Heat Transfer MCQ’s – Transient Heat Conduction in Infinite Thick Solids
“An infinite solid is one which extends itself infinitely in all directions of space”. Identify the correct option
Explanation: If the infinite solid is split in the middle by the plane, each half is known as semi-infinite solid.
The boundary conditions in case of transient heat conduction in infinite thick solids are (i) t (x = 0) = t i (ii) t (0, T) = t a for T greater than zero (iii) t (infinity, T) = t i for T greater than zero Identify the correct statements
a) i and ii
b) i and iii
c) ii and iii
d) i, ii and iii
Explanation: Boundary conditions are those by which we could find out the values of constant.
The perturbation time varies as
a) d 2/α
b) 2 d 2/α
c) 3 d 2/α
d) 4 d 2/α
Explanation: At penetration depth d, there will be 1% perturbation.
The temperature perturbation at all the surface has penetrated to the depth
a) 1.6 (α T) 1/2
b) 2.6 (α T) 1/2
c) 3.6 (α T) 1/2
d) 4.6 (α T) ½
Explanation: At penetration depth d, there will be 1% perturbation at a time t.
A water line is buried underground in dry soil that has an assumed initial temperature of 4.5 degree Celsius. The pipe may have no flow through it for long period of time, yet it will not be drained in order that no freezing occurs, the pipe must be kept at a temperature not lower than 0 degree Celsius. The pipe is to be designed for a 30 hour period at the beginning of which the soil surface temperature suddenly drops to – 17 degree Celsius. Workout the minimum earth covering needed above the water pipe so as to prevent the possibility of freezing during 36 hour cold spell. The soil in which the pipe is buried has the following properties
Density = 640 kg/m3 Specific heat = 1843J/kg degree Thermal conductivity = 0.345 W/m degree
a) 0.25 m
b) 0.35 m
c) 0.45 m
d) 0.55 m
Explanation: t – t a/t i – t a = erf [x/2 (α T) ½].
At the penetration depth d, there will be 1% perturbation at a time given by
a) 4 d 2/13 α
b) 3 d 2/13 α
c) 2 d 2/13 α
d) d 2/13 α
Explanation: d/2 (α t) ½ = 1.8.
A large steel ingot, which has been uniformly heated to 750 degree Celsius, is hardened by quenching it in an oil bath that is maintained at 25 degree Celsius. What length of time is required for the temperature to reach 600 degree Celsius at a depth of 1 cm? Thermal diffusivity for the steel ingot is 1.21 * 10 -5 m2/s. The ingot may be approximated as a flat plate
a) 4.55 seconds
b) 3.55 seconds
c) 2.55 seconds
d) 1.55 seconds
Explanation: t – t a/t I – t a = erf [x/2 (α T) ½].
A mild steel plate 5 cm thick and initially at 40 degree Celsius temperature is suddenly exposed on one side to a fluid which causes the surface temperature to increase to and remain at 90 degree Celsius. Calculate maximum time that the slab be treated as a semi-infinitely body For steel, thermal diffusivity = 1.25 * 10 –5 m2/s
a) 100 seconds
b) 200 seconds
c) 300 seconds
d) 400 seconds
Explanation: T MAX = δ 2/4 α (0.5)2 = 200 seconds.
Consider the above problem, find the temperature at the center of the slab one minute after the change in surface temperature
a) 66 degree Celsius
b) 76 degree Celsius
c) 86 degree Celsius
d) 96 degree Celsius
Explanation: t – t a/t I – t a = erf [x/2 (α T) ½]. Therefore, temperature at the center of the slab is t a + 0.48 (t I – t a).
Water pipes are to be buried underground in a wet soil (thermal diffusivity = 2.78 * 10 -3 m2/s) which is initially at 4.5 degree Celsius. The soil surface temperature suddenly drops to -5 degree Celsius and remains at this value for 10 hours. Calculate the minimum depth at which the pipe be laid if the surrounding soil temperature is to remain above 0 degree Celsius. The soil may be considered as semi-infinite solid
a) 0.467 m
b) 0.367 m
c) 0.267 m
d) 0.167 m
Explanation: t – t a/t I – t a = erf [x/2 (α T) ½]. Thus x = 2 (0.50) (α T) ½ = 0.167 m.