Question 1 |
Consider a hydrodynamically and thermally
fully-developed, steady fluid flow of 1 kg/s in a
uniformly heated pipe with diameter of 0.1 m and
length of 40 m. A constant heat flux of magnitude 15000 W/m^2
is imposed on the outer surface of the
pipe. The bulk-mean temperature of the fluid at the
entrance to the pipe is 200^{\circ}C.
The Reynolds number (R_e) of the flow is 85000, and the Prandtl number (P_r) of the fluid is 5. The thermal conductivity and the specific heat of the fluid are 0.08 W\dot m^{-1} \dot K^{-1} and 2600 J \dot kg^{-1} \dot K^{-1}, respectively. The correlation N_u = 0.023 Re^{0.8} Pr^{0.4} is applicable, where the Nusselt Number (N_u) is defined on the basis of the pipe diameter. The pipe surface temperature at the exit is ________ ^{\circ}C (round off to the nearest integer).
The Reynolds number (R_e) of the flow is 85000, and the Prandtl number (P_r) of the fluid is 5. The thermal conductivity and the specific heat of the fluid are 0.08 W\dot m^{-1} \dot K^{-1} and 2600 J \dot kg^{-1} \dot K^{-1}, respectively. The correlation N_u = 0.023 Re^{0.8} Pr^{0.4} is applicable, where the Nusselt Number (N_u) is defined on the basis of the pipe diameter. The pipe surface temperature at the exit is ________ ^{\circ}C (round off to the nearest integer).
125 | |
321 | |
652 | |
228 |
Question 1 Explanation:
\begin{aligned} \dot{m}&=1kg/sec\\ Re&=85000\\ Pr&=5\\ K&=0.08 W/mK\\ C&=2600J/kgK\\ Nu&=0.023R_e^{0.8}Pr^{0.4}\\ Nu&=\frac{hD}{k}\\ T_s&=?\\ Nu&=\frac{hD}{k}=0.023R_e^{0.8}Pr^{0.4}\\ \frac{h \times 0.1}{0.08}&=0.023(85000)^{0.8}(5)^{0.4}\\ h&=307.56 W/m^2K\\ Q&=qA_s=\dot{m}c_p(T_o-T_i)\\ q \pi DL&=\dot{m}c_p(T_o-T_i)\\ 15000 \times \pi \times0.1 \times 40&=1 \times 2600 \times (T_o-200)\\ T_o&=272.5^{\circ}C\\ &\text{Newtons law of cooling}\\ q&=h(T_s-T_m)\\ T_s-T-m=\frac{q}{h}=constant \end{aligned}
At outlet, x=LT_m=T_o. T=T_s
\begin{aligned} q&=h(T_s-T_o)\\ T_s-T_o&=\frac{q}{h}\\ T_s&=T_o+\frac{q}{h}\\ T_s&=272.5+\frac{15000}{307.56}\\ T_s&=321.27^{\circ}C\\ T_s&=321^{\circ}C \end{aligned}
Question 2 |
Which of the following non-dimensional terms is
an estimate of Nusselt number?
Ratio of internal thermal resistance of a solid to the boundary layer thermal resistance | |
Ratio of the rate at which internal energy is advected to the rate of conduction heat transfer | |
Non-dimensional temperature gradient | |
Non-dimensional velocity gradient multiplied by Prandtl number |
Question 2 Explanation:
Let,
\begin{aligned} \theta ^*&=\frac{T_s-T}{T_s-T_\infty }\\ y^*&=\frac{y}{L}\\ \frac{dy^*}{dy}&=\frac{1}{L}\\ \frac{d\theta ^*}{dy}&=-\frac{dT}{dy} \end{aligned}
T_s \text{ and } T_\infty are constant
At boundary
\begin{aligned} q_{\text{conduction}}&=q_{\text{convection}}\\ -K_f\left.\begin{matrix} \frac{dT}{dy} \end{matrix}\right|_{y=0}&=h(T_s-T_\infty )\\ -K_f\frac{d}{dy}\left [ \frac{T_s-T}{T_s-T_\infty }h \right ]&=h\\ \frac{d\theta ^*}{dy}&=\frac{h}{K_f}\\ \end{aligned}
Applying chain rule
\begin{aligned} \frac{d\theta ^*}{dy^*}\frac{dy^*}{dy}&=\frac{h}{K_f}\\ \frac{d\theta ^*}{dy^*}\frac{1}{L}=\frac{h}{K_f}\\ \frac{d\theta ^*}{dy^*}=\frac{hL}{K_f}=\text{Nusselt Number}\\ Nu&=\frac{d\theta ^*}{dy^*} \end{aligned}
Nu represents non dimensional temperature gradient.
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 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 4 Explanation:
Unsteady state
Question 5 |
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 5 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 6 |
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 6 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 7 |
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 7 Explanation:
Grashof number is ratio of buoyancy force to viscous force.
Question 8 |
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 8 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 9 |
The ratio of momentum diffusivity (v) to thermal diffusivity (\alpha), is called
Prandtl number | |
Nusselt number | |
Biot number | |
Lewis number |
Question 9 Explanation:
Pr=\frac{v}{\alpha }
Question 10 |
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 10 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}
There are 10 questions to complete.