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.

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yemmy yemmy · Answer 1 · 2020-04-16T07:53:41+02:00

Answer:

Explanation:

Mean temperature is given by

[tex]T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}[/tex]

Tmean = (Ti + T∞)/2

[tex]T_mean = 107.5^{0}[/tex]

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

[tex]\mu[/tex] = 221.6 × 10⁻⁷N.s/m²

[tex]\kappa[/tex] = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

= 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/[tex]\mu[/tex]

Substituting for Pv

Re = 4m/(πD[tex]\mu[/tex])

= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

= 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K