Question

A fluid $\left(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, \mu=1.4 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right.\(, \)c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\(, and \)k=0.58 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( ) flows with an average velocity of \)0.3 \mathrm{~m} / \mathrm{s}$ through a \(14-\mathrm{m}\)-long tube with inside diameter of $0.01 \mathrm{~m}\(. Heat is uniformly added to the entire tube at the rate of \)1500 \mathrm{~W} / \mathrm{m}^{2}\(. Determine \)(a)$ the value of convection heat transfer coefficient at the exit, \((b)\) the value of \(T_{s}-T_{\text {m }}\), and (c) the value of \(T_{e}-T_{i}\).

Short answer

a) 580.69 W/m²⋅K

Step-by-step solution

(1) Calculate the Reynolds number

The formula for Reynolds number (Re) is: Re = (ρ * V_avg * D) / μ Given values: ρ = 1000 kg/m³ V_avg = 0.3 m/s D = 0.01 m μ = 1.4 * 10^(-3) kg/m⋅s Substituting given values into the formula, we can find the Reynolds number as follows: Re = (1000 * 0.3 * 0.01) / (1.4 * 10^(-3)) Re = 2142.857

(2) Calculate the Nusselt number using Dittus-Boelter correlation

The Dittus-Boelter correlation for turbulent flow in a heating pipe: Nu = 0.023 * Re^(4/5) * Pr^(1/3) We need to calculate the Prandtl number (Pr): Pr = cp * μ / k Given values: cp = 4.2 kJ/kg⋅K = 4200 J/kg⋅K k = 0.58 W/m⋅K Calculating Prandtl number (Pr): Pr = (4200 * 1.4 * 10^(-3)) / 0.58 Pr = 10.086 Now, we can calculate the Nusselt number using the Dittus-Boelter correlation: Nu = 0.023 * (2142.857)^(4/5) * (10.086)^(1/3) Nu = 100.119

(3) Calculate the convection heat transfer coefficient (h) at the exit

The formula for the convection heat transfer coefficient (h) is: h = Nu * k / D Substitute the calculated values of Nu, k, and D: h = 100.119 * 0.58 / 0.01 h = 580.69 W/m²⋅K

(4) Calculate the value of Ts - Tm

The formula for q" (heat flux) is: q" = h * (Ts - Tm) Therefore, we can find Ts - Tm as follows: Ts - Tm = q" / h Given values: q" = 1500 W/m² Substitute the values: Ts - Tm = 1500 / 580.69 Ts - Tm = 2.583 K

(5) Calculate the value of Te - Ti

We need to determine the total heat transfer (Q) along the tube, which can be calculated by multiplying the heat transfer rate by the tube surface area: Q = q" * π * D * L Given values: L = 14 m Substitute the values: Q = 1500 * π * 0.01 * 14 Q = 659.73 W Now, we can calculate the temperature difference for the fluid (Te - Ti) using the formula: ΔT = Q / (ρ * V_avg * A * cp) Where A is the cross-sectional area of the tube: A = π * (D/2)^2 Substitute the values: A = π * (0.01/2)^2 A = 7.854 * 10^(-5) m² Now, calculate the temperature difference: Te - Ti = 659.73 / (1000 * 0.3 * 7.854 * 10^(-5) * 4200) Te - Ti = 4.215 K #Summary# (a) The convection heat transfer coefficient at the exit (h_e) is 580.69 W/m²⋅K. (b) The value of Ts - Tm is 2.583 K. (c) The value of Te - Ti is 4.215 K.