This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Effect Of Variable Conductivity”.

1. With variable thermal conductivity, Fourier law of heat conduction through a plane wall can be expressed as

a) Q = -k_{0} (1 + β t) A d t/d x

b) Q = k_{0} (1 + β t) A d t/d x

c) Q = – (1 + β t) A d t/d x

d) Q = (1 + β t) A d t/d x

View Answer

Explanation: Here k

_{0}is thermal conductivity at zero degree Celsius.

2. The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation

K = (1.45 + 0.5 * 10^{-5} t^{2}) KJ/m hr deg

Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?

a) 1355.3 kJ/m^{2} hr

b) 2345.8 kJ/m^{2} hr

c) 1745.8 kJ/m^{2} hr

d) 7895.9 kJ/m^{2} hr

View Answer

Explanation: Q = -k A d t/d x, Q d x = – k A d t = – (1.45 + 0.5 * 10

^{-5}t

^{2}) A d t. Integrating over the wall thickness δ, we get Q = 436.45/0.25 = 1745.8 kJ/m

^{2}hr.

3. A plane wall of thickness δ has its surfaces maintained at temperatures T_{1} and T_{2}. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k_{0} T^{2}. Find the expression to work out the steady state heat conduction through the wall?

a) Q = 2A k_{0} (T _{1 }^{3} – T _{2 }^{3})/3 δ

b) Q = A k_{0} (T _{1 }^{3} – T _{2 }^{3})/3 δ

c) Q = A k_{0} (T _{1 }^{2} – T _{2 }^{2})/3 δ

d) Q = A k_{0} (T _{1} – T _{2})/3 δ

View Answer

Explanation: Q = -k A d t/d x = k

_{0}T

^{2 }A d t/d x. Separating the variables and integrating within the prescribed boundary conditions, we get Q = A k

_{0}(T

_{1 }

^{3}– T

_{2 }

^{3})/3 δ.

4. The mean thermal conductivity evaluated at the arithmetic mean temperature is represented by

a) k_{m} = k_{0} [1 + β (t_{1} – t_{2})/2].

b) k_{m} = k_{0} [1 + (t_{1} + t_{2})/2].

c) k_{m} = k_{0} [1 + β (t_{1} + t_{2})/3].

d) k_{m} = k_{0} [1 + β (t_{1} + t_{2})/2].

View Answer

Explanation: At arithmetic mean temperatures i.e. (t

_{1}+ t

_{2})/2.

5. With respect to the equation k = k_{0} (1 +β t) which is true if we put β = 0?

a) Slope of temperature curve is constant

b) Slope of temperature curve does not change

c) Slope of temperature curve increases

d) Slope of temperature curve is decreases

View Answer

Explanation: As temperature profile is linear so it is constant.

6. 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

Explanation: Q = k A (t

_{1}– t

_{2})/δ. So, k = 0.71 W/m K.

7. If β is greater than zero, then choose the correct statement with respect to given relation

k = k_{0} (1 +β t)

a) k doesn’t depend on temperature

b) k depends on temperature

c) k is directly proportional to t

d) Data is insufficient

View Answer

Explanation: k increases with increases temperature.

8. The unit of thermal conductivity doesn’t contain which parameter?

a) Watt

b) Pascal

c) Meter

d) Kelvin

View Answer

Explanation: Its unit is W/m K.

9. The temperatures on the two sides of a plane wall are t_{1} and t_{2} and thermal conductivity of the wall material is prescribed by the relation

K = k_{0 }e ^{(-x/δ)}

Where, k_{0} is constant and δ is the wall thickness. Find the relation for temperature distribution in the wall?

a) t _{1} – t _{x }/ t _{1} – t _{2} = x

b) t _{1} – t _{x }/ t _{1} – t _{2} = δ

c) t _{1} – t _{x }/ t _{1} – t _{2} = δ/x

d) t _{1} – t _{x }/ t _{1} – t _{2} = x/δ

View Answer

Explanation: Q = -k A d t/d x = -k

_{0 }e

^{(-x/δ)}d t/d x. Separating the variables and upon integration, we get Q/k

_{0 }A = (t

_{1}– t

_{2})/ δ (e – 1). Therefore heat transfer through the wall, Q = k

_{0 }A (t

_{1}– t

_{2})/ δ (e – 1). At x = x and t = t

_{x }we get the answer.

10. “If β is less than zero, then with respect to the relation k = k_{0} (1 + β t), conductivity depends on surface area”.

a) True

b) False

View Answer

Explanation: k decreases with increasing temperature.

**Sanfoundry Global Education & Learning Series – Heat Transfer.**

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