1. The heat transfer process between body and its ambient is governed by an Internal Conductive Resistance (ICR) and an External Conductive Resistance (ECR). The body can be considered to be a lumped heat capacity system is
(a) ICR > ECR
(b) ICR is marginally smaller than ECR
(c) ICR = ECR
(d) ICR is negligible
(1 Mark, 1989)

Ans: d
2. Biot number signifies
(a) The ratio of heat conducted to heat convected
(b) The ratio of heat convected to heat conducted
(c) The ratio of external convective resistance to internal conductive resistance
(d) The ratio of internal conductive resistance to external convective resistance
(1 Mark, 1991)

Ans: d
3. Two rods, one of length L and the other of length 2L are made of the same material and have the same diameter. The two ends of the longer rod are maintained at 100°C. One end of the shorter end is maintained at 100°C while the other end is insulated. Both the rods are exposed to the same environment at 40°C. The temperature at the insulated end of the shorter rod is measured to be 55°C. The temperature at the midpoint of the longer rod would be
(a) 40°C
(b) 50°C
(c) 55°C
(d) 100°C
(2 Mark, 1992)

Ans: c
4. When the fluid velocity is doubled, the thermal time constant of a thermometer used for measuring the fluid temperature reduces by a factor of 2. (T/F)
(1 Mark, 1994)

Ans: F
Explanation:
We know, Time constant = \frac { \rho Vc }{ hA }
From the above relation, we can not tell the relationship between velocity and time constant.
So, we can not tell the change in time constant with velocity.
5. Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different temperature is valid when
(a) Biot number < 0.1
(b) Biot number > 0.1
(c) Fourier number < 0.1
(d) Fourier number > 0.1
(1 Mark, 2001)

Ans: a
Explanation:
6. The value of Biot number is very small (less than 0.1) when
(a) The convective resistance of the fluid is negligible
(b) The conductive resistance of the fluid is negligible
(c) The conductive resistance of the solid is negligible
(d) None of these
(1 Mark, 2002)

Ans: c
7. A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer coefficient on the bead surface is 400 W/m^{2} Thermophysical properties of thermocouple material are k = 20 W/m^{2}K, C = 400 J/kgK and ρ = 8500 kg/m^{3}. If the thermocouple initially at 30^{0}C is placed in a hot stream of 300^{0}C, the time taken by the bead to reach 298^{0}C, is
(a) 2.35 s
(b) 4.9 s
(c) 14.7 s
(d) 29.4 s
(2 Mark, 2004)

Ans: b
Explanation:
We know: \frac { { TT }_{ \infty } }{ { T }_{ i }{ T }_{ \infty } } ={ e }^{ \frac { hA }{ \rho Vc } t }
=> \frac { { TT }_{ \infty } }{ { T }_{ i }{ T }_{ \infty } } ={ e }^{ \frac { h\times \left( \pi { D }^{ 2 } \right) }{ \rho \times \frac { \pi }{ 6 } { D }^{ 3 }\times c } \times t }
=> \frac { 298300 }{ 30300 } ={ e }^{ \frac { 400\times \left( \pi { \times 0.706 }^{ 2 } \right) }{ 8500\times \frac { \pi }{ 6 } { 0.706 }^{ 3 }\times 400 } \times t }
=> t = 4.9 s
8. A small copper ball of 5 mm diameter at 500K is dropped into an oil bath whose temperature is 300K. the thermal conductivity of copper is 400 W/m.K, its density 9000 kg/m^{3} and its specific heat 385 J/kg.K. if the heat transfer coefficient is 250 W/m^{2}.K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be, in K/s
(a) 8.7
(b) 13.9
(c) 17.3
(d) 27.7
(2 Mark, 2005)

Ans: c
Explanation:
Mass of the copper ball = \rho V = \rho \times \frac { 4 }{ 3 } \times \pi \times { r }^{ 3 }
=> m = 9000 \times \frac { 4 }{ 3 } \times \pi \times { 0.0025 }^{ 3 } = 0.589\times { 10 }^{ 4 } kg
We know: hA(T – T_{0}) = mc\frac { dT }{ dt }
At the beginning of the cooling, i.e. at t = 0, T = T_{i} = 500K
Now, hA(T_{i} – T_{o}) = mc{ \left( \frac { dT }{ dt } \right) }_{ t=0 }
=> { \left( \frac { dT }{ dt } \right) }_{ t=0 } =\frac { hA\left( { T }_{ i}{ T }_{ o } \right) }{ mc }
=> { \left( \frac { dT }{ dt } \right) }_{ t=0 }=\frac { 250\times \pi \times { 0.005 }^{ 2 }\left( 500300 \right) }{ 0.589\times { 10 }^{ 4 }\times 385 }
=> { \left( \frac { dT }{ dt } \right) }_{ t=0 } = 17.3 s
[ve sign shows that the temperature of body is decreasing with increase in time.]
9. The average heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assume the plate temperature to be uniform at any instant of time and radiation heat exchange with the surroundings negligible. The ambient temperature is 25^{0}C, the plate has a total surface area of 0.1 m^{2} and a mass of 4 kg. The specific heat of the plate material is 2.5 kJ/kgK. The convective heat transfer coefficient in W/m^{2}K, at the instant when the plate temperature is 225^{0}C and the change in plate temperature with time dT/dt = – 0.02 K/s, is
(a) 200
(b) 20
(c) 15
(d) 10
(2 Mark, 2007)

Ans: d
Explanation:
Heat transfer by convection = Rate of decrease of internal energy
=> hA(T_{2} – T_{1}) = mc\frac { dT }{ dt }
=> h * 0.1 * (225 – 25) = – 4 * 2500 * (0.02)
=> h = 10 W/m^{2}K
10. Which one of the following configurations has the highest fin effectiveness?
(a) Thin, closely spaced fins
(b) Thin, widely spaced fins
(c) Thick, widely spaced fins
(d) Thick, closely spaced fins
(1 Mark, 2012)

Ans: a
11. A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient (h) = 20 W/m^{2}. The thermophysical properties of steel are: density (ρ) = 7800 kg/m^{3}, conductivity (k) = 40 W/mK and specific heat (c) = 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is
(a) 519
(b) 931
(c) 1195
(d) 2144
(2 Mark, 2013)

Ans: d
Explanation:
We know: \frac { { TT }_{ \infty } }{ { T }_{ i }{ T }_{ \infty } } ={ e }^{ \frac { hA }{ \rho Vc } t }
=> \frac { 43030 }{ 103030 } ={ e }^{ \frac { h\times \left( \pi { D }^{ 2 } \right) }{ \rho \times \frac { \pi }{ 6 } { D }^{ 3 }\times c } \times t }
=> \frac { 400 }{ 1000 } ={ e }^{ \frac { 20\times \left( \pi { \times 0.06 }^{ 2 } \right) }{ 7800\times \frac { \pi }{ 6 } \times 0.06^{ 3 }\times 600 } \times t }
=> t = 2144 second
12. Biot number signifies the ratio of
(a) Convective resistance in the fluid to conductive resistance in the solid
(b) Conductive resistance in the solid to convective resistance in the fluid
(c) Ineritia force to viscous force in the fluid
(d) Buoyancy force to viscous force in the fluid
(1 Mark, 2014[1])

Ans: b
13. A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in a water environment at 300 K. The convective heat transfer coefficient is 1000 W/m^{2}K. Thermal conductivity of steel is 40 W/mK. The time constant for the cooling process is 16 s. The time required (in s) to reach the final temperature is _______.
(2 Mark, 2016[1])

Ans: 42.23
Explanation:
Given, Time constant for the cooling process = \frac { \rho Vc }{ hA } = 16 s
Now, \frac { { TT }_{ \infty } }{ { T }_{ i }{ T }_{ \infty } } ={ e }^{ \frac { hA }{ \rho Vc } t }
=> \frac { { 350300 } }{ 1000300 } ={ e }^{ \frac { 1 }{ 16 } t }
=> t = 42.23 s
14. Two cylindrical shafts A and B at the same initial temperature are simultaneously placed in a furnace. The surfaces of the shafts remain at the furnace gas temperature at all times after they are introduced into the furnace. The temperature variation in the axial direction of the shafts can be assumed to be negligible. The data related to shafts A and B is given in the following Table.
The temperature at the centerline of the shaft A reaches 400^{o}C after two hours. The time required (in hours) for the centerline of the shaft B to attain the temperature of 400^{o }C is _______.
(2 Mark, 2016[2])

Ans: 2.5
Explanation:
Thermal diffusivity for both the shaft:
{ \alpha }_{ A }={ \left( \frac { k }{ \rho { c }_{ p } } \right) }_{ A }=\frac { 40 }{ 2\times { 10 }^{ 6 } } =20\times { 10 }^{ 6 }Again, { \alpha }_{ B }={ \left( \frac { k }{ \rho { c }_{ p } } \right) }_{ B }=\frac { 20 }{ 2\times { 10 }^{ 7 } } =10\times { 10 }^{ 7 }
To get the answer of this question, we must equate the Fourier number for both the shaft.
So, { \left( \frac { \alpha t }{ { L }^{ 2 } } \right) }_{ A }={ \left( \frac { \alpha t }{ { L }^{ 2 } } \right) }_{ B }
=> \frac { 20\times { 10 }^{ 6 }\times 2 }{ { 0.4 }^{ 2 } } =\frac { 10\times { 10 }^{ 7 }\times { t }_{ B } }{ { 0.1 }^{ 2 } }
=> t_{B }= 2.5 hours
15. A cylindrical steel rod, 0.01 m in diameter and 0.2 m in length is first heated to 750^{0 }C and then immersed in a water bath at 100^{0}C. The heat transfer coefficient is 250 W/m^{2}K. The density, specific heat and thermal conductivity of steel are ρ = 7801 kg/m^{3}, c = 473 J/kgK, and k = 43 W/mK, respectively. The time required for the rod to reach 300^{0} C is ________ seconds.
(2 Mark, 2016[3])

Ans: 42.42
Explanation:
We know: \frac { { TT }_{ \infty } }{ { T }_{ i }{ T }_{ \infty } } ={ e }^{ \frac { hA }{ \rho Vc } t }
=> \frac { 300100 }{ 750100 } ={ e }^{ \frac { 250\times \left( \pi DL+2\times \frac { \pi }{ 4 } { D }^{ 2 } \right) }{ 7801\times \frac { \pi }{ 4 } { D }^{ 2 }L\times 473 } \times t }
=> \frac { 300100 }{ 750100 } ={ e }^{ \frac { 250\times \left( \pi \times 0.01\times 0.2+2\times \frac { \pi }{ 4 } \times { 0.01 }^{ 2 } \right) }{ 7801\times \frac { \pi }{ 4 } \times 0.01^{ 2 }\times 0.2\times 473 } \times t }
=> t = 42.42 seconds
16. A metal ball of diameter 60 mm is initially at 220^{o} The ball is suddenly cooled by an air jet of 20^{o}C. The heat transfer coefficient is 200 W/m^{2}K. The specific heat, thermal conductivity and density of the metal ball are 400 J/kgK, 400 W/mK and 9000 kg/m^{3} respectively. The ball temperature (in ^{o}C) after 90 seconds will be approximately
(a) 141
(b) 163
(c) 189
(d) 210
(2 Mark, 2017[2])

Ans: a
Explanation:
\frac { V }{ A } =\frac { D }{ 6 } =\frac { 0.06 }{ 6 } = 0.01 mm
\frac { hA }{ \rho V{ c }_{ P } } =\frac { h }{ \rho (V/A){ c }_{ P } } =\frac { 200 }{ 9000\times 0.01\times 400 } =\frac { 1 }{ 180 }The ball temperature after 90 second:
\frac { { TT }_{ \infty } }{ { T }_{ i }{ T }_{ \infty } } ={ e }^{ \frac { hA }{ \rho Vc } t }=> \frac { T20 }{ 22020 } ={ e }^{ \frac { 1 }{ 180 } \times 90 }
=> T = 141.306^{0}C