Answer:
The efficiency of fin is 64.20%
Heat transfer rate 543.03249W
Effectiveness of each fin is 21.6996
Explanation:
Given
Surface temperature = Tb = 350°C
Thermal Conductivity = k = 235 W/m·K
Ambient air temperature = T = 25°C
Convective heat transfer coefficient of air = h = 154 W/m^2·K.
Length of fin = L = 82 mm
Base of fin = t = 5 mm
Width of fin = w = 100 mm
The mass of fin is calculated by;
m = √(2h/kt)
m = √(2 * 154/235*0.005)
m = 16.19
Then the critical length is calculated;
Lc. = L + ½t
Lc = 0.082 + ½ * 0.005.
Lc = 0.0845m
Therefore, the efficiency of the fin can be calculated as follows:
tanh (mLc)/mLc
= tanh(16.19 * 0.0845)/((16.19 * 0.0845)
tanh(1.368055)/1.368055
= 0.878248/1.368055
= 0.6420
Eff = 64.20%
Area of the fin is calculated as follows
A = 2wLc
A = 2 * 0.1 * 0.0845
A = 0.0169m²
Rate of heat transfer is calculated by;
Eff * h * A(Tb - T)
= 0.642 * 154 * 0.0169 * (350 - 25)
E = 543.03249W
The base area of the fin is calculated by;
A = wt
A = 0.1 * 0.005
A = 0.0005m²
Lastly, the effectiveness of each fin is calculated as follows;
Eff = E/(hA(TB - T))
Eff = 543.03249/(154 * 0.0005(350 - 25))
Eff = 21.6996
Therefore, the efficiency of fin is 64.20%
, heat transfer rate 543.03249W
and effectiveness of each fin is 21.6996
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