A 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

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁ = 0.05m

r₂ = 0.08m

Tn = Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg = 1 Wm-1K-1

Z = 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ - T₁/ln (r2/r1) / 2πKi = 2πKi T0 - T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 - 77/ln 0.08/0.05

⇒ T₀ - 77 = 231.72

T₀ = 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 * 2 / 2 * 0.08 = 1.606m

The heat gained in the pipe is = S * Kg * (Tg - T₀)

(10 * 1) = 1.606 * 1 * (Tg - 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K