2. Determine the heat transfer coefficient and the linear heat flux in the transverse air flow for a pipe with a diameter of d=80mm if its surface temperature is t_(s)= 80^circ C the air temperature is t_(a)=8^circ C and the wind speed is w=3m/s

To determine the heat transfer coefficient and the linear heat flux in the transverse airflow for a pipe, we can use the Dittus-Boelter equation, which is a commonly used empirical correlation for convective heat transfer. The Dittus-Boelter equation is given by:<br /><br />\[ h = 0.023 \cdot Re^{0.8} \cdot Pr^{0.4} \cdot \frac{L}{d} \]<br /><br />where:<br />- \( h \) is the heat transfer coefficient,<br />- \( Re \) is the Reynolds number,<br />- \( Pr \) is the Prandtl number,<br />- \( L \) is the characteristic length (in this case, the diameter of the pipe),<br />- \( d \) is the diameter of the pipe.<br /><br />First, we need to calculate the Reynolds number (\( Re \)) and the Prandtl number (\( Pr \)):<br /><br />\[ Re = \frac{w \cdot d}{\nu} \]<br /><br />where \( \nu \) is the kinematic viscosity of air at the film temperature (which is approximately the average of the air and surface temperatures). At \( t_{film} = \frac{80 + 8}{2} = 44^{\circ}C \), the kinematic viscosity of air is approximately \( \nu = 1.5 \times 10^{-5} m^2/s \).<br /><br />\[ Re = \frac{3 \cdot 0.08}{1.5 \times 10^{-5}} = 400,000 \]<br /><br />Next, we calculate the Prandtl number:<br /><br />\[ Pr = \frac{c_p \cdot \nu}{k} \]<br /><br />where \( c_p \) is the specific heat capacity of air at the film temperature, and \( k \) is the thermal conductivity of air at the film temperature. At \( 44^{\circ}C \), the specific heat capacity of air is approximately \( c_p = 1005 J/(kg \cdot K) \), and the thermal conductivity of air is approximately \( k = 0.03 W/(m \cdot K) \).<br /><br />\[ Pr = \frac{1005 \cdot 1.5 \times 10^{-5}}{0.03} = 0.75 \]<br /><br />Now, we can calculate the heat transfer coefficient:<br /><br />\[ h = 0.023 \cdot 400,000^{0.8} \cdot 0.75^{0.4} \cdot \frac{0.08}{0.08} \]<br /><br />\[ h = 0.023 \cdot 1.68 \times 10^{6} \cdot 0.75^{0.4} \]<br /><br />\[ h = 0.023 \cdot 1.68 \times 10^{6} \cdot 0.767 \]<br /><br />\[ h = 38,000 W/(m^2 \cdot K) \]<br /><br />Finally, we can calculate the linear heat flux:<br /><br />\[ q = h \cdot (t_s - t_a) \]<br /><br />\[ q = 38,000 \cdot (80 - 8) \]<br /><br />\[ q = 38,000 \cdot 72 \]<br /><br />\[ q = 2,736,000 W/m^2 \]<br /><br />Therefore, the heat transfer coefficient is approximately \( 38,000 W/(m^2 \cdot K) \), and the linear heat flux is approximately \( 2,736,000 W/m^2 \).