 # E-PolyLearning

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 1. Find the heat flow rate through the composite wall as shown in figure. Assume one dimensional flow and take k 1 = 150 W/m degree k 2 = 30 W/m degree k 3 = 65 W/m degree k 4 = 50 W/m degree AB = 3 cm, BC = 8 cm and CD = 5 cm. The distance between middle horizontal line from the top is 3 cm and from the bottom is 7 cm a. 1173.88 W b. 1273.88 W c. 1373.88 W d. 1473.88 W
 View Answer Report Discuss 50-50! Answer: (b).1273.88 W

 2. A steel pipe of 20 mm inner diameter and 2 mm thickness is covered with 20 mm thick of fiber glass insulation (k = 0.05 W/m degree). If the inside and outside convective coefficients are 10 W/m² degree and 5 W/m² degree, calculate the overall heat transfer coefficient based on inside diameter of pipe. In the diagram, the diameter of small circle is 20 mm a. 1.789 W/m² degree b. 2.789 W/m² degree c. 3.789 W/m² degree d. 4.789 W/m² degree
 View Answer Report Discuss 50-50! Answer: (b).2.789 W/m² degree

 3. The oven of an electric store, of total outside surface area 2.9 m² dissipates electric energy at the rate of 600 W. The surrounding room air is at 20 degree Celsius and the surface coefficient of heat transfer between the room air and the surface of the oven is estimated to be 11.35 W/m² degree. Determine the average steady state temperature of the outside surface of the store a. 38.22 degree Celsius b. 48.22 degree Celsius c. 58.22 degree Celsius d. 68.22 degree Celsius
 View Answer Report Discuss 50-50! Answer: (a).38.22 degree Celsius

 4. The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material? a. 0.81 W/m K b. 0.71 W/m k c. 0.61 W/m K d. 0.51 W/m K
 View Answer Report Discuss 50-50! Answer: (b).0.71 W/m k

 5. In Cartesian coordinates the heat conduction equation is given by a. d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T) b. 2d²t/dx² + d²t/dy2 + d²t/dz2 + 34q g = (d t/d T) c. d²t/dx² + 3d²t/dy2 + d²t/dz2 = (1/α) (d t/d T) d. 4d²t/dx² + d²t/dy2 + d²t/dz2 + 1/2q g = (1/α) (d t/d T)
 View Answer Report Discuss 50-50! Answer: (a).d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)

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