This set of Thermal Engineering Multiple Choice Questions & Answers (MCQs) focuses on “ Heat Transfer – Fins”.

1. A finned surface consists of root or base area of 1m^{2} and fin surface area of 2m^{2}. The average heat transfer coefficient for finned surface is 20W/m^{2}K effectiveness of fins provided is 0.75. If finned surface with surface with root or base temperature of 50°C is transferring heat to fluid of 30°C, then the rate of heat transfer without using fin is _______

a) 400W

b) 800W

c) 1000W

d) 1200W

View Answer

Explanation: (Q

_{without}) = h

_{A}(T

_{h}– T

_{infinity})

Q = 20×2×(50 – 30)

Q = 800W.

2. Heat pipe is widely used nowadays because _________

a) It acts as an insulator

b) It acts as conductor and insulator

c) It acts as a superconductor

d) It acts as a fin

View Answer

Explanation: Heat pipe is a heat transfer device. The thermal conductance of heat pipe may be several hundred 500times than that best available metal conductor, hence they act as super conductor.

3. Which the following statements pertaining to heat transfer through fins?

a) Fins are equally effective irrespective of whether they are on the hot side or cold side of the fluid

b) The temperature along the fin is variable and hence the rate of heat transfer varies along the elements of the fin

c) The fins may be made of materials that have a lower conductivity than the material of the wall

d) Fins must be arranged at right angles to the direction of the flow of the working fluid

View Answer

Explanation: Effectiveness of fin = \(\sqrt{\frac{kP}{hA}}\)

Fins are made up of material as that of base metal is integral but higher thermal conductivity metal are preferable as temperature drop will be close to uniform temperature.

4. Addition of fin to the surface increases the heat transfer if \(\sqrt{hA/kP}\) is __________

a) Equal to one

b) Greater than one

c) Less than one

d) Greater than one but less than two

View Answer

Explanation: ε = \(\sqrt{\frac{kP}{hA}}\)

For increased heat transfer, εfin should be higher. Hence εfin should be greater than one \(\sqrt{\frac{kP}{hA}}\) should be less than one.

5. The insulated tip temperature of a rectangular longitudinal fin having an excess (over ambient) root temperature of θ_{0}is _____

a) θ_{0}tanh(ml)

b) θ_{0}/sinh(ml)

c) θ_{0} tanh(ml)/(ml)

d) θ_{0}/cosh(ml)

View Answer

Explanation: Temperature distribution for insulated tip

\(\frac{θ}{θ_0} = \frac{coshm(l-x)}{cosh(ml)} \)

X=l, θ=\(\frac{θ_0}{cosh(ml)}\).

6. A fin length l, protrudes from a surface held at temperature T_{0}; it being higher than the ambient temperature T_{a}. The heat dissipation from the free end of the fin is stated to be negligibly small. What is the temperature gradient (dT/dx)_{x=l} at the tip of the fin?

a) Zero

b) (T_{0}-T_{l})/l

c) h(T_{0}-T_{a})

d) (T_{l}-T_{a})/(T_{0}-T_{a})

View Answer

Explanation: It is mentioned that heat dissipation for the free end of fin is stated to be negligibly small

(dT/dx)

_{x=l}= 0.

7. On heat transfer surface, fins are provided?

a) To increase temperature gradient so as to enhance heat transfer

b) To increase turbulence in flow for enhancing

c) To increase surface area to promote the rate of heat transfer

d) To decrease the pressure drop of the fluid

View Answer

Explanation: Fins are the extended surface of a body. Fins are provided to increase the surface area to promote the rate of heat transfer.

8. Extended surface are used to increase the rate of heat transfer. When the convective heat transfer coefficient h=mk, the addition of extended surface will:

a) Increase the rate of heat transfer

b) Decrease the rate of heat transfer

c) Not increase the rate of heat transfer

d) Increase the rate of heat transfer when the length of the fin is very large

View Answer

Explanation: For the fins, Effectiveness = \(\sqrt{\frac{kP}{hA}}=\frac{km}{h} \) And when h = mk effectiveness will become unity and by providing fin, heat transfer rate will be unity.

9. Three fins of equal length and diameter but made of aluminum, brass and cast iron are heated to 2000C at one end. If the fins dissipate heat to the surrounding air is 250C the temperature at the end will be least in case of ______

a) Aluminum fin

b) Brass fin

c) Cast-iron fin

d) Each fin have the same temperature at the free end

View Answer

Explanation: Effectiveness = \(\sqrt{\frac{kP}{hA}}\)

Effectiveness is proportional to thermal conductivity. In the given material thermal conductivity is minimum for cast iron. Therefore of cast iron is least effective.

10. An increase in pin fin effectiveness is caused by high value of _________

a) Convective coefficient

b) Change in temperature

c) Thermal conductivity

d) Sectional Area

View Answer

Explanation: Effectiveness α \(\sqrt{\frac{kP}{hA}}\)

For high value of effectiveness k (thermal conductivity) should be high. Effectiveness is directly proportional to thermal conductivity.

11. Which one of the following configuration has the highest fin effectiveness?

a) Thin, closely spaced fins

b) Thin, widely spaced fins

c) Thick, widely spaced fins

d) Thick, closely spaced fins

View Answer

Explanation: Effectiveness = \(\sqrt{\frac{kP}{hA}}\)

To keep effectiveness high, the term \(\sqrt{\frac{P}{A}}\) should be high. For that we use fins with larger perimeter (P) and with small cross sectional area (A). Hence instead of using 1fin with large cross section area we can use number of thin fins to keep the term \(\sqrt{\frac{P}{A}}\) high i.e. to keep high effectiveness.

12. A fin will be necessary and effective only when _____

a) k is small and h is large

b) k is large and h is also large

c) k is small and h is also small

d) k is large and h is small

View Answer

Explanation: Effectiveness α \(\sqrt{\frac{kP}{hA}}\)

A fin will be necessary and effective only when k thermal conductivity of fin material is large. Also h conductivity heat transfer coefficient between the surface and environment temperature is small. Effectiveness is directly proportional to k (thermal conductivity) and inversely proportional to h ( heat transfer coefficient).

**Sanfoundry Global Education & Learning Series – Thermal Engineering**

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