Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
