A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

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A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.001 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 5 m/s. Assume the thermophysical properties of the exhaust gas are those of air. (a) Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube. (b) Estimate the heat transfer coefficient for the air flowing across the outside of the tube. (c) Estimate the overall heat transfer coefficient U and the temperature of the exhaust gas when it reaches the laboratory.

Engineering

1 answer:

Kisachek [356] · Answer 1 · 2026-04-02T20:33:44+03:00

The mean temperature is calculated as Tmean = (Ti + T∞)/2, resulting in Tmean = 107.5⁰C. Furthermore, converting this to Kelvin gives Tmean = 107.5 + 273 = 380.5K. The corresponding properties of air at this average temperature are: v = 24.2689 × 10⁻⁶m²/s and α = 35.024 × 10⁻⁶m²/s, with viscosity = 221.6 × 10⁻⁷N.s/m² and thermal conductivity = 0.0323 W/m.K. The specific heat capacity is Cp = 1012 J/kg.K. The Prandtl number is found using Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶, equating to 0.693. To find the Reynolds number, we utilize the formula Pv = 4m/πD², and apply it: Re = (Pv * D). Substituting gives us Re = 4m/(πD), simplifying to Re = (4 x 0.003)/(π × 6 ×10⁻³ × 221.6 × 10⁻⁷) = 28728.3. Since Re is greater than 2000, the flow is classified as turbulent. For turbulent flow, we employ the Dittus - Doeltr correlation with n = 0.03, expressed as Nu = 0.023Re⁰⁸Pr⁰³, leading to the equation: (h₁D)/k = 0.023(28728.3)⁰⁸(0.693)⁰³. Resolving this gives us: (h₁ × 0.006)/0.0323 = 75.962. Thus, h₁ = (75.962 × 0.0323)/0.006, resulting in h₁ = 408.93 W/m².K.