DESIGN OF TRANSFORMERS

1. EE6604
DESIGNOF ELECTRICAL MACHINES
ELECTRICAL AND ELECTRONICS ENGINEERING
SRI KRISHNA KUMAR S
HTS953
AP/EEE

2. UNITIII
TRANSFORMERS
Output Equations
1 2
11
Main Dimensions
kVA output for single and three phase
transformers
1
Window space factor
1 2Design of core and winding
Overall dimensions
Operating characteristics
1
No load current
Temperature rise in Transformers
1
1Design of Tank
Methods of cooling of Transformers. 1

3. 3.7 Ratio of Iron loss to copper loss

4. 3.8 No-load current of a transformer
The phasor sum of the magnetizing current (Im) and the loss component of current (I1) ; Im is
calculated using the MMF/m required for the core and yoke and their respective length of flux
path. I1 is determined using the iron loss curve of the material used for the core and yoke and
the flux density employed and their weight.
The no-load current I is the vectorial sum of the magnetizing current Im and core loss or
working component current Ic. [Function of Im is to produce flux fm in the magnetic circuit and
the function of Ic is to satisfy the no load losses of the transformer].

5. 3.8 No-load current of a transformer
Since the copper loss under no load condition is almost negligible, the no load losses can
entirely be taken as due to core loss only. Thus the core loss component of the no load current
The magnetic circuit of a transformer consists of both iron and air path. The iron path is due to
legs and yokes and air path is due to the unavoidable joints created by the core composed of
different shaped stampings. If all the joints are assumed to be equivalent to an air gap of l , then
the total ampere turns for the transformer magnetic circuit is equal to ATfor iron + 800,000lg Bg .

6. 3.9 Temperature rise in Transformers
Losses dissipated in transformers in the core and windings get converted into thermal energy
and cause heating of the corresponding transformer parts. The heat dissipation occurs as
follows:
i) from the internal heated parts to the outer surface in contact with oil by conduction
ii) from oil to the tank walls by convection and
iii) from the walls of the tank to the atmosphere by radiation and convection.

7. 3.9 Temperature rise in Transformers
The specific heat dissipation due to convection of oil
= λconv = 40.3 (θ /H) ¼ W/m2- oC ; = temp difference of the surface relative to the oil and
H = height of the dissipating surface.
Experimentally found that a plain tank surface dissipates 6.0 W/m2-oC by radiation and 6.5 W/m2-oc by
convection (for a temp rise of 40 oC above an ambient temp of 20 oc).
Thus a total of 12.5 W/m2- oC is taken.
The temp rise θ = total loss/ (λ St) = (Pi + Pc) / (λ St)
Where St = heat dissipating surface area of the tank.
For small transformers , plain walled tank is enough to dissipate the losses. As the rating of the
transformer increases, the volume increases as he cube of the linier dimensions but the heat dissipating
surface area increases only as square of the linesr dimensions. So above certain rating, plain tank becomes
inadequate to dissipate losses and the area is increased by providing tubes. For larger ratings forced air
cooling is used.
3.10 DESIGN OF TANK

8. If tubing is provided, the oil circulation is improved due to the head of the oil, and this causes
an additional dissipation by convection of about 35 % .
Let x St be the area of the cooling tubes. Then
Loss dissipated by the tank surface = 12.5 St W/0C
Loss dissipated by the tubes (1.35 x 6.5) x St W/0 C = 8.8 x St W/0 C
Total loss dissipated by the tank and oil tubes
= (12.5 St + 8.8 x St ) W/0C
Hence θ = ( Pi + Pc)/ (12.5 St + 8.8 x St )
Total tube area x St
= (1/8.8) [ {(Pi + Pc)/ θ} – 12.5 St ]
The number of tubes = nt = Total tube area /(π dt lt )
The arrangement of the tubes on tank side walls should be made uniformly with a spacing of
usually 75 mm. Examples of calculation of nt and the arrangement of the tubes should be studied.
3.10 DESIGN OF TANK

9. 3.10 DESIGN OF TANK