# A 3 cm long, 2 mm × 2 mm rectangular cross-section aluminium fin [k = 237 W/m°c] is attached to a surface. If the fin efficiency is 65%, the effectiveness of this single fin is:

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TNTRB 2017 ME Official Question Paper
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1. 30
2. 24
3. 8
4. 39

Option 4 : 39
Free
ST 1: Logical reasoning
5280
20 Questions 20 Marks 20 Mins

## Detailed Solution

Concept:

The relation between the efficiency of the fin and the effectiveness of the fin is given by,

$$\frac{\eta }{\varepsilon } = \frac{{Cross~Section~area~of~the~fin~\left( {Ac} \right)}}{{Surface~~area~~of~the~fin~\left( {As} \right)}}$$

Calculation

Given

Length of the fin ,L = 3 cm = 30 mm, Side of the square cross-section, a = b = 2 mm, Efficiency of the fin, η = 65% = 0.65

As = p × L = 2 × (a + b) × L = 2 × (2 + 2) × 30 = 240 mm2

Ac = a × b = 2 × 2 = 4 mm2

$$Thermal~conductivity,k = 237~\frac{W}{{m^\circ C}}$$

Therefore,
$$⇒ \frac{{0.65}}{{\rm{\varepsilon }}} = \frac{4}{{240}}$$

⇒ ε = 39

Important Points

• The Biot number of the fin with good effectiveness should be less than 1.
• Fin material should have convection resistance higher than conduction resistance.