A mild steel tank with a wall thickness of 12 mm contains water at 85ºC, while the atmospheric temperature is 15ºC. The thermal conductivity of mild steel (\(\lambda\)) is 50 W/mK. The heat transfer coefficients for the inside and outside of the tank are 2500 W/m\(^2\)K and 14 W/m\(^2\)K, respectively.

Find:

(i) The heat transfer per unit area.

(ii) The temperature of the outside surface of the tank.

A is the surface area of the cylinder, and Q is the heat transferred. For calculating the heat transfer per unit area Q/A, we use the formula given above. Radius of the cylinder, Given data:

Mild steel tank of wall thickness = 12mm

Water temperature = 85°C

Atmospheric temperature = 15°C

Thermal conductivity of mild steel (λ) = 50 W/m

K Heat transfer coefficient inside tank (h1) = 2500 W/m

Heat transfer coefficient outside tank (h2) = 14 W/2K

Formula used:

Rate of heat transfer through the cylindrical wall per unit area = Q/A = kL (T1 - T2)/ln(R2/R1) Temperature distribution through cylindrical wall is given by the following formula:

Where T = temperature at any radius r,

R1 = inner radious of cylinder,

R2 = outer radius of cylinder,

r = radius at any from the center of the cylinder,

k = thermal conductivity of the cylinder,

L = length of cylinder Heat transfer coefficient

(h) = Q/(A * ΔT)

[tex](ln(R2/r) × (T1 - T2) × h2/2k) + T2T = (ln(0.018/0.006) × (85 - 15) × 14/2 × 50) + 15T = 50.88°C[/tex]

The temperature of the outside surface of the tank is 50.88°C.

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