph2 850 = Ke (220)2 + Kh 850 = 48400 Ke + 811.17 Kh ------- ②
Substitute eq ① in ② 850 = 48400 Ke + 811.17 Kh 850 = 48400 Ke + 811.17 [ 1.795 – 114.87 Ke ] 850 = 48400 Ke + 1456 – 93,179.1 Ke
Example Problem: A 60 hz, 200 KVA, three winding transformer is rated 2400 volts primary and their two secondary windings, one rated 600 V and the other at 240 volts. There are 200 primary turns and the rating of each secondary winding is 100 KVA. Calculate the current in primary when rated current flows @ PF = 1. In the 240 V winding and also rated current flows @ 0.707 lagging in the 600 volt windings. Given: f = 60 Hz KVA = 200 KVA KVA1 = 100 KVA
KVA2 = 100 KVA
VS1 = 600 V
PF2 = 1
VS2 = 240 V
1 = 00
PF1 = 0.707 lag 1 = -450
Req’d:
Ip = ?
Solution:
load 1
From the diagram:
load 2
Example Problem: A transformer consist of a primary winding with 500 turns and two secondary windings of 125 turns and 36 turns. The 125turn secondary windings has 60 ohms. Connected to its terminals and the 36-turn secondary winding has 3 ohms connected to its terminal. If the primary is connected to a 120v, 60 hz source, determine the current in primary windings. Given: Np = 500 turns
Z₁ = 60 ohms
N₁ = 125 turns
Z₂ = 3 ohms
N₂ = 36 turns
Vp = 120 v
f
= 60 hz
FIGURE:
load 1
For currents:
From the figure:
load 2
Three-Phase Transformer Connections: There are only 4 possible transformer combinations: 1. Delta to Delta - use: industrial applications 2. Delta to Wye - use : most common; commercial and industrial 3. Wye to Delta - use : high voltage transmissions 4. Wye to Wye - use : rare, don't use causes harmonics and balancing problems. A three- phase transformer is made of three of sets of primary and secondary windings. Each set wound around one leg of an iron core assembly. Essentially it looks like three single –phase transformer sharing a joined core as in fig. below.
Three phase transformer core has three sets of windings.
Those sets of primary and secondary windings will be connected in either Δ or Y configurations to form a complete unit. Whether the winding sets share a common core assembly or each winding pair is a separate transformer, the winding connection options are the same:
• Primary - Secondary • Y Y • Y Δ • Δ Y • Δ Δ
•
The reasons for choosing a Y or Δ configuration for transformer winding connections are the same as for any other three-phase application: Y connections provide the opportunity for multiple voltages, while Δ connections enjoy a higher level of reliability (if one winding fails open, the other two can still maintain full line voltages to the load).
•
Probably the most important aspect of connecting three sets of primary and secondary windings together to form a threephase transformer bank is paying attention to proper winding phasing (the dots used to denote “polarity” of windings). Remember the proper phase relationships between the phase windings of Δ and Y: (Figure below)
•
(Y) The center point of the “Y” must tie either all the “-” or all the “+” winding points together. (Δ) The winding polarities must stack together in a complementary manner ( + to -).
•
Getting this phasing correct when the windings aren't shown in regular Y or Δ configuration can be tricky. Let me illustrate, starting with Figure below.
Three individual transformers are to be connected together to transform power from one three-phase system to another. First, I'll show the wiring connections for a Y-Y configuration: Figure below.
Phase wiring for “Y-Y” transformer.
ADVANTAGE OF (3ф) THREE PHASE TRANSFORMER vs. THREE OF SINGLE PHASE TRANSFORMER. a) It is cheaper and economical to operate. b) It requires lesser space to mount the transformer. DISADVANTAGE OF THREE PHASE (3ф) TRANSFORMER vs. SINGLE PHASE TRANSFORMER. a) If one winding becomes disabled the entire operation is affected, thereby power is interrupted. b) In 3 of 1ф, when one transformer becames disabled, the remaining two transformer can be re-bank in open-delta (V) so that service can be continued although at reduce capacity (57.7%) while the damage transformer is being repaired.
Phase wiring for “Y-Δ” transformer.
Equivalent Circuit Diagram: Ip = Iф
X
Is
Ap
Y
Cp
1
As
Bp
Cs
Z
Bs 2
3
This connection is most economical for small high voltage transformer because the number of turns per phase and the amount of insulation is minimum. The transformer works satisfactorily only if the load is BALANCE.
Phase wiring for “Δ-Y” transformer.
EQUIVALENT CIRCUIT DIAGRAM: Ip = Iф
X Y Bp
2
Ap As
Cs Cp
Bs
Z
2 3
The main used of this is at substation and of the transmission where the voltage is to be stepped down. The primary is always Y-connected with grounded neutral to allow the flow of zero sequence current in the even of SLGF, source line to ground fault and DLGF, double line to ground fault
Phase wiring for “Δ-Δ” transformer.
EQUIVALENT CIRCUIT DIAGRAM Iф = Is
1 2
X
As Bs Ap
Cp Bp
Y Z
Cs 3
This transformer is generally where it is necessary to stepup the voltage at the beginning at high transmission lines.
“V” or “open-Δ” provides 2-φ power with only two transformers.
This configuration is called “V” or “Open-Δ.” Of course, each of the two transformers have to be oversized to handle the same amount of power as three in a standard Δ configuration, but the overall size, weight, and cost advantages are often worth it. Bear in mind, however, that with one winding set missing from the Δ shape, this system no longer provides the fault tolerance of a normal Δ-Δ system. If one of the two transformers were to fail, the load voltage and current would definitely be affected.
EQUIVALENT CIRCUIT DIAGRAM:
2
X Ap
Cp Bp
As
Cs Bs
Y
2
Z
3
This connection is economical for large low voltage transformer which insulation problem is not so urgent because it increases the number of turns per phase. No difficulty is experienced from unbalanced load.
Figure 1-1: Wye-Delta connection
Figure 1-2: Delta-Wye connection
Delta Connections: •
A delta system is a good short-distance distribution system. It is used for neighborhood and small commercial loads close to the supplying substation. Only one voltage is available between any two wires in a delta system. The delta system can be illustrated by a simple triangle. A wire from each point of the triangle would represent a three-phase, three-wire delta system. The voltage would be the same between any two wires (see figure 1-3).
Figure 1-3:
Wye Connections: Figure 1-4:
Connecting Single-Phase Transformers into a Three-Phase Bank: •
If three phase transformation is need and a three phase transformer of the proper size and turns ratio is not available, three single phase transformers can be connected to form a three phase bank. When three single phase transformers are used to make a three phase transformer bank, their primary and secondary windings are connected in a wye or delta connection. The three transformer windings in figure 1-5 are labeled H1 and the other end is labeled H2. One end of each secondary lead is labeled X1 and the other end is labeled X2.
Figure 1-5:
Figure 1-6 shows three single phase transformers labeled A, B, and C. The primary leads of each transformer are labeled H1 and H2 and the secondary leads are labeled X1 and X2. The schematic diagram of figure 1-5 will be used to connect the three single phase transformers into a three phase wye-delta connection as shown in figure 1-7.
Figure 1-6:
Figure 1-7:
Open Delta Connection: The open delta transformer connection can be made with only two transformers instead of three (figure 1-8). This connection is often used when the amount of three phase power needed is not excessive, such as a small business. It should be noted that the output power of an open delta connection is only 57% of the rated power of the two transformers. For example, assume two transformers, each having a capacity of 25 kVA, are connected in an open delta connection. The total output power of this connection is 43.5 kVA (50 kVA x 0.87 = 43.5 kVA).
Figure 1-8: Open Delta Connection
•
Another
figure
percentage
given
assumes
for a
this
closed
calculation delta
bank
is
87%.
This
containing
3
transformers. If three 25 kVA transformers were connected to form a closed delta connection, the total output would be 75 kVA (3 x 25 = 75 kVA). If one of these transformers were removed and the transformer bank operated as an open delta connection, the output power would be reduced to 58% of its original capacity of 75 kVA. The output capacity of the open delta bank is 43.5 kVA (75 kVA x .58% = 43.5 kVA).
The voltage and current values of an open delta connection are computed in the same manner as a standard delta-delta connection when three transformers are employed. The voltage and current rules for a delta connection must be used when determining line and phase values of voltage current. Closing a Delta: •
When closing a delta system, connections should be checked for proper polarity before making the final connection and applying power. If the phase winding of one transformer is reversed, an extremely high current will flow when power is applied. Proper phasing can be checked with a voltmeter at delta opening.
If power is applied to the transformer bank before the delta connection is closed, the voltmeter should indicate 0 volts. If one phase winding has been reversed, however, the voltmeter will indicate double the amount of voltage. It should be noted that a voltmeter is a high impedance device. It is not unusual for a voltmeter to indicate some amount of voltage before the delta is closed, especially if the primary has been connected as a wye and the secondary as a delta. When this is the case, the voltmeter will generally indicate close to the normal output voltage if the connection is correct and double the output voltage if the connection is incorrect.
Over current Protection for the Primary: • Electrical Code Article 450-3(b) states that each transformer 600 volts, nominal or less, shall be protected by an individual over current device on the primary side, rated or set at not more
than
125%
of
the
rated
primary
current
of
the
transformer. Where the primary current of a transformer is 9 amps or more and 125% of this current does not correspond to a standard rating of a fuse or nonadjustable circuit breaker, the next higher standard rating shall be permitted. Where the primary current is less than 9 amps, an over current device rated or set at not more than 167% of the primary current shall be permitted. Where the primary current is less than 2 amps, an overcurrent device rated or set at not more than 300% shall be permitted.
Example #1: What size fuses is needed on the primary side to protect a 3 phase 480v to 208v 112.5 kVA transformer? •
Important when dealing with 3 phase applications always use 1.732 (square root of 3).
To solve: P / I x E •
112.5 kVA X 1000 = 112500 VA
•
112500 VA divided by 831 (480 x 1.732) = 135.4 amps
•
Since the transformer is more than 9 amps you have to use 125 %.
•
135.4 X 1.25 = 169 amps.
Answer: 175 amp fuses (the next higher standard, Electrical Code 240-6).
Example #2: •
What size breaker is needed on the primary side to protect a 3 phase 208v to 480v 3kVA transformer?
To solve: P / I x E •
3kVA X 1000 = 3000 VA
•
3000 VA divided by 360 (208 x 1.732) = 8.3 amps
•
Since the transformer is 9 amps or less you have to use 167%.
•
8.3 X 1.67 = 13.8 amps
Answer: 15 amp breaker (preferably a 20 amp breaker)
•
Electrical Code Article 450-3(b)(2) states if a transformer 600 v, nominal, or less, having a an overcurrent device on the secondary side rated or set at not more than 125% of the rated secondary current of the transformer shall not be required to have an individual overcurrent device on the primary side if the primary feeder overcurrent device is rated or set at a current value not more than 250% of the rated primary current of the transformer.
Over current Protection for the Secondary: •
Electrical Code Article 450-3(b)(2) states that a transformer 600 v, nominal, or less, shall be protected by an individual over current device on the secondary side, rated or set at not more than 125% of the rated secondary current of the transformer. Where the secondary current of a transformer is 9 amps or more and 125% of this current does not correspond to a standard rating of a fuse or nonadjustable circuit breaker, the next higher standard rating shall be permitted. Where the secondary current is less than 9 amps, an overcurrent device rated or set at not more than 167% of the secondary current shall be permitted.
Example: •
What size breaker is needed on the secondary side to protect a 3 phase 480v/208v 112.5 kVA transformer?
To solve : P / I x E •
112.5 kVA x 1000 = 112500 VA
•
112500 divided by 360 (208 x 1.732) = 312.5 amps
•
312.5 X 1.25 = 390.6 amps
EQUIVALENT CIRCUIT DIAGRAM
X
1 As
Ap Cp Y Z
Cs
2 3
Note:
This is a method of transformer 3ф power by means only “Two Transformer.
It is employed when: a) Three phase load is too small to warrant to installation of full phase transformer. b) When one of the transformer in a delta-delta bank (∆∆) is disabled, so that us continued at reduced capacity until the faulty transformer is repaired or a new one is substituted. c) When it is anticipated that in the future, the load will increase necessitating the closing of open delta.
F.) SCOTT or T-CONNECTION: Equivalent Circuit Diagram
C
A
86.6%
4
3
50%
1
B
2
This connection is used to transform 3ф system to 2ф system or vice versa. It consist of 2 identical 1ф x
4mer one having a
50% tap & the other an 86.6% tap on their primary windings.
CLOSED DELTA BANK
3ф L O A D
Let:
Note:
OPEN-DELTA BANK
3 ф L O A D
Example Problem: two transformer are connected open delta delivering a load of 100 KVA at a power factor of 0.8 lagging. Calculate the power delivered by each x’4mer? Given: SL = 100 KVA L = 0.8 lagging = -36.870 Req’d:
3ф
T-1
T-2
L O A D
100 KVA
Example Problem: two single phase distribution transformer connected in open delta will supply power to a 200 Hp, 3ф induction motor operating at 0.707 OF and 90% efficiency. Solve the minimum size in KVA of each transformer needed to supply the power without being over loaded. Given:
Example Problem: An electrical utility company is supplied by two single phase x’4mers bank in rated 75 KVA capacity. What is the maximum 3ф load in kw that the bank can carry without overloading considering that the load has a lagging PF of 0.8. Given:
Example Problem: Three single phase (3ф) transformer each rated 75 KVA are bank in delta supplying a 3ф load drawing 160 KVA at 0.8 lagging PF. If one x’4mer has burned out and is removed for repair, solve for the amount of overloading of the remaining units. Given:
Example Problem: An open-delta bank, consisting of two single phase x’4mers is operating with a balance 3ф load of 50 KVA, 440 V at 0.8 PF lag and a single phase load of 10 kw resistive connected across AC leg. Determine minimum ratings of the two x’4mers. Assume a phase sequence a-b-c. Given:
10 kw
a C
T-1
Icc
T-2 N b
Ibb
440 V
50 kva 0.8 lag L O A D
Ia
a
L 10kw
C
440 v PF = 1
N
Ic
Example Problem: a 3ф transformer connected on delta on the primary side step down. The voltage from 13200 v – 460 v and delivers 750 kva, 0.8 PF lagging to the load. Calculate: a) Transformation ratio b) The current delivered to the load c) The current in primary line wires d) The current that flows in the transformer winding Given: ∆ Connected
SL = 750 kva
Vp = 13200 V PFL = 0.8 lag Vs = 460 V Req’d: a, ISL, IPL and IpФ = ?
Example Problem: A power transformer rated 50,000 KVA, 34.5/13.8 kvolts is connected Y-Y. what are the line currents at full load. Given:
Example Problem: A power transformer is rated 50,000 KVA 34.5/13.8 kvolts is connected Y-Grounded on the primary and delta on the secondary. Determine the full load phase currents on the secondary side. Given: Y (grounde) = ∆ connection
IpL Ipф
N
Example Problem: What should be the turns ratio of the 3ф transformers that transform from 230 kvolts to 4160 volts if the transformer is to be connected Y-∆ & ∆-Y Given:
PARALLEL OPERATION OF TRANSFORMER requirements to be met for ideal operation of two or more transformers in parallel. a)The voltage rating of both primary and secondary windings must be identical. Note: if this condition is not fulfilled or met, there will be a current that will circulate through the transformer secondaries which contribute to the transformer loss. b.)
The transformer must be properly connected with regards
to their polarity.
Note: if this condition is not met, there will be a large current that will circulate in the transformer secondaries and will damage the windings. c.) the equivalent impedance must be inversely proportional to their respective KVA rating.
Note: if this condition is not met, the transformer will not share the load in proportion to their respective KVA ratings. A possibility that the large transformer operates at a underload condition while smaller transformer operates at overload conditions.
d.) The ratio of the equivalent resistances (Res) and reactances (Xes) referred to the secondary side must be the same.
Note: if this condition is not met each transformer will operate at different power factor with that of the connected load. e.)
in the case of 3ф transformer to be connected in parallel, it
should be of the same winding connections: ∆-∆ to be connected to ∆-∆ or ∆-Y to be connected to ∆-Y
Note: if this condition is not met there will be a current that will circulate will be a current that will circulate through the secondary windings and will contribute to the transformer losses. ADDITIONAL INFORMATION 1. All circuit diagrams for convenient purposes must be referred to the secondary winding.
OR
2.)
Always perform vector addition for apparent power and
current.
When connected in parallel:
CASE 1: (Equal turn Ration)
5. KVA rating of each transformer is not necessary.
Example Problem: two transformer are connected in parallel supplying a common load of 175 KVA @ 0.8 PF lag. Both transformer has a voltage rating of 2300/230 V single phase and each rated 100 KVA. Transformer 1 has equivalent impedance of 16 ohms and that of transformer 2 is 13 ohms. Referred both to the primary. Find KVA load in each transformer. Given:
CASE2: (equal turn ratio)
Example Problem: Determine the KW ratio of the KW output of transformer T-1 to that transformer T-2 when they are connected in parallel supplying a load of 150 KW @ 0.8 PF lagging. Given:
CASE 3: (Equal Turn Ratio) 1. If given is percent (%) or per unit impedance, either in absolute or with equal angle.
2. Use algebraic addition
Example Problem: A 125 KVA distribution transformer w/ 4% impedance is connected in parallel w/ another transformer rated 75 KVA and 3%, both have the same voltage ratio. Neglect the resistance of each transformer, the total load is 140 KVA at 80% PF lagging, how much load does it carry?
CASE IV: (Equal turn ration) 1. If given is percent (%) or per unit impedance but of different angle 2. Use vector addition.
4. KVA rating of each transformer is necessary.
MODIFIED FORMULAS:
Sample Problem: A 500 KVA single phase transformer “A” w/ percentage impedance of 0.01 + j 0.05 is to be connected in parallel w/ a 250 KVA transformer “B” w/ percentage impedance of 0.015 + j 0.04. if they are serving single phase load rated 800 KVA @ 0.8
PF
lagging,
determine
the
power
factor
of
each
transformer. Assuming the transformer have the same turn ration.
For both case 3 and 4: Note: if the individual rated KVA rating of each transformer is not given
used this formula. With percent (%) or per
unit impedance given,
Example Problem: Meralco has two single phase transformer w/equal turns ratio and ratings are operating at in parallel to supply a load of 280 kw @ 0.8 PF lagging. Transformer “A” has resistance of 2% and a reactance 8% while transformer “B: has a resistance of 1% and reactance of 6%. Determine the power delivered by transformer A and B to the load.
LOAD OPERATIONS FOR TRANSFORMER’S W/ UNEQUAL TURN’S RATIO:
Where:
AUTO-TRANSFORMER defined as a transformer in which part of the winding is common to both primary and secondary. its and electrically continuous winding w/one or more taps on a magnetic core. One circuit is connected to the end of terminals while the other is connected to one end terminal and to a part way along the winding. in general, an autotransformer is both magnetically and electrically linked.
Notes: a) Always assumed auto-transformer efficiency of 100% if not specified. b) If efficiency is equal to 100%
Example Problem: An
auto-transformer
designed
for
4,000
to
2300
operation supplies a load of 32 kw at a power factor of 0.8 Calculate: a) The transform power b) The power conducted
volts
Example Problem: A
5KVA,
2300/460
V
distribution
transformer
is
to
be
connected as an auto transformer to step-up. The voltage from 2300V to 2760V. When used to transform 5KVA. Calculate the kilovolt-ampere load output.
Example Problem: A 20 kva, 500 V load is to be supplied by an ideal step-up auto-transformer from 400v source. Find the current in the common winding.
l o a d
AC MOTORS 1. Induction Motor (3ф) – is simply a rotating electrical machines which convert electrical power (energy output). a) Synchronous Speed (Ns) – is defined at which the rotating flux rotates. It is the speed of synchronous generator (as generator or alternator) that supplies all the load connected to it or it is called the speed of incoming source. WHERE: Ns = synchronous speed in RPM f = frequency of the incoming source supplying the induction motor. Stator frequency.
P = number of poles of the machine.
Is known as the actual speed of induction motor. It is the speed written on its nameplate in RPM. It is measured by means of instrument “Tachometer” c. Slip (S) – rotor of an induction motor rotates somewhat less than the synchronous speed
of the rotating magnetic
flux. The difference of the synchronous speed and the actual speed is known as SLIP.
2 MAIN PARTS OF INDUCTION MOTOR 1. STATOR – stationary part; It is the one that receives electrical power from the AC SOURCE by means of conduction.
When
compared
to
a
transformer, it
is
considered as the primary winding. 2. ROTOR – rotating part; it is the one that delivers the mechanical power output to the load and takes power from the stator via the airgap by means of a electromagnetic induction. When compared to a transformer it is considered to be the secondary winding.
TYPES OF INDUCTION MOTOR a) Squirrel cage type b) Wound rotor type
air gap STATOR
air gap SPI
RPD STATOR
ROTOR
RPO
RPI
SCL
COL
RCL
F&W
ROTOR
Example Problem A 6 pole, 50 hz, 3ф induction motor runs at 960 RPM while delivering a shaft torque of 120 n-m. if friction and windage loss amount to 180 watts, determine the rotor copper loss.
SPI
STATOR
SCL
CL
RPI
RPD ROTOR
RCL
RPO
F&W
Example Problem: A 3ф, 220 v, 60 Hz, 4 pole induction motor drives a “Fan” which provides 40 m³/sec of air requirement to furnace. Assume an efficiency of 60% and 150mm of H ₂0 gage pressure. Determine the current drawn by the induction motor if eff. and power factor is 80% and 70% respectively.
FPO ŋ-60%
ŋ-80% SPI
STATOR
RPI
ROTOR
RPO
(RPO=FPI)
air
TEST PERFORMED IN AN INDUCTION MOTOR a) No Load Test – this test is similar to open ckt test performed in a transformer. Purpose:
To be able to determine the value of constant “stry
power loss (SPL)” @ No load condition:
b.)
Blocked Rotor Test this test is similar to short circuit test performed in a
transformer. During this test the rotor is blocked and the rotor windings are short circuited, cut slip rings if the motor has a wound rotor. A reduced voltage is applied to the stator terminals and is so adjusted that full load current flows in the stator. The values of the current voltage and power input are measured. Purpose: To determine the equivalent resistance per phase referred on the stator side
c.)
Load Test - Motor is loaded with its normal load or visual load. - During this test, the motor is loaded. The problem should clearly specified what is the specified load on the motor. - This is not an assurance that the motor is tested at full load.
Example Problem: A 5 hp, 6 pole, 3ф, 60 Hz induction motor operates with a slip of 2% and requires 11 amp and 3500 watts, when during its visual load. When the rotor is blocked, 440 watts at 52 volts are required to circulate a current of 14 amp. Calculate the following when motor is driving its visual load. a) Horsepower output b) Torque Excerted c) Efficiency (Assume Y-connected stator) Given:
SPI
STATOR
SCL
CL
RPI
RPD ROTOR
RCL
RPO
F&W
Example Problem: A 440 v, 60 Hz, 4 pole, 3ф wound rotor induction motor is directly connected to a pump which delivers 1000 cfm of water against an effective head of 8.7 ft. under this load, the motor drains 15.62 kw at a power factor of 0.92 when rated at no load. When the motor drains 803 watts, the rotor resistance per phase is 0.022 ohm while stator resistance per phase is 0.202 ohm. The effective turn ration between stator and rotor is 4:1. calculate the pump efficiency? Given:
H20
8.7 ft Pop
M
P
Example Problem: A 4 pole, 60 Hz, 3ф induction motor draws 8200 watts from the line. The losses in the machines are: SCL
= 300 watts
RCL
= 160 watts
CoL
= 400 watts
F & W = 130 watts Calculate the torque output: SPI
STATOR
SCL
CL
RPI
RPD ROTOR
RCL
RPO
F&W
CASE 2: Note: When the given data regarding core loss
and friction
and windage (F & W) are incorporated as one in the stray power loss (SPL).
air gap SPI
RPD STATOR
SCL
SPL=(F&W)+CL
ROTOR
RCL
Example Problem: A 6 pole, 3ф, 60 Hz induction motor takes 48 kw in power at 1140 RPM. The stator copper loss is 1.6 kw and stray power losses are 1 kw. Find the eff.
SPI
STATOR
SCL
SPL
RPI
RPD = RPO ROTOR
RCL
Example Problem: A 230 v, 8ф, 4 pole, 60 Hz squirrel cage induction motor operating at 90% efficiency & 85% PF and 2.5% slips drives a pump for reservoir. Determine the current taken from the line if the discharge rate is 475 GPM and the total head of which the pump is working 200 ft. Assume pump efficiency of 80%.
water reservior
SPI
Pom Pip
M
P
Pop
200 ft
DC RESISTANCE TEST OR OHMIC TEST. – This is exactly the same test applied on the alternator to solve for value of ohmic resistance of the stator per phase. – When this test is applied blocked rotor test is no longer necessary. Blocked Rotor Test Resф – effective or AC equivalent resistance per phase referred to the stator. (Rotor is included) During DC Resistance Test Resф – effective or AC resistance per phase of the stator alone. (Rotor is not included)
Open Terminal Stator Winding
Stator Winding
Example Problem: A 400 V, 60 Hz, 6 Pole, Y connected 3ф induction motor draws 75 kw with a line current of 100 amp, the core loss is 2 kw, friction and windage loss is 1.2 kw, the stator resistance between two terminals is 0.35 ohm. What is the efficiency if the motor runs at a slip of 2.5%
Solution: SPI
STATOR
SCL
+CL
RPI
RPD ROTOR
RCL
RPO
F&W
RPI (electrical)
RPD (electro mechanical) ROTOR RPO
RCL (electrical)
F&W
(mechanical)
Example Problem: A 40 Hp, 230 v, 8 pole, 25 Hz 3ф induction motor is running at 355 RPM @ rated load. The torque lost by friction is 24 n-m. if the total stator loss is 1000 watts. Calculate the eff.of the motor at this load.
Solution: RPI
ROTOR
RCL
RPO
STARTING AN INDUCTION MOTOR 1. All formulas derived are applicable to all type of AC motors and DC motors. 2. During staring, the current by the motor is too high about (5 to 7times its full load value) simply because at the instant of starting, the motor is not rotating and the slip is 1.0 and the counter emf or back emf is zero. 3. Motors should be protected at the instant of starting to prevent damage on the windings and to lessen the effect on the other equipment connected on the same line wires. Note: for small motors (usually 10 hp below) use across the line, magnetic starter (full voltage starting method)
The starting line current is always proportional to the starting line voltage: The starting torque (Ts) is always proportional to the square of the starting line voltage: For two condition problem: 1st condition: use subscript “1”
2nd condition: use subscript “2” Ratio
Where: Vs – starting line voltage Ts – starting torque Is - starting current Td – rated or full load torque developed IL – rated or full load current
Example Problem: A 25 hp, 230 volt, 3ф induction motor with 85% PF has a starting current of 5.5 times its rated current. To reduced the starting current, at “Y-∆” starter is installed. What will be the new starting current? Given:
Example Problem: A 10 hp, 550v, 3ф induction motor has a starting torque of 160% of full load torque and a starting current of 425% of full load current. If the motor is used on 440v, 60hz system. What will be the starting torque express in percent of full load value?
Example Problem An induction motor of 30 hp, 220v, 3ф, draws 450% of rated current with rated voltage and delivers during the starting period of 130% of the normal torque. The full load efficiency of the motor and full load PF is 80% and 70% respectively. If the auto-transformer unit is used as a starting unit
and the
starting torque of the loads is only 50% of the rated torque of the motor. Find the starting current of the line.
REDUCED VOLTAGE METHOD FOR LARGE 3ф MOTORS 1. Y-∆ starter (Y-start ; ∆-run) 2. Part winding method 3. Auto-transformer method 4. Line resistance method 5. Line reactance method 6. Wound rotor starting
FULL VOLTAGE STARTING METHOD Line 1 Line 2 Line 3
Start OLR
Circuit Breaker
Normally Open Overload Relay
Motor
Stop
Coil
SYNCHRONOUS MOTOR TYPES OF S.M. 1. Under-Excited Synchronous M. – normally operates at a lagging power factor. 2. Normally-Excited Synchronous M. – operates at unity power factor. 3. Over-Excited Synchronous M. – operates at leading power factor.
CHARACTERISTICS OF SYNCHRONOUS MOTOR 1. It runs a definite constant speed called synchronous speed. 2. It is not self-starting, it needs amortisseur winding. 3. It can be operated over a wide range of power factors hence it can be used for power factor correction. AMORTISSEUR WINDING – it is case winding with bars embedded in the rotor pole faces of a synchronous and shorted at each end rings to make the rotor self starting in dumping out hunting.
Field Rheostat
ίƒ field circuit separately Vf
Vф
excited by DC source
Supply armature circuit
Voltage
Notations: If
– DC shield current taken from the exciter
Vf
– DC field exciter voltage
Reф – effective or AC resistance per phase of the armature Xsф – synchronous reactance per phase of the armature winding. Zsф – synchronous impedance per phase of the armature winding.
Vф – supply AC voltage per phase (rated or nameplate voltage) Iф
– supply armature phase current
Ecф – back or counter emf developed in the armature per phase.
VOLTAGE EQUATION (in general):
Where: - deflection angle - torque angle
Electrical
Pin
Pd
Mechanical
Pd ARMATURE
SPL
Po
Notes: 1. If the types of motor is not specified to be 3ф or 1ф, “always consider the motor as 1ф”. 2. If the synchronous motor is used for power factor correction purposes. It is understood and considered to operate always at leading power factor (over excited sm) 3. If the field winding is not mentioned, then it is considered as ZERO.
Example Problem: The output of a 200 v synchronous motor taking a current of 20 amp is 4 hp. Effective armature armature resistance is 0.5 ohms. The iron and friction losses amount to 400 watts. What is the power factor of the machine.
Pin
Pd
Pd ARMATURE
Example Problem: A 20hp, 440v, 3ф, star connected sm has an armature effective resistance per phase of 0.4 ohm at full load output, the power factor is 0.9 leading, the iron and friction losses amount to 500w. What is the value of armature line current?
Pin
Pd
ARMATURE
Pd
Example Problem: A 500v, 1ф sm gives a net mechanical power of 7.46 kw and operates at 0.9 lagging PF, its effective resistance is 0.8 ohm. If the iron and friction losses are 500 w and excitation losses are 800 w. estimate the armature current. Calculate the commercial efficiency.
Pin
Pd
ARMATURE
Example Problem: A 2300-v, 3ф star connected synchronous motor has a resistance of 0.2 ohm per phase and a sm reactance of 2.2 ohm per phase. The motor is operating at 0.5 PF leading with a line current of 200 amp. Determine the value of the generated emf per phase.
Another Solution:
Example Problem: A 100 v, synchronous motor having 40% reactance and a negligible resistance is to be operated at rated load a) Unity b) 0.8 lag c) 0.8 leading What are the values of induced emf is?