Question 1 |

Ambient air flows over a heated slab having flat, top surface at y=0. The local temperature (in Kelvin) profile within the thermal boundary layer is given by T(y)=300+200\; exp(-5y), where y is the distance measured from the slab surface in meter. If the thermal conductivity of air is 1.0 W/m.K and that of the slab is 100 W/m.K, then the magnitude of temperature gradient |dT/dy| within the slab at y=0 is ________K/m (round off to the nearest integer).

5 | |

10 | |

15 | |

20 |

Question 1 Explanation:

Given that

Temp. variation of air, T=300+200 e^{-5 y}

\begin{aligned} K_{\text {air }} &=1 \mathrm{~W} / \mathrm{m}-\mathrm{K} \\ K_{\text {slab }} &=100 \mathrm{~W} / \mathrm{m}-\mathrm{K} \\ \left.\frac{d T}{d y}\right|_{y=0, \mathrm{slab}} &=? \end{aligned}

Applying surface energy balance at interface (y=0)

heat flux leaving from slab at interface by conduction = heat flux received from air at interface by conduction

\begin{aligned} -\left.k_{\text {slab }} \frac{d T}{d y}\right|_{y=0, \text { slab }}&=-\left.k_{\text {air }} \frac{d T}{d y}\right|_{y=0, \text { air }} \\ \left.\frac{d T}{d y}\right|_{y=0, \text { air }}&=0+200 \times-5 \times 1=-1000 \mathrm{~K} / \mathrm{m} \\ -\left.100 \frac{d T}{d y}\right|_{y=0, \text { slab }}&=-1 \times-1000 \\ \left.\frac{d T}{d y}\right|_{y=0, \text { slab }}&=-10 \mathrm{~K} / \mathrm{m}\\ \text { Magnitude of }\left.\frac{d T}{d y}\right|_{y=0, \text { slab }}&=10 \mathrm{~K} / \mathrm{m} \end{aligned}

Question 2 |

A shell and tube heat exchanger is used as a steam condenser. Coolant water enters the tube at 300 K at a rate of 100 kg/s. The overall heat transfer coefficient is 1500 W/m^2.K, and total heat transfer area is 400 m^2. Steam condenses at a saturation temperature of 350 K. Assume that the specific heat of coolant water is 4000 J/kg.K. The temperature of the coolant water coming out of the condenser is _______K (round off to the nearest integer).

254 | |

339 | |

254 | |

654 |

Question 2 Explanation:

\begin{aligned} \mathrm{NTU}&=\frac{U A}{\left(\dot{m} c_{p}\right)_{\text {small }}}=\frac{1500 \times 400}{100 \times 4000}=1.5\\ \epsilon_{H E}&=1-e^{-N T U}=\frac{T_{c e}-T_{c i}}{T_{h i}-T_{c i}}=\frac{T_{C e}-300}{350-300}=0.7768\\ T_{c e}&=338.84 \mathrm{~K} \simeq 339 \mathrm{~K} \end{aligned}

Question 3 |

In forced convective heat transfer, Stanton number (St), Nusselt number (Nu), Reynolds number (Re) and Prandtl number (Pr) are related as

\text{St}=\frac{\text{Nu}}{\text{Re Pr}} | |

\text{St}=\frac{\text{Nu Pr}}{\text{Re}} | |

\text{St}=\text{Nu Pr Re} | |

\text{St}=\frac{\text{Nu Re}}{\text{Pr}} |

Question 3 Explanation:

S t=\frac{N u}{R e \times P r}

Question 4 |

A solid sphere of radius 10 mm is placed at the centroid of a hollow cubical enclosure of side length 30 mm. The outer surface of the sphere is denoted by 1 and the inner surface of the cube is denoted by 2. The view factor F_{22} for radiation heat transfer is ________ (rounded off to two decimal places).

0.12 | |

0.45 | |

0.88 | |

0.77 |

Question 4 Explanation:

\begin{aligned} r_{1} &=10 \mathrm{~mm} \\ A_{1} &=4 \pi r_{1}^{2} \\ A_{2} &=6 \times\left(30^{2}\right) \\ F_{12} &=1 \\ A_{1} F_{12} &=A_{2} F_{21} \quad \Rightarrow \quad F_{21}=\frac{A_{1}}{A_{2}}=\frac{4 \pi \times(10)^{2}}{6 \times(30)^{2}}=0.2327 \\ F_{22} &=1-F_{21}=0.7672 \simeq 0.77 \end{aligned}

Question 5 |

An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 75 ^{\circ}C when placed in air at 25 ^{\circ}C. When the wire is coated with PVC of thickness
1.0 mm, the temperature of the surface of the wire reduces to 55 ^{\circ}C. Assume that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______W/m.K (round off to two decimal places).

0.45 | |

0.32 | |

0.11 | |

0.05 |

Question 5 Explanation:

For uninsulated wire:

Given: T_{\infty}=25^{\circ} \mathrm{C}, R_{1}=1 \mathrm{~mm}, \dot{q}_{\text {gen }}=5 \mathrm{~W} / \mathrm{m}, T_{s_{1}}=7.5^{\circ} \mathrm{C}

\begin{aligned} q &=\dot{q}_{g e n} \times L=\frac{T_{s 1}-T_{\infty}}{\frac{1}{h \times 2 \pi R_{1} L}} \\ 5 \times L &=h \times(2 \pi \times 0.001) L \times(75-25) \\ h &=15.91 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} \end{aligned}

For Insulated wire:

R_{2}=2 \mathrm{~mm}

Heat transfer rate after insulation kept on wire,

q=q_{\text {gen, wire }} \times L=5 \times L

\begin{aligned} q &=5 \times L=\frac{55-25}{\frac{\ln (2 / 1)}{2 \pi k_{\text {PVC }} L}+\frac{1}{15.91 \times 2 \times \pi \times 0.002 \times L}} \\ k_{P V C} &=0.1103 \mathrm{~W} / \mathrm{m}-\mathrm{K} \\ &=0.11 \mathrm{~W} / \mathrm{m}-\mathrm{K} \end{aligned}

Given: T_{\infty}=25^{\circ} \mathrm{C}, R_{1}=1 \mathrm{~mm}, \dot{q}_{\text {gen }}=5 \mathrm{~W} / \mathrm{m}, T_{s_{1}}=7.5^{\circ} \mathrm{C}

\begin{aligned} q &=\dot{q}_{g e n} \times L=\frac{T_{s 1}-T_{\infty}}{\frac{1}{h \times 2 \pi R_{1} L}} \\ 5 \times L &=h \times(2 \pi \times 0.001) L \times(75-25) \\ h &=15.91 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} \end{aligned}

For Insulated wire:

R_{2}=2 \mathrm{~mm}

Heat transfer rate after insulation kept on wire,

q=q_{\text {gen, wire }} \times L=5 \times L

\begin{aligned} q &=5 \times L=\frac{55-25}{\frac{\ln (2 / 1)}{2 \pi k_{\text {PVC }} L}+\frac{1}{15.91 \times 2 \times \pi \times 0.002 \times L}} \\ k_{P V C} &=0.1103 \mathrm{~W} / \mathrm{m}-\mathrm{K} \\ &=0.11 \mathrm{~W} / \mathrm{m}-\mathrm{K} \end{aligned}

Question 6 |

An infinitely long pin fin, attached to an isothermal hot surface, transfers heat at a steady rate of Q_1 to the ambient air. If the thermal conductivity of the fin material is doubled, while keeping everything else constant, the rate of steady- state heat transfer from the fin becomes Q_2. The ratio Q_2/Q_1 is

\sqrt{2} | |

2 | |

\frac{1}{\sqrt{2}} | |

\frac{1}{2} |

Question 6 Explanation:

Fin problem:

q=\sqrt{h P k A}\left(T_{o}-T_{\infty}\right) \text { wait }

If k gets doubled q increases by \sqrt{2} times.

q=\sqrt{h P k A}\left(T_{o}-T_{\infty}\right) \text { wait }

If k gets doubled q increases by \sqrt{2} times.

Question 7 |

A hot steel spherical ball is suddenly dipped into a low temperature oil bath.

Which of the following dimensionless parameters are required to determine instantaneous center temperature of the ball using a Heisler chart?

Which of the following dimensionless parameters are required to determine instantaneous center temperature of the ball using a Heisler chart?

Biot number and Fourier number | |

Reynolds number and Prandtl number | |

Biot number and Froude number | |

Nusselt number and Grashoff number |

Question 7 Explanation:

Unsteady state

Question 8 |

Water flows through a tube of 3 cm internal diameter and length 20 m, The outside surface
of the tube is heated electrically so that it is subjected to uniform heat flux circumferentially
and axially. The mean inlet and exit temperatures of the water are 10^{\circ}C \; and \; 70^{\circ}C,
respectively. The mass flow rate of the water is 720 kg/h. Disregard the thermal resistance
of the tube wall. The internal heat transfer coefficient is 1697 W/m^2\cdot K. Take specific heat
C_p of water as 4.179 kJ/kg.K. The inner surface temperature at the exit section of the
tube is __________ ^{\circ}C (round off to one decimal place).

125.4 | |

25.6 | |

48.8 | |

85.7 |

Question 8 Explanation:

h=1697 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}

From energy balance equation,

\begin{aligned} \text{Heat flux }\times \text{Area of HT}&=\dot{m}_{w} \times C_{p w}\left(T_{w}-T_{w}\right) q^{\prime \prime} \times \pi D L \\ &=\dot{m}_{w} \times C_{p w}(70-10) \\ q^{\prime \prime} &=\frac{720}{3600} \times \frac{4.179 \times 10^{3} \times 60}{\pi \times\left(\frac{3}{100}\right) \times 20} \mathrm{W} / \mathrm{m}^{2} \\ q^{\prime \prime} &=26604.34 \mathrm{W} / \mathrm{m}^{2} \end{aligned}

Applying Newton's law of cooling at exit

\begin{aligned} q^{\prime \prime}&=h \times\left(T_{\text {tube at exit}}-T_{\text {water at exit }}\right) W / m^{2}\\ 26604.34&=1697 \times\left(T_{\text {tube at exit }}-70\right) \mathrm{W} / \mathrm{m}^{2} \\ T_{\text {tube at exit }}&=85.67^{\circ} \mathrm{C} \end{aligned}

Question 9 |

The spectral distribution of radiation from a black body at T_1=3000 K has a maximum
at wavelength\lambda _{max}. The body cools down to a temperature T_2. If the wavelength
corresponding to the maximum of the spectral distribution at T_2 is 1.2 times of the original
wavelength \lambda _{max}, then the temperature T_2 is ________ K (round off to the nearest integer).

3000 | |

4500 | |

2500 | |

1800 |

Question 9 Explanation:

From Wien's Displacement law,

\begin{aligned} \lambda_{m} T &=\text { constant } \Rightarrow \lambda_{m 1} T_{1}=\lambda_{m 2} T_{2} \\ \lambda_{m 1} \times 3000 &=1.2 \lambda_{m 1} \times T_{2} \\ T_{2} &=\left(\frac{3000}{1.2}\right) K=2500 K \end{aligned}

\begin{aligned} \lambda_{m} T &=\text { constant } \Rightarrow \lambda_{m 1} T_{1}=\lambda_{m 2} T_{2} \\ \lambda_{m 1} \times 3000 &=1.2 \lambda_{m 1} \times T_{2} \\ T_{2} &=\left(\frac{3000}{1.2}\right) K=2500 K \end{aligned}

Question 10 |

In a furnace, the inner and outer sides of the brick wall (k_1 = 2.5 W/mK) are maintained
at 1100^{\circ}C and 700^{\circ}C respectively as shown in figure.

The brick wall is covered by an insulating material of thermal conductivity k_2. The thickness of the insulation is 1/4^{th} of the thickness of the brick wall. The outer surface of the insulation is at 200^{\circ}C. The heat flux through the composite walls is 2500 W/m^2.

The value of k_2 is ________ W/mK (round off to 2 decimal places).

The brick wall is covered by an insulating material of thermal conductivity k_2. The thickness of the insulation is 1/4^{th} of the thickness of the brick wall. The outer surface of the insulation is at 200^{\circ}C. The heat flux through the composite walls is 2500 W/m^2.

The value of k_2 is ________ W/mK (round off to 2 decimal places).

0.25 | |

0.5 | |

0.75 | |

1 |

Question 10 Explanation:

Given, L_{2}=\frac{L_{1}}{4}

Assuming steady state, one-dimensional conduction heat transfer through composite slab,

Thermal circuit:

\begin{array}{l} \Rightarrow \quad q=\frac{1100-700}{\frac{L_{1}}{k_{1} A}}=\frac{700-200}{\frac{L_{2}}{k_{2} A}} \\ \Rightarrow \quad \frac{400}{\frac{L_{2}}{2.5}}=\frac{500}{\frac{L_{2}}{k_{2}}}\\ \Rightarrow \quad k_{2}=0.5 \mathrm{W}-\mathrm{mK} \end{array}

Assuming steady state, one-dimensional conduction heat transfer through composite slab,

Thermal circuit:

\begin{array}{l} \Rightarrow \quad q=\frac{1100-700}{\frac{L_{1}}{k_{1} A}}=\frac{700-200}{\frac{L_{2}}{k_{2} A}} \\ \Rightarrow \quad \frac{400}{\frac{L_{2}}{2.5}}=\frac{500}{\frac{L_{2}}{k_{2}}}\\ \Rightarrow \quad k_{2}=0.5 \mathrm{W}-\mathrm{mK} \end{array}

There are 10 questions to complete.