Condition for minimum cost transformer design

A transformer has minimum total cost of active material (iron + copper) when the cost of iron equals the cost of copper.

Derivation

Let

• $\phi_m$ = maximum flux, $AT$ = ampere-turns per limb
• $c_i$, $c_c$ = cost per kg of iron and copper
• $\rho_i$, $\rho_c$ = densities; $l_i$ = mean length of flux path; $L_{mt}$ = mean length of turn
• $B_m$, $\delta$ = flux density and current density (fixed by losses and heating)

For a single-phase transformer:

$$Q = 4.44f\phi_m\,(IT)\times10^{-3} = 4.44f\,\phi_m\,AT\times10^{-3}$$

so for a given rating the product $\phi_m \cdot AT$ is constant:

$$\phi_m\cdot AT = \frac{Q\times10^3}{4.44f} = \text{constant} = C$$

Weight of iron $G_i = \rho_il_iA_i = \rho_il_i\dfrac{\phi_m}{B_m}$.

Copper area (both windings) $= \dfrac{2AT}{\delta}$, so weight of copper $G_c = \rho_cL_{mt}\dfrac{2AT}{\delta}$.

Total cost:

$$C_t = c_iG_i + c_cG_c = \underbrace{\frac{c_i\rho_il_i}{B_m}}_{k_1}\phi_m + \underbrace{\frac{2c_c\rho_cL_{mt}}{\delta}}_{k_2}AT$$

Put $AT = C/\phi_m$:

$$C_t = k_1\phi_m + \frac{k_2C}{\phi_m}$$

For minimum cost:

$$\begin{aligned} \frac{dC_t}{d\phi_m} &= k_1 - \frac{k_2C}{\phi_m^2} = 0\\ k_1\phi_m &= \frac{k_2C}{\phi_m} = k_2\,AT \end{aligned}$$

that is,

$$c_iG_i = c_cG_c \quad\Longrightarrow\quad \text{cost of iron} = \text{cost of copper}$$

Optimum ratio of magnetic to electric loading

From $k_1\phi_m = k_2AT$:

$$r = \frac{\phi_m}{AT} = \frac{k_2}{k_1} = \frac{2c_c\rho_cL_{mt}B_m}{c_i\rho_il_i\delta}$$

This ratio fixes the voltage per turn, $E_t = K\sqrt{Q}$ with $K = \sqrt{4.44fr\times10^3}$.

Notes

• In the same way, minimum total loss (maximum efficiency) at a load occurs when iron loss = copper loss.
• In practice designers use the "minimum cost" ratio only as a guide, since loss capitalisation, regulation and cooling also matter.