Air at 207 kPa and 200◦C enters a 2.5-cm-ID tube at 6 m/s. The tube is constructed ofcopper with a thickness of 0.8 mm and a length of 3 m. Atmospheric air at 1 atm and 20◦Cflows normal to the outside of the tube with a free-stream velocity of 12 m/s. Calculate theair temperature at exit from the tube. What would be the effect of reducing the hot-air flowing half?

Temperature of air at exit = 24.32 C, After reducing hot air the temperature of the exit air becomes = 20.11 C

Explanation:

ρ = P/R (Ti) where ρ is the density of air at the entry, P is pressure of air at entrance, R is the gas constant, Ti is the temperature at entry

ρ = (2.07 x 10⁵) / (287) (473) = 1.525 kg/m³

Calculate the mass flow rate given by

m (flow rate) = (ρ x u (i) x A (i)) where u (i) is the speed of air, A (i) is the area of the tube (πr²) of the tube

m (flow rate) = 1.525 x (π x 0.0125²) x 6 = 4.491 x 10⁻³ kg/s

The Reynold's Number for the air inside the tube is given by

R (i) = (ρ x u (i) x d) / μ where d is the inner diameter of the tube and μ is the dynamic viscosity of air (found from the table at Temp = 473 K)

R (i) = (1.525) x (6) x 0.025/2.58 x 10⁻⁵ = 8866

Calculate the convection heat transfer Coefficient as

h (i) = (k/d) (R (i) ^0.8) (Pr^0.3) where k is the thermal conductivity constant known from table and Pr is the Prandtl's Number which can also be found from the table at Temperature = 473 K

h (i) = (0.0383/0.025) x (8866^0.8) x (0.681^0.3) = 1965.1 W/m². C

The fluid temperature is given by T (f) = (T (i) + T (o)) / 2 where T (i) is the temperature of entry and T (o) is the temperature of air at exit

T (f) = (200 + 20) / 2 = 110 C = 383 K

Now calculate the Reynold's Number and the Convection heat transfer Coefficient for the outside

R (o) = (μ∞ x do) / V (f) where μ∞ is the speed of the air outside, do is the outer diameter of the tube and V (f) is the kinematic viscosity which can be known from the table at temperature = 383 K

R (o) = (12 x 0.0266) / (25.15 x 10⁻⁶) = 12692

h (o) = K (f) / d (o) (0.193 x Ro^0.618) (∛Pr) where K (f) is the Thermal conductivity of air on the outside known from the table along with the Prandtl's Number (Pr) from the table at temperature = 383 K

h (o) = (0.0324/0.0266) x (0.193 x 12692^0.618) x (0.69^1/3) = 71.36 W/m². C

Calculate the overall heat transfer coefficient given by

U = 1/{ (1/h (i)) + A (i) / (A (o) x h (o)) } simplifying the equation we get

U = 1/{ (1/h (i) + (πd (i) L) / (πd (o) L) x h (o) } = 1/{ (1/h (i) + di / (d (o) x h (o)) }

U = 1/{ (1/1965.1) + 0.025 / (0.0266 x 71.36) } = 73.1 W/m². C

Find out the minimum capacity rate by

C (min) = m (flow rate) x C (a) where C (a) is the specific heat of air known from the table at temperature = 473 K

C (min) = (4.491 x 10⁻³) x (1030) = 4.626 W / C

hence the Number of Units Transferred may be calculated by

NTU = U x A (i) / C (min) = (73.1 x π x 0.025 x 3) / 4.626 = 3.723

Calculate the effectiveness of heat ex-changer using

∈ = 1 - е^ (-NTU) = 1 - e^ (-3.723) = 0.976

Use the following equation to find the exit temperature of the air

(Ti - Te) = ∈ (Ti - To) where Te is the exit temperature

(200 - Te) = (0.976) x (200 - 20)

Te = 24.32 C

The effect of reducing the hot air flow by half, we need to calculate a new value of Number of Units transferred followed by the new Effectiveness of heat ex-changer and finally the exit temperature under these new conditions.

Since the new NTU is half of the previous NTU we can say that

NTU (new) = 2 x NTU = 2 x 3.723 = 7.446

∈ (new) = 1 - e^ (-7.446) = 0.999

(200 - Te (new)) = (0.999) x (200 - 20)

Te (new) = 20.11 C