# Heat Transfer Questions and Answers – Conduction Through a Plane Wall

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This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Conduction Through A Plane Wall”.

1. 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)

Explanation: This is one dimensional heat conduction through a homogenous, isotropic wall with constant thermal conductivity.

2. The temperature distribution in a large thin plate with uniform surface temperature will be
(Assume steady state condition)
a) Logarithmic
b) Hyperbolic
c) Parabolic
d) Linear

Explanation: The temperature increases with increasing value of x. Temperature gradient will be positive i.e. linear.

3. Let us assume two walls of same thickness and cross-sectional area having thermal conductivities in the ratio 1/2. Let us say there is same temperature difference across the wall faces, the ratio of heat flow will be
a) 1
b) 1/2
c) 2
d) 4

Explanation: Q1 = k1 A1 d t11 and Q2 = k2A2 d t22
Now, δ1 = δ2 and A1 = A2 and d t1 = d t2
So, Q1/Q2 = ½.
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4. The interior of an oven is maintained at a temperature of 850 degree Celsius by means of a suitable control apparatus. The oven walls are 500 mm thick and are fabricated from a material of thermal conductivity 0.3 W/m degree. For an outside wall temperature of 250 degree Celsius, workout the resistance to heat flow
a) 0.667 degree/W
b) 1.667 degree/W
c) 2.667 degree/W
d) 3.667 degree/W

Explanation: R t = 0.5/0.3 = 1.667 degree/W.

5. A plane slab of thickness 60 cm is made of a material of thermal conductivity k = 17.45 W/m K. Let us assume that one side of the slab absorbs a net amount of radiant energy at the rate q = 530.5 watt/m2. If the other face of the slab is at a constant temperature t2 = 38 degree Celsius. Comment on the temperature with respect to the slab?
a) 87.5 degree Celsius
b) 32 degree Celsius
c) 47.08 degree Celsius
d) 32.87 degree Celsius

Explanation: Heat flux, q = k (t s – t f) / Thickness. So, t s = 56.17 degree Celsius. Now, t = t s + (t f – t s) x/Thickness.

6. The rate of heat transfer for a plane wall of homogenous material with constant thermal conductivity is given by
a) Q = kA (t1-t2)/δ
b) Q = 2kAx/ δ
c) Q = 2kAδx
d) Q = 2k/δ x

Explanation: Computations for heat flow can be made by substituting the value of temperature gradient into the general equation. The heat flow somehow doesn’t depend on x.

7. In case of homogeneous plane wall, there is a linear temperature distribution given by
a) t = t1 + (t2-t1) δ/x
b) t = t2 – (t2-t1) x/ δ
c) t = t1 + (t2-t1) x
d) t = t1 + (t2-t1) x/ δ

Explanation: The expression for steady state temperature distribution can be set up by integrating the Fourier rate equation.

8. The rate of convective heat transfer between a solid boundary and adjacent fluid is given by
a) Q = h A (t s – t f)
b) Q = h A
c) Q = (t s – t f)
d) Q = h (t s – t f)

Explanation: Here, h is heat transfer coefficient i.e. convective.

9. A homogeneous wall of area A and thickness δ has left and right hand surface temperatures of 0 degree Celsius and 40 degree Celsius. Determine the temperature at the center of the wall
a) 10 degree Celsius
b) 20 degree Celsius
c) 30 degree Celsius
d) 40 degree Celsius

Explanation: At the midpoint x = δ/2. So, temperature = 40 + (0 – 40)/2 = 20 degree Celsius.

10. A rod of 3 cm diameter and 20 cm length is maintained at 100 degree Celsius at one end and 10 degree Celsius at the other end. These temperature conditions are attained when there is heat flow rate of 6 W. If cylindrical surface of the rod is completely insulated, determine the thermal conductivity of the rod material
a) 21.87 W/m degree
b) 20.87 W/m degree
c) 19.87 W/m degree
d) 18.87 W/m degree

Explanation: Q = k A C (t 1 – t 2)/δ = 0.318 k.

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