Consider a \(10-\mathrm{m}\)-long smooth rectangular tube, with \(a=50 \mathrm{~mm}\) and \(b=25 \mathrm{~mm}\), that is maintained at a constant surface temperature. Liquid water enters the tube at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.01 \mathrm{~kg} / \mathrm{s}\). Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\).

First, we need to determine the volume flow rate (Q) and the average velocity (V_avg) of water in the tube. To find Q, we use: Q = (mass flow rate) / (density of water) For water at 20°C, the density is approximately 998 kg/m³. Thus, Q = 0.01 kg/s ÷ 998 kg/m³ = 0.00001 m³/s Now, let's calculate the average velocity (V_avg). The cross-sectional area (A) of the rectangular tube is: A = a * b = (0.05 m)(0.025 m) = 0.00125 m² And the average velocity is: V_avg = Q / A = 0.00001 m³/s ÷ 0.00125 m² = 0.008 m/s

Next, we will calculate the Reynolds number (Re) to characterize the flow regime. The Reynolds number is given by: Re = (density * V_avg * hydraulic diameter) / dynamic viscosity The hydraulic diameter (D_h) for a rectangular tube is: D_h = 2ab/(a+b) = 2(0.05 m)(0.025 m) / (0.05 m + 0.025 m) = 0.0333 m For water at 20°C, the dynamic viscosity is approximately 0.001002 kg/m·s. Therefore, Re = (998 kg/m³)(0.008 m/s)(0.0333 m) / 0.001002 kg/m·s = 265.9 Since Re < 2000, the flow is considered laminar.

For laminar flow in a rectangular tube, the Nusselt number (Nu) is estimated using the following equation: Nu = 3.66 Next, we calculate the convective heat transfer coefficient (h) using: h = (Nu * thermal conductivity of water) / D_h For water at average temperature of 50°C, the thermal conductivity is approximately 0.617 W/m·K. Thus, h = (3.66)(0.617 W/m·K) / 0.0333 m = 67.6 W/m²·K

Now, let's find the heat transfer rate (Q_total) required to heat the water from 20°C to 80°C. We use: Q_total = (mass flow rate) * (specific heat of water) * (temperature difference) For water at 20°C, the specific heat is approximately 4182 J/kg·K. So, Q_total = (0.01 kg/s)(4182 J/kg·K)(80°C - 20°C) = 2510 J/s or 2510 W

Finally, we can find the tube surface temperature (T_s) using the following equation for the heat transfer rate: Q_total = h * A_total * (T_s - T_water_avg) Where A_total is the total surface area of the tube, and T_water_avg is the average water temperature, which is (20°C + 80°C) / 2 = 50°C. The total surface area (A_total) of the tube is: A_total = 2(ab + ac) * tube_length = 2(0.05 m * 0.025 m + 0.05 m * 0.025 m) * 10 m = 0.05 m² By plugging the values into the equation, we get: 2510 W = (67.6 W/m²·K)(0.05 m²)(T_s - 50°C) Now we only need to solve for T_s: T_s = (2510 W ÷ (67.6 W/m²·K)(0.05 m²)) + 50°C = 113.5°C The tube surface temperature necessary to heat the water to the desired outlet temperature of 80°C is 113.5°C.