# Free and Forced Convection

 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).
 A 125 B 321 C 652 D 228
GATE ME 2022 SET-2   Heat Transfer
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?
 A Ratio of internal thermal resistance of a solid to the boundary layer thermal resistance B Ratio of the rate at which internal energy is advected to the rate of conduction heat transfer C Non-dimensional temperature gradient D Non-dimensional velocity gradient multiplied by Prandtl number
GATE ME 2022 SET-2   Heat Transfer
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
 A $\text{St}=\frac{\text{Nu}}{\text{Re Pr}}$ B $\text{St}=\frac{\text{Nu Pr}}{\text{Re}}$ C $\text{St}=\text{Nu Pr Re}$ D $\text{St}=\frac{\text{Nu Re}}{\text{Pr}}$
GATE ME 2021 SET-2   Heat Transfer
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?
 A Biot number and Fourier number B Reynolds number and Prandtl number C Biot number and Froude number D Nusselt number and Grashoff number
GATE ME 2021 SET-1   Heat Transfer
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).
 A 125.4 B 25.6 C 48.8 D 85.7
GATE ME 2020 SET-2   Heat Transfer
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
 A 283.3 B 299.9 C 149.9 D 449.7
GATE ME 2020 SET-1   Heat Transfer
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}
 Question 7
Grashof number signifies the ratio of
 A inertia force to viscous force B buoyancy force to viscous force C buoyancy force to inertia force D inertia force to surface tension force
GATE ME 2016 SET-3   Heat Transfer
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 __________
 A 6.123mm B 5.269mm C 8.2654mm D 4.9mm
GATE ME 2016 SET-1   Heat Transfer
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}
 Question 9
The ratio of momentum diffusivity (v) to thermal diffusivity ($\alpha$), is called
 A Prandtl number B Nusselt number C Biot number D Lewis number
GATE ME 2015 SET-3   Heat Transfer
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
 A $\delta = \delta_{r}$ B $\delta \gt \delta_{r}$ C $\delta \lt \delta_{r}$ D $\delta = 0$ but $\delta_{r} \neq 0$
GATE ME 2015 SET-2   Heat Transfer
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}
There are 10 questions to complete.