Heat Transfer MCQ’s – Periodic Variation
1 - Question
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
View Answer
Explanation: Any type of waveform can be analyzed and resolved into an infinite number of sine and cosine waves.
2 - Question
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)br/> c) α S,T – α S,A = sin (2 π n T)br/> d) α S,T – α S,A = 3 sin (2 π n T)
View Answer
Explanation: α S,T = t S,T – t M.
3 - Question
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
View Answer
Explanation: d T = x/2 (1/α π n) ½ where x = 0.25 m and n = frequency.
4 - Question
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
View Answer
Explanation: α X,A = α S,A exponential [-x (π n/α) ½] where frequency = 1500 * 60.
5 - Question
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
View Answer
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 - Question
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
View Answer
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 - Question
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
View Answer
Explanation: d t/d T = α d 2t/d x 2 = 0.888 degree Celsius/hour.
8 - Question
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
View Answer
Explanation: Q = – k A (d t/d r), Rate of heat storage = Q IN – Q OUT = – 3.672 * 10 5 W.
9 - Question
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
View Answer
Explanation: d t/d T = α [d 2t/d r2 + d t/r d r] = – 55 degree Celsius/hour.
10 - Question
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].
View Answer
Explanation: The time interval between the two instants is called the time lag.