Heat Transfer MCQ’s – Heat Dissipation from a Fin Insulated at the Tip
The relevant boundary conditions in case of heat dissipation from a fin insulated at the tip are
a) t = t 0 at x = 0 and d t/d x = 0 at x = 0
b) t = t 0 at x = 0 and d t/d x = 0 at x = 1
c) t = t 0 at x = 1 and d t/d x = 0 at x = 1
d) t = t 0 at x = infinity and d t/d x = 0 at x = infinity
Explanation: It should be at x = 0 and x = 1 respectively.
The temperature distribution in case of fin insulated at the tip is given by
a) t – t0/t0 – t a = cos h m (3 – x)/cos ml
b) t – t0/t0 – t a = cos h m (2 – x)/sin h ml
c) t – t0/t0 – t a = cos h m (l – x)/cos h ml
d) t – t0/t0 – t a = cos m (l – x)/sin ml
Explanation: It should contain cos h term and (1 – x) term.
The rate of heat transfer from the fin in case of fin insulated at the tip is
a) (h P k A)1/2 (t 0 – t a) tan h ml
b) (h k A)1/2 (t 0 – t a) tan h ml
c) (h P A)1/2 (t 0 – t a) tan h ml
d) (h P k)1/2 (t 0 – t a) tan h ml
Explanation: It should contains all the terms i.e. h, A, P, k.
“Fin is insulated at the tip”. What does that mean?
a) Less heat is transferred from the tip
b) Heat will transferred from tip only
c) More heat is transferred from the tip
d) No heat is transferred from the tip
Explanation: The fin is of finite length with the tip insulated and so no heat is transferred from the tip.
Find the heat transfer rate from a hot surface for 6 fins of 10 cm length? The base temperature of the fin is maintained at 200 degree Celsius and the film is exposed to a convection environment at 15 degree Celsius with convective coefficient 25W/square m K. Each fin has cross-sectional area 2.5 square centimeter and is made of a material having thermal conductivity 250W/m K
Explanation: n = 6 and l = 10 cm = 0.1 m, ml = 0.4472, Q = 6[(250) (2.5 * 10– 4) (4.472) (200 – 15) tan h (0.4472) = 130.18W.
An array of 10 fins of anodized aluminum (k = 180W/m K) is used to cool a transistor operating at a location where the ambient conditions correspond to temperature 35 degree Celsius and convective coefficient 12W/square m K. The distance AB is 3 mm, EF is 0.4 mm. The length of the fin is 5 mm and has its base at 60 degree Celsius. Find the power dissipated by the fin array?
Explanation: P = 2(3 + 0.4) = 6.8 mm, A = (3) (0.4) = 1.2 square meter, m = 19.44 per meter. So, Q = k A m (t 0 –ta) tan h ml = 0.0786, therefore heat loss from the array of 10 fins = (0.0786) (10) = 0.786W.
An electronic semiconductor device generates 0.16 k J/hr of heat. To keep the surface temperature at the upper safe limit of 75 degree Celsius, it is desired that the heat generated should be dissipated to the surrounding environment which is at 30 degree Celsius. The task is accomplished by attaching aluminum fins, 0.5 square mm and 10 mm to the surface. Work out the number of fins if thermal conductivity of fin material is 690W/m K and the heat transfer coefficient is 45k J/square m hr K. Neglect the heat loss from the tip of the fin
Explanation: P = 2(0.5 + 0.5) = 2 mm, A = (0.5) (0.5) = 0.25 square meter. m = 22.85 per meter, so Q = k A m (t 0 –ta) tan h ml = 39.77 * 10-3 k J/hr per fin. So number of fins = 0.16/39.77 * 10 -3 = 4.02.
A rod of 10 mm diameter and 80 mm length with thermal conductivity 16W/ m K protrudes from a surface at 160 degree Celsius. The rod is exposed to air at 30 degree Celsius with a convective coefficient of 25W/square m K. How does the heat flow from this rod get affected if the same material volume is used for two fins of the same length? Assume short fin with insulated end
a) 12.25 %
b) 25.6 %
c) 23.4 %
d) 21.2 %
Explanation: Case 1 – m1 = 25 per meter, m1 l = 25 * 0.08 = 2. Therefore, Q1 = 3.935W Case 2 – d =0.00707 m, m2 = 29.73 per meter, m2 l = 2.378. Therefore, Q2 = 2.385W % increase in heat flow = 4.77 – 3.935/3.935 = 0.21.
Two rods A and B of the same length and diameter protrude from a surface at 120 degree Celsius and are exposed at air at 25 degree Celsius. The temperatures measured at the end of the rods are 50 degree Celsius and 75 degree Celsius. If thermal conductivity of material A is 20W/ m K, calculate it for B
a) 31.13W/m K
b) 41.13W/m K
c) 51.13W/m K
d) 61.13W/m K
Explanation: α/ α0 = t – t a/t 0– ta = 1/cos h ml. For rod A, cos h m1 l =3.8. Similarly for rod 2, cos h m2 l = 1.9, m1/m2 = 1.599. So k2 = k1 (1.599)2 = 51.13W/m K.
A centrifugal pump which circulates a hot liquid metal at 500 degree Celsius is driven by a 3600 rpm electric motor. The motor is coupled to the pump impeller by a horizontal steel shaft of dia 25 mm. Let us assume the motor temperature as 60 degree Celsius with the ambient air at 25 degree Celsius, what length of shaft should be specified between the motor and the pump? It may be presumed that the thermal conductivity of the shaft material is 35W/m K and that the convective film coefficient between the steel shaft and the ambient air is 15.7W/square m K
a) 38.96 cm
b) 54.76 cm
c) 23.76 cm
d) 87.43 cm
Explanation: 60 – 25/500 – 25 = cos h m (l – l)/cos h ml = 1/cos h ml, so ml = 3.3. For a circular shaft of diameter d, P/A = 4/d, m = 8.47 per meter. So, l = 3.3/8.47 = 38.96 cm.