A plane wall with surface temperature of 350°C is attached with straight rectangular fins (k = 235 W/m·K). The fins are exposed to an ambient air condition of 25°C and the convection heat transfer coefficient is 154 W/m^2·K. Each fin has a length of 82 mm, a base of 5 mm thick and a width of 100 mm. Determine the efficiency, heat transfer rate, and effectiveness of each fin

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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