Question

An iron bar (L₁ = 0.1 m, A₁ = 0.02 m², K₁ = 79 wm^-1 K^-1). and a brass bar (L₂ = 0.1 m, A₂ = 0.02 m², K₂ = 109 Wm^-1 K^-1) are soldered end to end as shown in the figure below. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K    respectively. Computer (a) the heat Current through the compound bar, (b) the equivalent thermal conductivity of the compound bar, (c) the temperature of the junction of the two bars.

Answer 1

Given,

L = L₂ = L = 0.1 m.

A₁ = A₂ = A = 0.02 m²

K₁ = 79 Wm^-1 K^-1

K₂ = 109 Wm^-1 K^-1

T₁ = 373 K and T₂ = 27

3 K.

Under the steady state condition the heat current (H₁) through iron bar is equal to the heat current (H₂) through brass bar.

So, H = H₁ = H₂

⇒ (K₁A₁(T₁ - T₀)) / L₁ = (K₂A₂(T₀ - T₂)) / L₂

where the junction temperature is

T₀ = (K₁A₁ + K₂T₂ / K₁ + K₂)

Since, A₁ = A₂ = A and  L₁ = L₂ = L, we have

If K is the equivalent thermal conductivity of the compound bar, then by using equation. we have