AC MACHINERY FORMULAS
ALTERNATORS
𝑷𝑵
𝒇=
𝟏𝟐𝟎
𝑬𝒑 = 𝟐. 𝟐𝟐𝒌𝒑 𝒌𝒃 𝒇∅𝒁 𝒁 = # 𝒐𝒇 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔 𝒊𝒏 𝒔𝒆𝒓𝒊𝒆𝒔 𝒑𝒆𝒓 𝒑𝒉𝒂𝒔𝒆
𝒄𝒐𝒊𝒍 𝒔𝒑𝒂𝒏 𝒊𝒏 # 𝒔𝒍𝒐𝒕𝒔
𝒌𝒑 = 𝒔𝒊𝒏[(𝟗𝟎)(𝒑𝒊𝒕𝒄𝒉)] 𝒑𝒊𝒕𝒄𝒉 =
𝒔𝒍𝒐𝒕𝒔/𝒑𝒐𝒍𝒆
𝒏𝜹
𝒔𝒊𝒏[ ] 𝟏𝟖𝟎𝒐 𝒔𝒍𝒐𝒕𝒔
𝟐
𝒌𝒃 = 𝜹 𝜹= 𝒏=
𝒏𝒔𝒊𝒏[ ] 𝒔𝒍𝒐𝒕𝒔/𝒑𝒐𝒍𝒆 𝒑𝒐𝒍𝒆/𝒑𝒉𝒂𝒔𝒆
𝟐
𝑬𝒑 = √(𝑽𝒑 𝒄𝒐𝒔𝜽 + 𝑰𝒑 𝑹𝒂 )𝟐 + (𝑽𝒑 𝒔𝒊𝒏𝜽 ± 𝑰𝒑 𝑿𝒍𝒂 𝟐 ∴ +𝑰𝑿 𝒊𝒇 𝒑. 𝒇. 𝒊𝒔 𝒍𝒂𝒈𝒈𝒊𝒏𝒈
𝑬𝒑 ∠ ± 𝜹 = 𝑽𝒑 ∠𝟎 + (𝑰𝒑 ∠ ± 𝜽)(𝑹𝒂 + 𝒋𝑿𝒍𝒂 ) ∴ +𝜽 𝒊𝒇 𝒑. 𝒇. 𝒊𝒔 𝒍𝒆𝒂𝒅𝒊𝒏𝒈
Open Circuit Test
𝑽
𝒀 ⟹ 𝑬𝒐𝒄 = 𝚫 ⟹ 𝑬𝒐𝒄 = 𝑽 : V = voltmeter reading
√𝟑
Short Circuit Test
𝑨
𝒀 ⟹ 𝑰𝒔𝒄 = 𝑨 𝚫 ⟹ 𝑰𝒔𝒄 = : A = ammeter reading
√𝟑
DC Resistance Test
𝟏 𝟑
𝒀 ⟹ 𝑹𝒂 = 𝑹𝟏 𝚫 ⟹ 𝑹𝒂 = 𝑹𝟏
𝟐 𝟐
: R1 = DC resistance of the alternator between any two terminals; Ohmmeter reading
Parameters derived from tests
𝑬𝒐𝒄
𝒁𝒔 = 𝑿𝒔 = √𝒁𝒔 𝟐 − 𝑹𝒂 𝟐 : Z,E,I,X,R – per phase values
𝑰𝒔𝒄
Per unit values
𝑺𝒓𝒂𝒕𝒆𝒅 𝑽𝒓𝒂𝒕𝒆𝒅 𝑽𝟐 𝒓𝒂𝒕𝒆𝒅
𝑰𝒓𝒂𝒕𝒆𝒅 = 𝒁𝒓𝒂𝒕𝒆𝒅 = = : per phase values
𝑽𝒓𝒂𝒕𝒆𝒅 𝑰𝒓𝒂𝒕𝒆𝒅 𝑺𝒓𝒂𝒕𝒆𝒅
𝑹𝒂 𝑿𝒂
𝑹𝒑𝒖 = 𝑿𝒑𝒖 = 𝒁𝒑𝒖 = 𝑹𝒑𝒖 ± 𝒋𝑿𝒑𝒖
𝒁𝒓𝒂𝒕𝒆𝒅 𝒁𝒓𝒂𝒕𝒆𝒅
Voltage Regulation
𝑬𝒑 −𝑽𝒑
%𝑽𝑹 = %𝑽𝑹 = √(𝒄𝒐𝒔𝜽 + %𝑰𝑹)𝟐 + (𝒔𝒊𝒏𝜽 ± %𝑰𝑿)𝟐 − 𝟏 : +%IX if lagging pf
𝑽𝒑
Power Developed in the armature per phase
𝑬𝒑 𝑽𝒑 𝑬𝒑 𝑽𝒑
𝑷𝒅 = 𝒔𝒊𝒏𝜹 𝑷𝒅𝒎𝒂𝒙 =
𝑿𝒔 𝑿𝒔
𝑷𝒅𝒎𝒂𝒙 𝒘𝒉𝒆𝒏 𝜹 = 𝟗𝟎
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AC MACHINERY FORMULAS
Power losses and efficiency
𝑷𝒐𝒖𝒕
𝜼= 𝑷𝒍𝒐𝒔𝒔𝒆𝒔 = 𝑷𝒄𝒖 + 𝑷𝒄𝒐𝒓𝒆 + 𝑷𝒇𝒘 + 𝑷𝒇 + 𝑷𝒗
𝑷𝒐𝒖𝒕 +𝑷𝒍𝒐𝒔𝒔𝒆𝒔
𝑷𝒐𝒖𝒕(𝟏−𝝓) = 𝑽𝑳 𝑰𝑳 𝒄𝒐𝒔𝜽 𝑷𝒐𝒖𝒕(𝟑−𝝓) = √𝟑𝑽𝑳 𝑰𝑳 𝒄𝒐𝒔𝜽
𝑷𝒄𝒖(𝟏𝝓) = 𝑰𝒑 𝟐 𝑹𝒂 𝑷𝒄𝒖(𝟑𝝓) = 𝟑𝑰𝒑 𝟐 𝑹𝒂 ………………………….: 𝑰𝒑 − 𝒑𝒉𝒂𝒔𝒆 𝒄𝒖𝒓𝒓𝒆𝒏𝒕, 𝑹𝒂 − 𝒑𝒆𝒓 𝒑𝒉𝒂𝒔𝒆
𝑽𝒇 𝟐
𝑷𝒇 = 𝑰𝒇 𝟐 𝑹𝒇 = 𝑰𝒇 𝑽𝒇 =
𝑹𝒇
𝑷𝒄𝒐𝒓𝒆 , 𝑷𝒇𝒘 , 𝑷𝒇 , 𝑷𝒗 − 𝒂𝒓𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒍𝒐𝒔𝒔𝒆𝒔, 𝒖𝒏𝒍𝒆𝒔𝒔 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅
Copper loss at any size of load: 𝑷𝒄𝒖𝒙 = 𝒙𝟐 𝑷𝒄𝒖𝒇𝒍
Governor Speed Regulation
𝑵𝑵𝑳 − 𝑵𝑭𝑳 𝒇𝑵𝑳 − 𝒇𝑭𝑳
𝑮𝑺𝑹 = =
𝑵𝑭𝑳 𝒇𝑭𝑳
𝚫𝐏 𝑷𝑭𝑳 𝚫𝐟 𝑮𝑺𝑹 𝒙 𝒇𝑭𝑳
= 𝑮𝑫 = =
fNL 𝚫𝐟 𝒇𝑵𝑳 −𝒇𝑭𝑳 𝚫𝐏 𝑷𝑭𝑳
PFL-rating of the alternator
fNew GD-governor’s droop
𝚫𝐟
fFL
𝚫𝐏
PNEW PFL
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AC MACHINERY FORMULAS
TRANSFORMERS
Induced emf:
𝑬 = 𝟒. 𝟒𝟒𝑵𝒇∅
𝑬𝟏 𝑵𝟏 𝑰𝟐
= = = 𝒂 (𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒓𝒂𝒕𝒊𝒐)
𝑬𝟐 𝑵𝟐 𝑰𝟏
For an ideal transformer, there’s no leakage flux, no copper loss, and core is loss less.
For practical transformer:
𝑬𝟏 𝟐 𝑬𝟏 𝟐
𝑺 𝒎 = 𝑬𝟏 𝑰 𝒐 𝑸𝒎 = √𝑺𝒎 𝟐 − 𝑷𝒎 𝟐 𝑹𝒎 = 𝑿𝒎 =
𝑷𝒎 𝑸𝒎
Shifting parameters
𝑹𝟏
𝑹𝒆𝟏 = 𝑹𝟏 + 𝑹𝟐 (𝒂𝟐 ) 𝑹𝒆𝟐 = 𝑹𝟐 +
(𝒂𝟐 )
𝑿𝟏
𝑿𝒆𝟏 = 𝑿𝟏 + 𝑿𝟐 (𝒂𝟐 ) 𝑿𝒆𝟐 = 𝑿𝟐 +
(𝒂𝟐 )
𝒁𝒆𝟏 = √𝑹𝒆𝟏 𝟐 + 𝑿𝒆𝟏 𝟐 𝒁𝒆𝟐 = √𝑹𝒆𝟐 𝟐 + 𝑿𝒆𝟐 𝟐
Referred to secondary side: 𝑬𝟐 = √(𝑽𝟐 𝒄𝒐𝒔𝜽 + 𝑰𝟐 𝑹𝒆𝟐 )𝟐 + (𝑽𝟐 𝒔𝒊𝒏𝜽 ± 𝑰𝟐 𝑿𝒆𝟐 )𝟐
𝑰𝟐 − 𝒍𝒐𝒂𝒅 𝒄𝒖𝒓𝒓𝒆𝒏𝒕, 𝑽𝟐 − 𝒍𝒐𝒂𝒅 𝒗𝒐𝒍𝒕𝒂𝒈𝒆 +𝑰𝑿 𝒊𝒇 𝑷𝒐𝒘𝒆𝒓𝑭𝒂𝒄𝒕𝒐𝒓 𝒍𝒂𝒈𝒈𝒊𝒏𝒈, −𝑰𝑿 𝒊𝒇 𝒍𝒆𝒂𝒅𝒊𝒏𝒈
Using complex number format:
𝑬𝟐 ∠ ± 𝜹 = 𝑽𝟐 ∠𝟎 + (𝑰𝟐 ∠ ± 𝜽)(𝑹𝒆𝟐 + 𝒋𝑿𝒆𝟐 ) +𝜽 𝒊𝒇 𝒍𝒆𝒂𝒅𝒊𝒏𝒈 𝒑. 𝒇. , −𝜽 𝒊𝒇 𝒍𝒂𝒈𝒈𝒊𝒏𝒈 𝒑. 𝒇.
Open Circuit Test
𝑷𝒐𝒄 = 𝑷𝒄𝒐𝒓𝒆
Short Circuit Test
𝑬𝒔𝒄 𝑷𝒔𝒄
𝑷𝒔𝒄 = 𝑷𝒄𝒖𝒇𝒍 𝒁𝒆 = 𝑹𝒆 = 𝑿𝒆 = √𝒁𝒆 𝟐 − 𝑹𝒆 𝟐
𝑰𝒔𝒄 𝑰𝒔𝒄 𝟐
Power losses and Efficiency
𝑷𝒐𝒖𝒕
𝜼= 𝒙 𝟏𝟎𝟎% 𝑷𝒊𝒏 = 𝑷𝒐𝒖𝒕 + 𝑷𝒍𝒐𝒔𝒔𝒆𝒔 𝑷𝒐𝒖𝒕 = 𝑽𝑳 𝑰𝑳 𝒑. 𝒇. 𝑷𝒍𝒐𝒔𝒔𝒆𝒔 = 𝑷𝒄𝒐𝒓𝒆 + 𝑷𝒄𝒖
𝑷𝒊𝒏
𝑷𝒄𝒖 = 𝑰𝟏 𝟐 𝑹𝟏 + 𝑰𝟐 𝟐 𝑹𝟐 = 𝑰𝟏 𝟐 𝑹𝒆𝟏 = 𝑰𝟐 𝟐 𝑹𝒆𝟐 𝑷𝒄𝒐𝒓𝒆 = 𝑷𝒆𝒅𝒅𝒚 + 𝑷𝒉𝒚𝒔𝒕𝒆𝒓𝒆𝒔𝒊𝒔
𝑬𝒈 𝟏,𝟔
𝑷𝒆𝒅𝒅𝒚 = 𝒌𝒆 (𝒇𝜷)𝟐 = 𝒌𝒆 𝑬𝒈 𝟐 𝑷𝒉𝒚𝒔𝒕𝒆𝒓𝒆𝒔𝒊𝒔 = 𝒌𝒉 𝒇𝜷𝟏.𝟔 = 𝒌𝒉
𝒇𝟎.𝟔
Copper loss at any size of load: 𝑷𝒄𝒖𝒙 = 𝒙𝟐 𝑷𝒄𝒖𝒇𝒍
𝑷𝒐𝒖𝒕
Maximum efficiency: 𝑷𝒄𝒐𝒓𝒆 = 𝑷𝒄𝒖 𝜼= 𝒙 𝟏𝟎𝟎%
𝑷𝒐𝒖𝒕 +𝟐𝑷𝒄𝒐𝒓𝒆
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AC MACHINERY FORMULAS
𝑾𝒐𝒖𝒕
All Day efficiency: 𝜼𝒂𝒍𝒍 𝒅𝒂𝒚 = 𝒙 𝟏𝟎𝟎%
𝑾𝒐𝒖𝒕 +𝑾𝒍𝒐𝒔𝒔𝒆𝒔
𝑾𝒐𝒖𝒕 = 𝚺𝑷𝒐𝒖𝒕 (𝒕) = 𝚺(𝒙)(𝑽𝑨 𝒓𝒂𝒕𝒊𝒏𝒈)(𝒑𝒇)(𝒕)………….: x - size of load
𝑾𝒄𝒐𝒓𝒆 = 𝑷𝒄𝒐𝒓𝒆 (𝟐𝟒𝒉𝒓𝒔) 𝑾𝒄𝒖 = 𝚺(𝐱)𝟐 (𝑷𝒄𝒖𝒇𝒍 )(𝒕)
Per Unit Values
𝑰𝟐 𝑹𝒆𝟐 𝑷𝒄𝒖𝒇𝒍 𝑷𝒔𝒄 𝑰𝟐 𝑿𝒆𝟐 𝑬𝒔𝒄
%𝑰𝑹 = = = %𝑰𝑿 = %𝑰𝒁 = %𝑰𝑿 = √%𝑰𝒁𝟐 − %𝑰𝑹𝟐
𝑽𝟐 𝑽𝑨𝒓𝒂𝒕𝒊𝒏𝒈 𝑽𝑨𝒓𝒂𝒕𝒊𝒏𝒈 𝑽𝟐 𝑽𝒓𝒂𝒕𝒆𝒅
Voltage Regulation – percentage rise in the terminal voltage of the transformer when the transformer load is removed
𝑬𝟐 −𝑽𝟐
%𝑽𝑹 = %𝑽𝑹 = √(𝒄𝒐𝒔𝜽 + %𝑰𝑹)𝟐 + (𝒔𝒊𝒏𝜽 ± %𝑰𝑿)𝟐 − 𝟏
𝑽𝟐
Autotransformer – is a transformer with only one winding common to both the primary and the secondary sides.
𝑰𝟏 𝑵𝟐 𝟏 𝑽𝟏 𝑵𝟏
= = = =𝒂 𝑷𝒊𝒏 = 𝑽𝟏 𝑰𝟏
𝑰𝟐 𝑵! 𝒂 𝑽𝟐 𝑵𝟐
𝟏
𝑷𝒕𝒓𝒂𝒏𝒔 = 𝑷𝒊𝒏 (𝟏 − ) 𝑷𝒄𝒐𝒏𝒅 = 𝑷𝒊𝒏 − 𝑷𝒕𝒓𝒂𝒏𝒔
𝒂
3-phase transformers
Delta 𝑽𝑳𝒊𝒏𝒆 = 𝑽𝝓 𝑰𝑳𝒊𝒏𝒆 = √𝟑𝑰𝝓
Wye 𝑽𝑳𝒊𝒏𝒆 = √𝟑𝑽𝝓 𝑰𝑳𝒊𝒏𝒆 = 𝑰𝝓
𝑺𝒗 = √𝟑𝑽𝑨𝒓𝒂𝒕𝒊𝒏𝒈 𝒐𝒇 𝟏 − 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒓………….apparent power that an open-delta bank can deliver.
𝑺𝚫 = 𝟑𝑽𝑨 𝒐𝒇 𝒐𝒏𝒆 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒓…………apparent power that a closed-delta bank can deliver.
Scott (T-connected transformer bank) – uses to transform 3-Φ to 2-Φ or vice versa
Parallel operation of transformers
Voltage rating of primaries and secondaries must be identical.
Properly connected with regards to polarity.
The ratio of equivalent resistance to reactance (Re:Xe) of all transformers should be the same.
(𝒂𝟐 −𝒂𝟏 )𝑽𝟐 +(𝒂𝟐 𝟐𝒁𝒆𝟐 𝒐𝑰𝒕 ) (𝒂𝟏 −𝒂𝟐 )𝑽𝟐 +(𝒂𝟏 𝟏𝒁𝒆𝟐 𝒐𝑰𝒕)
UNEQUAL TURNS RATIO: 𝟏𝑰𝟐 = 𝟐𝑰𝟐 =
𝒂𝟏 𝟐𝒁𝒆𝟐 +𝒂𝟐 𝟐𝒁𝒆𝟐 𝒂𝟏 𝟏𝒁𝒆𝟐 +𝒂𝟐 𝟐𝒁𝒆𝟐
𝟐𝒁𝒆𝟐 𝟏𝒁𝒆𝟐
EQUAL TURNS RATIO: 𝟏𝑰𝟐 = [ ]𝒐𝑰𝒕 𝟐𝑰𝟐 = [ ]𝒐𝑰𝒕
𝟏𝒁𝒆𝟐 +𝟐𝒁𝒆𝟐 𝟏𝒁𝒆𝟐 +𝟐𝒁𝒆𝟐
𝑺𝒕 𝟐𝒁𝒆𝟐 𝑺𝒕 𝟏𝒁𝒆𝟐
𝟏𝑺𝟐 = [ ] 𝟐𝑺𝟐 = [ ] 𝒐𝑰𝒕 - load current
𝟏𝒁𝒆𝟐 +𝟐𝒁𝒆𝟐 𝟏𝒁𝒆𝟐 +𝟐𝒁𝒆𝟐
Circulating current (load current, oIt=0):
(𝒂𝟐 −𝒂𝟏 )𝟏𝑽𝟐 (𝒂𝟏 −𝒂𝟐 )𝟏𝑽𝟐
𝟏𝑰𝒄 = 𝟐𝑰𝒄 = ……….: 1V2 = load voltage
𝒂𝟏 𝟏𝒁𝒆𝟐 +𝒂𝟐 𝟐𝒁𝒆𝟐 𝒂𝟏 𝟏𝒁𝒆𝟐 +𝒂𝟐 𝟐𝒁𝒆𝟐
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