# Heat Transfer in Flow Over Plates and Pipes

 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).
 A 5 B 10 C 15 D 20
GATE ME 2021 SET-2   Heat Transfer
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
For a hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant cross-section, the Nusselt number at constant wall heat flux ($Nu_q$) and that at constant wall temperature ($Nu_T$) are related as
 A $Nu_q \gt Nu_T$ B $Nu_q \lt Nu_T$ C $Nu_q = Nu_T$ D $Nu_q = ( Nu_T)^2$
GATE ME 2019 SET-1   Heat Transfer
Question 2 Explanation:
For laminar flow through circular tube:
$\mathrm{Nu}_{\mathrm{q}}=4.36$ (For constant heat flux)
$\mathrm{Nu}_{\mathrm{T}}=3.66$ (For constant wall temperature)
$\mathrm{Nu}_{\mathrm{q}} \gt \mathrm{Nu}_{\mathrm{T}}$
 Question 3
For flow through a pipe of radius R, the velocity and temperature distribution are as follows $u(r,x)=C_{1}$ and $T(r,x)=C_{2}[1-(\frac{r}{R})^{3}],$ where $C_{1}$ and $C_{2}$ are constants. The bulk mean 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 $\frac{0.5C_{2}}{U_{m}}$ B $0.5C_{2}$ C $0.6C_{2}$ D $\frac{0.6C_{2}}{U_{m}}$
GATE ME 2015 SET-1   Heat Transfer
Question 3 Explanation:
\begin{aligned} u(r, x)&=C_{1} \\ T(r, x)&=C_{2}\left[1-\left(\frac{r}{R}\right)^{3}\right] \\ T_{m}&=\frac{2}{U_{m} R^{2}} \int_{0}^{R} u(r, x) T(r, x) r d r \\ &=\frac{2}{U_{m} R^{2}} \int_{0}^{R} C_{1} \times C_{2}\left[1-\left(\frac{r}{R}\right)^{3}\right] r d r \\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}} \int_{0}^{R}\left[1-\left(\frac{r}{R}\right)^{3}\right] r d r \\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}} \int_{0}^{R}\left(r-\frac{r^{4}}{R^{3}}\right) d r \\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}}\left[\frac{r^{2}}{2}-\frac{r^{5}}{5 R^{3}}\right]_{0}^{R}\\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}}\left[\frac{R^{2}}{2}-\frac{R^{5}}{5 R^{3}}\right] \\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}}\left[\frac{R^{2}}{2}-\frac{R^{2}}{5}\right] \\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}}\left[\frac{5 R^{2}-2 R^{2}}{10}\right] \\ &=\frac{2 C_{1} C_{2}}{U_{m} R^{2}} \times \frac{3}{10} R^{2}=\frac{0.6 C_{1} C_{2}}{U_{m}} \quad\ldots(i) \end{aligned}
since $u(r, x)=C_{1}$, constant $U_{m}=C_{1}$
Substituting $U_{m}=C_{1}$ in above Eq. (i), we get
$T_{m}=\frac{0.6 C_{1} C_{2}}{C_{1}}=0.6 C_{2}$
 Question 4
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
GATE ME 2015 SET-1   Heat Transfer
Question 4 Explanation:
For the flow of viscous fluid over a flat plate if the fluid temperature is the same as the plate temperature the thermal boundary layer is not formed at all because boundary layer formation took place if there is some difference in fluid property of no slip layer and remaining fluid.
As here given that fluid is viscous and flowing over a flat plate. Here, definitely a kinematic boundary layer will be formed but as there is no temperature difference so no formation of thermal boundary layer.
 Question 5
The Blasius equation related to boundary layer theory is a
 A third-order linear partial differential equation B third-order nonlinear partial differential equation C second-order nonlinear ordinary differential equation D third-order nonlinear ordinary differential equation
GATE ME 2015 SET-1   Heat Transfer
Question 5 Explanation:
$2 \frac{d^{3} f}{d \eta^{3}}+f \frac{d^{2} f}{d \eta^{2}}=0$
third-order non-linear differential equation.
 Question 6
Water flows through a tube of diameter 25 mm at an average velocity of 1.0 m/s. The properties of water are $\rho =1000 kg/m^{3}$ , $\mu =7.25\times 10^{-4} N.s/m^{2}$ , $k=0.625 W/m.K, P_{r}=4.85$. Using $Nu=0.023 Re^{0.8} Pr^{0.4}$, the convective heat transfer coefficient (in $W/m^{2}.K$ )is______
 A 1458.25 B 6589.54 C 4613.66 D 8754.65
GATE ME 2014 SET-4   Heat Transfer
Question 6 Explanation:
Reynolds number,
\begin{aligned} \operatorname{Re} &=\frac{\rho V d}{\mu}=\frac{1000 \times 1 \times 25 \times 10^{-3}}{7.25 \times 10^{-4}} \\ &=34482.75 \\ \operatorname{Pr} &=4.85 \\ \mathrm{Nu} &=0.023 \mathrm{Re}^{0.8} \mathrm{Pr} .^{0.4} \\ \mathrm{Nu} &=0.023 \times(34482.758)^{0.8}(4.85)^{0.4} \\ &=184.546 \\ \text { also, } \quad \mathrm{Nu} &=\frac{h d}{k} \\ \text { or } \quad h &=\frac{\mathrm{Nuk}}{d}=\frac{184.546 \times 0.625}{25 \times 10^{-3}}\\ &=4613.66 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \end{aligned}
 Question 7
The non-dimensional 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 $N_u=\frac{q''}{T_W-T_\infty }\frac{L}{k}$ is equal to
 A a B b C 2c D (b+2c)
GATE ME 2014 SET-1   Heat Transfer
Question 7 Explanation:
\begin{aligned} q^{\prime \prime} &=-\left.k \frac{\partial T}{\partial y}\right|_{y=0} \\ &=h\left(T_{w}-T_{\infty}\right) \\ T_{w}-T &=\left(T_{w}-T_{\infty}\right)\left[a+\frac{b y}{L}+c\left(\frac{y}{L}\right)^{2}\right] \\ \frac{\partial T}{\partial y} & =\left(T_{w}-T_{\infty}\right)\left[\frac{b}{L}+\frac{2 c}{L^{2}} y\right] \\ -\left.\frac{\partial T}{\partial y}\right|_{y=0} &=\left(T_{w}-T_{\infty}\right) \frac{b}{L} \\ \text{or} \quad q^{\prime \prime}&=k\left(T_{w}-T_{\infty}\right) \frac{b}{L} \\ \therefore \quad N u &=\frac{q^{\prime \prime}}{T_{w}-T_{\infty}} \frac{L}{k}=b \end{aligned}
 Question 8
Consider a two-dimensional 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
GATE ME 2014 SET-1   Heat Transfer
Question 8 Explanation: \begin{aligned} &1.\mathrm{Re}=500\\ &2.\mathrm{Re}=10^{4}\\ &3.\mathrm{Re}=2 \times 10^{5}\\ &4.\mathrm{Re}=2 \times 10^{6}\\ \end{aligned}
Experimental results for the variation of the local Nusselt number with $\theta$ are shown in above figure for the cylinder in cross flow of air. The results are strongly influenced by the nature of boundary layer development on the surface. Starting at the stagnation point, Nu decreases with increasing $\theta$ as a result of laminar boundary layer development. Nu as a result of laminar boundary layer development. Nu reaches minimum at $80^{\circ}$ where separation occurs. So, at point 2, the local heat transfer coefficient is minimum.
 Question 9
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 $\frac{1}{2}$ 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 $\frac{1}{8}$ 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
GATE ME 2011   Heat Transfer
Question 9 Explanation:
Given data:
For fluid P
\begin{aligned} \frac{\delta}{\delta_{t}} &=\frac{1}{2} \\ (\mathrm{Re})_{P} &=1000 \\ (\mathrm{Pr})_{P} &=\frac{1}{8} \\ (\mathrm{Nu})_{P} &=35 \\ (\mathrm{Nu})_{P} &=0.664(\mathrm{Re})_{P}^{0.5}(\mathrm{Pr})_{P}^{0.333} \end{aligned}
For fluid Q
\begin{aligned} \frac{\delta}{\delta_{t}} &=2 \\ (\mathrm{Re})_{a} &=10000 \\ (\mathrm{Pe})_{a} &=? \\ (\mathrm{Nu})_{a} &=? \\ \frac{\delta}{\delta_{t}} &=(\mathrm{Pr})_{a}^{1 / 3} \\ 2 &=(\mathrm{Pr})_{a}^{1 / 3} \\ \text{or} \quad (\mathrm{Pr}) &=(2)^{3}=8 \end{aligned}
For laminar boundary layer on a flat plate,
\begin{aligned} (P r)_{Q} &=(P r)_{Q}^{0.5}(P r)_{C}^{0.333} \\ \text{Ratio} \quad \frac{(N u)_{P}}{(N u)_{Q}} &=\frac{0.664(R e)_{P}^{0.5}(P r)_{P}^{0.333}}{0.664(R e)_{Q}^{0.5}(P r)_{0}^{0.333}} \\ \frac{(N u)_{P}}{(N u)_{Q}} &=\frac{(R e)_{P}^{0.5}(P r)_{P}^{0.333}}{(R e)_{Q}^{0.5}(P r)_{Q}^{0.333}} \\ \frac{35}{(N u)_{Q}} &=\frac{(10000)^{0.5} \times\left(\frac{1}{8}\right)^{0.333}}{(10000)^{0.5} \times(8)^{0.333}} \\ \frac{35}{(N u)_{Q}} &=\frac{1}{(8)^{0.333} \times(8)^{0.333}} \\ &=\frac{1}{(8)^{0.666}}=\frac{1}{4} \\ \left(N_{u}\right)_{Q} &=35 \times 4=140 \end{aligned}
 Question 10
A pipe of 25mm outer diameter carries steam. The heat transfer coefficient between the cylinder and surroundings is 5W /$m^{2}K$. It is proposed to reduce the heat loss from the pipe by adding insulation having a thermal conductivity of 0.05W/mK. Which one of the following statements is TRUE?
 A The outer radius of the pipe is equal to the critical radius B The outer radius of the pipe is less than the critical radius C Adding the insulation will reduce the heat loss D Adding the insulation will increase the heat loss
GATE ME 2011   Heat Transfer
Question 10 Explanation:
Given data:
Outer diameter of pipe,
\begin{aligned} d_{0}&=25 \mathrm{mm}\\ \therefore \text{Radius: }r_{0}&=\frac{d_{0}}{2}=\frac{25}{2}=12.5 \mathrm{mm} \\ h&=5 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \\ k&=0.05 \mathrm{W} / \mathrm{mK} \end{aligned}
$r_{c}=\frac{k}{h}=\frac{0.05}{5}=0.01 \mathrm{m}=10 \mathrm{mm}$ As $r_{0}>r_{c}$ , then adding the insulation will reduce the heat loss. 