Sure, let's work through this problem step-by-step to determine the value of the heat transfer coefficient [tex]\( h \)[/tex].
Given:
- Diameter of the cylinder, [tex]\( D = 30 \)[/tex] meters
- Air temperature, [tex]\( T_{\text{air}} = 23^\circ \text{C} \)[/tex]
- Surface temperature of the cylinder, [tex]\( T_{\text{surface}} = 90^\circ \text{C} \)[/tex]
- Heat output, [tex]\( Q = 40 \)[/tex] watts per meter
First, we need to determine the temperature difference between the surface of the cylinder and the air:
[tex]\[ \Delta T = T_{\text{surface}} - T_{\text{air}} \][/tex]
[tex]\[ \Delta T = 90^\circ \text{C} - 23^\circ \text{C} \][/tex]
[tex]\[ \Delta T = 67^\circ \text{C} \][/tex]
Next, we calculate the surface area of the curved part of the cylinder per meter of its length. The cylinder is very long, so we consider the side surface of a unit length (1 meter). The formula for the surface area of a cylinder is:
[tex]\[ A = \pi D L \][/tex]
where [tex]\( D \)[/tex] is the diameter, and [tex]\( L \)[/tex] is the length. For a unit length (1 meter):
[tex]\[ L = 1 \text{ meter} \][/tex]
Therefore,
[tex]\[ A = \pi \times 30 \text{ meters} \times 1 \text{ meter} \][/tex]
[tex]\[ A \approx 94.2478 \text{ square meters} \][/tex]
Now, we use the formula for heat transfer:
[tex]\[ Q = h \times A \times \Delta T \][/tex]
We need to isolate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{Q}{A \times \Delta T} \][/tex]
Substituting the values:
[tex]\[ h = \frac{40 \text{ watts}}{94.2478 \text{ square meters} \times 67^\circ \text{C}} \][/tex]
[tex]\[ h \approx 0.00633452509818489 \text{ watts per square meter per degree Celsius} \][/tex]
Thus, the heat transfer coefficient [tex]\( h \)[/tex] is approximately [tex]\( 0.0063 \ \text{W/m}^2 \cdot ^\circ C \)[/tex].
