Arun
Last Activity: 6 Years ago
50°C 45°C 35°C
Surrounding temperature is ‘T’°C Avg.
t = 50 + 45/2 = 47.5 Avg. temp.
Variation from surrounding T = 47.5 – T
Degree of drop in temp = 50 – 45/5 = 1 °C/mm
By Network’s Law 1°C/mm = bA * t ⇒ bA = 1/t = 1/47.5 – T …(1)
In second case,
Avg, temp = 35 + 45/2 = 57.5 Avg. temp.
Variation from surrounding t’ = 57.5 – t
rate of fall of temp = 45 – 35/8 = 40.625 °C/mm
By Network’s Law ⇒ 5/B = 1/(47.5 - T) * (57.5 - T)
By Componendo & Dividendo method
We get,
T = 34.1°C
