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Heat Transfer From Extended Surfaces Online Exam Quiz

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A very long copper rod 20 mm in diameter extends horizontally from a plane heated wall maintained at 100 degree Celsius. The surface of the rod is exposed to an air environment at 20 degree Celsius with convective heat transfer coefficient of 8.5 W/m² degree. Workout the heat loss if the thermal conductivity of copper is 400 W/m degree

Options

A : a. 10.71 W

B : b. 20.71 W

C : c. 30.71 W

D : d. 40.71 W

Common applications of finned surfaces are with (i) Electrical motors (ii) Economizers for steam power plant (iii) Convectors for steam and cold water heating systems (iv) Cooling coils Identify the correct option

Options

A : a. 1, 2 and 4

B : b. 1, 2 and 3

C : c. 1, 2, 3 and 4

D : d. 1 and 2

The extended surface used for the enhancement of heat dissipation is

Options

A : a. Convective coefficient

B : b. Fourier number

C : c. Fin

D : d. No finned surface

It is said that fins can take a variety of forms (i) Longitudinal fins of rectangular cross section attached to a wall (ii) Cylindrical tubes with radial fins (iii) Conical rod protruding from a wall Identify the correct statement

Options

A : a. 1 only

B : b. 1 and 2

C : c. 2 and 3

D : d. 1, 2 and 3

A steel rod (k = 30 W/m degree) 1 cm in diameter and 5 cm long protrudes from a wall which is maintained at 10 degree Celsius. The rod is insulated at its tip and is exposed to an environment with h = 50 W/m² degree and t a = 30 degree Celsius. Calculate the fin efficiency

Options

A : a. 56.57%

B : b. 66.57%

C : c. 76.57%

D : d. 86.57%

If fin is sufficiently thin, so heat flows pertains to

Options

A : a. One dimensional heat conduction

B : b. Two dimensional heat conduction

C : c. Three dimensional heat conduction

D : d. No heat flow is there

If heat dissipation for one fin is given by 377.45 k J/hour, then what is the heat dissipation for 12 fins?

Options

A : a. 7529.4 k J/hour

B : b. 6529.4 k J/hour

C : c. 5529.4 k J/hour

D : d. 4529.4 k J/hour

In order to achieve maximum heat dissipation the fin should be designed in such a way that has a

Options

A : a. Maximum lateral surface towards the tip side of fin

B : b. Minimum lateral surface near the center line

C : c. Maximum lateral surface at the root side of fin

D : d. Maximum lateral surface near the center of fin

On a heat transfer surface, fins are provided to

Options

A : a. Increase turbulence in flow for enhancing heat transfer

B : b. Increase temperature gradient so as to enhance heat transfer

C : c. Pressure drop of the fluid should be minimized

D : d. Surface area is maximum to promote the rate of heat transfer

In heat dissipation from an infinitely long fin, the boundary conditions are

Options

A : a. t = t0 at x = infinity and t = ta at x = 0

B : b. t = t0 at x = 0 and t = ta at x = infinity

C : c. t = t0 at x = 0 and t = ta at x = 0

D : d. t = t0 at x = infinity and t = ta at x = infinity

The temperature distribution in case of infinitely long fin is

Options

A : a. t – t a/t? – t a = mx

B : b. t – t a/t? – t a = -mx

C : c. t – t a/t? – t a = e-m x

D : d. t – t a/t? – t a = log (m x)

The rate of heat transfer in case of infinitely long fin is given by

Options

A : a. (h P k A)^1/2 (t? – t a)

B : b. (h P A)^1/2 (t? – t a)

C : c. (P k A)^1/2 (t? – t a)

D : d. (h k A)^1/2 (t? – t a)

Let us say there are two rods having same dimensions, one made of brass (k = 85W/m K) and the other of copper (k = 375W/m K), having one of their ends inserted into a furnace. At a section 10.5 cm away from the furnace, the temperature of brass rod is 120 degree Celsius. Find the distance at which the same temperature would be reached in the copper rod? Both ends are exposed to the same environment

Options

A : a. 12.54 cm

B : b. 45.87 cm

C : c. 12.34 cm

D : d. 22.05 cm

Three rods, one made of silver (l = 420W/m K), second made of aluminum (k = 210W/m K) and the third made of iron (k = 70W/m K) are coated with a uniform layer of wax all around. The rods are placed vertically in a boiling water bath with 250 mm length of each rod projecting outside. If all the rods are having following dimensions i.e. diameter = 15 mm and length = 300 mm and have identical surface coefficient 12.5W/ m² K, work out the ratio of lengths up to which wax will melt on each rod

Options

A : a. 2.45:1:1.732

B : b. 1.732:1:2.45

C : c. 2.45:1.732:1

D : d. 1.732:1:2.45

A rod of 10 mm square section and 160 mm length with thermal conductivity of 50W/m K protrudes from a furnace wall at 200 degree Celsius with convective coefficient 20 W/ square m K. Make calculations for the heat convective up to 80 mm length

Options

A : a. 6.84W

B : b. 7.34W

C : c. 4.54W

D : d. 5.47W

A fin protrudes from a surface which is held at a temperature higher than that of its environment. The heat transferred away from the fin is

Options

A : a. Heat escaping from the tip of the fin

B : b. Heat conducted along the fin length

C : c. Convective heat transfer from the fin surface

D : d. Sum of heat conducted along the fin length and that convected from the surface

The value of correction length for equilateral fin is

Options

A : a. L C = 2 L + a/4(3)^1/2

B : b. L C = L + a/4(3)^1/2

C : c. L C = 3 L + a/4(3)^1/2

D : d. L C = 6 L + a/4(3)^1/2

The parameter m = (h P/k A C)^1/2 has been stated to increase in a long fin. If all other parameters are constant, then

Options

A : a. Profile of temperature will remain the same

B : b. Along the length temperature drop will be less

C : c. The parameter influences the heat flow only

D : d. The temperature drop along the length will be steeper

The relevant boundary conditions in case of heat dissipation from a fin insulated at the tip are

Options

A : a. t = t? at x = 0 and d t/d x = 0 at x = 0

B : b. t = t? at x = 0 and d t/d x = 0 at x = 1

C : c. t = t? at x = 1 and d t/d x = 0 at x = 1

D : d. t = t? at x = infinity and d t/d x = 0 at x = infinity

The temperature distribution in case of fin insulated at the tip is given by

Options

A : a. t – t?/t? – t a = cos h m (3 – x)/cos ml

B : b. t – t?/t? – t a = cos h m (2 – x)/sin h ml

C : c. t – t?/t? – t a = cos h m (l – x)/cos h ml

D : d. t – t?/t? – t a = cos m (l – x)/sin ml

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