Very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitrogen. The top surface of the ground is flat and has a uniform temperature. The pipe wall has a very low thermal conductivity of 0.0035 Wm-1K-1 and receives a cooling power of 10 W/m to keep the liquid nitrogen at 77 K. If the thermal conductivity of the ground is 1 Wm-1K-1, what is the surface temperature of the ground

Therefore, the surface temperature of the ground is 352.2 K

Explanation:

Recall that

r₁ = 0.05 m

r₂ =0.08 m

T₁ =77K

Ki = 0.0035 Wm-1K-1

Kg = Wm-1K-1

z = 2m

φ =10 W/m

Now,

The heat transfer will occur in two resistance

Ln (r₂/r₁)/ 2π.kI =  z/K * A

so,

φ =  - (Ti - Tg)/ln(r₂/r₁)/2πr.kI + z/kg * 2r₂* I

which is

10 = (77-Tg/ ln(0.08/0.05)/ 2π * 0.0035 + 2/1*2*0.16

10 = Δ T/2.137 + 6.25

So,

276.2 = (Tg - 77)

Therefore Tg = 353.2K

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explanation:

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