Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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}
Question 7 |
A hollow cylinder has length L, inner radius r_{1}, outer radius r_{2}, and thermal conductivity k. The thermal resistance of the cylinder for radial conduction is
\frac{\ln( r_{2}/r_{1} )}{2\pi kL} | |
\frac{\ln( r_{1}/r_{2} )}{2\pi kL} | |
\frac{2\pi kL}{\ln( r_{2}/r_{1})} | |
\frac{2\pi kL}{\ln( r_{1}/r_{2})} |
Question 7 Explanation:
Q=\frac{\Delta T}{\text { Thermal resistance }}
Thermal resistance =\int_{t_{1}}^{t_{2}} \frac{L}{k A}
\begin{aligned} \quad&=\int_{t_{1}}^{r_{2}} \frac{d r}{k \times 2 \pi r l} \\ \quad&=\frac{1}{2 \pi / k} \int_{r_{1}}^{r_{2}} \frac{d r}{r} \\ \quad&=\frac{1}{2 \pi l k}[\ln r]_{r_{1}}^{r_{2}} \\ \quad&=\frac{1}{2 \pi l k} \ln \frac{r_{2}}{r_{1}} \\ \end{aligned}
Thermal resistance =\int_{t_{1}}^{t_{2}} \frac{L}{k A}
\begin{aligned} \quad&=\int_{t_{1}}^{r_{2}} \frac{d r}{k \times 2 \pi r l} \\ \quad&=\frac{1}{2 \pi / k} \int_{r_{1}}^{r_{2}} \frac{d r}{r} \\ \quad&=\frac{1}{2 \pi l k}[\ln r]_{r_{1}}^{r_{2}} \\ \quad&=\frac{1}{2 \pi l k} \ln \frac{r_{2}}{r_{1}} \\ \end{aligned}
Question 8 |
A plastic sleeve of outer radius r_{0}=1mm covers a wire (radius r = 0.5 mm) carrying electric current. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 W/m^{2}-K. Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will
increase | |
remain the same | |
decrease | |
be zero |
Question 8 Explanation:
\begin{aligned} \text { Given data: } r_{o} &=1 \mathrm{mm} \\ \text { Inner radius: } r_{i} &=0.5 \mathrm{mm} \\ k &=0.15 \mathrm{W} / \mathrm{mK} \\ h_{o} &=25 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}\\ \text{Critical radius,}\\ r_{c} &=\frac{k}{h_{0}}=\frac{0.15}{25} \\ &=6 \times 10^{-3} \mathrm{m}=6 \mathrm{mm} \end{aligned}
As r_{o} \lt r_{c^{\prime}} due to addition of plastic cover heat transfer will increase till r_{o}=r_{c} beyond that heat transfer will decrease.
As r_{o} \lt r_{c^{\prime}} due to addition of plastic cover heat transfer will increase till r_{o}=r_{c} beyond that heat transfer will decrease.
Question 9 |
A brick wall (k= 0.9\frac{W}{m.K}) of thickness 0.18m separate the warm air in a room from the cold ambient air. On a particular winter day, the outside air temperature is -5^{\circ}C and the room needs to be maintained at 27^{\circ}C. The heat transfer coefficient associated with outside air is 20\frac{W}{m^{2} K} . Neglecting the convective resistance of the air inside the room, the heat loss, in\frac{W}{m^{2}}, is
88 | |
110 | |
128 | |
160 |
Question 9 Explanation:
Given data:
\begin{aligned} k&=0.9 \mathrm{W} / \mathrm{mK} \\ \delta&=0.18 \mathrm{m} \end{aligned}
T_{0}=-5^{\circ} \mathrm{C} ; \quad T_{1}=27^{\circ} \mathrm{C} ; \quad h_{0}=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}
\begin{aligned} \text{Heat loss: }Q&=\frac{T_{1}-T_{0}}{\frac{\delta}{\mathrm{k} A}+\frac{1}{h_{0} A}} \\ \frac{Q}{A} &=\frac{T_{1}-T_{0}}{\frac{\delta}{k}+\frac{1}{h_{0}}} \\ q&=\frac{27-(-5)}{0.18+\frac{1}{20}} \quad \because q=\frac{Q}{A} \\ &=\frac{32}{0.2+0.05} = \frac{32}{0.25}\\ &=128 \mathrm{W} / \mathrm{m}^{2} \end{aligned}
\begin{aligned} k&=0.9 \mathrm{W} / \mathrm{mK} \\ \delta&=0.18 \mathrm{m} \end{aligned}
T_{0}=-5^{\circ} \mathrm{C} ; \quad T_{1}=27^{\circ} \mathrm{C} ; \quad h_{0}=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}
\begin{aligned} \text{Heat loss: }Q&=\frac{T_{1}-T_{0}}{\frac{\delta}{\mathrm{k} A}+\frac{1}{h_{0} A}} \\ \frac{Q}{A} &=\frac{T_{1}-T_{0}}{\frac{\delta}{k}+\frac{1}{h_{0}}} \\ q&=\frac{27-(-5)}{0.18+\frac{1}{20}} \quad \because q=\frac{Q}{A} \\ &=\frac{32}{0.2+0.05} = \frac{32}{0.25}\\ &=128 \mathrm{W} / \mathrm{m}^{2} \end{aligned}
Question 10 |
A cylindrical uranium fuel rod of radius 5 mm in a nuclear reactor is generating heat at the rate of 4\times 10^{7} Wm^{3} . The rod is cooled by a liquid (convective heat transfer coefficient 1000 W/m^{2}-K) at 25^{\circ}C. At steady state, the surface temperature (in K) of the rod is
308 | |
398 | |
418 | |
448 |
Question 10 Explanation:
Given data:
\begin{aligned} r&=5 \mathrm{mm}=0.005 \mathrm{m}\\ \therefore \quad &=2 r=2 \times 0.005=0.010 \mathrm{m} \\ q_{G} &=4 \times 10^{7} \mathrm{W} / \mathrm{m}^{3} \\ h_{0} &=1000 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \\ T_{0} &=25^{\circ} \mathrm{C} \end{aligned}
For steady state heat transfer,
Rate of heat generation in the rod
= rate of convection heat transfer from rod surface to fluid
\begin{aligned} q_{G} \times \text { volume of } \operatorname{rod} &=h_{0} A\left(T_{s}-T_{0}\right) \\ q_{G} \times \frac{\pi}{4} d^{2} l^{\prime} &=h_{0} \times \pi d l\left(T_{s}-T_{0}\right) \\ \frac{d q_{G}}{4} &=h_{0}\left(T_{s}-T_{0}\right) \\ \frac{0.010 \times 4 \times 10^{7}}{4} &=1000\left(T_{s}-25\right) \\ \text{or }\quad T_{s}-25 &=100\\ \text{or }\quad T_{s}&=100+25=125^{\circ} \mathrm{C}\\ &=(125+273) \mathrm{K}=398 \mathrm{K} \end{aligned}
\begin{aligned} r&=5 \mathrm{mm}=0.005 \mathrm{m}\\ \therefore \quad &=2 r=2 \times 0.005=0.010 \mathrm{m} \\ q_{G} &=4 \times 10^{7} \mathrm{W} / \mathrm{m}^{3} \\ h_{0} &=1000 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \\ T_{0} &=25^{\circ} \mathrm{C} \end{aligned}
For steady state heat transfer,
Rate of heat generation in the rod
= rate of convection heat transfer from rod surface to fluid
\begin{aligned} q_{G} \times \text { volume of } \operatorname{rod} &=h_{0} A\left(T_{s}-T_{0}\right) \\ q_{G} \times \frac{\pi}{4} d^{2} l^{\prime} &=h_{0} \times \pi d l\left(T_{s}-T_{0}\right) \\ \frac{d q_{G}}{4} &=h_{0}\left(T_{s}-T_{0}\right) \\ \frac{0.010 \times 4 \times 10^{7}}{4} &=1000\left(T_{s}-25\right) \\ \text{or }\quad T_{s}-25 &=100\\ \text{or }\quad T_{s}&=100+25=125^{\circ} \mathrm{C}\\ &=(125+273) \mathrm{K}=398 \mathrm{K} \end{aligned}
There are 10 questions to complete.
9 number answer A hoga
As per Official answer key answer is C.