Heat Transfer


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 in Flow Over Plates and Pipes
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
A cylindrical rod of length h and diameter d is placed inside a cubic enclosure of side length L . S denotes the inner surface of the cube. The view-factor F_{S-S} is
A
0
B
1
C
\frac{(\pi d h+\pi d^2/2)}{6L^2}
D
1-\frac{(\pi d h+\pi d^2/2)}{6L^2}
GATE ME 2023      Radiation
Question 2 Explanation: 


Surface area of centrifugal rod =\pi d h+\frac{2 \pi d^{2}}{4}
Or A_{1}=\pi\left(d h+\frac{d^{2}}{2}\right)
Surface area of cube A_{2}=6 \times \mathrm{L}^{2}

For surface 1 (cylinder), \mathrm{F}_{11}=0
So \mathrm{F}_{12}=1 \quad \text {...(i) }

For surface 2(cube)
\begin{aligned} \mathrm{F}_{21}+\mathrm{F}_{22} & =1 \\ \mathrm{F}_{22} & =1-\mathrm{F}_{21} \\ \mathrm{F}_{21} & =1-\mathrm{F}_{22}\;\;...(ii) \end{aligned}
By reciprocity theorem for surface 1 and 2
\begin{aligned} A_{1} F_{12} & =A_{2} F_{21} \\ F_{21} & =\frac{A_{1}}{A_{2}} \end{aligned}
\left[F_{12}=1\right. from equation (i)]
So, By equation (ii)
F_{22}=F_{s S}=1-\left(\frac{\pi \mathrm{dh}+\frac{\pi d^{2}}{2}}{6 L^{2}}\right)


Question 3
Consider a counter-flow heat exchanger with the inlet temperatures of two fluids (1 and 2) being T_{1,in}=300K and T_{2,in}=350K. The heat capacity rates of the two fluids are C_1=1000 W/K and C_2=400 W/K, and the effectiveness of the heat exchanger is 0.5. The actual heat transfer rate is _____ kW. (Answer in integer)
A
6
B
10
C
12
D
18
GATE ME 2023      Heat Exchanger
Question 3 Explanation: 


Data given
\begin{aligned} \mathrm{C}_{1} & =1000 \mathrm{W} / \mathrm{K} \\ \mathrm{C}_{2} & =400 \mathrm{W} / \mathrm{K} \\ \epsilon & =\frac{\mathrm{q}_{\text {actual }}}{\mathrm{q}_{\max }} \end{aligned}
for q_{\max }, C should be minimum
So, \quad \mathrm{C}_{\min }=\mathrm{C}_{2}
So, \quad \in=\frac{\mathrm{q}_{\text {actual }}}{\mathrm{C}_{1}\left(\mathrm{T}_{\mathrm{h}_{1}}-\mathrm{T}_{\mathrm{c}_{1}}\right)}=0.5
or
\begin{aligned} q_{\text {actual }} & =0.5 \times 400(350-300) \\ & =10 \mathrm{kW} \end{aligned}
Question 4
Consider a laterally insulated rod of length L and constant thermal conductivity. Assuming one-dimensional heat conduction in the rod, which of the following steady-state temperature profile(s) can occur without internal heat generation?

A
A
B
B
C
C
D
D
GATE ME 2023      Conduction
Question 4 Explanation: 


This is simply heat transfer through plane surface along it's length.
So, Governing equation for it is
\frac{d^{2} T}{d x^{2}}=0
or \frac{\mathrm{d} T}{\mathrm{dx}}=4
or \mathrm{T}=\mathrm{C}_{1}+\mathrm{C}_{2}
a straight line equation} So, both option (A) and (B) are correct.
Question 5
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 in Flow Over Plates and Pipes
Question 5 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)




There are 5 questions to complete.

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