Heat Transfer Questions and Answers – Dielectric Heating

This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Dielectric Heating”.

1. Which one of the following materials are quickly heated by applying high frequency?
a) Textiles
b) Engines
c) Rubber
d) Coal
View Answer

Answer: a
Explanation: They can be heated at high voltage alternating current to the plated of the condenser.

2. Generally heat generated depends on some parameters. It is directly proportional to
a) Time
b) Conductivity
c) Voltage
d) Distance between plates
View Answer

Answer: c
Explanation: It generally depends on the voltage as directly proportional.

3. Consider a 1.2 m thick slab of poured concrete (k = 1.148 W/m degree) with both of side surfaces maintained at a temperature of 20 degree Celsius. During its curing, chemical energy is released at the rate of 80 W/m3. Workout the maximum temperature of concrete
a) 30.73 degree celsius
b) 29.73 degree celsius
c) 28.73 degree celsius
d) 27.73 degree celsius
View Answer

Answer: b
Explanation: t = q g (δ – x) x/2 k + t w = 29.73 degree celsius.
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4. The insulating material used in dielectric heating is
a) Coal
b) Silver
c) Coin
d) Wool
View Answer

Answer: d
Explanation: Wool is good for heat conduction and from a dielectric heating point of view.

5. A composite slab consists of 5 cm thick layer of steel (k = 146 kJ/m hr degree) on the left side and a 6 cm thick layer of brass (k = 276 kJ/m hr degree) on the right hand side. The outer surfaces of the steel and brass are maintained at 100 degree Celsius and 50 degree Celsius. The contact between the two slabs is perfect and heat is generated at the rate of 4.2 * 10 5 k J/m2 hr at the plane of contact. The heat thus generated is dissipated from both sides of composite slab for steady state conditions. Calculate the temperature at the interface
Find the temperature at the interface if heat is generated at rate of 4.2 * 10 5 k J/m2
a) 115.26 degree celsius
b) 125.26 degree celsius
c) 135.26 degree celsius
d) 145.26 degree celsius
View Answer

Answer: b
Explanation: Q 1 + Q 2 = Q g. Q 1 = k 1 A 1 t i – t 1)/δ 1 and Q 2 = k 2 A 2 t i – t 2)/δ 2.
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6. Unit of specific resistance is
a) Ohm mm2/m
b) Ohm mm
c) Ohm/m
d) Ohm mm/m
View Answer

Answer: a
Explanation: Specific resistance is resistance per unit length.

7. What maximum thickness of concrete can be poured without causing the temperature gradient to exceed 98.5 degree Celsius per meter anywhere in the slab? Consider a 1.2 m thick slab of poured concrete (k = 1.148 W/m degree) with both of side surfaces maintained at a temperature of 20 degree Celsius. During its curing, chemical energy is released at the rate of 80 W/m3. Workout the maximum temperature of concrete
a) 2.64 m
b) 3.64 m
c) 4.64 m
d) 5.64 m
View Answer

Answer: b
Explanation: d t/d x = q g (δ – 2 x)/2 k. The temperature is largest at x = 0.
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8. Dielectric heating apparatus consists of
a) 4 electrodes
b) Elemental strip
c) No Insulating material
d) 4 plates
View Answer

Answer: b
Explanation: It consists of an elemental strip in the middle of the system.

9. The given expression can be used to solve the electrode temperature t w1 and t w2
q g δ = h 1 α 1 + h 2 α 2
Where, α 1 = A (t w 2 – t a) and α 2 = (t w1 – t a).
a) True
b) False
View Answer

Answer: b
Explanation: α 1 = A (t w 1 – t a) and α 2 = (t w2 – t a).
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10. A slab of insulating material of thickness 6 cm and thermal conductivity 1.4kJ/m hr deg is placed between and is in contact with two parallel electrodes, and is then subjected to high frequency dielectric heating at a uniform rate of 140,000kJ/m3 hr. At steady state coefficients of combined radiation and convection are 42 and 48 kJ/m2 hr deg. If atmospheric temperature is 25 degree Celsius, find surface temperatures?
a) 144.10 degree Celsius and 134.47 degree Celsius
b) 123.50 degree Celsius and 154.34 degree Celsius
c) 121.60 degree Celsius and 115.45 degree Celsius
d) 165.40 degree Celsius and 165.45 degree Celsius
View Answer

Answer: c
Explanation: α = -q g x2 /2k + h 1 α 1/k + α 1. At x =0.06 m and α = α 2, α 2 = -180 + 2.8 α 1. Also q g A δ = h 1 α 1 + h 2 α 2.

Sanfoundry Global Education & Learning Series – Heat Transfer.

To practice all areas of Heat Transfer, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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