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4 years ago

11

A 50 mm diameter thin walled pipe is covered with an insulation layer with thicknessof 25mm and thermal conductivity of0.075W/mK

. The inner pipe carries a superheated vapor at atmospheric pressure. The steam temperature entering the pipe is 120°Candthe air temperature is 20°C. The convection heat transfer coefficient on top of the insulation layer is 15W/m(K. If the velocity of the steam is 10 m/s, at what point along the pipe the steam starts to condense?

1 answer:

Wittaler [7]4 years ago

3 0

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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In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

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