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Transformer heating and cooling time constants (25°C/40°C)

Short answer

Solve heating equation: a = e^(-1/τ) from 25=θm(1-a), 40=θm(1-a²) gives a=0.6, θm=62.5°C, τh=1.958h (117.5min). Cooling from rise 62.5°C to 15°C in 1.5h yields τc=1.051h (63.1min). Final steady temperature rise 62.5°C, heating time constant 1.958h, cooling time constant 1.051h.

The rise in temperature of a transformer after one hour and two hours of starting from cold conditions are 25°C and 40°C respectively. Determine its final steady temperature rise and the heating time constant. If its temperature falls from the final steady value to 45°C in 90 minutes when disconnected from the operation, determine its cooling time constant. The ambient temperature is 30°C.

Given

  • From cold: θ1=25\theta_1 = 25 °C after 1 h, θ2=40\theta_2 = 40 °C after 2 h
  • On disconnection, temperature falls from its final steady value to 45 °C in 90 min (1.5 h); ambient = 30 °C

Heating: final rise and heating time constant

θ=θm(1−e−t/τh)\theta = \theta_m(1 - e^{-t/\tau_h}). Let a=e−1/τha = e^{-1/\tau_h}:

25=θm(1−a),40=θm(1−a2)1+a=4025=1.6  ⇒  a=0.6θm=251−0.6=62.5 ∘Cτh=1ln⁡(1/0.6)=10.5108=1.958 h≈117.5 min\begin{aligned} 25 &= \theta_m(1 - a), \qquad 40 = \theta_m(1 - a^2) \\ 1 + a &= \frac{40}{25} = 1.6 \;\Rightarrow\; a = 0.6 \\ \theta_m &= \frac{25}{1 - 0.6} = 62.5\ ^\circ\text{C} \\ \tau_h &= \frac{1}{\ln(1/0.6)} = \frac{1}{0.5108} = 1.958\ \text{h} \approx 117.5\ \text{min} \end{aligned}

Cooling time constant

Final temperature = 30 + 62.5 = 92.5 °C (rise 62.5 °C). After 1.5 h the temperature is 45 °C, a rise of 45 − 30 = 15 °C.

15=62.5 e−1.5/τcτc=1.5ln⁡(62.5/15)=1.5ln⁡4.1667=1.51.4271=1.051 h≈63.1 min\begin{aligned} 15 &= 62.5\,e^{-1.5/\tau_c} \\ \tau_c &= \frac{1.5}{\ln(62.5/15)} = \frac{1.5}{\ln 4.1667} = \frac{1.5}{1.4271} = 1.051\ \text{h} \approx 63.1\ \text{min} \end{aligned}

Answer: final steady temperature rise = 62.5 °C; heating time constant = 1.958 h (117.5 min); cooling time constant = 1.051 h (63.1 min).

✓ Answer checked by Claude Sonnet 5.5

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