1. Thermal conductivity is lower for
(a) Wood
(b) Air
(c) Water at 100^{0}C
(d) Steam at 1 bar
(1 Mark, 1990)

Ans: b
Explanation:
Air has the lowest thermal conductivity.
Thermal conductivity of air at 20^{0}C is = 0.022 W/mk.
2. Match the property with their units.
(a) A1, B2, C6, D5
(b) A2, B5, C6, D1
(c) A2, B 6, C4, D1
(d) A1, B5, C3, D2
(2 Mark, 1991)

Ans: b
3. For a current carrying wire of 20 mm diameter exposed to air (h = 25 W/m^{2}K), maximum heat dissipation occurs when the thickness of insulation (k = 0.5 W/mK), is:
(a) 20 mm
(b) 10 mm
(c) 2.5 mm
(d) 0 mm
(2 Mark, 1993)

Ans: b
Explanation:
Given: Diameter of current carrying wire = 20 mm
radius of current carrying wire(r) = 20/2 = 10 mm
maximum heat dissipation occurs when radius of wire reaches to critical radius of insulation.
Critical radius (r_{c}) = k/h = 0.5/25 = 20 mm
Thickness of insulation = r_{c}– r = 20 – 10 = 10 mm
4. Two insulating materials of thermal conductivity K and 2K are available for lagging a pipe carrying a hot fluid. If the radial thickness of each material is the same.
(a) Material with higher thermal conductivity should be used for the inner layer and one with lower thermal conductivity for the outer.
(b) Material with lower thermal conductivity should be used for the inner layer and one with higher thermal conductivity for the outer.
(c) It is immaterial in which sequence the insulating materials are used
(d) It is not possible to judge unless numerical values of dimensions are given
(1 Mark, 1994)

Ans: b
5. For a given heat flow and for the same thickness, the temperature drop across the material will be maximum for
(a) Copper
(b) Steel
(c) Glasswool
(d) Refractory brick
(1 Mark, 1996)
6. The temperature variation under steady heat conduction across a composite slab of two materials with thermal conductivities K_{1} and K_{2} is shown in Figure. Then, which one of the following statements holds?
(a) K_{1} > K_{2}
(b) K_{1} = K_{2}
(c) K_{1} = 0
(d) K_{1} < K_{2}
(2 Mark, 1998)
7. It is proposed to coat a 1 mm diameter wire with enamel paint (k = 0.1 W/mK) to increase heat transfer with air. If the air side heat transfer coefficient is 100 W/m^{2}K, the optimum thickness of enamel paint should be
(a) 0.25 mm
(b) 0.5 mm
(c) 1 mm
(d) 2 mm
(2 Mark, 1999)
8. In descending order of magnitude, the thermal conductivity of (a) Pure iron (b) Liquid water (c) Saturated water vapour and (d) aluminum can be arranged as
(a) abcd
(b) bcad
(c) dabc
(d) dcba
(1 Mark, 2001)
Linked Answer Questions Q.9 – 10.
Heat is being transferred by convection from water at 48°C to a glass plate whose surface that is exposed to the water is at 40°C. The thermal conductivity of water is 0.6 W/mK and the thermal conductivity of glass is 1.2 W/mK. The spatial gradient of temperature in the water at the waterglass interface is dT/dy=1×10^{4} K/m.
9. The value of the temperature gradient in the glass at the waterglass interface in K/m is
(a) 2 × 10^{4}
(b) 0.0
(c) 0.5 × 10^{4}
(d) 2 × 10^{4}
(2 Mark, 2003)
10. The heat transfer coefficient h in W/m^{2} K is
(a) 0.0
(b) 4.8
(c) 6
(d) 750
(2 Mark, 2003)
11. One dimensional unsteady state heat transfer equation for a sphere with heat generation at the rate of ‘q’ can be written as
(a) \frac { 1 }{ r } \frac { \partial }{ \partial r } \left( r\frac { \partial T }{ \partial r } \right) +\frac { q }{ k } =\frac { 1 }{ \alpha } \frac { \partial T }{ \partial r }
(b) \frac { 1 }{ { r }^{ 2 } } \frac { \partial }{ \partial r } \left( { r }^{ 2 }\frac { \partial T }{ \partial r } \right) +\frac { q }{ k } =\frac { 1 }{ \alpha } \frac { \partial T }{ \partial t }
(c) \frac { { \partial }^{ 2 }T }{ \partial { r }^{ 2 } } +\frac { q }{ k } =\frac { 1 }{ \alpha } \frac { \partial T }{ \partial t }
(d) \frac { { \partial }^{ 2 } }{ \partial { r }^{ 2 } } \left( rT \right) +\frac { q }{ k } =\frac { 1 }{ \alpha } \frac { \partial T }{ \partial t }
(1 Mark, 2004)

Ans: b
12. A stainless steel tube (k_{s} = 19 W/mK) of 2 cm ID and 5 cm OD is insulated with 3 cm thick asbestos (k_{a} = 0.2 W/mK). If the temperature difference between the innermost and outermost surfaces is 600^{0}C, the heat transfer rate per unit length is
(a) 0.94 W/m
(b) 9.44 W/m
(c) 944.72 W/m
(d) 9447.21 W/m
(2 Mark, 2004)
13. A well machined steel plate of thickness L is kept such that the wall temperatures are T_{h} and T_{c} as seen in the figure below. A smooth copper plate of the same thickness L is now attached to the steel plate without any gap as indicated in the figure below. The temperature at the interface is T_{i}. The temperatures of the outer walls are still the same at T_{h }and T_{c}. The heat transfer rates are q_{1} and q_{2} per unit area in the two cases respectively in the direction shown. Which of the following statements is correct?
(a) T_{h }> T_{i }> T_{c }and q_{1 }< q_{2}
(b) T_{h }< T_{i }< T_{c }and q_{1 }= q_{2}
(c) T_{h }= (T_{i }+ T_{c})/2 and q_{1 }> q_{2}
(d) T_{i }< (T_{h }+ T_{c})/2 and q_{1 }> q_{2}
(1 Mark, 2005)
14. In a case of one dimensional heat conduction in a medium with constant properties, T is the temperature at position x, at time t. then \frac { \partial T }{ \partial t } is proportional to
(a) \frac { T }{ x }
(b) \frac { \partial T }{ \partial x }
(c) \frac { { \partial }^{ 2 }T }{ \partial x\partial t }
(d) \frac { { \partial }^{ 2 }T }{ \partial { x }^{ 2 } }
(1 Mark, 2005)

Ans: d
15. Heat flows through a composite slab, as shown below. The depth of the slab is 1m. The k values are in W/mK. The overall thermal resistance in K/W is
(a) 17.2
(b) 21.9
(c) 28.6
(d) 39.2
(2 Mark, 2005)
16. In a composite slab, the temperature at the interface (T_{inter}) between two materials is equal to the average of the temperatures at the two ends. Assuming steady onedimensional heat conduction, which of the following statements is true about the respective thermal conductivities?
(a) 2k_{1} = k_{2}
(b) k_{1} = k_{2}
(c) 2k_{1} = 3k_{2}
(d) k_{1} = 2k_{2}
(1 Mark, 2006)
17. With an increase in thickness of insulation around a circular pipe, heat loss to surroundings due to
(a) Convection increases, while that due to conduction decreases
(b) Convection decreases, while that due to conduction increases
(c) Convection and conduction decreases
(d) Convection and conduction increases
(2 Mark, 2006)

Ans: a
Linked Answer Questions Q.18 – 19.
Consider steady onedimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m^{3}. The left and right faces are kept at constant temperatures of 160^{0}C and 120^{0}C respectively. The plate has a constant thermal conductivity of 200 W/mK.
18. The location of maximum temperature within the plate from its left face is
(a) 15 mm
(b) 10 mm
(c) 5 mm
(d) 0 mm
(2 Mark, 2007)
19. The maximum temperature within the plate in ^{0}C is
(a) 160
(b) 165
(c) 200
(d) 250
(2 Mark, 2007)
20. Steady two dimensional heat conduction takes place in the body shown in the figure below. The normal temperature gradients over surfaces P and Q can be considered to be uniform. The temperature gradient \frac { \partial T }{ \partial x } at surface Q is equal to 10 K/m. Surfaces P and Q are maintained at constant temperatures as shown in the figure, while the remaining part of the boundary is insulated. The body has a constant thermal conductivity of 0.1 W/m.K. The values of \frac { \partial T }{ \partial x } and \frac { \partial T }{ \partial y } at surface P are
(a) \frac { \partial T }{ \partial x } = 20 K/m, \frac { \partial T }{ \partial y } = 0 K/m
(b) \frac { \partial T }{ \partial x } = 0K/m, \frac { \partial T }{ \partial y } =10K/m
(c) \frac { \partial T }{ \partial x } = 10 K/m, \frac { \partial T }{ \partial y } =10 K/m
(d) \frac { \partial T }{ \partial x } = 0K/m, \frac { \partial T }{ \partial y } = 20 K/m
(2 Mark, 2008)

Ans: d
Explanation:
Heat always flows perpendicular to the crosssection area of any surface.
So, at surface p: \frac { \partial T }{ \partial x } = 0 K/m
Again, Q_{P }= Q_{Q}
=> { k }_{ p }\times { A }_{ P }\times { \left( \frac { \partial T }{ \partial y } \right) }_{ P }={ k }_{ Q }\times { A }_{ Q }\times { \left( \frac { \partial T }{ \partial x } \right) }_{ Q }
=> { \left( \frac { \partial T }{ \partial y } \right) }_{ P }=\frac { 0.1\times 2\times { 10 } }{ 0.1\times 1 }
=> { \left( \frac { \partial T }{ \partial y } \right) }_{ P } = 20 K/m
21. A coolant fluid at 30°C flows over a heated flat plate maintained at a constant temperature of 100°C. The boundary layer temperature distribution at a given location on the plate may be approximated as T = 30 + 70 exp(−y) where y (in m) is the distance normal to the plate and T is in °C. If thermal conductivity of the fluid is 1.0 W/mK, the local convective heat transfer coefficient (in W/m^{2}K) at that location will be
(a) 0.2
(b) 1
(c) 5
(d) 10
(1 Mark, 2009)

Ans: b
Explanation:
Energy balance in steady state condition at the surface:
kA{ \left( \frac { dT }{ dy } \right) }_{ y=0 }=hA\Delta T=> h=\frac { kA{ \left( \frac { dT }{ dy } \right) }_{ y=0 } }{ A\Delta T }
=> h=\frac { kA{ \left( \frac { d }{ dy } (30+70exp(y)) \right) }_{ y=0 } }{ A\Delta T }
=> h=\frac { kA{ \left( 70{ e }^{ y } \right) }_{ y=0 } }{ A\Delta T }
=> h=\frac { 70\times 1 }{ 10030 } = 1 W/m^{2}K
22. Consider steadystate heat conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides.
Given: h_{i} = 20 W/m^{2}K, h_{o} = 50 W/m^{2}K,
{ T }_{ \infty ,i } = 20 ^{0}C, { T }_{ \infty ,o } = 2 ^{0}C,
k_{1} = 20 W/mK, k_{2} = 50 W/mK,
L_{1} = 0.30 m and L_{2} = 0.15 m.
Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (in °C), of the two walls will be
(a) 0.50
(b) 2.75
(c) 3.75
(d) 4.50
(2 Mark, 2009)

Ans: c
Explanation:
Rate of heat flux (q) = \frac { { T }_{ \infty ,i }{ T }_{ \infty ,o } }{ \frac { 1 }{ { h }_{ i } } +\frac { { L }_{ 1 } }{ { k }_{ 1 } } +\frac { { L }_{ 2 } }{ { k }_{ 2 } } +\frac { 1 }{ { h }_{ o } } }
=> q = \frac { 20\left( 2 \right) }{ \frac { 1 }{ 20 } +\frac { 0.3 }{ 20 } +\frac { 0.15 }{ 50 } +\frac { 1 }{ 50 } } = 250 W/m^{2}
Again, q = { h }_{ i }\left( { T }_{ \infty ,i }{ T }_{ 1 } \right)
=> Surface temperature of wall 1 (T_{1}) = 7.5^{0}C
Again, q = \frac { { k }_{ 1 }({ T }_{ 1 }{ T }_{ i }) }{ { L }_{ 1 } } =\frac { 20(7.5{ T }_{ i }) }{ 0.3 }
=> T_{i }= 3.75^{0}C
23. A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400Wm^{−1}K^{−1}. One end of the fin is maintained at 130ºC and its remaining surface is exposed to ambient air at 30ºC. if the convective heat transfer coefficient is 40 Wm^{2}K^{1}, the heat loss (in W) from the fin is
(a) 0.08
(b) 5.0
(c) 7.0
(d) 7.8
(2 Mark, 2010)

Ans: b
Explanation:
Considering finite length of fin with insulated tip:
m = \sqrt { \frac { hP }{ kA } } =\quad \sqrt { \frac { h\times 2\pi r }{ k\times \pi { r }^{ 2 } } } =\sqrt { \frac { 2h }{ k{ r } } } =\sqrt { \frac { 2\times 40 }{ 400\times 2.5\times { 10 }^{ 3 } } } = 8.94
Heat loss by fin (Q_{fin}) = \sqrt { hPkA } \left( { T }_{ o }{ T }_{ \infty } \right) tanhmL
=> Q_{fin }= \sqrt { 40\times \left( 2\pi \times 2.5\times { 10 }^{ 3 } \right) \times 400\times \pi { \left( 2.5\times { 10 }^{ 3 } \right) }^{ 2 } } \left( 13030 \right) tanh\left( 8.94\times 0.1 \right)
=> Q_{fin }= 5 W
24. A pipe of 25 mm outer diameter carries steam. The heat transfer coefficient between the cylinder and surroundings is 5 W/m^{2}K . It is proposed to reduce the heat loss from the pipe by adding insulation having a thermal conductivity of 0.05 W/mK. Which one of the following statements is TRUE?
(a) The outer radius of the pipe is equal to the critical radius
(b) The outer radius of the pipe is less than the critical radius
(c) Adding the insulation will reduce the heat loss
(d) Adding the insulation will increase the heat loss
(2 Mark, 2011)

Ans: c
Explanation:
Critical radius of insulation (r_{c}) = \frac { k }{ h } =\frac { 0.05 }{ 5 } = 0.01 m = 10 mm
r_{c }< Outer radius of pipe (12.5 mm)
So, Adding the insulation will reduce the heat loss.
25. A spherical steel ball of 12mm diameter is initially at 1000K. It is slowly cooled in a surrounding of 300K. The heat transfer coefficient between the steel ball and the surrounding is 5 W/m^{2} The thermal conductivity of steel is 20W/mK. The temperature difference between the centre and the surface of the steel ball is
(a) Large because conduction resistance is far higher than the convective resistance
(b) Large because conduction resistance is far less than the convective resistance
(c) Small because conduction resistance is far higher than the convective resistance
(d) Small because conduction resistance is far less than the convective resistance
(2 Mark, 2011)

Ans: d
Explanation:
For the steel ball,
Characteristic length (L) = \frac { d }{ 6 } =\frac { 12 }{ 6 } = 2 mm = 0.002 m
Bi = \frac { hL }{ k } =\frac { 5\times 0.002 }{ 20 } = 0.0005
Again, Bi = \frac { Conduction\quad resistance }{ Covection\quad resistance } = 0.0005
So, Conduction resistance is very small than convective resistance.
26. Consider onedimensional steady state heat conduction along xaxis (0 ≤ x ≤ L), through a plane wall with the boundary surfaces (x = 0 and x = L) maintained at temperatures of 0°C and 100°C. Heat is generated uniformly throughout the wall. Choose the CORRECT statement.
(a) The direction of heat transfer will be from the surface at 100°C to the surface at 0°C.
(b) The maximum temperature inside the wall must be greater than 100°C.
(c) The temperature distribution is linear within the wall.
(d) The temperature distribution is symmetric about the midplane of the wall.
(1 Mark, 2013)

Ans: b
Explanation:
Option ‘b’ is correct because if heat is generated uniformly through out the wall, then the maximum temperature is always inside the wall and direction of heat transfer is from the point of maximum temperature to the wall.
27. Consider onedimensional steady state heat conduction, without heat generation, in a plane wall; with boundary conditions as shown in the figure below. The conductivity of the wall is given by k = k_{0} + bT; where k_{0} and b are positive constants, and T is temperature.
As x increases, the temperature gradient (dT/dx) will
(a) Remain constant
(b) Be zero
(c) Increase
(d) Decrease
(1 Mark, 2013)

Ans: d
Explanation:
Given, k = k_{o} + bT
Along x direction, the value of ‘k’ increases because the value of ‘T’ increases. (Given: T_{2 }> T_{1} )
We know: Q = kA \frac { dT }{ dx }
Q is constant. The value of \frac { dT }{ dx } decreases to compensate the increase of k.
28. A material P of thickness 1 mm is sandwiched between two steel slabs, as shown in the figure below. A heat flux 10 kW/m^{2} is supplied to one of the steel slabs as shown. The boundary temperatures of the slabs are indicated in the figure. Assume thermal conductivity of this steel is 10 W/m.K. considering onedimensional steady state heat conduction for the configuration, the thermal conductivity (k, in W/m.K) of material P is _____.

Ans: 0.1
Explanation:
We know: q” = \frac { { T }_{ 1 }{ T }_{ 2 } }{ \frac { { l }_{ 1 } }{ { k }_{ s } } +\frac { l }{ { k } } +\frac { { l }_{ 2 } }{ { k }_{ s } } }
=> 10 * 10^{3 }= \frac { 500360 }{ \frac { 20\times { 10 }^{ 3 } }{ 10 } +\frac { 1\times { 10 }^{ 3 } }{ { k } } +\frac { 20\times { 10 }^{ 3 } }{ 10 } }
=> k = 0.1 W/mK
29. Heat transfer through a composite wall is shown in figure. Both the sections of the wall have equal thickness (l). The conductivity of one section is k and that of the other is 2k. The left face of the wall is at 600 K and the right face is at 300 K. The interface temperature T_{i} (in K) of the composite wall is _____.

Ans: 400
Explanation:
Heat transfer (Q) = \frac { kA\left( { T }_{ 1 }{ T }_{ i } \right) }{ l } =\frac { 2kA\left( { T }_{ i }{ T }_{ 2 } \right) }{ l }
=> 3T_{i} = T_{1} + 2T_{2}
=> 3T_{i} = 600 + 2 * 300 = 1200K
=> T_{i} = 400K
30. Consider a long cylindrical tube of inner and outer radii, r_{i} and r_{o} , respectively, length, L and thermal conductivity, k. Its inner and outer surfaces are maintained at T_{i} and T_{o}, respectively (T_{i} > T_{o}). Assuming onedimensional steady state heat conduction in the radial direction, the thermal resistance in the wall of the tube is
(a) \frac { 1 }{ 2\pi kL}ln\left( \frac { { r }_{ i } }{ { r }_{ o } } \right)
(b) \frac { L }{ 2\pi { r }_{ i }L }
(c) \frac { 1 }{ 2\pi kL }ln\left( \frac { { r }_{ o } }{ { r }_{ i } } \right)
(d) \frac { 1 }{ 4\pi kL }ln\left( \frac { { r }_{ o } }{ { r }_{ i } } \right)
(1 Mark, 2014[3])

Ans: c
31. A plane wall has a thermal conductivity of 1.15 W/m.K. If the inner surface is at 1100^{0}C and the outer surface is at 350^{0}C, then the design thickness (in meter) of the wall to maintain a steady heat flux of 2500 W/m^{2} should be _____.
(2 Mark, 2014[4])

Ans: 0.345
Explanation:
We know, q=k\frac { \left( { T }_{ o }{ T }_{ i } \right) }{ L }
=> 2500=1.15\times \frac { \left( 3501100 \right) }{ L }
=> Thickness of the wall (L) = 0.345 m
32. As the temperature increases, the thermal conductivity of a gas
(a) increases
(b) decreases
(c) remains constant
(d) increases up to a certain temperature and then decreases
(2 Mark, 2014[4])

Ans: a
Explanation:
If temperature increases of a gas
Then. the molecules move faster
Then, molecules transport energy faster
thus, the the transport property called thermal conductivity increases.
33. A 10 mm diameter electrical conductor is covered by an insulation of 2 mm thickness. The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 10 W/m^{2}K. Addition of further insulation of the same material will
(a) increase heat loss continuously
(b) decrease heat loss continuously
(c) increase heat loss to a maximum and then decrease heat loss
(d) decrease heat loss to a minimum and then increase heat loss
(2 Mark, 2015[1])

Ans: c
Explanation:
Critical radius of the wire (r_{c}) = \frac { k }{ h } =\frac { 0.08 }{ 10 } = 8 mm
radius of wire = radius of electrical conductor + insulation thickness
=> r = 5 + 2 = 7 mm
As r < r_{c}, So further insulation of same material, increase the heat loss up to critical radius, then decreases.
34. If foam insulation is added to a 4 cm outer diameter pipe as shown in the figure, the critical radius of insulation (in cm) is __.
(1 Mark, 2015[2])

Ans: 5
Explanation:
Critical radius of insulation (r_{c}) = \frac { { k }_{ foam } }{ { h }_{ o } } =\frac { 0.1 }{ 2 } = 5 cm
35. A cylindrical uranium fuel rod of radius 5 mm in a nuclear is generating heat at the rate of 4\times 10^{7} W/m^{3}. The rod cooled by a liquid (convective heat transfer coefficient 1000 W/m^{2}k) at 25^{o} In steady state, the surface temperature (in K) of the rod is
(a) 308
(b) 398
(c) 418
(d) 448
(2 Mark, 2015[2])

Ans: b
Explanation:
We know: Rate of heat generation inside the rod = Heat transfer by convection
=> { \dot { Q } }_{ gen }\times volume=h{ A }_{ s }\left( { T }_{ s }{ T }_{ \infty } \right)
=> { \dot { Q } }_{ gen }\times \pi { r }^{ 2 }L=h\left( 2\pi rL \right) \left( { T }_{ s }{ T }_{ \infty } \right)
=> { \dot { Q } }_{ gen }\times { r }=2\times h\left( { T }_{ s }{ T }_{ \infty } \right)
=> 4\times { 10 }^{ 7 }\times { 0.005 }=2\times 1000\left( { T }_{ s }25 \right)
=> T_{s }= 125^{0}C = 398 K
36. A brick wall (k = 0.9W/mK) of thickness 0.18 m separates the warm air in a room from the cold ambient air. On a particular winter day, the outside air temperature is −5°C and the room needs to be maintained at 27°C. The heat transfer coefficient associated with outside air is 20 W/m^{2}K. Neglecting the convective resistance of the air inside the room, the heat loss, in (W/m^{2}), is
(a) 88
(b) 110
(c) 128
(d) 160
(2 Mark, 2015[3])

Ans: c
Explanation:
Heat loss (q) = q=\frac { \left( { T }_{ i }{ T }_{ o } \right) }{ \frac { L }{ k } +\frac { 1 }{ { h }_{ o } } }
=> q=\frac { \left( 27+5 \right) }{ \frac { 0.18 }{ 0.9 } +\frac { 1 }{ 20 } } = 128 W/m^{2}
37. A plastic sleeve of outer radius r_{0} = 1 mm covers a wire (radius r = 0.5 mm) carrying electric current. Thermal conductivity of the plastic is 0.15 W/mK. The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 W/m^{2}K. Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will
(a) Increase
(b) Remain the same
(c) Decrease
(d) Be zero
(2 Mark, 2016[1])

Ans: a
Explanation:
Critical radius of the wire (r_{c}) = \frac { k }{ h } =\frac { 0.15 }{ 25 } = 6 mm
Given, r_{o} = 1 mm
As r_{o} < r_{c}, So further addition of plastic cover, the heat transfer from the wire to the ambient will increase.
38. A hollow cylinder has length L, inner radius r_{1}, outer radius r_{2}, and thermal conductivity k. The thermal resistance of the cylinder for radial conduction is
(a) \frac { ln({ r }_{ 2 }/{ r }_{ 1 }) }{ 2\pi kL }
(b) \frac { ln({ r }_{ 1 }/{ r }_{ 2 }) }{ 2\pi kL }
(c) \frac { 2\pi kL }{ ln({ r }_{ 2 }/{ r }_{ 1 }) }
(d) \frac { 2\pi kL }{ ln({ r }_{ 1 }/{ r }_{ 2 }) }
(2 Mark, 2016[1])

Ans: a
39. Steady onedimensional heat conduction takes place across the faces 1 and 3 of a composite slab consisting of slabs A and B in perfect contact as shown in the figure, where k_{A} , k_{B} denote the respective thermal conductivities. Using the data as given in the figure, the interface temperature T_{2} (in °C) is _____.
(1 mark, 2016[3])

Ans: 67.5
Explanation:
The heat flow through the walls A and B are same.
q_{A }=_{ }q_{B}
=> \frac { { k }_{ A }\left( { T }_{ 1 }{ T }_{ 2 } \right) }{ { L }_{ 1 } } =\frac { { k }_{ B }\left( { T }_{ 2 }{ T }_{ 3 } \right) }{ { L }_{ 2 } }
=> \frac { 20\left( 130{ T }_{ 2 } \right) }{ 0.1 } =\frac { 100\left( { T }_{ 2 }30 \right) }{ 0.3 }
=> T_{2 }= 67.5^{0}C
40. Equal amounts of a liquid metal at the same temperature are poured into three moulds made of steel, copper and aluminum. The shape of the cavity is a cylinder with 15 mm diameter. The size of the moulds are such that the outside temperature of the moulds do not increase appreciably beyond the atmospheric temperature during solidification. The sequence of solidification in the mould from the fastest to slowest is (Thermal conductivities of steel, copper and aluminum are 60.5, 401 and 237 W/mK, respectively. Specific heats of steel, copper and aluminum are 434, 385 and 903 J/kgK, respectively. Densities of steel, copper and aluminum are 7854, 8933 and 2700 kg/m^{3}, respectively.)
(a) Copper – Steel – Aluminum
(b) Aluminum – Steel – Copper
(c) Copper – Aluminum – Steel
(d) Steel – Copper – Aluminum
(1 Mark, 2016[3])

Ans: c
Explanation:
Higher the value of thermal diffusivity(\alpha ), faster will be the colling.
For copper = { \alpha }_{ copper }={ \left( \frac { k }{ \rho c } \right) }_{ c }=\frac { 401 }{ 8933\times 385 } =1.16\times { 10 }^{ 4 }
For steel = { \alpha }_{ steel }={ \left( \frac { k }{ \rho c } \right) }_{ s }=\frac { 60.5 }{ 7854\times 434 } =1.77\times { 10 }^{ 5 }
For Aluminium = { \alpha }_{ Alu }={ \left( \frac { k }{ \rho c } \right) }_{ A }=\frac { 237 }{ 2700\times 903 } =9.72\times { 10 }^{ 5 }
As, { \alpha }_{ copper } > { \alpha }_{ Alu } > { \alpha }_{ steel }
So, option ‘c’ is correct.
41. Heat is generated uniformly in a long solid cylindrical rod (diameter = 10mm) at the rate of 4\times10^{7} W/m^{3}.The thermal conductivity of the rod material is 25 W/mK. Under steady state conditions, the temperature difference between the centre and the surface of the rod is _____^{o}C.
(2 Mark, 2017[1])

Ans: 10
Explanation:
We know: { T }_{ c }{ T }_{ S }=\frac { { \dot { Q } }_{ gen }R }{ 4k }
=> { T }_{ c }{ T }_{ S }=\frac { 4\times { 10 }^{ 7 }\times { 0.005 }^{ 2 } }{ 4\times 25 } = 10^{0}C
42. The heat loss from a fin is 6 W. The effectiveness and efficiency of the fin are 3 and 0.75 respectively. The heat loss (in W) from the fin, keeping the entire fin surface at base temperature is __________.
(1 Mark, 2017[2])

Ans: 8
Explanation:
We know: \eta =\frac { { Q }_{ act } }{ { Q }_{ max } }
=> 0.75 =\frac { 6 }{ { Q }_{ max } }
=> { Q }_{ max }=8W
The heat loss from the fin keeping the entire fin surface at base temperature = Q_{max }= 8 W
43. A plane slab of thickness L and thermal conductivity k is heated with a fluid on one side (P) and the other side (Q) is maintained at a constant temperature, T_{Q} of 25^{0}C, as shown in the figure. The fluid is at 45^{0}C and the surface heat transfer coefficient, h, is 10 W/m^{2}K. The steady state temperature, T_{P}, (in ^{0}C) of the side which is exposed to the fluid is ____. (Correct to two decimal places).
(2 Mark, 2018[1])

Ans: 33.89
Explanation:
Here, q_{convection} = q_{conduction}
=> \frac { { T }_{ \infty }{ T }_{ P } }{ \frac { 1 }{ h } } =\frac { { T }_{ P }{ T }_{ Q } }{ \frac { L }{ k } }
=> \frac { 45{ T }_{ P } }{ \frac { 1 }{ 10 } } =\frac { { T }_{ P }25 }{ \frac { 0.2 }{ 2.5 } }
=> T_{P }= 33.89^{0}C
44. A slender rod of length L, diameter d (L >> d) and thermal conductivity k_{1} is joined with another rod of identical dimensions, but of thermal conductivity k_{2}, to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is
(a) k_{1} + k_{2}
(b) \sqrt { { k }_{ 1 }+{ k }_{ 2 } }
(c) k_{1}k_{2}/(k_{1} + k_{2})
(d) 2k_{1}k_{2}/(k_{1} + k_{2})
(1 Mark, 2019[1])
45. Three slabs are joined together as shown in the figure. There is no thermal contact resistance at the interfaces. The centre slab experience a nonuniform internal heat generation with an average value equal to 10000 Wm^{–3}, while the left and right slabs have no internal heat generation. All slabs have thickness equal to 1 m and thermal conductivity of each slab is equal to 5 Wm^{–1}K^{–1}. The two extreme faces are exposed to fluid with heat transfer coefficient 100 Wm^{–2}K^{–1} and bulk temperature 30°C as shown. The heat transfer in the slabs is assumed to be one dimensional and steady, and all properties are constant. If the left extreme face temperature T_{1} is measured to be 100°C, the right extreme faced temperature T_{2} is _____ °C.
(2 Mark, 2019[1])

Ans: 60
Explanation:
Let the crosssection area(A) = 1 m^{2}
Heat generation (Q_{gen}) = 10,000 * A * width = 10000 * 1 * 1 = 10000 W
Heat convected from slab 1 = (Q_{conv})_{1 }= hA( T_{1} – T_{0}) = 100 * 1 * (100 – 30) = 7000 W
Heat generated in the slab is convected from slab 1 and 3.
i.e. Q_{gen }= (Q_{conv})_{1} + (Q_{conv})_{3}
=> 10000 = 70000 + (Q_{conv})_{3}
=> (Q_{conv})_{3 }= 3000 W
=> hA(T_{3 }– T_{0}) = 3000
=> 100 * 1 * (T_{3 }– 30) = 3000
=> T_{3 }= 60^{0}C
46. Onedimensional steady state heat conduction takes place through a solid whose crosssectional area varies linearly in the direction of heat transfer. Assume there is no heat generation in the solid and the thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is
(a) Linear
(b) Quadratic
(c) Logarithmic
(d) Exponential
(1 Mark, 2019[2])

Ans: c
Explanation:
Crosssection area varies linearly in the direction of heat transfer.
i.e. A = a + bx
One dimensional steady state heat conduction through a solid:
Q = kA\frac { dT }{ dx }
=> \frac { dT }{ dx }=\frac { Q }{ kA }
=> dT=\frac { Q }{ k(a+bx) } dx
=> \int _{ 1 }^{ 2 }{ dT } =\int _{ 1 }^{ 2 }{ \frac { Q }{ k(a+bx) } dx }
=> { T }_{ 2 }{ T }_{ 1 }=\frac { Q }{ k } ln\left( \frac { a+b{ x }_{ 2 } }{ a+b{ x }_{ 1 } } \right)
So, the temperature distribution is logarithmic.