Question 1 |
Consider steady state, one-dimensional heat
conduction in an infinite slab of thickness
2L (L = 1 m) as shown in the figure. The conductivity
(k) of the material varies with temperature as
k = CT, where T is the temperature in K, and C is a
constant equal to 2 W\dot m^{-1} \dot K^{-2}. There is a uniform
heat generation of 1280 kW/m^3
in the slab. If both
faces of the slab are maintained at 600 K, then the
temperature at x = 0 is ________ K (in integer).
825 | |
1126 | |
1000 | |
1256 |
Question 1 Explanation:
General Heat conduction equation in Cartesian
coordinates is
\frac{\partial }{\partial x}\left ( K_x\frac{\partial T}{\partial x} \right )+\frac{\partial }{\partial y}\left ( K_y\frac{\partial T}{\partial y} \right )\frac{\partial }{\partial z}\left ( K_z\frac{\partial T}{\partial z} \right )+q_g=\rho c\frac{\partial T}{\partial t}
For one dimensional steady state heat generation with variable thermal conductivity, the above equation can be written as
\begin{aligned} \frac{\partial }{\partial x}\left ( CT\frac{\partial T}{\partial x} \right )+q_g&=0\\ C\frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+q_g&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+\frac{q_g}{C}&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\frac{q_g}{C}\\ \int \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\int \frac{q_g}{C}\\ \text{For 1-D heat flow}\\ \frac{\partial T}{\partial x}&=\frac{dT }{\partial x}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}+C_1\\ x=0\;\;\frac{dT}{dx}&=0\;\;C_1=0 \text{ (At centerline)}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}\\ \int TdT&=\int -\frac{q_gx}{C}dx\\ \frac{T^2}{2}&=\frac{-q_gx^2}{2C}+C_2 . . . (1)\\ T=T_s&=\text{Outside surface temperature}\\ \frac{T_S^2}{2}&=\frac{-q_gL^2}{2C}+C_2\\ C_2&=\frac{T-s^2}{2}+\frac{q_gL^2}{2C}\\ \text{Put }&c_2 \text{ in equation (1)}\\ \frac{T^2}{2}&=\frac{-q_ax^2}{2C}+\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \text{At }&x=0,T=T_{max}\\ \frac{T_{max}^2}{2}&=\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \frac{T_{max}^2}{2}&=\frac{600^2}{2}+\frac{128010^3 \times 1^2}{2 \times 2}\\ T_{max}&=1000K \end{aligned}
\frac{\partial }{\partial x}\left ( K_x\frac{\partial T}{\partial x} \right )+\frac{\partial }{\partial y}\left ( K_y\frac{\partial T}{\partial y} \right )\frac{\partial }{\partial z}\left ( K_z\frac{\partial T}{\partial z} \right )+q_g=\rho c\frac{\partial T}{\partial t}
For one dimensional steady state heat generation with variable thermal conductivity, the above equation can be written as
\begin{aligned} \frac{\partial }{\partial x}\left ( CT\frac{\partial T}{\partial x} \right )+q_g&=0\\ C\frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+q_g&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+\frac{q_g}{C}&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\frac{q_g}{C}\\ \int \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\int \frac{q_g}{C}\\ \text{For 1-D heat flow}\\ \frac{\partial T}{\partial x}&=\frac{dT }{\partial x}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}+C_1\\ x=0\;\;\frac{dT}{dx}&=0\;\;C_1=0 \text{ (At centerline)}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}\\ \int TdT&=\int -\frac{q_gx}{C}dx\\ \frac{T^2}{2}&=\frac{-q_gx^2}{2C}+C_2 . . . (1)\\ T=T_s&=\text{Outside surface temperature}\\ \frac{T_S^2}{2}&=\frac{-q_gL^2}{2C}+C_2\\ C_2&=\frac{T-s^2}{2}+\frac{q_gL^2}{2C}\\ \text{Put }&c_2 \text{ in equation (1)}\\ \frac{T^2}{2}&=\frac{-q_ax^2}{2C}+\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \text{At }&x=0,T=T_{max}\\ \frac{T_{max}^2}{2}&=\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \frac{T_{max}^2}{2}&=\frac{600^2}{2}+\frac{128010^3 \times 1^2}{2 \times 2}\\ T_{max}&=1000K \end{aligned}
Question 2 |
Consider a solid slab (thermal conductivity, k =10 W \dot m^{-1}\dot K^{-1}) with thickness 0.2 m and of infinite extent in the other two directions as shown in the
figure. Surface 2, at 300 K, is exposed to a fluid
flow at a free stream temperature ( T_\infty ) of 293 K,
with a convective heat transfer coefficient (h) of
100 W\dot m^{-2} \dot K^{-1} . Surface 2 is opaque, diffuse and
gray with an emissivity (\varepsilon ) of 0.5 and exchanges
heat by radiation with very large surroundings at
0 K. Radiative heat transfer inside the solid slab
is neglected. The Stefan-Boltzmann constant is
5.67 \times 10^{-8} W \dot m^{-2}\dot K^{-4} . The temperature T_1
of Surface
1 of the slab, under steady-state conditions, is
_________ K (round off to the nearest integer).
254 | |
632 | |
985 | |
319 |
Question 2 Explanation:
T_\infty =293K,T_{surrounding}=0K, h=100W/m^2K, K_{slab}=10W/mK
Under steady state
\begin{aligned} Q_{stored}&=0\\ Q_{in}&=Q_{out}\\ Q_{cond}&=Q_{conv}+Q_{rad}\\ K\left [ \frac{T_1-T_2}{L} \right ]&=h(T_2-T_\infty )+\varepsilon \sigma (T_2^4-T_{surr}^4)\\ 10\left [ \frac{T_1-300}{0.2} \right ]&=100(300-293 )+0.5 \times 5.67 \times 10^{-8} (300^4-0^4)\\ T_1-300&=\frac{929.635}{50}\\ T_1&=18.59+300\\ T_1&=318.59\\ T_1&=318.6K=319K \end{aligned}
Under steady state
\begin{aligned} Q_{stored}&=0\\ Q_{in}&=Q_{out}\\ Q_{cond}&=Q_{conv}+Q_{rad}\\ K\left [ \frac{T_1-T_2}{L} \right ]&=h(T_2-T_\infty )+\varepsilon \sigma (T_2^4-T_{surr}^4)\\ 10\left [ \frac{T_1-300}{0.2} \right ]&=100(300-293 )+0.5 \times 5.67 \times 10^{-8} (300^4-0^4)\\ T_1-300&=\frac{929.635}{50}\\ T_1&=18.59+300\\ T_1&=318.59\\ T_1&=318.6K=319K \end{aligned}
Question 3 |
Consider a one-dimensional steady heat conduction
process through a solid slab of thickness 0.1 m. The
higher temperature side A has a surface temperature
of 80^{\circ}C , and the heat transfer rate per unit area to
low temperature side B is 4.5 \; kW/m^2 . The thermal
conductivity of the slab is 15 \; W/m.K . The rate of
entropy generation per unit area during the heat
transfer process is ________ W/m^2.K (round off to
2 decimal places).
2.25 | |
3.36 | |
1.18 | |
1.85 |
Question 3 Explanation:
1D Steady heat conduction
\begin{aligned} \frac{Q}{A} &=4.5\frac{kW}{m^2}=4500 \frac{W}{m^2}\\ k&=15W/mk \\ S_{gen}&=? \\ \frac{Q}{A} &= k\left [ \frac{T_1-T_2}{dx} \right ]\\ 4500&= 15\left [ \frac{80-T_2}{0.1} \right ]\\ T_2 &=50^{\circ}C=323K \\ T_1 &=80^{\circ}C=353K \\ \therefore \;\; \frac{S_{gen}}{Area}&=\frac{-Q}{AT_1}+\frac{Q}{AT_2} \\ &=\frac{-4500}{353}+\frac{4500}{323} \\ S_{gen} &= 1.18\frac{W}{m^2K} \end{aligned}
\begin{aligned} \frac{Q}{A} &=4.5\frac{kW}{m^2}=4500 \frac{W}{m^2}\\ k&=15W/mk \\ S_{gen}&=? \\ \frac{Q}{A} &= k\left [ \frac{T_1-T_2}{dx} \right ]\\ 4500&= 15\left [ \frac{80-T_2}{0.1} \right ]\\ T_2 &=50^{\circ}C=323K \\ T_1 &=80^{\circ}C=353K \\ \therefore \;\; \frac{S_{gen}}{Area}&=\frac{-Q}{AT_1}+\frac{Q}{AT_2} \\ &=\frac{-4500}{353}+\frac{4500}{323} \\ S_{gen} &= 1.18\frac{W}{m^2K} \end{aligned}
Question 4 |
A tiny temperature probe is fully immersed in a
flowing fluid and is moving with zero relative
velocity with respect to the fluid. The velocity
field in the fluid is \vec{V}=2x\hat{i}+(y+3t)\hat{j},and the
temperature field in the fluid is T=2x^2+xy+4t where x and y are the spatial coordinates, and t is
the time. The time rate of change of temperature
recorded by the probe at (x=,y=1,t=1) is
_______.
4 | |
0 | |
18 | |
14 |
Question 4 Explanation:
\begin{aligned}
\frac{dT}{dt}&=? \text{ at }x=1,y=1,t=1\\
\frac{dT}{dt}&=\frac{\partial T}{\partial t}\frac{dt}{dt}+\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{\partial T}{\partial y}\frac{dy}{dt}\\
&=\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}\\
&=4+(2x)(4x+y)+(y+3t)(x)\\
&=4+8x^2+3xy+3xt\\
&\text{at }x=1,y=1,t=1\\
\frac{dT}{dt}&=4+8(1)^2+3(1)+3(1)\\
&=18
\end{aligned}
Question 5 |
In a furnace, the inner and outer sides of the brick wall (k_1 = 2.5 W/mK) are maintained
at 1100^{\circ}C and 700^{\circ}C respectively as shown in figure.
The brick wall is covered by an insulating material of thermal conductivity k_2. The thickness of the insulation is 1/4^{th} of the thickness of the brick wall. The outer surface of the insulation is at 200^{\circ}C. The heat flux through the composite walls is 2500 W/m^2.
The value of k_2 is ________ W/mK (round off to 2 decimal places).
The brick wall is covered by an insulating material of thermal conductivity k_2. The thickness of the insulation is 1/4^{th} of the thickness of the brick wall. The outer surface of the insulation is at 200^{\circ}C. The heat flux through the composite walls is 2500 W/m^2.
The value of k_2 is ________ W/mK (round off to 2 decimal places).
0.25 | |
0.5 | |
0.75 | |
1 |
Question 5 Explanation:
Given, L_{2}=\frac{L_{1}}{4}
Assuming steady state, one-dimensional conduction heat transfer through composite slab,
Thermal circuit:
\begin{array}{l} \Rightarrow \quad q=\frac{1100-700}{\frac{L_{1}}{k_{1} A}}=\frac{700-200}{\frac{L_{2}}{k_{2} A}} \\ \Rightarrow \quad \frac{400}{\frac{L_{2}}{2.5}}=\frac{500}{\frac{L_{2}}{k_{2}}}\\ \Rightarrow \quad k_{2}=0.5 \mathrm{W}-\mathrm{mK} \end{array}
Assuming steady state, one-dimensional conduction heat transfer through composite slab,
Thermal circuit:
\begin{array}{l} \Rightarrow \quad q=\frac{1100-700}{\frac{L_{1}}{k_{1} A}}=\frac{700-200}{\frac{L_{2}}{k_{2} A}} \\ \Rightarrow \quad \frac{400}{\frac{L_{2}}{2.5}}=\frac{500}{\frac{L_{2}}{k_{2}}}\\ \Rightarrow \quad k_{2}=0.5 \mathrm{W}-\mathrm{mK} \end{array}
Question 6 |
One-dimensional steady state heat conduction takes place through a solid whose cross-sectional area varies linearly in the direction of heat transfer. Assume there is no heat generation in the solid and the thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is
Linear | |
Quadratic | |
Logarithmic | |
Exponential |
Question 6 Explanation:
\begin{array}{c} \text { Area }(\mathrm{A}) \propto \mathrm{x} \\ \mathrm{A}=\mathrm{c} \mathrm{x} \end{array}
According to Fourier's law of heat conduction
\mathrm{Q}=-\mathrm{k} \mathrm{A} \frac{\mathrm{d} \mathrm{T}}{\mathrm{d} \mathrm{x}}
\begin{array}{l} \mathrm{Q}=-\mathrm{k} \cdot \mathrm{c} \mathrm{x} \frac{\mathrm{d} \mathrm{T}}{\mathrm{d} \mathrm{x}} \\ \mathrm{Q} \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}=-\mathrm{k} \mathrm{c} \mathrm{d} \mathrm{T} \\ \int \mathrm{d} \mathrm{T}=-\frac{\mathrm{Q}}{\mathrm{k} \mathrm{c}} \int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}} \\ \mathrm{T}=-\frac{\mathrm{Q}}{\mathrm{k}_{\mathrm{r}}} \ln \mathrm{x}+\mathrm{c}_{1} \end{array}
Temperature distribution is logarithmic.
(or)
In hollow cylinder, area is linearly proportional to radius.
A = 2\pi rL
Temperature profile is logarithmic in case of hollow cylinder with no heat generation.
According to Fourier's law of heat conduction
\mathrm{Q}=-\mathrm{k} \mathrm{A} \frac{\mathrm{d} \mathrm{T}}{\mathrm{d} \mathrm{x}}
\begin{array}{l} \mathrm{Q}=-\mathrm{k} \cdot \mathrm{c} \mathrm{x} \frac{\mathrm{d} \mathrm{T}}{\mathrm{d} \mathrm{x}} \\ \mathrm{Q} \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}=-\mathrm{k} \mathrm{c} \mathrm{d} \mathrm{T} \\ \int \mathrm{d} \mathrm{T}=-\frac{\mathrm{Q}}{\mathrm{k} \mathrm{c}} \int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}} \\ \mathrm{T}=-\frac{\mathrm{Q}}{\mathrm{k}_{\mathrm{r}}} \ln \mathrm{x}+\mathrm{c}_{1} \end{array}
Temperature distribution is logarithmic.
(or)
In hollow cylinder, area is linearly proportional to radius.
A = 2\pi rL
Temperature profile is logarithmic in case of hollow cylinder with no heat generation.
Question 7 |
A slender rod of length L, diameter d (L \gt \gt d) and thermal conductivity k_1 is joined with another rod of identical dimensions, but of thermal conductivity k_2, to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is
k_1+k_2 | |
\sqrt{k_1k_2} | |
\frac{k_1k_2}{k_1+k_2} | |
\frac{2k_1k_2}{k_1+k_2} |
Question 7 Explanation:
Area is constant in the direction of heat flow. (similar like slab)
Thermal circuit :
\begin{array}{l} \text { Heat transfer rate, }(Q)=\frac{T_{1}-T_{3}}{\frac{L}{k_{1} A}+\frac{L}{k_{2} A}}=\frac{T_{1}-T_{3}}{\frac{2 L}{k_{e q} A}} \\ \frac{L}{k_{1} A}+\frac{L}{k_{2} A}=\frac{2 L}{k_{e q} A} \\ \frac{2}{k_{e q}}=\frac{1}{k_{1}}+\frac{1}{k_{2}} \\ k_{e q}=\frac{2 k_{1} k_{2}}{k_{1}+k_{2}} \end{array}
Question 8 |
A 0.2 m thick infinite black plate having a thermal conductivity of 3.96 W/m-K is exposed
to two infinite black surfaces at 300 K and 400 K as shown in the figure. At steady state,
the surface temperature of the plate facing the cold side is 350 K. The value of Stefan-
Boltzmann constant, \sigma , is 5.67 \times 10^{-8} \: W/m^{2}\: K^{4} Assuming 1-D heat conduction, the magnitude of heat flux through the plate (in W/m^{2}) is ________ (correct to two decimal places).
258.78 | |
963.25 | |
391.61 | |
147.93 |
Question 8 Explanation:
Under steady state condition, all rate of heat transfer i.e. from surface at 400 K to black
plate (via radiation), inside black plate (via conduction) and from black plate to surface
at 300 K (via radiation) are equal.
So, heat flux through wall= flux from wall to surface at 300 K
=\frac{\sigma\left(350^{4}-300^{4}\right)}{\frac{1}{1}+\frac{1}{1}-1}=391.612 \text { watt/m }^{2}
So, heat flux through wall= flux from wall to surface at 300 K
=\frac{\sigma\left(350^{4}-300^{4}\right)}{\frac{1}{1}+\frac{1}{1}-1}=391.612 \text { watt/m }^{2}
Question 9 |
Heat is generated uniformly in a long solid cylindrical rod (diameter = 10mm) at the rate of 4\,\times\,10^{7} W/m^{3}. The thermal conductivity of the rod material is 25W/m.K. Under steady state conditions, the temperature difference between the centre and the surface of the rod is _________ ^{\circ}C.
7 | |
8 | |
9 | |
10 |
Question 9 Explanation:
\begin{aligned} T_{0}&= \text{ Temp. of cylinder at the axis}\\ T_{S} &=\text { Surface Temp. of Rod (cylindel) } \\ T_{O}-T_{S} &=\frac{\dot{q}}{4 k} R^{2} \end{aligned}
\dot{q} uniform heat generation rate per unit volumne
\left(w / m^{3}\right)
T_{0}-T_{S}=\frac{4 \times 10^{7}}{4 \times 25} \times\left(\frac{5}{1000}\right)^{2}=10^{\circ} \mathrm{C}
Question 10 |
Steady one-dimensional heat conduction takes place across the faces 1 and 3 of a composite slab consisting of slabs A and B in perfect contact as shown in the figure, where k_{A},k_{B} denote the respective thermal conductivities. Using the data as given in the figure, the interface temperature T2 (in ^{\circ}C) is __________
67.5 C | |
68.5 C | |
70.6 C | |
54.6 C |
Question 10 Explanation:
At steady state
\begin{aligned} Q_{A} &=Q_{B} \\ \frac{k_{A}\left[T_{1}-T_{2}\right] A}{L_{1}} &=\frac{k_{B}\left[T_{2}-T_{3}\right] A}{L_{2}} \\ \frac{20 \times\left[130-T_{2}\right]}{0.1} &=\frac{100 \times\left[T_{2}-30\right]}{0.3} \\ 3\left[130-T_{2}\right] &=5 T_{2}-150 \\ 390-3 T_{2} &=5 T_{2}-150 \\ 540 &=8 T_{2} \\ T_{2} &=67.5^{\circ} \mathrm{C} \end{aligned}
There are 10 questions to complete.
9 number answer A hoga
As per Official answer key answer is C.
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