1. For the fluid flowing over a flat plate with Prandtl number greater than unity, the thermal boundary layer for laminar forced convection
(a) Is thinner than the hydrodynamic boundary layer
(b) Has thickness equal to zero
(c) Is of same thickness as hydrodynamic boundary layer
(d) Is thicker than the hydrodynamic boundary layer
(1 Mark, 1988)

Ans: a
Explanation:
We know: \left( \frac { { \delta }_{ v } }{ { \delta }_{ t } } \right) ={ Pr }^{ 1/3 }
If Pr > 1
=> { Pr }^{ 1/3 } > 1
=> \frac { { \delta }_{ v } }{ { \delta }_{ t } } > 1
=> { \delta }_{ v } > { \delta }_{ t }
2. A fluid flowing over a flat plate has the following properties: Dynamic viscosity: 25\times 10^{6 }kg/ms, Specific heat: 2.0 kJ/kgK, Thermal conductivity: 0.05 W/mk. The hydrodynamic boundary layer thickness is measured to be 0.5 mm. The thickness of thermal boundary layer would be
(a) 0.1 mm
(b) 0.5 mm
(c) 1.0 mm
(d) None of the above
(2 Mark, 1992)

Ans: b
Explanation:
We know: Pr = \frac { \mu { c }_{ p } }{ k }
=> { \left( \frac { \delta }{ { \delta }_{ t } } \right) }^{ 3 }=\frac { \mu { c }_{ p } }{ k }
=> { \left( \frac { 0.5 }{ { \delta }_{ t } } \right) }^{ 3 }=\frac { \left( 25\times { 10 }^{ 6 } \right) \times 2000 }{ 0.05 }
=> { \delta }_{ t }=0.5mm
3. For air near atmospheric conditions flowing over a flat plate, the laminar thermal boundary layer is thicker than the hydrodynamic boundary layer. (T/F)
(1 Mark, 1994)

Ans: T
Explanation:
For air, Pr = 0.7
Now, \left( \frac { { \delta }_{ v } }{ { \delta }_{ t } } \right) ={ Pr }^{ 1/3 }={ 0.7 }^{ 1/3 } = 0.89
=> { \delta }_{ v }=0.89{ \delta }_{ t }
=> { \delta }_{ v } < { \delta }_{ t }
4. Heat transfer coefficient for free convection in gases, forced convection in gases and vapours, and for boiling water lie, respectively in the ranges of
(a) 5 – 15, 20 – 200 and 3000 – 50,000 W/m^{2}K.
(b) 20 – 50, 200 – 500 and 50,000 – 10^{5} W/m^{2}K.
(c) 50 – 100, 500 – 1000 and 10^{5} – 10^{6} W/m^{2}K.
(d) 20 – 100, 200 – 1000, and a constant 10^{6} W/m^{2}K.
(1 Mark, 1998)

Ans: a
5. Water (Prandtl number = 6) flows over a flat plate which is heated over the entire length. Which one of the following relationship between the hydrodynamic boundary layer thickness (\delta ) and the thermal boundary layer thickness ({ \delta }_{ t } ) is true?
(a) { \delta }_{ t } = \delta
(b) { \delta }_{ t } < \delta
(c) { \delta }_{ t } > \delta
(d) Cannot be predicted
(2 Mark, 2001)

Ans: b
Explanation:
We know: \left( \frac { { \delta } }{ { \delta }_{ t } } \right) ={ Pr }^{ 1/3 }
=> \left( \frac { { \delta } }{ { \delta }_{ t } } \right) ={ 6 }^{ 1/3 } = 1.81
=> \delta =1.81\times { \delta }_{ t }
=> { \delta }_{ t } < \delta
6. The properties of mercury at 300 K are: density = 13529 kg/m^{3}, specific heat at constant pressure = 0.1393 kJ/kgK, dynamic viscosity =0.1523×10^{2}s/m^{2} and thermal conductivity = 8.540 W/mK. The Prandtl number of the mercury at 300 K is
(a) 0.0248
(b) 2.48
(c) 24.8
(d) 248
(2 Mark, 2002)

Ans: a
Explanation:
We know: Pr = \frac { \mu { c }_{ p } }{ k } =\frac { \left( 0.1523\times { 10 }^{ 2 } \right) \times 139.3 }{ 8.54 } = 0.0248
7. Consider a laminar boundary layer over a heated flat plate. The free stream velocity is { U }_{ \infty }. At some distance x from the leading edge the velocity boundary layer thickness is { \delta }_{ V } and the thermal boundary layer thickness is { \delta }_{ T }. If the Prandtl number is greater than 1, then
(a) { \delta }_{ V } > { \delta }_{ T }
(b) { \delta }_{ T } > { \delta }_{ V }
(c) { \delta }_{ V } = { \delta }_{ T }\sim { \left( { U }_{ \infty }x \right) }^{ 1/2 }
(d) { \delta }_{ V } = { \delta }_{ T }\sim { \left( x \right) }^{ 1/2 }
(2 Mark, 2003)

Ans: a
Explanation:
We know: \left( \frac { { \delta }_{ v } }{ { \delta }_{ t } } \right) ={ Pr }^{ 1/3 }
If Pr > 1
=> { Pr }^{ 1/3 } > 1
=> \frac { { \delta }_{ v } }{ { \delta }_{ t } } > 1
=> { \delta }_{ v } > { \delta }_{ t }
Statement for Linked Questions 8 and 9.
An uninsulated air conditioning duct of rectangular cross section 1 m × 0.5 m, carrying air at 20°C with a velocity of 10 m/s, is exposed to an ambient of 30°C. Neglect the effect of duct construction material. For air in the range of 20 – 30°C, data are as follows: thermal conductivity =0.025 W/m.K; viscosity = 18 μPa.s; Prandtl number =0.73; density = 1.2 kg/m^{3}. The laminar flow Nusselt number is 3.4 for constant wall temperature conditions and, for turbulent flow, Nu = 0.023 Re^{0.8 }Pr^{0.33}.
8. The Reynolds number for the flow is
(a) 444
(b) 890
(c) 4.44×10^{5}
(d) 5.33×10^{5}
(2 Mark, 2005)

Ans: c
Explanation:
Re=\frac { \rho v{ L }_{ c } }{ \mu }Where, L_{c }= Characterstic length
For Rectangular crosssection:
L_{c }= \frac { 4A }{ P } =\frac { 4ab }{ 2(a+b) } =\frac { 4\times 1\times 0.5 }{ 2(1+0.5) } = 0.66 m
Now, Re=\frac { \rho v{ L }_{ c } }{ \mu } =\frac { 1.2\times 10\times 0.66 }{ 18\times { 10 }^{ 6 } } =
=> Re = 4.44\times { 10 }^{ 5 }
9. The heat transfer per metre length of the duct, in watts, is:
(a) 3.8
(b) 5.3
(c) 89
(d) 769
(2 Mark, 2005)

Ans: d
Explanation:
Given, Nu = 0.023 Re^{0.8 }Pr^{0.33}
=> \frac { h{ L }_{ c } }{ k } =0.023\times { \left( 4.44\times { 10 }^{ 5 } \right) }^{ 0.8 }\times { 0.73 }^{ 0.33 }
=> \frac { h\times 0.66 }{ 0.025 } = 683.173
=> h = 25.64 W/m^{2}
Total surface area(A) = 2(1 + 0.5)L = 3L
Now, Heat transfer by convection (Q) = h(3L)(30 – 20)
=>Q/L = 25.64 * 3 * 10 = 769 W
10. The temperature distribution within the thermal boundary layer over a heated isothermal flat plate is given by \frac { T{ T }_{ w } }{ { T }_{ \infty }{ T }_{ w } } =\frac { 3 }{ 2 } \left( \frac { y }{ { \delta }_{ t } } \right) \frac { 1 }{ 2 } { \left( \frac { y }{ { \delta }_{ t } } \right) }^{ 3 }, where T_{w} and { T }_{ \infty } are the temperature of plate and free stream respectively, and y is the normal distance measured from the plate. The ratio of average to the local Nusselt number based on the thermal boundary layer thickness { \delta }_{ t } is given by
(a) 1.33
(b) 1.50
(c) 2.0
(d) 4.64
(2 Mark, 2007)

Ans: c
Explanation:
\frac { { Nu }_{ avg } }{ { Nu }_{ loc } } =2
11. For flow of fluid over a heated plate, the following fluid properties are known Viscosity = 0.001 Pa.s; specific heat at constant pressure = 1kJ/kg.K; thermal conductivity = 1 W/mk. The hydrodynamic boundary layer thickness at a specified location on the plate is 1 mm. The thermal boundary layer thickness at the same location is
(a) 0.001 mm
(b) 0.01 mm
(c) 1 mm
(d) 1000 mm
(1 Mark, 2008)

Ans: c
Explanation:
We know: Pr = \frac { \mu { c }_{ p } }{ k }
=> { \left( \frac { \delta }{ { \delta }_{ t } } \right) }^{ 3 }=\frac { \mu { c }_{ p } }{ k }
=> { \left( \frac { 1 }{ { \delta }_{ t } } \right) }^{ 3 }=\frac { 0.001\times { 10 }^{ 3 } }{ 1 }
=> { \delta }_{ t }=1mm
12.For the threedimensional object shown in the figure below, five faces are insulated. The sixth face (PQRS), which is not insulated, interacts thermally with the ambient, with a convective heat transfer coefficient of 10 W/m^{2}.K. The ambient temperature is 30^{0}C. Heat is uniformly generated inside the object at the rate of 100 W/m^{3}. Assuming the face PQRS to be at uniform temperature, its steady state temperature is
(a) 10^{0}C
(b) 20^{0}C
(c) 30^{0}C
(d) 40^{0}C
(2 Mark, 2008)

Ans: d
Explanation:
T = Temperature of the face PQRS
T_{0 }= Ambient temperature
Total heat generated = Heat transfer by convection through PQRS
=> Rate of heat generation * Volume = hA(T – T_{0})
=> 100(1*2*2) = (2*2)(T – 30)
=> T = 40^{0}C
13. The ratios of the laminar hydrodynamic boundary layer thickness to thermal boundary layer thickness of flows of two fluids P and Q on a flat plate are ½ and 2 respectively. The Reynolds number based on the plate length for both the flows is 10^{4} . The Prandtl and Nusselt numbers for P are 1/8 and 35 respectively. The Prandtl and Nusselt numbers for Q are respectively
(a) 8 and 140
(b) 8 and 70
(c) 4 and 70
(d) 4 and 35
(2 Mark, 2011)

Ans: a
Explanation:
Given: { \left( \frac { \delta }{ { \delta }_{ t } } \right) }_{ Q } = 2
=> { \left( { Pr }_{ Q } \right) }^{ 1/3 } = 2
=> { Pr }_{ Q } = 8
For laminar boundary layer on flat plate:
Nu\quad \alpha \quad { Re }^{ 1/2 }{ Pr }^{ 1/3 }=> \frac { { Nu }_{ Q } }{ { Nu }_{ P } } ={ \left( \frac { { Pr }_{ Q } }{ { Pr }_{ P } } \right) }^{ 1/3 } [As, Re is same for both]
=> { Nu }_{ Q }=35{ \left( \frac { 8 }{ 1/8 } \right) }^{ 1/3 } = 140
Statement for Linked Questions 14 & 15.
Water (specific heat, c_{p} = 4.18 kJ/kgK) enters a pipe at a rate of 0.01 kg/s and a temperature of 20°C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux q_{w}” W/m^{2}.
14. If q_{w}“= 2500x, where x is in m and in the direction of flow (x = 0 at the inlet), the bulk mean temperature of the water leaving the pipe in °C is
(a) 42
(b) 62
(c) 74
(d) 104
(2 Mark, 2013)

Ans: b
Explanation:
Heat transfer to the pipe:
Q = \int _{ 0 }^{ 3 }{ { q }_{ w }^{ " } } .\pi d.dx
=> Q = \int _{ 0 }^{ 3 }{ 2500x } \times 0.05\pi \times dx
=> Q = 392.5{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 } = 1766 W
Q = mc_{p}(T_{0} – T_{i})
=> 1766 = 0.01 * 4180 * (T_{0} – 20)
=> T_{0} = 80^{0}C
15. If q_{w}” = 5000 and the convection heat transfer coefficient at the pipe outlet is 1000 W/m^{2}K, the temperature in °C at the inner surface of the pipe at the outlet is
(a) 71
(b) 76
(c) 79
(d) 81
(2 Mark, 2013)

Ans: d
Explanation:
Here, T_{1} = Temperature at inner surface of pipe at outlet
T_{i }= Bulk mean temperature at inlet of pipe
T_{0 }= Bulk mean temperature at outlet of pipe
Total heat into the pipe (Q) = { q }_{ w }^{ " }\times A = 5000\times \pi dL = 2355 W
Q = mc_{p}(T_{0} – T_{i})
=> 2355 = 0.01 * 4180 * (T_{0} – 20)
=> T_{0 }= 76.33^{0}C
Heat transfer taking place at the end of the pipe:
{ q }_{ w }^{ " } = h(T_{1} – T_{i})
=> 5000 = 1000(T_{1} – 76.33)
=> T_{1 }= 81^{0}C
16. Consider a twodimensional laminar flow over a long cylinder as shown in the figure below.
The free stream velocity is { U }_{ \infty } and the free stream temperature { T }_{ \infty } is lower than the cylinder surface temperature T_{S}. The local heat transfer coefficient is minimum at point
(a) 1
(b) 2
(c) 3
(d) 4
(1 Mark, 2014[1])
17. The nondimensional fluid temperature profile near the surface of a convectively cooled flat plate is given by \frac { { T }_{ w }T }{ { T }_{ w }{ T }_{ \infty } } =a+b\frac { y }{ L } +c{ \left( \frac { y }{ L } \right) }^{ 2 }, where y is measured perpendicular to the plate, L is the plate length and a, b and c are arbitrary constants. T_{w} and { T }_{ \infty } are wall and ambient temperatures respectively. If the thermal conductivity of the fluid is k and the wall heat flux is q, the Nusselt number Nu=\frac { q" }{ { T }_{ w }{ T }_{ \infty } } \frac { L }{ k } is equal to
(a) a
(b) b
(c) 2c
(d) (b + 2c)
(2 Mark, 2014[1])

Ans: b
Explanation:
We know:
Heat flux (q”) = k{ \left( \frac { dT }{ dy } \right) }_{ y=0 }=h\left( { T }_{ w }{ T }_{ \infty } \right)
Given: \frac { { T }_{ w }T }{ { T }_{ w }{ T }_{ \infty } } =a+b\frac { y }{ L } +c{ \left( \frac { y }{ L } \right) }^{ 2 }
=> { T }_{ w }T=\left( { T }_{ w }{ T }_{ \infty } \right) \left[ a+b\frac { y }{ L } +c{ \left( \frac { y }{ L } \right) }^{ 2 } \right]
=> { \left( \frac { dT }{ dy } \right) }_{ y=0 }=\left( { T }_{ w }{ T }_{ \infty } \right) \left[ \frac { b }{ L } \right]
=> \frac { q" }{ k } =\left( { T }_{ w }{ T }_{ \infty } \right) \left[ \frac { b }{ L } \right]
=> \frac { q" }{ \left( { T }_{ w }{ T }_{ \infty } \right) } .\frac { L }{ k } =b
=> Nu = b
18. For laminar forced convection over a flat plate, if the free stream velocity increases by a factor of 2, the average heat transfer coefficient
(a) Remains same
(b) Decreases by a factor of \sqrt { 2 }
(c) Rises by a factor of \sqrt { 2 }
(d) Rises by a factor of 4
(1 Mark, 2014[2])

Ans: c
Explanation:
Nu = 0.664 Re^{0.5 }Pr^{0.33}
=> \frac { h{ L } }{ k } =0.664\times { \left( \frac { \rho vL }{ \mu } \right) }^{ 0.5 }\times { \frac { \mu { c }_{ p } }{ k } }^{ 0.33 }
=> h\quad \alpha \quad { v }^{ 0.5 }
=> \frac { { h }_{ 2 } }{ { h }_{ 1 } } ={ \left( \frac { { v }_{ 2 } }{ { v }_{ 1 } } \right) }^{ 0.5 }
=> \frac { { h }_{ 2 } }{ { h }_{ 1 } } ={ \left( \frac { { 2v }_{ 1 } }{ { v }_{ 1 } } \right) }^{ 0.5 }
=> { h }_{ 2 }=\sqrt { 2 } { h }_{ 1 }
19. Water flows through a tube of diameter 25 mm at an average velocity of 1.0m/s. The properties of water are, ρ = 1000 kg/m^{3}, μ = 7.25 × 10^{4} N.s/m^{2}, k = 0.625 W/mK, Pr = 4.85. Using Nu = 0.023Re^{0.8}Pr^{0.4}, the convective heat transfer coefficient (in W/m^{2}.K) is ______.
(2 Mark, 2014[4])

Ans: 4613.66
Explanation:
Pr = 4.85
Re = \frac { \rho Vd }{ \mu } =\frac { 1000\times 1\times 25\times { 10 }^{ 3 } }{ 7.25\times { 10 }^{ 4 } } = 34482.75
Now, Nu = 0.023Re^{0.8}Pr^{0.4}
\frac { h{ L }_{ c } }{ k } = 0.023Re^{0.8}Pr^{0.4}
=> \frac { h\times 25\times { 10 }^{ 3 } }{ 0.625 } =0.023\times { 34482.75 }^{ 0.8 }\times { 4.85 }^{ 0.4 } [Here, L_{c }= d]
=> h = 4613.66 W/m^{2}K
20. In the laminar flow of air (Pr = 0.7) over a heated plate, if \delta \quad and\quad { \delta }_{ T } denote respectively, the hydrodynamic and thermal boundary layer thickness, then

Ans: c
Explanation:
\frac { \delta }{ { \delta }_{ T } } ={ Pr }^{ 1/3 }=> \frac { \delta }{ { \delta }_{ T } } ={ 0.7 }^{ 1/3 } = 0.887
=> \frac { \delta }{ { \delta }_{ T } } < 1
=> \delta < { \delta }_{ T }
21. For flow of viscous fluid over a flat plate, if the fluid temperature is the same as the plate temperature, the thermal boundary layer is
(a) Thinner than the velocity boundary layer
(b) Thicker than the velocity boundary layer
(c) Of the same thickness as the velocity boundary layer
(d) not formed at all
(1 Mark, 2015[1])

Ans: d
Explanation:
No temperature difference between wall and fluid means no thermal boundary layer will form.
22. For flow through a pipe radius R, the velocity and temperature distribution are as follows:
u(r, x) = C_{1}, and T(r, x) ={ C }_{ 2 }\left[ 1{ \left( \frac { r }{ R } \right) }^{ 3 } \right], Where C_{1 }and C_{2} are constants. The bulk temperature is given by { T }_{ m }=\frac { 2 }{ { U }_{ m }{ R }^{ 2 } } \int _{ 0 }^{ R }{ u(r,x) } T(r,x)rdr with U_{m} being the mean velocity of flow. The value of T_{m} is
(a) \frac { 0.5{ C }_{ 2 } }{ { U }_{ m } }
(b) 0.5C_{2}
(c) 0.6C_{2}
(d) \frac { 0.6{ C }_{ 2 } }{ { U }_{ m } }
(2 Mark, 2015[1])
23. A fluid (Prandtl number, Pr = 1) at 500 K flows over a flat plate of 1.5 m length, maintained at 300 K. The velocity of the fluid is 10 m/s. Assuming kinematic viscosity, υ = 30 * 10^{6 }m^{2}/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is ____.
(2 Mark, 2016[1])

Ans: 6.123
Explanation:
Reynold number at 0.5 m from the leading edge:
Re=\frac { Ux }{ \nu } =\frac { 10\times 0.5 }{ 30\times { 10 }^{ 6 } } =166,666.67
As, Re < 5\times { 10 }^{ 5 }
So, \delta =\frac { 5x }{ \sqrt { { Re }_{ x } } } =\frac { 5\times 0.5 }{ \sqrt { 166666.67 } } = 6.123 mm
Given, Pr = 1.
We know: \frac { \delta }{ { \delta }_{ T } } ={ Pr }^{ 1/3 }={ 1 }^{ 1/3 } = 1
=> \delta ={ \delta }_{ T }=6.123mm
24. For a hydro dynamically and thermally fully developed laminar flow through a circular pipe of constant crosssection, the Nusselt number at constant wall heat flux (Nu_{q}) and that at constant wall temperature (Nu_{T}) are related as
(a) Nu_{q} < Nu_{T}
(b) Nu_{q} < (Nu_{T})^{2}
(c) Nu_{q} = Nu_{T}
(d) Nu_{q} > Nu_{T}
(1 Mark, 2019[1])

Ans: d
Explanation:
Nu = 4.36 (for constant heat flux)
Nu = 3.66 (for constant wall temperature)
So, Nu_{q} > Nu_{T}
25. The wall of a constant diameter pipe of length 1 m is heated uniformly with flux q” by wrapping a heater coil around it. The flow at the inlet to the pipe is hydrodynamically fully developed. The fluid is incompressible and the flow is assumed to be laminar and steady all through the pipe. The bulk temperature of the fluid is equal to 0°C at the inlet and 50°C at the exit. The wall temperatures are measured at three locations, P, Q and R, as shown in the figure. The flow thermally develops after some distance from the inlet. The following measurements are made:
Among the locations P, Q and R, the flow is thermally developed at:
(a) P and Q only
(b) P, Q and R
(c) R only
(d) Q and R only
(2 Mark, 2019[1])

Ans: d
Explanation:
In case of uniform heat flux, bulk mean temperature varies linearly.
So, Bulk mean temperature (T_{b}) = A + Bx …..(1)
According to the question,
At x = 0, T = 0^{0}C => A= 0^{0}C ……..(2)
Again, At x= 1, T = 50^{0}C => B= 50^{0}C …..(3)
From equation (1), (2) and (3), T_{b} = 50x
The value of (T_{w} – T_{b}) is remain constant in thermal developed region.
So, Q and R in thermally developed region.
26. A thin vertical flat plate of height L, and infinite width perpendicular to the plane of the figure, is losing heat to the surroundings by natural convection. The temperature of the plate and the surroundings, and the properties of the surrounding fluid, are constant. The relationship between the average Nusselt and Rayleigh numbers is given as Nu = KRa^{1/4 } , where K is a constant. The length scales for Nusselt and Rayleigh numbers are the height of the plate. The height of the plate is increased to 16L keeping all other factors constant
If the average heat transfer coefficient for the first plate is h_{1} and that for the second plate is h_{2}, the value of the ratio h_{1}/ h_{2} is_____.
(1 Mark, 2019[2])

Ans: 2
Explanation:
Given: Nu = KRa^{1/4}
We know: Ra = f(Gr, Pr)
So, Nu = K\times { Gr }^{ 1/4 }\times { Pr }^{ 1/4 }
=> \frac { hL }{ k } =K\times \left( \frac { g\beta { L }^{ 3 }\Delta T }{ { \nu }^{ 2 } } \right) ^{ 1/4 }\times { \left( \frac { \mu { c }_{ p } }{ k } \right) }^{ 1/4 }
=> h\alpha \frac { { L }^{ 3/4 } }{ L }
=> h\alpha \frac { 1 }{ { L }^{ 1/4 } }
=> \frac { { h }_{ 1 } }{ { h }_{ 2 } } = { \left( \frac { { L }_{ 2 } }{ { L }_{ 1 } } \right) }^{ 1/4 }
=> \frac { { h }_{ 1 } }{ { h }_{ 2 } } = { \left( \frac { { 16L } }{ { L } } \right) }^{ 1/4 }
=> \frac { { h }_{ 1 } }{ { h }_{ 2 } } =2