Heat Transfer Questions and Answers – Newton- Rikhman Law

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This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Newton- Rikhman Law”.

1. Newton-Rikhman law is given by
a) Q = h A (t s – t f)
b) Q = 2 h A (t s – t f)
c) Q = 3 h A (t s – t f)
d) Q = 4 h A (t s – t f)
View Answer

Answer: a
Explanation: Regardless of the particular nature, the appropriate rate equation for the convective heat transfer is prescribed by Newton’s law of cooling.

2. The value of film coefficient is dependent upon
(i) Boundary layer configuration
(ii) Geometry and orientation of the surface
(iii) Surface conditions
a) i and ii
b) ii and iii
c) i and ii
d) i, ii and iii
View Answer

Answer: d
Explanation: It depends upon surface conditions i.e. roughness and cleanliness, geometry and orientation of the surface i.e. plate, tube and cylinder placed vertically or horizontally.

3. The convection coefficients for boiling and condensation lie in the range
a) 5000-12500 W/m 2 K
b) 2500-100000 W/m 2 K
c) 2500-5000 W/m 2 K
d) 2500-12500 W/m 2 K
View Answer

Answer: b
Explanation: Convection mechanisms involving phase changes lead to important field of boiling and condensatio.

4. Forced air flows over a convection heat exchanger in a room heater, resulting in a convective heat transfer coefficient 1.136 k W/m2 K. The surface temperature of heat exchanger may be considered constant at 65 degree Celsius, and the air is at 20 degree Celsius. Determine the heat exchanger surface area required for 8.8 k W of heating
a) 0.272 m2
b) 0.472 m2
c) 0.172 m2
d) 0.672 m2
View Answer

Answer: c
Explanation: Q = h A (t s – t f). So. A = 0.172 m2.
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5. A region of fluid motion near a plate in which temperature gradient exist is
a) Thermal boundary layer
b) Diathermia boundary layer
c) Turbulent flow
d) Laminar flow
View Answer

Answer: a
Explanation: The fluid velocity decreases as it approaches the solid surface reaching to zero in the fluid layer immediately next to the surface. The thin layer of stagnated fluid is called thermal boundary layer.
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6. Thermo-physical properties of the fluid are represented by
(i) Density
(ii) Viscosity
(iii) Specific heat
(iv) Thermal conductivity
Identify the correct option
a) i and ii
b) i, ii, iii and iv
c) ii, iii and iv
d) i, ii and iii
View Answer

Answer: b
Explanation: The value of film coefficient is dependent upon thermos-physical properties of he fluid i.e. density, viscosity, specific heat, coefficient of expansion and thermal conductivity.

7. A motor cycle cylinder consists of ten fins, each 150 mm outside diameter and 75 mm inside diameter. The average fin temperature is 500 degree Celsius and the surrounding air is at 20 degree Celsius temperature. Make calculations for the rate of heat dissipation from the cylinder fins by convection when motor cycle is stationary and convective coefficient is 6 W/m2 K
a) 432.2 W
b) 532.2 W
c) 632.2 W
d) 763.2 W
View Answer

Answer: d
Explanation: A = 0.265 m2 and Q = (6) (0.265) (500 – 20) = 763.2 W.
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8. Consider the above problem, make calculations for the rate of heat dissipation from the cylinder fins by convection when motor cycle is moving at 60 km/hr and convective coefficient is 75 W/m2 K
a) 9640 W
b) 9540 W
c) 9440 W
d) 9340 W
View Answer

Answer: b
Explanation: A = 0.265 m2 and Q = (75) (0.265) (500 – 20) = 9540 W.

9. The temperature profile at a particular location on a surface is prescribed by the identity
(t s – t) / (t s – t infinity) = (1/2) (y/0.0075) 3 + (3/2) (y/0.0075)
If thermal conductivity of air is stated to be 0.03 W/m K, determine the value of convective heat transfer coefficient
a) 4 W/m2 K
b) 5 W/m2 K
c) 6 W/m2 K
d) 7 W/m2 K
View Answer

Answer: c
Explanation: h = – k/ (t s – t infinity) [d t/d y] y = 0.
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10. Air at 20 degree Celsius flows over a flat surface maintained at 80 degree Celsius. The local heat flow at a point was measured as 1250 W/m2 .Take thermal conductivity of air as 0.028 W/m K, calculate the temperature at a distance 0.5 mm from the surface
a) 57.682 degree celsius
b) 67.682 degree celsius
c) 77.682 degree celsius
d) 87.682 degree celsius
View Answer

Answer: a
Explanation: Temperature at 0.5 mm from the surface is 80 + (d t/d y) y = 0 (0.0005) = 57.682 degree celsius.

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.

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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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