## 2 Theoretical Background
### Stray Losses  
Stray loss at tank walls accounts for approximately 70-80% of the total stray loss in large power transformers. Eddy currents cause power loss in the transformer tank and are caused by leakage flux in the winding. The power loss in the transformer tank is known as stray loss and can be calculated using the Poynting vector. Stray loss is calculated using [PERSON]'s equations [34; 35].  
\[\
abla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial\mathbf{t}} \tag{3}\]  
\[\
abla\times\mathbf{H}=\mathbf{J} \tag{4}\]  
\[\
abla\cdot\mathbf{B}=0 \tag{5}\]  
The two fundamental relationships between the magnetic flux density and the magnetic field strength and between the current density and the electric field are  
\[\mathbf{B}=\mathrm{\mu}\mathbf{H} \tag{6}\]  
\[\mathbf{J}=\mathrm{\sigma}\mathbf{E} \tag{7}\]  
where \(\mathbf{H}\) represents the strength of the magnetic field in joules per meter, \(\mathbf{E}\) represents the strength of the electric field in volts per meter, \(\mathbf{B}\) represents the magnetic flux density in watts per square meter, and \(\mathbf{J}\) represents the current density in amps per square meter.  
Equation (4) can be simplified with vector algebra by taking the curl of both sides.  
\[\
abla(\
abla\cdot\mathbf{H})-\
abla^{2}\mathbf{H}=\
abla\times\mathbf{J} \tag{8}\]Using Equations (3) and (5)-(7), Equation (8) is expressed as follows:  
\[\
abla^{2}\mathbf{H}-\mu\sigma\frac{\partial\mathbf{H}}{\partial\mathbf{t}}=0 \tag{9}\]  
A structural component is assumed as shown in Figure 1. The magnetic field strength \(\mathrm{Hy}\) and current density \(\mathrm{J}\mathbf{x}\) are considered as functions of z. The complex permeability can be written as Equation (9) for this problem.  
\[\frac{\mathrm{d}^{2}\mathrm{H}_{\mathrm{y}}}{\mathrm{d}z^{2}}=\mathrm{j}\omega \sigma\mu\mathrm{H}_{\mathrm{y}} \tag{10}\]  
\[\mathrm{H}_{\mathrm{y}}=\mathrm{H}_{\mathrm{a}}\text{ for }\mathrm{z}=0 \text{ \ \ \ }\mathrm{H}_{\mathrm{y}}=0\text{ for }\mathrm{z}=\infty \tag{11}\]  
By simplifying Equation (10) from the above conditions,  
\[\mathrm{H}_{\mathrm{y}}=\mathrm{H}_{\mathrm{a}}\mathrm{e}^{-\mathrm{m}z} \tag{12}\]  
\[\mathrm{m}=\sqrt{\mathrm{j}\omega\sigma\mu}=(1+\mathrm{j})\sqrt{\frac{\omega \sigma\mu}{2}} \tag{13}\]  
where \(\mathrm{H}_{\mathrm{a}}\) is a complex number, and \(\mathrm{m}\) is the complex propagation constant that represents propagation.  
By solving Equations (10) and (11),  
\[\mathrm{H}_{\mathrm{y}}=\mathrm{H}_{\mathrm{a}}\mathrm{e}^{-\frac{(1+\mathrm{ j})z}{\sigma}} \tag{14}\]  
Equation (14) is substituted into Equation (4) and simplified using vector algebra.  
\[\mathbf{J}_{\mathbf{x}}=\frac{(1+\mathrm{j})}{\delta}\mathrm{e}^{-\frac{(1+ \mathrm{j})z}{\delta}} \tag{15}\]  
The time-averaged intensity of stray loss in the transformer tank can be obtained by calculating the real part of the complex Poynting vector on the surface.  
\[\mathrm{P}=\frac{1}{2}\mathrm{Re}(\mathbf{E}\times\mathbf{H}^{*}) \tag{16}\]  
On the surface (z = 0), Equations (14) and (15) are substituted into Equation (16). The stray loss per unit surface area is  
\[\mathrm{P}=\mathrm{Re}(\frac{\mathrm{H}_{\mathrm{a}}}{\sigma\delta}(1+\mathrm{ j})\mathrm{H}_{\mathrm{a}}\mathrm{z})=\frac{1}{2}\mathrm{Re}\Big{(}Z_{\mathrm{S}} \mathrm{H}_{\mathrm{a}}^{2}\Big{)}\mathrm{z} \tag{17}\]  
\[\mathrm{P}=\frac{1}{2}\frac{\mathrm{H}_{\mathrm{a}}^{2}}{\sigma\delta}=\sqrt{ \frac{\omega\mu}{2\sigma}}\frac{\mathrm{H}_{\mathrm{a}}^{2}}{2} \tag{18}\]  
Figure 1: Indicative representation of the dimensions of the structure that is considered for the stray loss.  
Therefore, the transformer tank's total power loss is  
\[\mathrm{P}=\sqrt{\frac{\omega\upmu}{8\sigma}}\int_{\mathrm{surface}}\mathrm{H}_{ \mathrm{a}}^{2}\,\mathrm{ds} \tag{19}\]