A thin-walled tube with a diameter of 12 mm and length of 25 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 0.006 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 2.5 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

(a)  h₁   = 204.45 W/m²k

(b) h₀ = 46.80 W/m².k

(c) T = T = 15.50°C

Explanation:

Given Data;

Diameter = 12mm

Length = 25 m

Entry temperature = 200°C

Flow rate = 0.006 kg/s

velocity = 2.5 m/s.

Step 1: Calculating the mean temperature;

(200 + 15)/2

Mean temperature = 107.5°C = 380.5 K

The properties of air at mean temperature 380.5 K are given as:

v = 24.2689*10⁻⁶m²/s

a = 35.024*10⁻⁶m²/s

μ    = 221.6 *10⁻⁷Ns/m²

k = 0.0323 W/m.k

Cp = 1012 J/kg.k

Step 2: Calculating the prantl number using the formula;

Pr = v/a

   = 24.2689*10⁻⁶/ 35.024*10⁻⁶

   = 0.693

Step3: Calculating the reynolds number using the formula;

Re = 4m/πDμ

    = 4 *0.006/π*12*10⁻³ * 221.6 *10⁻⁷

    = 0.024/8.355*10⁻⁷

    = 28725

Since Re is greater than 2000, the flow is turbulent. Nu becomes;

Nu = 0.023Re^0.8 *Pr^0.3

Nu = 0.023 * 28725^0.8 * 0.693^0.3

     = 75.955

(a) calculating the heat transfer coefficient:

Nu = hD/k

h = Nu *k/D

  = (75.955 * 0.0323)/12*10^-3

h   = 204.45 W/m²k

(b)

Properties of air at 15°C

v = 14.82 *10⁻⁶m²/s

k = 0.0253 W/m.k

a = 20.873 *10⁻⁶m²/s

Pr(outside) = v/a

                  = 14.82 *10⁻⁶/20.873 *10⁻⁶

                 = 0.71

Re(outside) = VD/v

                   = 2.5 * 12*10⁻³/14.82*10⁻⁶

                    =2024.29

Using Zakauskus correlation,

Nu = 0.26Re^0.6 * Pr^0.37 * (Pr(outside)/Pr)^1/4

    = 0.26 * 2024.29^0.6 *  0.71^0.37 * (0.71/0.693)^1/4

    = 22.199

Nu = h₀D/k

h₀ = Nu*k/D

     = 22.199* 0.0253/12*10⁻³

h₀ = 46.80 W/m².k

 (c)

Calculating the overall heat transfer coefficient using the formula;

1/U =1/h₁ +1/h₀

1/U = 1/204.45 + 1/46.80

1/U = 0.026259

U = 1/0.026259

U = 38.08

Calculating the temperature of the exhaust using the formula;

T -T₀/T₁-T₀ = e^-[uπDL/Cpm]

T - 15/200-15 = e^-[38.08*π*12*10⁻³*25/1012*0.006]

T - 15/185 = e^-5.911

T -15 = 185 * 0.002709

T = 15+0.50

T = 15.50°C

jesus loves you

explanation: