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A wall has two layers of materials A and B; each made of a different material. Both the layers have the same thickness. The thermal conductivity of materialA is twice that of B. Under the equilibrium, the temperature difference across the wall is 36°C. The temperature difference across the layer A is __________ °C.
A
6
B
12
C
18
D
24
Correct Answer:
12
A furnace wall consists of two layers. The inside layer of 450 mm is made of light weight bricks of thermal conductivity IW/m.K and outside layer of 900 mm is made of refractory of thermal conductivity 2W/m.K. The hot face of the inside layer is at temperature 1300 K and the cold face of the outer layer is at 400 K. The temperature at the interface between the two layers is . . . . . . . . K.
A
1000
B
850
C
700
D
600
A composite flat wall of a furnace is made of two materials 'A' and 'B'. The thermal conductivity of 'A' is twice of that of material 'B', while the thickness of layer of 'A' is half that of B. If the temperature at the two sides of the wall are 400 and 1200°K, then the temperature drop (in °K) across the layer of material 'A' is
A
125
B
133
C
150
D
160
A furnace is made of a refractory brick wall of thickness 0.5 metre and thermal conductivity 0.7 W/m.°K For the same temperature drop and heat loss, this refractory wall can be replaced by a layer of diatomaceous earth of thermal conductivity 0.14 W/m.K and thickness __________ metre.
A
0.01
B
0.1
C
0.25
D
0.5
A composite slab has two layers of different materials with thermal conductivities k
1
and k
2
. If each layer has the same thickness, then the equivalent thermal conductivity of the slab will be
A
$${{\text{k}}_1}{{\text{k}}_2}$$
B
$${{\text{k}}_1} + {{\text{k}}_2}$$
C
$$\frac{{{{\text{k}}_1} + {{\text{k}}_2}}}{{{{\text{k}}_1}{{\text{k}}_2}}}$$
D
$$\frac{{2{{\text{k}}_1}{{\text{k}}_2}}}{{{{\text{k}}_1} + {{\text{k}}_2}}}$$
A furnace wall is made of three materials (I, II and III) of equal thickness and having thermal conductivity k
1
k
2
and k
3
respectively. The steady state temperature profile inside each material is shown in the figure. Thermal conductivities of the materials will vary as
A
k<sub>1</sub> > k<sub>2</sub> > k<sub>3</sub>
B
k<sub>2</sub> > k<sub>1</sub> > k<sub>3</sub>
C
k<sub>3</sub> > k<sub>1</sub> > k<sub>2</sub>
D
k<sub>3</sub> > k<sub>2</sub> > k<sub>1</sub>
A furnace wall consists of four layers of different materials, M
1
, M
2
, M
3
and M
4
. If the layers are of equal thickness and the steady state temperature profiles is as shown below, then the material with the lowest thermal conductivity is
A
M<sub>1</sub>
B
M<sub>2</sub>
C
M<sub>3</sub>
D
M<sub>4</sub>
A cylinder of radius r and made of material of thermal conductivity k 1 is surrounded by a cylindrical shell of inner radius r and outer radius 2r. This outer shell is made of a material of thermal conductivity k 2. Net conductivity would be
A
k 1 + 3 k 2/4
B
k 1 + k 2/4
C
k 1 + 3k 2
D
k 1 + k 2
A composite wall is made of two layers of thickness δ1 and δ2 having thermal conductivities k and 2k and equal surface area normal to the direction of heat flow. The outer surface of composite wall are at 100 degree Celsius and 200 degree Celsius. The minimum surface temperature at the junction is 150 degree Celsius. What will be the ratio of wall thickness?
A
1:1
B
2:1
C
1:2
D
2:3
The temperatures on the two sides of a plane wall are t1 and t2 and thermal conductivity of the wall material is prescribed by the relation K = k0 e (-x/δ) Where, k0 is constant and δ is the wall thickness. Find the relation for temperature distribution in the wall?
A
t 1 – t x / t 1 – t 2 = x
B
t 1 – t x / t 1 – t 2 = δ
C
t 1 – t x / t 1 – t 2 = δ/x
D
t 1 – t x / t 1 – t 2 = x/δ
A composite slab has two layers having thermal conductivities in the ratio of 1:2. If the thickness is the same for each layer then the equivalent thermal conductivity of the slab would be
A
1/3
B
2/3
C
2
D
4/3