Question

Water vapor at \(250^{\circ} \mathrm{C}\) is flowing with a velocity of $5 \mathrm{~m} / \mathrm{s}\( in parallel over a \)2-\mathrm{m}$-long flat plate where there is an unheated starting length of \(0.5 \mathrm{~m}\). The heated section of the flat plate is maintained at a constant temperature of \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at the trailing edge, \((b)\) the average convection heat transfer coefficient for the heated section, and \((c)\) the rate of heat transfer per unit width for the heated section.

Short answer

Based on the given problem with a flow of water vapor over a flat plate with an unheated and heated section, we calculated the following values: 1. The local convection heat transfer coefficient at the trailing edge is 18.75 W/(m^2⋅K). 2. The average convection heat transfer coefficient for the heated section is 22.0 W/(m^2⋅K). 3. The rate of heat transfer per unit width for the heated section is 6600 W/m.

Step-by-step solution

Calculate the Reynolds number at the trailing edge

To determine the type of flow (laminar or turbulent) and the appropriate Nusselt number correlations, we first need to calculate the Reynolds number at the trailing edge. The Reynolds number is determined by the following formula: $$Re_{x} = \frac{\rho u x}{\mu}$$ Where \(Re_{x}\) is the Reynolds number, \(\rho\) is the density of the water vapor, \(u\) is the velocity, \(x\) is the distance from the leading edge, and \(\mu\) is the dynamic viscosity of the water vapor. Given the problem's temperature and velocity, we can obtain the properties of water vapor from a table. At \(250^{\circ}\mathrm{C}\), the density \(\rho = 0.621 \mathrm{~kg/m^3}\), and the dynamic viscosity \(\mu = 2.481 \times 10^{-5}\mathrm{ kg/(m \cdot s)}\). With these values, calculate the Reynolds number at the trailing edge: $$Re_{2} = \frac{0.621 \mathrm{~kg/m^3} \cdot 5 \mathrm{~m/s} \cdot 2 \mathrm{~m}}{2.481 \times 10^{-5}\mathrm{ kg/(m \cdot s)}} \approx 250066$$

Determine the Nusselt number

Since the Reynolds number at the trailing edge is larger than \(5 \times 10^5\), we can assume turbulent flow. For turbulent flow over a flat plate, we use the Colburn equation for the Nusselt number: $$Nu_{x} = 0.0296 Re_{x}^{4/5} Pr^{1/3}$$ Where \(Nu_{x}\) is the Nusselt number, \(Re_{x}\) is the Reynolds number calculated in step 1, and \(Pr\) is the Prandtl number. The Prandtl number for the water vapor at \(250^{\circ}\mathrm{C}\) can be found in a table, and is equal to \(Pr = 0.676\). Calculate the Nusselt number at the trailing edge: $$Nu_{2} = 0.0296 (250066)^{4/5} (0.676)^{1/3} \approx 888.25$$

Calculate the local convection heat transfer coefficient

Now that we have the Nusselt number, we can determine the local convection heat transfer coefficient at the trailing edge using the following formula: $$h_{x} = \frac{Nu_{x} \cdot k}{x}$$ Where \(h_{x}\) is the local convection heat transfer coefficient, \(Nu_{x}\) is the Nusselt number, \(k\) is the thermal conductivity of the water vapor, and \(x\) is the distance from the leading edge. At \(250^{\circ}\mathrm{C}\), the thermal conductivity of water vapor is \(k = 0.0422 \mathrm{~W/(m \cdot K)}\). Calculate the local convection heat transfer coefficient at the trailing edge: $$h_{2} = \frac{888.25 \cdot 0.0422 \mathrm{~W/(m \cdot K)}}{2\mathrm{~m}} \approx 18.75\mathrm{~W/(m^2 \cdot K)}$$

Calculate the average convection heat transfer coefficient

We can determine the average convection heat transfer coefficient for the heated section by calculating the Nusselt number for the entire length of the flat plate and then using the formula for the convection heat transfer coefficient. Calculate the average Nusselt number for the heated section: $$Nu_{L} = 0.0296 Re_{L}^{4/5} Pr^{1/3}$$ $$Nu_{L} = 0.0296 (250066)^{4/5} (0.676)^{1/3} \approx 1041.4$$ Now calculate the average convection heat transfer coefficient for the heated section: $$\overline{h} = \frac{Nu_{L} \cdot k}{L}$$ $$\overline{h} = \frac{1041.4 \cdot 0.0422 \mathrm{~W/(m \cdot K)}}{2 \mathrm{~m}} \approx 22.0\mathrm{~W/(m^2 \cdot K)}$$

Calculate the rate of heat transfer per unit width

Finally, we can determine the rate of heat transfer per unit width for the heated section using the average convection heat transfer coefficient and the difference in temperature between the water vapor and the heated section: $$q = \overline{h} \cdot L \cdot \Delta T$$ Where \(q\) is the heat transfer rate, \(\overline{h}\) is the average convection heat transfer coefficient, \(L\) is the heated section length, and \(\Delta T\) is the difference in temperature. Calculate the rate of heat transfer per unit width for the heated section: $$q = 22.0\mathrm{~W/(m^2 \cdot K)} \cdot (2\mathrm{~m} - 0.5\mathrm{~m}) \cdot (250^{\circ}\mathrm{C} - 50^{\circ}\mathrm{C}) \approx 6600\mathrm{~W/m}$$ The final results are: - Local convection heat transfer coefficient at the trailing edge: \(h_{2} = 18.75\mathrm{~W/(m^2 \cdot K)}\) - Average convection heat transfer coefficient for the heated section: \(\overline{h} = 22.0\mathrm{~W/(m^2 \cdot K)}\) - Rate of heat transfer per unit width for the heated section: \(q = 6600\mathrm{~W/m}\)