# Heat Transfer in Flow Over Plates and Pipes

 Question 1
A very large metal plate of thickness $d$ and thermal conductivity $k$ is cooled by a stream of air at temperature $T_{\infty }=300K$ with a heat transfer coefficient $h$, as shown in the figure. The centerline temperature of the plate is $T_P$. In which of the following case(s) can the lumped parameter model be used to study the heat transfer in the metal plate?

 A $h=10Wm^{-2}K^{-1},k=100Wm^{-1}K^{-1},d=1mm,T_P=350K$ B $h=100Wm^{-2}K^{-1},k=100Wm^{-1}K^{-1},d=1mm,T_P=325K$ C $h=100Wm^{-2}K^{-1},k=1000Wm^{-1}K^{-1},d=1mm,T_P=325K$ D $h=1000Wm^{-2}K^{-1},k=1Wm^{-1}K^{-1},d=1mm,T_P=350K$
GATE ME 2023   Heat Transfer
Question 1 Explanation:

For lumped heat analysis
$\mathrm{B}_{\mathrm{i}} \leq 0.1$
$L_{c} =\frac{\text { volume }}{\text { surface area }} =\frac{d \times L \times L}{L \times L+L \times L}=\frac{d}{2}$

(a) $B_{i}=\frac{100 \times 0.5 \times 10^{-3}}{100}=0.5 \times 10^{-4} \leq 0.5$

(c) $B_{i}=\frac{h L_{c}}{\mathrm{~K}}=\frac{100 \times 0.5 \times 10^{-3}}{1000}=0.5 \times 10^{-4} \leq 0.1$

(b) $\mathrm{B}_{\mathrm{i}}=\frac{\mathrm{hL}_{\mathrm{c}}}{\mathrm{K}}=\frac{100 \times 0.5}{100}=0.5 \gt 0.1$

(d) $B_{i}=\frac{1000 \times 0.5}{1}=500 \gt 0.1$
Hence, answer is $(A, C)$
 Question 2
Consider incompressible laminar flow of a constant property Newtonian fluid in an isothermal circular tube. The flow is steady with fully-developed temperature and velocity profiles. The Nusselt number for this flow depends on
 A neither the Reynolds number nor the Prandtl number B both the Reynolds and Prandtl numbers C the Reynolds number but not the Prandtl number D the Prandtl number but not the Reynolds number
GATE ME 2023   Heat Transfer
Question 2 Explanation:
Nusselt no. for convection with uniform temperature circular tubes
$Nu_D = 3.66$

Convection with uniform heat flux for circular tubes
$Nu_D = 4.36$

Dittus-Boelter equation is an explicit function for calculating the nusselt number for rough tubes
$Nu_D = 0.023 {Re_D}^{4/5}Pr^n$

Flat plate in laminar flow
$N_{u_x} = 0.332R{e_x}^{1/2}Pr^{1/3} \;\;\; (Pr \gt 0.6)$

 Question 3
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 3 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 4
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 4 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 5
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 5 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
$T_{m}=\frac{0.6 C_{1} C_{2}}{C_{1}}=0.6 C_{2}$

There are 5 questions to complete.