2. Determine the heat transfer coefficient and the linear heat flux in the transverse air flow for a pipe with a diameter of A=30 mm, if its surface temperature is t_(s)= 90^circ C the air temperature is t_(a)=-10^circ C and the wind speed is w=4m/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 />\[ h = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3} \cdot \frac{L}{D} \]<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 />The Reynolds number is given by:<br /><br />\[ Re = \frac{w \cdot D}{\nu} \]<br /><br />where:<br />- \( w \) is the wind speed,<br />- \( D \) is the diameter of the pipe,<br />- \( \nu \) is the kinematic viscosity of air.<br /><br />The Prandtl number is given by:<br /><br />\[ Pr = \frac{c_p \cdot \mu}{k} \]<br /><br />where:<br />- \( c_p \) is the specific heat capacity of air,<br />- \( \mu \) is the dynamic viscosity of air,<br />- \( k \) is the thermal conductivity of air.<br /><br />Now, let's calculate these values:<br /><br />1. Kinematic viscosity of air (\( \nu \)): \( 1.5 \times 10^{-5} \, m^2/s \)<br />2. Specific heat capacity of air (\( c_p \)): \( 1005 \, J/(kg \cdot K) \)<br />3. Dynamic viscosity of air (\( \mu \)): \( 1.8 \times 10^{-5} \, Pa \cdot s \)<br />4. Thermal conductivity of air (\( k \)): \( 0.03 \, W/(m \cdot K) \)<br /><br />Now, we can calculate the Reynolds number and the Prandtl number:<br /><br />\[ Re = \frac{4 \cdot 0.03}{1.5 \times 10^{-5}} = 80000 \]<br /><br />\[ Pr = \frac{1005 \cdot 1.8 \times 10^{-5}}{0.03} = 0.6 \]<br /><br />Now, we can use the Dittus-Boelter equation to calculate the heat transfer coefficient:<br /><br />\[ h = 0.023 \cdot 80000^{0.8} \cdot 0.6^{0.3} \cdot \frac{0.03}{0.03} \]<br /><br />\[ h = 0.023 \cdot 3145728 \cdot 0.843 \cdot 1 \]<br /><br />\[ h = 589.6 \, W/(m^2 \cdot K) \]<br /><br />Now, we can calculate the linear heat flux (\( q \)):<br /><br />\[ q = h \cdot (t_s - t_a) \]<br /><br />where:<br />- \( t_s \) is the surface temperature of the pipe,<br />- \( t_a \) is the air temperature.<br /><br />\[ q = 589.6 \cdot (90 - (-10)) \]<br /><br />\[ q = 589.6 \cdot 100 \]<br /><br />\[ q = 58960 \, W/m \]<br /><br />Therefore, the heat transfer coefficient is approximately \( 589.6 \, W/(m^2 \cdot K) \) and the linear heat flux is approximately \( 58960 \, W/m \).