# Heat Transfer Questions and Answers – Periodic Variation

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This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Periodic Variation”.

1. When the surface temperature variation inside a solid are periodic in nature, the profile of temperature variation with time may assume
a) Triangular
b) Linear
c) Parabolic
d) Hyperbolic

Explanation: Any type of waveform can be analyzed and resolved into an infinite number of sine and cosine waves.

2. The surface temperature oscillates about the mean temperature level in accordance with the relation
a) α S,T – α S,A = 2 sin (2 π n T)
b) α S,T – α S,A = 5 sin (2 π n T)
c) α S,T – α S,A = sin (2 π n T)
d) α S,T – α S,A = 3 sin (2 π n T)

Explanation: α S,T = t S,T – t M.

3. The temperature variation of a thick brick wall during periodic heating or cooling follows a sinusoidal waveform. During a period of 24 hours, the surface temperature ranges from 25 degree Celsius to 75 degree Celsius. Workout the time lag of the temperature wave corresponding to a point located at 25 cm from the wall surface. Thermo-physical properties of the wall material are; thermal conductivity = 0.62 W/m K; specific heat = 450 J/kg K and density = 1620 kg/m3
a) 3.980 hour
b) 6.245 hour
c) 2.648 hour
d) 3.850 hour

Explanation: d T = x/2 (1/α π n) ½ where x = 0.25 m and n = frequency.

4. A single cylinder 2-stroke engine operates at 1500 rpm. Calculate the depth where the temperature wave due to variation in cylinder is damped to 1% of its surface value. For the cylinder material, thermal diffusivity = 0.042 m2/hr
a) 0.1996 cm
b) 0.3887 cm
c) 0.2774 cm
d) 0.1775 cm

Explanation: α X,A = α S,A exponential [-x (π n/α) ½] where frequency = 1500 * 60.

5. The temperature distribution at a certain time instant through a 50 cm thick wall is prescribed by the relation
T = 300 – 500 x – 100 x2 + 140 x3
Where temperature t is in degree Celsius and the distance x in meters has been measured from the hot surface. If thermal conductivity of the wall material is 20 k J/m hr degree, calculate the heat energy stored per unit area of the wall
a) 4100 k J/hr
b) 4200 k J/hr
c) 4300 k J/hr
d) 4400 k J/hr

Explanation: d t/d x = -500 + 200 x + 420 x2. Now heat storage rate = Q IN – Q OUT = 10000 – 5900 = 4100 k J/hr.

6. A large plane wall, 40 cm thick and 8 m2 area, is heated from one side and temperature distribution at a certain time instant is approximately prescribed by the relation
T = 80 – 60 x +12 x2 + 25 x3 – 20 x4
Where temperature t is in degree Celsius and the distance x in meters. Make calculations for heat energy stored in the wall in unit time.
For wall material:
Thermal conductivity = 6 W/m K and thermal diffusivity = 0.02 m2/hr.
a) 870.4 W
b) 345.6 W
c) 791.04 W
d) 238.5 W

Explanation: Q IN = – k A (d t/d x)X = 0 = 2880 W and Q OUT = – k A (d t/d x)X = 0.4 = 2088.96 W.

7. Consider the above problem, calculate rate of temperature change at 20 cm distance from the side being heated
a) 0.777 degree Celsius/hour
b) 0.888 degree Celsius/hour
c) 0.999 degree Celsius/hour
d) 0.666 degree Celsius/hour

Explanation: d t/d T = α d 2t/d x 2 = 0.888 degree Celsius/hour.

8. At a certain time instant, the temperature distribution in a long cylindrical fire tube can be represented approximately by the relation
T = 650 + 800 r – 4250 r2
Where temperature t is in degree Celsius and radius r is in meter. Find the rate of heat flow such that the tube measures: inside radius 25 cm, outside radius 40 cm and length 1.5 m.
For the tube material
K = 5.5 W/m K
α = 0.004 m2/hr
a) 3.672 * 10 8 W
b) 3.672 * 10 2 W
c) 3.672 * 10 5 W
d) – 3.672 * 10 5 W

Explanation: Q = – k A (d t/d r), Rate of heat storage = Q IN – Q OUT = – 3.672 * 10 5 W.

9. Consider he above problem, find the rate of change of temperature at the inside surface of the tube
a) – 35 degree Celsius/hour
b) – 45 degree Celsius/hour
c) – 55 degree Celsius/hour
d) – 65 degree Celsius/hour

Explanation: d t/d T = α [d 2t/d r2 + d t/r d r] = – 55 degree Celsius/hour.

10. Time lag is given by the formula
a) x/2 [1/ (α π n) ½].
b) x/3 [1/ (α π n) ½].
c) x/4 [1/ (α π n) ½].
d) x/5 [1/ (α π n) ½].

Explanation: The time interval between the two instants is called the time lag.

Sanfoundry Global Education & Learning Series – Heat Transfer.

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