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Answer:
The steam will start to condense at 6.6 mm into the pipe
Explanation:
The volume flow rate =π×(50/1000)²/4×10 = 0.0196 m³/s
The specific volume of the steam = 1.769 m³/kg
Therefore;
The mass flow rate = 0.0196/1.769 = 0.011099 kg/s
The resistance of the insulation material = ln(0.075/0.05)/(2×π×0.075) = 0.860 K/W
The resistance of the outside film of the insulator = 1/(15×2×π×0.075×1) = 0.14147 K/W
The total resistance = 0.14147 + 0.860 = 1.00147 K/W
1/(UA) = 1.00147 K/W
A = 2×π×0.05×1
1/U = 0.3146
U = 3.178 W/m² K
We have;
T(x) = T₀ + (Tin - T₀) exp(-UπDx/mcp)
Therefore, when T(x) = 100°C, we have;
100 = 20 + (120 - 20)exp(-3.178×π×0.05x/(0.011099 × 1.33))
Solving, we get
x = 6.597× 10⁻³ m ≈ 6.6 mm
Therefore, the steam will start to condense at 10 mm into the pipe.
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