Distribution transformer efficiency and number of tubes

Why maximum efficiency at less than full load

A distribution transformer stays energised for 24 hours but its load changes through the day and is often light (night hours). So the important figure is all-day (energy) efficiency:

$$\eta_{all\ day} = \frac{\text{kWh output in 24 h}}{\text{kWh output} + \text{kWh losses in 24 h}}$$

• Iron loss $P_i$ occurs for all 24 hours, whatever the load.
• Copper loss $P_c$ occurs only in proportion to (load)², and is large only for a few peak hours.

Maximum efficiency occurs when copper loss = iron loss, i.e. at load fraction $x = \sqrt{P_i/P_{c,FL}}$. By keeping $P_i$ small (lower $B_m$, good-quality steel) the maximum efficiency is placed at about 50–70% of full load, where the transformer runs most of the time. This minimises total energy loss in a day.

Number of cooling tubes

Let

• $P_i + P_c$ = total full-load loss (W)
• $S_t$ = dissipating surface of plain tank walls (m²)
• $\theta$ = permissible mean temperature rise of oil (°C)

A plain tank wall dissipates by radiation ≈ 6 W/m²°C and by convection ≈ 6.5 W/m²°C, total 12.5 W/m²°C:

$$\theta = \frac{P_i + P_c}{12.5\,S_t}$$

If this $\theta$ is too high, tubes are added. Let the tube area be $xS_t$. Tubes do not add to radiation much (they screen each other), but they improve convection by about 35%:

$$6.5 \times 1.35 \approx 8.8\ \text{W/m}^2{}^\circ\text{C}$$

Heat dissipated with tubes:

$$\begin{aligned} P_i + P_c &= \theta\left[12.5\,S_t + 8.8\,xS_t\right]\\ xS_t &= \frac{1}{8.8}\left(\frac{P_i + P_c}{\theta} - 12.5\,S_t\right) \end{aligned}$$

If each tube has diameter $d_t$ and length $l_t$, its surface is $\pi d_tl_t$, so

$$n_t = \frac{xS_t}{\pi d_tl_t} = \frac{1}{8.8\,\pi d_tl_t}\left(\frac{P_i + P_c}{\theta} - 12.5\,S_t\right)$$

Round $n_t$ up to the next whole number. Usually $\theta$ = 35 °C (mean oil rise), tube diameter 50 mm and spacing 75 mm.