Question 1 |

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 | |

B | |

C | |

D |

Question 1 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 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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}

There are 5 questions to complete.

9 number answer A hoga

As per Official answer key answer is C.

can i get the answer key for this

You can press any options out of four given. It will give you true answer and your answer is correct or not.