# Heat Transfer Questions and Answers – Heat Flow Through Triangular and Parabolic Fins

This set of Heat Transfer quiz focuses on “Heat Flow Through Triangular And Parabolic Fins”.

1. Which fin yields the maximum heat flow per unit weight?
a) Straight triangular fin
b) Curved triangular fin
c) Parabolic fin
d) Hyperbolic fin

Explanation: In straight triangular fin, there is maximum heat flow.

2. Heat dissipation by every segment of the fin is
a) Sometimes same
b) Same
c) Not same
d) Sometimes same or sometimes not same

Explanation: It is always different as fins are not uniform with respect to cross-sectional area.

3. “If a fin of a constant cross section is used, there would be wastage of material”. Chose the correct option
a) True
b) False

Explanation: Cross section must vary to utilize the material.

4. Which one is true regarding parabolic fin?
a) It dissipates the minimum amount of heat at a minimum material cost
b) It dissipates the minimum amount of heat at a maximum material cost
c) It dissipates the maximum amount of heat at a maximum material cost
d) It dissipates the maximum amount of heat at a minimum material cost

Explanation: In this case, a parabolic fin is of great practical importance.

5. For parabolic fin, the curve follows which law?
a) y = C/x2
b) y = C x4
c) y = C x2
d) y = C x1/2

Explanation: Equation of parabola is y = 4 x2 or x = 4 y2.
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6. The correction length for cylindrical fin is
a) L C = L + d/4
b) L C = 2 L + d/4
c) L C = 3 L + d/4
d) L C = 4 L + d/4

Explanation: Area = π d2/4. Where, d is the diameter.

7. Provision of fins on a given heat transfer surface will be more effective if there is
a) Fewer but thick fins
b) Large number of thick fins
c) Fewer but thin fins
d) Large number of thin fins

Explanation: Increase in ratio of perimeter P to be cross sectional area A C brings about improvement in the effectiveness of fins.

8. The heat dissipation at any section of parabolic fin is given by
a) (t2 – t1) (b) (δ)
b) k (t2 – t1) (b) (δ)
c) k (t2 – t1) (δ)
d) k (t2 – t1) (b)

Explanation: Q = qx (A X) = k (t2 – t1) (b) (δ).

9. An air cooled cylindrical wall is to be fitted with triangular fins of 3 cm thickness at base and 12 cm in height. The fins are made from stainless steel with density 8000 kg/m3 and thermal conductivity 17.5 W/m K. The wall temperature is 600 degree Celsius and the fin is exposed to an environment with t a = 30 degree Celsius and h = 20 W/m2 K. What is the temperature distribution along the fin?
a) t = 10 + 250 I 0 [6.056 (x) 1/2].
b) t = 20 + 250 I 0 [6.056 (x) 1/2].
c) t = 30 + 250 I 0 [6.056 (x) 1/2].
d) t = 40 + 250 I 0 [6.056 (x) 1/2].

Explanation: α/α 0 = t – t 0/t 0 – t a = I 0 [2 B (x) ½]/ I 0 [2 B (l) ½]. Here B = (2 h l/k δ) ½ = 3.028.

10. Consider the above problem, make calculations for the rate of heat flow per unit mass of fin material used
a) 126.53 W/kg
b) 154.76 W/kg
c) 134.87 W/kg
d) 165.46 W/kg

Explanation: Q = b (2 h k δ) ½ α 0 I 1 [2 B (L) ½/ I 0 [2 B (L) ½ = 1822 W. Mass of fin per meter width = 14.4 kg. Therefore rate of heat flow per unit mass = 1822/14.4 = 126.53 W/kg.

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