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.

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