Heat Transfer MCQ’s – Heat Dissipation from an Infinitely Long Fin
In heat dissipation from an infinitely long fin, the boundary conditions are
a) t = t0 at x = infinity and t = ta at x = 0
b) t = t0 at x = 0 and t = ta at x = infinity
c) t = t0 at x = 0 and t = ta at x = 0
d) t = t0 at x = infinity and t = ta at x = infinity
Explanation: These conditions must be approached when ml is greater than 5.
The temperature distribution in case of infinitely long fin is
a) t – t a/t 0 – t a = mx
b) t – t a/t 0 – t a = -mx
c) t – t a/t 0 – t a = e-m x
d) t – t a/t 0 – t a = log (m x)
Explanation: Exponential curve should be here.
The rate of heat transfer in case of infinitely long fin is given by
a) (h P k A) 1/2 (t 0 – t a)
b) (h P A) 1/2 (t 0 – t a)
c) (P k A) 1/2 (t 0 – t a)
d) (h k A) 1/2 (t 0 – t a)
Explanation: It should contain all the terms i.e. h, p, k and A.
Let us say there are two rods having same dimensions, one made of brass (k = 85W/m K) and the other of copper (k = 375W/m K), having one of their ends inserted into a furnace. At a section 10.5 cm away from the furnace, the temperature of brass rod is 120 degree Celsius. Find the distance at which the same temperature would be reached in the copper rod? Both ends are exposed to the same environment
a) 12.54 cm
b) 45.87 cm
c) 12.34 cm
d) 22.05 cm
Explanation: For brass rod, 120 = t a + (t 0 – t a) e –m l and for copper rod, 120 = t a + (t 0 – t a) e –M L So L = L 0 (k 2/k 1) = 22.05 cm.
Three rods, one made of silver (l = 420W/m K), second made of aluminum (k = 210W/m K) and the third made of iron (k = 70W/m K) are coated with a uniform layer of wax all around. The rods are placed vertically in a boiling water bath with 250 mm length of each rod projecting outside. If all the rods are having following dimensions i.e. diameter = 15 mm and length = 300 mm and have identical surface coefficient 12.5W/ m2 K, work out the ratio of lengths up to which wax will melt on each rod
Explanation: k 1/l 12 = k 2/l 22 = k 3/l 32.
Let us assume there are two pieces of copper wire 0.1625 cm in diameter with a device that melts it at 195 degree Celsius. The wires are positioned vertically in air at 24 degree Celsius and the heat transfer coefficient of the wire is 17 W/ square m K. Let us say k = 335W/m K i.e. of wire. Find out the energy input?
Explanation: A C = π D 2/4 = 2.073 * 10 –6square meter. P = π D = 0.0051 m. Q FIN = k A C m (t2 – t 1) = 1.326W.
A rod of 10 mm square section and 160 mm length with thermal conductivity of 50W/m K protrudes from a furnace wall at 200 degree Celsius with convective coefficient 20 W/ square m K. Make calculations for the heat convective up to 80 mm length
Explanation: Q = k A C m (t 2 – t 1), m = (P h/k AC) 1/2 = 12.649 /m, so Q = 10.75W. At x = 80 mm, (m x) = 1.01192, so T – 30/200 – 30 = 0.3635. Therefore net heat is 10.75 – k A C m (t 0.08 – t a) = 6.84W.
A fin protrudes from a surface which is held at a temperature higher than that of its environment. The heat transferred away from the fin is
a) Heat escaping from the tip of the fin
b) Heat conducted along the fin length
c) Convective heat transfer from the fin surface
d) Sum of heat conducted along the fin length and that convected from the surface
Explanation: As the temperature is higher, so it’s convective.
The value of correction length for equilateral fin is
a) L C = 2 L + a/4(3)1/2
b) L C = L + a/4(3)1/2
c) L C = 3 L + a/4(3)1/2
d) L C = 6 L + a/4(3)1/2
Explanation: Area of triangle i.e. equilateral is (3)1/2/4.
The parameter m = (h P/k A C) 1/2 has been stated to increase in a long fin. If all other parameters are constant, then
a) Profile of temperature will remain the same
b) Along the length temperature drop will be less
c) The parameter influences the heat flow only
d) The temperature drop along the length will be steeper
Explanation: For an infinitely long fin t – t a/t 0 – t a = e – m x. Dimensionless temperature falls more with increase in factor m.