Final answer:
The heat loss per fin is calculated using the convective heat transfer formula and the given dimensions of the fin and the tube, resulting in a heat loss of approximately 22.18 watts per fin.
Step-by-step explanation:
The subject of this question is related to heat transfer, specifically addressing the calculation of heat loss from aluminum fins on a tube. To find the heat loss per fin, we'll use the concept of convection heat transfer rather than the equation Q = mcΔT, which is more commonly used for heat transfer involving mass and temperature changes in a substance. The information provided suggests a scenario of steady-state heat transfer from the fins to the surrounding fluid due to a temperature difference, and we'll use the convective heat transfer formula:
Q = hA(Tsurface - T∞)
where Q is the heat transfer per fin, h is the convective heat transfer coefficient, A is the surface area of the fin, Tsurface is the temperature of the tube surface, and T∞ is the ambient fluid temperature.
To calculate A, we need the dimensions of the fin, but we have to be careful to only consider the surface area that is exposed to the fluid and is able to dissipate heat, which is the perimeter times the length of the fin.
Given:
- Width of the fin (w) = 1.5 cm
- Thickness of the fin (t) = 1.0 mm = 0.1 cm
- Convective heat transfer coefficient (h) = 130 W/m²K
- Temperature of the tube surface (Tsurface) = 170°C
- Ambient fluid temperature (T∞) = 25°C
- First, we need to convert the width of the fin to meters:
w = 1.5 cm * (1 m / 100 cm) = 0.015 m
The surface area (A) of one side of the fin is then calculated by multiplying the perimeter that is in contact with the air by the width of the fin, assuming the length of the fin is equal to the perimeter of the tube:
A = 2 * π * (d/2) * w
With d = 2.5 cm = 0.025 m (the diameter of the tube), we get:
A = 2 * π * (0.025 m / 2) * 0.015 m = 2 * π * 0.0125 m * 0.015 m = 2 * 3.1416 * 0.0125 m * 0.015 m = 1.1781 x 10-3 m²
Finally, we can calculate the heat loss per fin as follows:
Q = hA(Tsurface - T∞) = 130 W/m²K * 1.1781 x 10-3 m² * (170°C - 25°C)
Q = 130 * 1.1781 x 10-3 * 145 = 22.18 W
Therefore, the heat loss per fin is approximately 22.18 watts.