Module 3. Dimensional analysis and heat transfer

Lesson 18

PROBLEMS ON FORCED CONVECTION

Example 18.1 Air flows inside a tube 60 mm in diameter (d) and 2.1 m long (l) at a velocity w = 5 m/sec. Find the heat-transfer coefficient α if the mean air temperature t_f = 100^oC.

At t_f = 100°C, λ_f = 0.0276 kcal/m-hr-°C; ν_f = 23.13 × 10^-6 m²/sec

Substituting these values in equation, we get

whence

 kcal/m-hr-°C

Since , it is necessary to introduce the correction factor ε_i ; we find = 1.04 (Table 17.2)

We finally have α = α’ = 16.2 × 1.04 = 16.8 kcal/m²-hr-°C.

Example 18.2 Water flows inside a tube 50 mm in diameter (d) and 3 m long (l) at a velocity ω = 0.8 m/sec. Determine the heat-transfer coefficient if the mean water temperature t_f = 50°C and wall temperature t_ω = 70°C

At t_f = 50°C, λ_f = 0.557 kcal/m²-hr-°C; ν = 5.56 × 10^-7 m²/sec and Pr_f = 3.54

At t_ω = 70°C, Pr_ω = 2.55

Substituting these values in formula, we get

From which

Since  = 60, the correction factor for the length of the tube = 1.

Example 18.3 Determine the mean heat-transfer coefficient and the quantity of heat transferred in water flowing in a horizontal tube 3 mm in diameter (d) and 2 m long (l), if ω = 0.3 m/sec; t_f = 60^°C and t_ω = 20°C.

We will solve the problem with the aid of formula

At t_f = 60°C, λ_f  = 0.567 kcal/m-hr-°C; ν_f  = 0.478 × 10^-6 m²/sec

β = 5.11 × 10-4 1/°C and Pr_f = 2.98.  At t_ω  = 20°C, Pr_𝜔 = 7.02

Employing the above values, we have

The temperature difference Δt = 60 – 20 = 40°C

We find

From which

Since the ratio it is not necessary to introduce the correction factor accounting for the length of the tube.

The quantity of transferred heat: