Uploaded by rono schngr

example

advertisement
Anand Jain, Govt. Polytechnic Hanumangarh
TRANSFORMER
DES
IGN
EXAMPLES-3
TRANSFORMER
DES IGN
EXAMPLES-3
Anand Jain, Govt.
Polytechnic Hanumangarh
Design magnetic frame and winding for a 50 KVA, 11000/400 V, 50Hz, three phase delta/star core type
distribution transformer. Assume flux density 1.2 T, current density (aluminum conductor) 1.5 A/mm2,
window space factor=0.2, having window area 10 times net iron area. Use three stepped cores.
Given Data :
kVA=50,
f = 50 Hz,
VHV=11000 V delta VLV=400 V star
Bm=1.2 T
d=1.5 A/mm2
Kw=0.2
Aw/Ai =10
Output equation of transformer is given as
𝑸 = πŸ‘. πŸ‘πŸ‘ 𝒇 π‘©π’Ž 𝑨𝑰 𝑲𝑾 𝑨𝑾 𝜹 × πŸπŸŽ−πŸ‘ here
So
π‘¨π’Š × π‘¨π‘Ύ = πŸπŸŽπ‘¨π’Š 𝟐 =
𝑨𝑾 = 𝟏𝟎 𝑨𝑰
𝑸×πŸπŸŽπŸ‘
πŸ‘.πŸ‘πŸ‘ 𝒇 π‘©π’Ž 𝑲𝑾 𝜹
=
πŸ“πŸŽπŸŽπŸŽπŸŽ
πŸ‘.πŸ‘πŸ‘ ×πŸ“πŸŽ×𝟏.𝟐×𝟎.𝟐×𝟏.πŸ“×πŸπŸŽπŸ”
Check πŸπŸŽπŸ” figure in current density value
π‘¨π’Š 𝟐 = πŸ–πŸ‘. πŸ‘πŸ‘πŸ‘ × πŸπŸŽ−πŸ”
or
π‘¨π’Š =. πŸŽπŸŽπŸ—πŸπŸπŸ–πŸ“ π’ŽπŸ = πŸ—πŸ. πŸπŸ– π’„π’ŽπŸ
Hence emf per turn
𝐸𝑑 = 4.44 π‘“π‘©π’Ž π‘¨π’Š = 4.44 × 50 × 1.2 ×. πŸŽπŸŽπŸ—πŸπŸπŸ–πŸ“ = 𝟐. πŸ’πŸ‘πŸ volt/turn
With such medium size 50KVA transformers three stepped or more stepped cores are used. When we use
higher number of steps in core, we utilize the iron space better. By increasing the number of steps, the
area of circumscribing circle is more effectively utilized for iron part. The relation between circumscribe
circle diameter “d” and core cross section “Ai” can be expressed as follows π‘¨π’Š = π’Œ π’…πŸ
The value of constant k depends upon the number of steps of core of the transformer, given as
Core type
Square
Cruciform
3 stepped
4 stepped
5 stepped
6 stepped
Constant k
0.45
0.56
0.6
0.62
0.64
0.65
for three stepped core circumscribing circle diameter 𝒅
=
The most economical dimensions of various steps for a
three-stepped core can be calculated. The results are
𝒂 = 𝟎. πŸ—πŸ 𝒅 = 𝟎. πŸ—πŸ × πŸπŸ. πŸ‘πŸ‘ = 𝟏𝟏. 𝟐𝟐 ≈ 𝟏𝟏. 𝟐 π’„π’Ž
𝒃 = 𝟎. πŸ•πŸŽπŸ•π’… = 𝟎. πŸ•πŸŽπŸ• × πŸπŸ. πŸ‘πŸ‘ = πŸ–. πŸ•πŸπŸ• ≈ πŸ–. πŸ• π’„π’Ž
𝒂𝒏𝒅 𝒄 = 𝟎. πŸ’πŸ 𝒅 = 𝟎. πŸ’πŸ × πŸπŸ. πŸ‘πŸ‘ = πŸ“. πŸπŸ•πŸ– ≈ πŸ“. 𝟐 π’„π’Ž
π‘¨π’Š
π’Œ
=
πŸ—πŸ.πŸπŸ–
𝟎.πŸ”
= 𝟏𝟐. πŸ‘πŸ‘ π’„π’Ž
It is given that window area is 10 times of net iron area so
π‘¨π’˜ = 𝟏𝟎 π‘¨π’Š = 𝟏𝟎 × πŸŽ. πŸŽπŸŽπŸ—πŸπŸπŸ–πŸ“ = 𝟎. πŸŽπŸ—πŸπŸπŸ–πŸ“ π’ŽπŸ = πŸ—πŸπŸ. πŸ–πŸ“ π’„π’ŽπŸ
Window area is determined as product of width and height of the window. The height
of the window, is usually taken 2 to 4 times of width of window.
π»π‘Š
π‘Šπ‘Š
= 2 π‘‘π‘œ 4
When height of window is taken high the mean turn length will decrease but leakage
flux will increase. Similarly, less height and more width of window will increase the
mean turn length but decrease leakage flux by some amount. Let us take height of
window 2.5 times of width of window.
𝑨𝑾 = πŸ—πŸπŸ. πŸ–πŸ“ π’„π’ŽπŸ = π»π‘Š × π‘Šπ‘Š = 2.5 π‘Šπ‘Š × π‘Šπ‘Š
or
π‘Šπ‘Š =
𝑨𝑾
2.5
=
so height of window
912.85
2.5
= 19.11 ≈ 19.1 π‘π‘š
π»π‘Š = 2.5 π‘Šπ‘Š = 2.5 × 19.11 = 47.8 π‘π‘š
or π‘Šπ‘Š = 19.1 π‘π‘š
so height of window
π»π‘Š = 2.5 π‘Šπ‘Š = 2.5 × 19.11 = 47.8 π‘π‘š
One significant dimension in core is center to
center limb distance
𝐷 = π‘Šπ‘€ + 𝑑 = 19.1 + 12.3 = 31.4 π‘π‘š
Overall width of core
π‘Š = 2𝐷 + π‘Ž = 2 × 31.4 + 11.2 = 74 π‘π‘š
Overall height of core
𝐻 = π»π‘Š + 2 π‘Ž = 47.8 + 2 × 11.2 = 70.2 π‘π‘š
Winding design
π‘‰π‘ƒβ„Ž =
𝑉𝐿
3
=
400
3
Low voltage winding 400 V star connection
𝑸 = πŸ‘. πŸ‘πŸ‘ 𝒇 π‘©π’Ž 𝑨𝑰 𝑲𝑾 𝑨𝑾 𝜹 × πŸπŸŽ−πŸ‘ here
= 231.2 𝑉
No of turns
𝑁𝐿𝑉 =
Current in LV
𝐼𝐿𝑉 =
𝑉
𝐸𝑑
=
𝑄
3π‘‰π‘ƒβ„Ž
231.2
2.432
=
= 95.06 ≈ 95 π‘‘π‘’π‘Ÿπ‘›π‘ 
50000
3×231.2
Cross section area of conductor π‘ŽπΏπ‘‰ =
𝑨𝑾 = 𝟏𝟎 𝑨𝑰
𝐸𝑑 = 4.44 π‘“π‘©π’Ž π‘¨π’Š = 𝟐. πŸ’πŸ‘πŸ volt/turn
= 72.088 𝐴
𝐼𝐿𝑉
𝛿
=
72.088
1.5
= 48.05867 π‘šπ‘š2
We will select rectangle conductor of next higher standard size from conductor table, one can also use foil winding.
High voltage winding 11KV Delta
π‘‰π‘ƒβ„Ž = 𝑉𝐿 = 11000 𝑉
𝑉
No of turns
𝑁𝐻𝑉 = 𝑁𝐿𝑉 𝑉𝐻𝑉 = 95 ×
Current in HV
𝐼𝐻𝑉 =
𝐿𝑉
𝑄
3π‘‰π‘ƒβ„Ž
=
50000
3×11000
Cross section area of conductor π‘Žπ»π‘‰ =
11000
231.2
= 4519.89 ≈ 4520 π‘‘π‘’π‘Ÿπ‘›π‘ 
= 1.5151 𝐴
𝐼𝐻𝑉
𝛿
=
1.5151
1.5
= 1.01 π‘šπ‘š2
We will select round conductor of next higher standard size from conductor table.
Design magnetic frame and winding for a 10 KVA 11000/230 V, 50Hz, single phase core type transformer.
Assume flux density 1.1 T, current density 2.2 A/mm2, window space factor=0.25, constant K=0.8 for single
phase core type transformer. Use cruciform two stepped core.
Given Data :
Hence emf per turn
kVA=10,
VHV=11000 V
f = 50 Hz,
VLV=230 V
Bm=1.1 T d=2.2 A/mm2
Kw=0.25 K=0.8
𝐸𝑑 = 𝐾 𝑄 = 0.8 10= 2.53 volt/turn
𝑬
The net core cross section area Ai is given as π‘¨π’Š = πŸ’.πŸ’πŸ’ 𝒇𝒕 𝑩
π’Ž
Bm – maximum flux density in the core of the transformer in Wb/ m2
f – Frequency of the power supply in Hz
With very small size transformers, rectangular core can be used with either circular or rectangular coils. With
small size transformers less than 5KVA, square core can be used, with circular coils. With medium and large
transformers, cruciform (two stepped) or more stepped cores are used. Stepped core utilizes the iron space
better.
π‘¨π’Š =
𝑬𝒕
𝟐. πŸ“πŸ‘
=
= 𝟎. πŸŽπŸπŸŽπŸ‘πŸ”πŸπŸ‘πŸ’ π’ŽπŸ = πŸπŸŽπŸ‘. πŸ” π’„π’ŽπŸ
πŸ’. πŸ’πŸ’ 𝒇 π‘©π’Ž πŸ’. πŸ’πŸ’ × πŸ“πŸŽ × πŸ. 𝟏
The relation between circumscribe circle diameter “d” and core cross section “Ai” can be
expressed as follows π‘¨π’Š = π’Œ π’…πŸ The value of constant k depends upon the number of
steps in core of the transformer, it is given as
Core type
Square
Cruciform
3 stepped
4 stepped
5 stepped
6 stepped
Constant k
0.45
0.56
0.6
0.62
0.64
0.65
for cruciform core circumscribing circle diameter
𝒅=
π‘¨π’Š
π’Œ
=
πŸπŸŽπŸ‘.πŸ”
𝟎.πŸ“πŸ”
= πŸπŸ‘. πŸ” π’„π’Ž
By increasing the number of steps, the area of
circumscribing circle is more effectively utilized.
The most economical dimensions of various steps
for a multi-stepped core can be calculated.
for two step size of core :
𝒂 = 𝟎. πŸ–πŸ“ 𝒅 = 𝟎. πŸ–πŸ“ × πŸπŸ‘. πŸ” = 𝟏𝟏. πŸ“πŸ” ≈ 𝟏𝟏. πŸ” π’„π’Ž
𝒂𝒏𝒅 𝒃 = 𝟎. πŸ“πŸ‘π’… = 𝟎. πŸ“πŸ‘ × πŸπŸ‘. πŸ” = πŸ•. 𝟐 π’„π’Ž
We can find out window area from output equation given as
𝑸 = 𝟐. 𝟐𝟐 𝒇 π‘©π’Ž 𝑨𝑰 𝑲𝑾 𝑨𝑾 𝜹 × πŸπŸŽ−πŸ‘
So
𝑸×πŸπŸŽπŸ‘
π’Ž 𝑨𝑰 𝑲𝑾 𝜹
𝑨𝑾 = 𝟐.𝟐𝟐 𝒇 𝑩
𝟏𝟎𝟎𝟎𝟎
= 𝟐.𝟐𝟐×πŸ“πŸŽ×𝟏.𝟏×𝟎.πŸŽπŸπŸŽπŸ‘πŸ”×𝟎.πŸπŸ“×𝟐.𝟐×πŸπŸŽπŸ” = 𝟎. πŸŽπŸπŸ’πŸ‘πŸ– π’ŽπŸ = πŸπŸ’πŸ’ π’„π’ŽπŸ
Check πŸπŸŽπŸ” figure in current density value
Window area is determined as product of width and height of the window.
The height of the window, is usually taken 2.5 to 4 times of width of window.
𝑯𝑾
𝑾𝑾
= 𝟐. πŸ“ 𝒕𝒐 πŸ’
When height of window is taken high the mean turn length will decrease but leakage flux will increase.
Similarly, less height and more width of window will increase the mean turn length but decrease
leakage flux by some amount. Let us take height of window three times of width of window.
Thus, from the above equations, the height of the window & the width of the window can be calculated.
𝑨𝑾 = πŸπŸ’πŸ’ π’„π’ŽπŸ = π»π‘Š × π‘Šπ‘Š = 3 π‘Šπ‘Š × π‘Šπ‘Š
or
π‘Šπ‘Š =
𝑨𝑾
3
=
144
3
= 6.93 ≈ 6.9 π‘π‘š
so height of window π»π‘Š = 3 π‘Šπ‘Š = 3 × 6.93 = 20.8 π‘π‘š
One significant dimension in
core is center to center limb
distance
𝐷 = π‘Šπ‘€ + 𝑑 = 6.9 +
13.6 = 20.5 π‘π‘š
Overall width of core
π‘Š = 𝐷 + π‘Ž = 20.5 +
11.6 = 32.1 π‘π‘š
Overall height of core
𝐻 = π»π‘Š + 2 π‘Ž
= 20.8 + 2 × 11.6 = 44 π‘π‘š
The transformer core frame
with dimensions is as
Winding design
Low voltage winding
No of turns
𝑁𝐿𝑉 =
Current in LV
𝐼𝐿𝑉 =
𝑉
𝐸𝑑
𝑄
𝑉
=
=
230
2.53
= 90.9 ≈ 91 π‘‘π‘’π‘Ÿπ‘›π‘ 
10000
230
= 43.48 𝐴
𝐼𝐿𝑉
𝛿
Cross section area of conductor π‘ŽπΏπ‘‰ =
=
43.48
2.2
= 19.763 π‘šπ‘š2
We will select rectangle conductor of next higher standard size from
conductor table, one can also use foil winding.
High voltage winding
No of turns
𝑁𝐻𝑉 = 𝑁𝐿𝑉
4352 π‘‘π‘’π‘Ÿπ‘›π‘ 
Current in LV
𝐼𝐻𝑉 =
𝑄
𝑉
=
𝑉𝐻𝑉
𝑉𝐿𝑉
= 91 ×
10000
11000
11000
230
= 4352.17 ≈
= 0.909 𝐴
Cross section area of conductor π‘Žπ»π‘‰ =
𝐼𝐻𝑉
𝛿
=
0.909
2.2
= 0.4132 π‘šπ‘š2
We will select round conductor of next higher standard size from
conductor table.
Design magnetic frame and winding for a 250 KVA, 11000/230 V, 50Hz, three phase delta/star core type distribution
transformer. Assume flux density 1.2 T, current density 2.5 A/mm2, window space factor=0.28, constant K=0.45 for
three phase core type distribution transformers. Use three or four stepped cores.
Given Data :
kVA=250,
Bm=1.2 T d=2.5 A/mm2
f = 50 Hz,
VHV=11000 V delta VLV=230 V star
Hence emf per turn
𝑬𝒕 = 𝑲 𝑸 = 𝟎. πŸ’πŸ“
The net core cross section area Ai is given as
π‘¨π’Š =
πŸπŸ“πŸŽ
πŸ‘
Kw=0.28 K=0.45
= πŸ’. 𝟏 volt/turn
𝑬𝒕
πŸ’.πŸ’πŸ’ 𝒇 π‘©π’Ž
Bm – maximum flux density in the core of the transformer in Wb/ m2
f – Frequency of the power supply in Hz
With very small size transformers, rectangular core can be used with either circular or rectangular coils. With small size
transformers less than 5KVA, square core can be used, with circular coils. With medium and large transformers,
cruciform (two stepped) or more stepped cores are used. Stepped core utilizes the iron space better.
π‘¨π’Š =
𝑬𝒕
πŸ’. 𝟏
=
= 𝟎. πŸŽπŸπŸ“πŸ’ π’ŽπŸ = πŸπŸ“πŸ’ π’„π’ŽπŸ
πŸ’. πŸ’πŸ’ 𝒇 π‘©π’Ž πŸ’. πŸ’πŸ’ × πŸ“πŸŽ × πŸ. 𝟐
The relation between circumscribe circle diameter “d” and core cross section “Ai” can be expressed as follows
π‘¨π’Š = π’Œ π’…πŸ
The value of constant k depends upon the number of steps of core of the transformer, given as
Square
0.45
Core type
Constant k
Cruciform
0.56
3 stepped
0.6
4 stepped
0.62
for four stepped core circumscribing circle diameter
𝒅=
π‘¨π’Š
π’Œ
=
πŸπŸ“πŸ’
𝟎.πŸ”πŸ
= πŸπŸ“. πŸ•πŸ” π’„π’Ž
By increasing the number of steps, the area of
circumscribing circle is more effectively utilized. The most
economical dimensions of various steps for a four-stepped
core can be calculated. The results are
a = 𝟎. πŸ—πŸ‘ 𝒅 = 𝟎. πŸ—πŸ‘ × πŸπŸ“. πŸ•πŸ” = πŸπŸ’. πŸ”πŸ” ≈ πŸπŸ’. πŸ• π’„π’Ž
𝒃 = 𝟎. πŸ– 𝒅 = 𝟎. πŸ– × πŸπŸ“. πŸ•πŸ” = 𝟏𝟐. πŸ” π’„π’Ž
𝒄 = 𝟎. πŸ” 𝒅 = 𝟎. πŸ” × πŸπŸ“. πŸ•πŸ” = πŸ—. πŸ’ π’„π’Ž
𝒂𝒏𝒅 𝒅 = 𝟎. πŸ‘πŸ” 𝒅 = 𝟎. πŸ‘πŸ” × πŸπŸ“. πŸ•πŸ” = πŸ“. πŸ• π’„π’Ž
5 stepped
0.64
6 stepped
0.65
We can find out window area from output equation given as
𝑸 = πŸ‘. πŸ‘πŸ‘ 𝒇 π‘©π’Ž 𝑨𝑰 𝑲𝑾 𝑨𝑾 𝜹 × πŸπŸŽ−πŸ‘
So
𝑸×πŸπŸŽπŸ‘
π’Ž 𝑨𝑰 𝑲𝑾 𝜹
𝑨𝑾 = πŸ‘.πŸ‘πŸ‘ 𝒇 𝑩
πŸπŸ“πŸŽπŸŽπŸŽπŸŽ
= πŸ‘.πŸ‘πŸ‘ ×πŸ“πŸŽ×𝟏.𝟐×𝟎.πŸŽπŸπŸ“πŸ’×𝟎.πŸπŸ–×𝟐.πŸ“×πŸπŸŽπŸ” = 𝟎. πŸπŸπŸ”πŸ π’ŽπŸ = πŸπŸπŸ”πŸ π’„π’ŽπŸ
Check πŸπŸŽπŸ” figure in current density value
Window area is determined as product of width and height of the window.
The height of the window, is usually taken 2 to 4 times of width of window.
π»π‘Š
π‘Šπ‘Š
= 2 π‘‘π‘œ 4
When height of window is taken high the mean turn length will decrease but leakage flux will increase.
Similarly, less height and more width of window will increase the mean turn length but decrease
leakage flux by some amount. Let us take height of window 2.5 times of width of window.
Thus, from the above equations, the height of the window & the width of the window can be calculated.
𝑨𝑾 = πŸπŸπŸ”πŸ π’„π’ŽπŸ = π»π‘Š × π‘Šπ‘Š = 2.5 π‘Šπ‘Š × π‘Šπ‘Š
or
π‘Šπ‘Š =
𝑨𝑾
2.5
so height of window
=
1161
2.5
= 21.55 ≈ 21.6 π‘π‘š
π»π‘Š = 2.5 π‘Šπ‘Š = 2.5 × 21.55 = 53.9 π‘π‘š
One significant dimension in core is
center to center limb distance
𝐷 = π‘Šπ‘€ + 𝑑 = 21.6 + 15.8 =
37.4 π‘π‘š
Overall width of core
π‘Š = 2𝐷 + π‘Ž = 2 × 37.4 + 14.7
= 89.5 π‘π‘š
Overall height of core
𝐻 = π»π‘Š + 2 π‘Ž = 53.9 + 2 × 14.7
= 83.3 π‘π‘š
The transformer core frame with
dimensions is as
Winding design: Low voltage winding 400 V star connection
𝑽𝑷𝒉 =
𝑽𝑳
πŸ‘
=
No of turns
𝑁𝐿𝑉 =
Current in LV
𝐼𝐿𝑉 =
πŸ’πŸŽπŸŽ
πŸ‘
𝑉
𝐸𝑑
= πŸπŸ‘πŸ. 𝟐 𝑽
=
𝑄
3π‘‰π‘ƒβ„Ž
231.2
4.1
=
= 56.39 ≈ 56 π‘‘π‘’π‘Ÿπ‘›π‘ 
250000
= 360.44 𝐴
3×231.2
Cross section area of conductor π‘ŽπΏπ‘‰ =
𝐼𝐿𝑉
𝛿
=
360.44
2.5
= 144.176 π‘šπ‘š2
We will select rectangle conductor of next higher standard size from
conductor table, one can also use foil winding.
High voltage winding 11KV Delta π‘‰π‘ƒβ„Ž = 𝑉𝐿 = 11000 𝑉
No of turns 𝑡𝑯𝑽 = 𝑡𝑳𝑽
Current in LV
𝑽𝑯𝑽
𝑽𝑳𝑽
𝐼𝐻𝑉 =
= πŸ“πŸ” ×
𝑄
3π‘‰π‘ƒβ„Ž
=
𝟏𝟏𝟎𝟎𝟎
πŸπŸ‘πŸ.𝟐
250000
3×11000
Cross section area of conductor π‘Žπ»π‘‰ =
= πŸπŸ”πŸ”πŸ’. πŸ‘πŸ“ ≈ πŸπŸ”πŸ”πŸ’ 𝒕𝒖𝒓𝒏𝒔
= 7.5757 𝐴
𝐼𝐻𝑉
𝛿
=
7.5757
2.5
= 3.03 π‘šπ‘š2
We will select round conductor of next higher standard size from
conductor table.
Download