Question 1 |
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 1 Explanation:
S t=\frac{N u}{R e \times P r}
Question 2 |
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 2 Explanation:
Unsteady state
Question 3 |
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 3 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 4 |
A small metal bead (radius 0.5 mm), initially at 100^{\circ}C, when placed in a stream of fluid
at 20^{\circ}C, attains a temperature of 28^{\circ}C in 4.35 seconds. The density and specific heat
of the metal are 8500 kg/m^3 and 400 J/kgK, respectively. If the bead is considered as
lumped system, the convective heat transfer coefficient (in W/m^2 K) between the metal
bead and the fluid stream is
283.3 | |
299.9 | |
149.9 | |
449.7 |
Question 4 Explanation:
r=0.5 \mathrm{mm} ; c_{p}=400 \mathrm{J} / \mathrm{kgk} ; \rho=8500 \mathrm{kg} / \mathrm{m}^{3} ; t=4.35 \mathrm{sec}
\begin{aligned} \therefore \quad \frac{V}{A}&=\frac{\frac{4}{3} \pi R^{3}}{4 \pi R^{2}}=\frac{R}{3} \\ \because \quad \frac{T-T_{\infty}}{T_{i}-T_{\infty}}&=e^{\frac{-h A \tau}{\rho C_{p} V}} \\ \Rightarrow \qquad \frac{28-20}{100-20}&=e^{-\left(\frac{h \cdot 3000}{0.5 \times 8500 \times 400}\right)} \times 4.35 \\ \Rightarrow \qquad h&=299.95 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \end{aligned}
\begin{aligned} \therefore \quad \frac{V}{A}&=\frac{\frac{4}{3} \pi R^{3}}{4 \pi R^{2}}=\frac{R}{3} \\ \because \quad \frac{T-T_{\infty}}{T_{i}-T_{\infty}}&=e^{\frac{-h A \tau}{\rho C_{p} V}} \\ \Rightarrow \qquad \frac{28-20}{100-20}&=e^{-\left(\frac{h \cdot 3000}{0.5 \times 8500 \times 400}\right)} \times 4.35 \\ \Rightarrow \qquad h&=299.95 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \end{aligned}
Question 5 |
Grashof number signifies the ratio of
inertia force to viscous force | |
buoyancy force to viscous force | |
buoyancy force to inertia force | |
inertia force to surface tension force |
Question 5 Explanation:
Grashof number is ratio of buoyancy force to viscous force.
Question 6 |
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, v=30 \times 10^{-6}m^{2}/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is __________
6.123mm | |
5.269mm | |
8.2654mm | |
4.9mm |
Question 6 Explanation:
Given
\begin{aligned} P r &=1 \\ T_{\infty} &=500 \mathrm{K} \\ \text { Length } &=1.5 \mathrm{m} \\ T_{\text {plate }} &=300 \mathrm{K} \\ U_{\infty} &=10 \mathrm{m} / \mathrm{s} \\ v &=30 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s} \end{aligned}
Thermal boundary layer at 0.5 \mathrm{m} from leadra edge =\delta_{t}
\text{Re}_{x=0.5}=\frac{U x}{v}=\frac{10 \times 0.5}{30 \times 10^{-6}}=166,666.67
since \text{Re}_{x=0.5} \lt 5 \times 10^{5}. Hence flow is laminar
\begin{aligned}\therefore \quad \delta &=\frac{5 x}{\sqrt{\text{Re}_{x}}} \\ &=\frac{5 \times 0.5}{\sqrt{166,666.67}} \\ &=6.1237 \times 10^{-3} \mathrm{m} \\ \frac{\delta_{t}}{\delta} &=\left(P_{f}\right)^{1 / 3} \\ \delta_{t} &=(1)^{1 / 3} \times \delta \\ &=1 \times 6.1237 \times 10^{-3} \mathrm{m} \\ &=6.1237 \mathrm{mm} \end{aligned}
\begin{aligned} P r &=1 \\ T_{\infty} &=500 \mathrm{K} \\ \text { Length } &=1.5 \mathrm{m} \\ T_{\text {plate }} &=300 \mathrm{K} \\ U_{\infty} &=10 \mathrm{m} / \mathrm{s} \\ v &=30 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s} \end{aligned}
Thermal boundary layer at 0.5 \mathrm{m} from leadra edge =\delta_{t}
\text{Re}_{x=0.5}=\frac{U x}{v}=\frac{10 \times 0.5}{30 \times 10^{-6}}=166,666.67
since \text{Re}_{x=0.5} \lt 5 \times 10^{5}. Hence flow is laminar
\begin{aligned}\therefore \quad \delta &=\frac{5 x}{\sqrt{\text{Re}_{x}}} \\ &=\frac{5 \times 0.5}{\sqrt{166,666.67}} \\ &=6.1237 \times 10^{-3} \mathrm{m} \\ \frac{\delta_{t}}{\delta} &=\left(P_{f}\right)^{1 / 3} \\ \delta_{t} &=(1)^{1 / 3} \times \delta \\ &=1 \times 6.1237 \times 10^{-3} \mathrm{m} \\ &=6.1237 \mathrm{mm} \end{aligned}
Question 7 |
The ratio of momentum diffusivity (v) to thermal diffusivity (\alpha), is called
Prandtl number | |
Nusselt number | |
Biot number | |
Lewis number |
Question 7 Explanation:
Pr=\frac{v}{\alpha }
Question 8 |
In the laminar flow of air (Pr = 0.7) over a heated plate, if \delta and \delta_{r} denote, respectively, the
hydrodynamic and thermal boundary layer thicknesses, then
\delta = \delta_{r} | |
\delta \gt \delta_{r} | |
\delta \lt \delta_{r} | |
\delta = 0 but \delta_{r} \neq 0 |
Question 8 Explanation:
Prandtl number,
\begin{aligned} P r&=0.7\\ \text{also}\quad \mathrm{Pr}&=\frac{\delta}{\delta_{r}}\\ \therefore 0.7&=\frac{\delta}{\delta_{r}}\\ \text{or }\quad \delta&=0.7 \delta_{r}\\ \therefore \delta& \lt \delta_{r}\\ \end{aligned}
\begin{aligned} P r&=0.7\\ \text{also}\quad \mathrm{Pr}&=\frac{\delta}{\delta_{r}}\\ \therefore 0.7&=\frac{\delta}{\delta_{r}}\\ \text{or }\quad \delta&=0.7 \delta_{r}\\ \therefore \delta& \lt \delta_{r}\\ \end{aligned}
Question 9 |
For laminar forced convection over a flat plate, if the free stream velocity increases by a factor of 2, the average heat transfer coefficient
remains same | |
decreases by a factor of \sqrt{2} | |
rises by a factor of \sqrt{2} | |
rises by a factor of 4 |
Question 9 Explanation:
For the laminar flow over flat plate,
\mathrm{Re} \lt 3 \times 10^{5}
Nusselt number,
\begin{aligned} & N_{u}=\frac{h l}{k}=0.664 \mathrm{Re}^{1 / 2} \mathrm{Pr}^{1 / 3} \\ \text { where } \mathrm{Re}&=\frac{U l}{\mathrm{v}} \\ \therefore \qquad \frac{h l}{k}&=0.664 \times\left(\frac{U l}{\mathrm{v}}\right)^{1 / 2} \times \mathrm{Pr}^{1 / 3}\\ h &\propto U^{1 / 2}\\ \frac{h_{2}}{h_{1}} &=\left(\frac{U_{2}}{U_{1}}\right)^{1 / 2} \\ &=\left(\frac{2 U_{2}}{U_{1}}\right)^{1 / 2}=\sqrt{2}\\ \text{or }\quad h_{2}&=\sqrt{2} h_{1}\\ \end{aligned}
\mathrm{Re} \lt 3 \times 10^{5}
Nusselt number,
\begin{aligned} & N_{u}=\frac{h l}{k}=0.664 \mathrm{Re}^{1 / 2} \mathrm{Pr}^{1 / 3} \\ \text { where } \mathrm{Re}&=\frac{U l}{\mathrm{v}} \\ \therefore \qquad \frac{h l}{k}&=0.664 \times\left(\frac{U l}{\mathrm{v}}\right)^{1 / 2} \times \mathrm{Pr}^{1 / 3}\\ h &\propto U^{1 / 2}\\ \frac{h_{2}}{h_{1}} &=\left(\frac{U_{2}}{U_{1}}\right)^{1 / 2} \\ &=\left(\frac{2 U_{2}}{U_{1}}\right)^{1 / 2}=\sqrt{2}\\ \text{or }\quad h_{2}&=\sqrt{2} h_{1}\\ \end{aligned}
Question 10 |
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^{\circ}C. The pipe, of diameter 50 mm and length 3 m, is subjected to a wall heat flux {q_{w}}^{''}
in W/m^{2}:
If {q_{w}}^{''}=5000 and the convection heat transfer coefficient at the pipe outlet is 1000 W/m^{2}K , the temperature in ^{\circ}C at the inner surface of the pipe at the outlet is
If {q_{w}}^{''}=5000 and the convection heat transfer coefficient at the pipe outlet is 1000 W/m^{2}K , the temperature in ^{\circ}C at the inner surface of the pipe at the outlet is
71 | |
76 | |
79 | |
81 |
Question 10 Explanation:
\begin{aligned} q_{w}^{\prime \prime} &=5000 \mathrm{W} / \mathrm{m}^{2} \\ h &=1000 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \end{aligned}
Heat transfer through pipe wall,
\begin{aligned} Q &=q_{w}^{\prime} \times \text { surface area }\\ &=5000 \times \pi d l \\ &=5000 \times 3.14 \times 0.05 \times 3 \\ &=2355 \mathrm{W}\\ \text{also }\quad Q&=m c_{p}\left(T_{0}-T_{i}\right), \\ & \text { heat gained by the water}\\ 2355 &=0.01 \times 4180\left(T_{0}-20\right) \\ 56.33 &=T_{0}-20 \\ T_{0} &=76.33^{\circ} \mathrm{C} \end{aligned}
Heat flux q_{w}^{\prime \prime} is same at any two sections,
\begin{aligned} f_{w}^{\prime \prime}&=h\left(T_{s^{\prime} 0}-T_{0}\right) \\ 5000 &=1000\left(T_{s^{\prime} 0}-76.33\right) \\ \text{or }\quad 5 &=T_{s^{\prime} 0}-76.33\\ \text{or }\quad T_{s^{\prime} \circ}&=76.33+5\\ &=81.33^{\circ} \mathrm{C} \simeq 81^{\circ} \mathrm{C} \end{aligned}
Heat transfer through pipe wall,
\begin{aligned} Q &=q_{w}^{\prime} \times \text { surface area }\\ &=5000 \times \pi d l \\ &=5000 \times 3.14 \times 0.05 \times 3 \\ &=2355 \mathrm{W}\\ \text{also }\quad Q&=m c_{p}\left(T_{0}-T_{i}\right), \\ & \text { heat gained by the water}\\ 2355 &=0.01 \times 4180\left(T_{0}-20\right) \\ 56.33 &=T_{0}-20 \\ T_{0} &=76.33^{\circ} \mathrm{C} \end{aligned}
Heat flux q_{w}^{\prime \prime} is same at any two sections,
\begin{aligned} f_{w}^{\prime \prime}&=h\left(T_{s^{\prime} 0}-T_{0}\right) \\ 5000 &=1000\left(T_{s^{\prime} 0}-76.33\right) \\ \text{or }\quad 5 &=T_{s^{\prime} 0}-76.33\\ \text{or }\quad T_{s^{\prime} \circ}&=76.33+5\\ &=81.33^{\circ} \mathrm{C} \simeq 81^{\circ} \mathrm{C} \end{aligned}
There are 10 questions to complete.
