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Question: Two walls of thickness \[{{d}_{1}}\] and \[{{d}_{2}}\] thermal conductivities \({{k}_{1}}\) and \({{...

Two walls of thickness d1{{d}_{1}} and d2{{d}_{2}} thermal conductivities k1{{k}_{1}} and k2{{k}_{2}} are in contact. In steady state if the temperature at the outer surface are T1{{T}_{1}} andT2{{T}_{2}}, temperature at the common wall will be:
A) T=k1T1+k2T2d1+d2T=\dfrac{{{k}_{1}}{{T}_{1}}+{{k}_{2}}{{T}_{2}}}{{{d}_{1}}+{{d}_{2}}}
B) T=k1T1d2+k2T2d1k1d2+k2d1T=\dfrac{{{k}_{1}}{{T}_{1}}{{d}_{2}}+{{k}_{2}}{{T}_{2}}{{d}_{1}}}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}
C) T=(k1d1+k2d2)T1T2T1+T2T=\dfrac{({{k}_{1}}{{d}_{1}}+{{k}_{2}}{{d}_{2}}){{T}_{1}}{{T}_{2}}}{{{T}_{1}}+{{T}_{2}}}
D) T=(k1T1d1+k2T2d2)k1d1+k2d2T=\dfrac{({{k}_{1}}{{T}_{1}}{{d}_{1}}+{{k}_{2}}{{T}_{2}}{{d}_{2}})}{{{k}_{1}}{{d}_{1}}+{{k}_{2}}{{d}_{2}}}

Explanation

Solution

This problem is solved by using heat rate formula .heat is a form of energy. As a result of temperature difference heat energy is transferred from one body to another body .Heat flows from hotter body to cooler body. Temperature (T) is the measure of the amount of heat energy present in a body.

Complete step by step solution:
Heat transfer is a branch of thermal engineering in that we are going to study the generation of heat transfer and conversion. Heat transfer is the flow of heat due a temperature difference between the bodies. There are three mechanisms of heat transfer and they are conduction, convection and radiation.
The walls are in contact so there will be same are area (A)
Consider the temperatures at the common surface as T
Heat rate at the first wall will be :Qt=k1A(T−T1)d1\dfrac{Q}{t}=\dfrac{{{k}_{1}}A(T-{{T}_{1}})}{{{d}_{1}}}
Heat rate at the second wall will be Qt=k2A(T2−T)d2\dfrac{Q}{t}=\dfrac{{{k}_{2}}A({{T}_{2}}-T)}{{{d}_{2}}}
Heat rate will be same throughout the length at steady state
Qt=k1A(T−T1)d1\dfrac{Q}{t}=\dfrac{{{k}_{1}}A(T-{{T}_{1}})}{{{d}_{1}}} = Qt=k2A(T2−T)d2\dfrac{Q}{t}=\dfrac{{{k}_{2}}A({{T}_{2}}-T)}{{{d}_{2}}}
T(k1d2+k2d1)=k2T2d1+k1T1d2T({{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}})={{k}_{2}}{{T}_{2}}{{d}_{1}}+{{k}_{1}}{{T}_{1}}{{d}_{2}}
T=k1T1d2+k2T2d1k1d2+k2d1T=\dfrac{{{k}_{1}}{{T}_{1}}{{d}_{2}}+{{k}_{2}}{{T}_{2}}{{d}_{1}}}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}
So the correct option is B.

Note: Students convection can be done naturally are by force and convection happens in gases and liquids. SI unit of heat is joule (J) and calorie is also used as the unit of heat. Steady state is a situation at which all the state variables are constant and thermal conductivity is usually measured in watts per meter-kelvin.