91影视

Given: - Gas temperature, \(T_{gas} = 1000\mathrm{~K}\) - Plate temperature, \(T_{plate} = 350\mathrm{~K}\) - Gas mean velocity, \(V_{gas} = 60\mathrm{~m/s}\) - Plate separation, \(d = 0.04\mathrm{~m\!}\) We are assuming: - The gas has the properties of atmospheric air. - The effects of radiation are negligible. Momentarily, we will need additional values for air properties at the mean temperature, such as the dynamic viscosity, thermal conductivity, and specific heat capacity at constant pressure.

To find the Reynolds number (Re), we use the formula: \(Re = \frac{V_{gas} \times d}{\nu}\) Where: - \(\nu\) is the kinematic viscosity of the air. For air at \(1000\mathrm{~K}\), we have: - Dynamic viscosity (\(渭\)), \(4.5 \times 10^{-5}\mathrm{~kg/(m\cdot s)}\) - Density (\(蟻\)), \(0.35\mathrm{~kg/m^3}\) Using the relationship \(谓 = 渭/蟻\), we find the kinematic viscosity: \(谓 = \frac{4.5 \times 10^{-5}}{0.35} = 1.29 \times 10^{-4}\mathrm{~m^2/s}\) Now, we can calculate the Reynolds number: \(Re = \frac{60 \times 0.04}{1.29 \times 10^{-4}} = 18671\) The Prandtl number (Pr) is calculated using the formula: \(Pr = \frac{渭 \times C_p}{k}\) Where: - \(C_p\) is the specific heat capacity at constant pressure. - \(k\) is the thermal conductivity. For air at \(1000\mathrm{~K}\), we have: - \(C_p\), \(1210\mathrm{~J/(kg\cdot K)}\) - Thermal conductivity (\(k\)), \(0.062\mathrm{~W/(m\cdot K}\!)\) Now, we can calculate the Prandtl number: \(Pr = \frac{4.5 \times 10^{-5} \times 1210}{0.062} = 0.7\)

For fully developed turbulent flow between parallel plates: \(Nu = 8.235Re^{0.5}Pr^{1/3}\) Now, substituting the values of Reynolds number and Prandtl number, we can calculate the Nusselt number: \(Nu = 8.235(18671)^{0.5}(0.7)^{1/3} = 398.41\) The convective heat transfer coefficient (h) is given by: \(h = \frac{Nu \times k}{d}\) Substituting the calculated values, we find the value for h: \(h = \frac{398.41 \times 0.062}{0.04} = 620.39\mathrm{~W/(m^2\cdot K}\!)\)

Using the convective heat transfer equation, we can now find the heat flux (q) for the original plates: \(q = h(T_{gas} - T_{plate})\) Substituting the values for h, gas temperature, and plate temperature: \(q = 620.39(1000 - 350)\mathrm{~W/m^2}\) \(q = 402.46 \times 10^3\mathrm{~W/m^2}\) Thus, the heat flux at the plate surface without the third plate is \(402.46kW/m^2\).

When the third plate is suspended midway between the original plates, the distance between the plates is halved, which decreases the residence time of the gas. Therefore, the convective heat transfer coefficient can be considered to remain the same, and the heat flux for the original plates will be as follows: \(q = h(T_{gas\_new} - T_{plate})\) Since the distance between the original plates is halved, the gas has half the time to exchange heat with the plates. Because of this, we can assume that the temperature drop will also be approximately halved. So, \(T_{gas\_new} = 350 + \frac{650}{2} = 675\mathrm{~K}\). Substituting the new gas temperature, we get the new heat flux for the original plates: \(q = 620.39(675 - 350)\mathrm{~W/m^2}\) \(q = 201.73 \times 10^3\mathrm{~W/m^2}\) Thus, the heat flux at the surface of the original plates with the third plate is \(201.73kW/m^2\). #Conclusion# (a) The heat flux at the plate surface for the original plates is \(402.46kW/m^2\). (b) When a third plate is suspended midway between the original plates, the heat flux at the surface of the original plates reduces to \(201.73kW/m^2\).