Qstnid -Iv-30-the Three Rods Shown In Figure (28-e7) Have Identical Geometrical Dimensions. Heat Flows From The Hot End At A Rate Of 40 W In The Arrangement (A) Find The Rates Of Heat Flow When The Rods Are Joined As In Arrangement (B) And In (C) Thermal Condcutivities Of Aluminium And Copper Are 200 W M-1°c-1 And 400 W M-1°c-1 Respectively. figure (A) Find The Rates Of Heat Flow When The Rods Are Joined As In Arrangement (B) And In (C) Thermal Condcutivities Of Aluminium And Copper Are 200 W M-1°c-1 And 400 W M-1°c-1 Respectively. figure - 28: Heat Transfer

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28: Heat Transfer

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#30
The three rods shown in figure (28-E7) have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a) Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c) Thermal condcutivities of aluminium and copper are 200 W m^-1°C^-1 and 400 W m^-1°C^-1 respectively.
Figure (a) Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c) Thermal condcutivities of aluminium and copper are 200 W m^-1°C^-1 and 400 W m^-1°C^-1 respectively.
Figure
Ans : For arrangement (a) ,

Temperature of the hot end ,T₁ = 100°C
Temperature of the cold end ,T₂ = 0°C
R_s = R₁ + R₂ + R₃
`` =\frac{l}{{K}_{\,\mathrm{\,Al\,}}\,\mathrm{\,A\,}}+\frac{l}{{K}_{\,\mathrm{\,cu\,}}\,\mathrm{\,A\,}}+\frac{l}{{K}_{\mathit{Al}}A}``
`` =\frac{\mathit{l}}{\mathit{A}}\left(\frac{1}{200}+\frac{1}{400}+\frac{1}{200}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{5}{400}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\times \frac{1}{80}``
`` \frac{\,\mathrm{\,d\,}Q}{dt}=q=\,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{{T}_{1}-{T}_{2}}{{R}_{\,\mathrm{\,S\,}}}``
`` =\frac{100-0}{{\displaystyle \frac{l}{\,\mathrm{\,A\,}}\times \frac{1}{80}}}``
`` ``
`` \,\mathrm{\,Given\,}:``
`` q=40\,\mathrm{\,W\,}``
`` 40=\frac{100}{{\displaystyle \frac{l}{\,\mathrm{\,A\,}}}\times {\displaystyle \frac{1}{80}}}``
`` \Rightarrow \frac{l}{\,\mathrm{\,A\,}}=200``
`` \Rightarrow \frac{\,\mathrm{\,A\,}}{l}=\frac{1}{200}``
`` ``
For arrangement (b) ,

`` {R}_{\,\mathrm{\,net\,}}={R}_{\,\mathrm{\,Al\,}}+\frac{{R}_{\,\mathrm{\,Cu\,}}\times {R}_{\,\mathrm{\,Al\,}}}{{\,\mathrm{\,R\,}}_{\,\mathrm{\,Cu\,}}+{\,\mathrm{\,R\,}}_{\,\mathrm{\,Al\,}}}``
`` =\frac{l}{{K}_{\,\mathrm{\,Al\,}}A}+\frac{{\displaystyle \frac{l}{{K}_{\,\mathrm{\,Cu\,}}A}}\times {\displaystyle \frac{l}{{K}_{\,\mathrm{\,Al\,}}l}}}{{\displaystyle \frac{l}{{K}_{\,\mathrm{\,Cu\,}}A}}+{\displaystyle \frac{l}{{K}_{\mathit{Al}}\,\mathrm{\,A\,}}}}``
`` =\frac{\mathit{l}}{\mathit{A}\mathit{·}{\mathit{K}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{l}}{\mathit{A}\left({\mathit{K}}_{\mathit{Al}}\mathit{+}{\mathit{K}}_{\mathit{Cu}}\right)}``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{1}{200}+\frac{1}{200+400}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{1}{200}+\frac{1}{600}\right)``
`` =\frac{4}{600}.\frac{l}{\,\mathrm{\,A\,}}``
Rate of flow of heat is given by
`` q=\frac{{\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}}{{\,\mathrm{\,R\,}}_{\,\mathrm{\,net\,}}}``
`` =\frac{\left(100-0\right)}{4l}600\,\mathrm{\,A\,}``
`` =\frac{100\times 600}{4}\times \frac{1}{200}``
`` =75\,\mathrm{\,W\,}``
For arrangement (c) ,

`` \frac{\mathit{1}}{{\mathit{R}}_{\mathit{net}}}\mathit{=}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Cu}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}``
`` =\frac{{K}_{\,\mathrm{\,Al\,}}A}{l}+\frac{{K}_{\,\mathrm{\,Cu\,}}A}{l}+\frac{{K}_{\,\mathrm{\,Al\,}}A}{l}``
`` \frac{1}{{R}_{\,\mathrm{\,net\,}}}=\left({K}_{\,\mathrm{\,Al\,}}+{K}_{\,\mathrm{\,Cu\,}}+{K}_{\,\mathrm{\,Al\,}}\right)\frac{\mathit{A}}{\mathit{l}}``
`` \frac{1}{{R}_{\,\mathrm{\,net\,}}}=\left(200+400+200\right)\times \frac{1}{200}``
`` \Rightarrow {R}_{\,\mathrm{\,net\,}}=\frac{200}{800}``
`` =\frac{1}{4}``
`` ``
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}\,\mathrm{\,flow\,}=\frac{\Delta T}{{R}_{\,\mathrm{\,net\,}}}``
`` =\frac{100}{{\displaystyle \frac{1}{4}}}``
`` =400\,\mathrm{\,W\,}``
Page No 100: (a) ,

Temperature of the hot end ,T₁ = 100°C
Temperature of the cold end ,T₂ = 0°C
R_s = R₁ + R₂ + R₃
`` =\frac{l}{{K}_{\,\mathrm{\,Al\,}}\,\mathrm{\,A\,}}+\frac{l}{{K}_{\,\mathrm{\,cu\,}}\,\mathrm{\,A\,}}+\frac{l}{{K}_{\mathit{Al}}A}``
`` =\frac{\mathit{l}}{\mathit{A}}\left(\frac{1}{200}+\frac{1}{400}+\frac{1}{200}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{5}{400}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\times \frac{1}{80}``
`` \frac{\,\mathrm{\,d\,}Q}{dt}=q=\,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{{T}_{1}-{T}_{2}}{{R}_{\,\mathrm{\,S\,}}}``
`` =\frac{100-0}{{\displaystyle \frac{l}{\,\mathrm{\,A\,}}\times \frac{1}{80}}}``
`` ``
`` \,\mathrm{\,Given\,}:``
`` q=40\,\mathrm{\,W\,}``
`` 40=\frac{100}{{\displaystyle \frac{l}{\,\mathrm{\,A\,}}}\times {\displaystyle \frac{1}{80}}}``
`` \Rightarrow \frac{l}{\,\mathrm{\,A\,}}=200``
`` \Rightarrow \frac{\,\mathrm{\,A\,}}{l}=\frac{1}{200}``
`` ``
For arrangement (b) ,

`` {R}_{\,\mathrm{\,net\,}}={R}_{\,\mathrm{\,Al\,}}+\frac{{R}_{\,\mathrm{\,Cu\,}}\times {R}_{\,\mathrm{\,Al\,}}}{{\,\mathrm{\,R\,}}_{\,\mathrm{\,Cu\,}}+{\,\mathrm{\,R\,}}_{\,\mathrm{\,Al\,}}}``
`` =\frac{l}{{K}_{\,\mathrm{\,Al\,}}A}+\frac{{\displaystyle \frac{l}{{K}_{\,\mathrm{\,Cu\,}}A}}\times {\displaystyle \frac{l}{{K}_{\,\mathrm{\,Al\,}}l}}}{{\displaystyle \frac{l}{{K}_{\,\mathrm{\,Cu\,}}A}}+{\displaystyle \frac{l}{{K}_{\mathit{Al}}\,\mathrm{\,A\,}}}}``
`` =\frac{\mathit{l}}{\mathit{A}\mathit{·}{\mathit{K}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{l}}{\mathit{A}\left({\mathit{K}}_{\mathit{Al}}\mathit{+}{\mathit{K}}_{\mathit{Cu}}\right)}``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{1}{200}+\frac{1}{200+400}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{1}{200}+\frac{1}{600}\right)``
`` =\frac{4}{600}.\frac{l}{\,\mathrm{\,A\,}}``
Rate of flow of heat is given by
`` q=\frac{{\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}}{{\,\mathrm{\,R\,}}_{\,\mathrm{\,net\,}}}``
`` =\frac{\left(100-0\right)}{4l}600\,\mathrm{\,A\,}``
`` =\frac{100\times 600}{4}\times \frac{1}{200}``
`` =75\,\mathrm{\,W\,}``
For arrangement (c) ,

`` \frac{\mathit{1}}{{\mathit{R}}_{\mathit{net}}}\mathit{=}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Cu}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}``
`` =\frac{{K}_{\,\mathrm{\,Al\,}}A}{l}+\frac{{K}_{\,\mathrm{\,Cu\,}}A}{l}+\frac{{K}_{\,\mathrm{\,Al\,}}A}{l}``
`` \frac{1}{{R}_{\,\mathrm{\,net\,}}}=\left({K}_{\,\mathrm{\,Al\,}}+{K}_{\,\mathrm{\,Cu\,}}+{K}_{\,\mathrm{\,Al\,}}\right)\frac{\mathit{A}}{\mathit{l}}``
`` \frac{1}{{R}_{\,\mathrm{\,net\,}}}=\left(200+400+200\right)\times \frac{1}{200}``
`` \Rightarrow {R}_{\,\mathrm{\,net\,}}=\frac{200}{800}``
`` =\frac{1}{4}``
`` ``
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}\,\mathrm{\,flow\,}=\frac{\Delta T}{{R}_{\,\mathrm{\,net\,}}}``
`` =\frac{100}{{\displaystyle \frac{1}{4}}}``
`` =400\,\mathrm{\,W\,}``
Page No 100:

#30-a
Find the rates of heat flow when the rods are joined as in arrangement
Ans : ,

Temperature of the hot end ,T₁ = 100°C
Temperature of the cold end ,T₂ = 0°C
R_s = R₁ + R₂ + R₃
`` =\frac{l}{{K}_{\,\mathrm{\,Al\,}}\,\mathrm{\,A\,}}+\frac{l}{{K}_{\,\mathrm{\,cu\,}}\,\mathrm{\,A\,}}+\frac{l}{{K}_{\mathit{Al}}A}``
`` =\frac{\mathit{l}}{\mathit{A}}\left(\frac{1}{200}+\frac{1}{400}+\frac{1}{200}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{5}{400}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\times \frac{1}{80}``
`` \frac{\,\mathrm{\,d\,}Q}{dt}=q=\,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{{T}_{1}-{T}_{2}}{{R}_{\,\mathrm{\,S\,}}}``
`` =\frac{100-0}{{\displaystyle \frac{l}{\,\mathrm{\,A\,}}\times \frac{1}{80}}}``
`` ``
`` \,\mathrm{\,Given\,}:``
`` q=40\,\mathrm{\,W\,}``
`` 40=\frac{100}{{\displaystyle \frac{l}{\,\mathrm{\,A\,}}}\times {\displaystyle \frac{1}{80}}}``
`` \Rightarrow \frac{l}{\,\mathrm{\,A\,}}=200``
`` \Rightarrow \frac{\,\mathrm{\,A\,}}{l}=\frac{1}{200}``
`` ``
For arrangement

#30-b
and in
Ans : ,

`` {R}_{\,\mathrm{\,net\,}}={R}_{\,\mathrm{\,Al\,}}+\frac{{R}_{\,\mathrm{\,Cu\,}}\times {R}_{\,\mathrm{\,Al\,}}}{{\,\mathrm{\,R\,}}_{\,\mathrm{\,Cu\,}}+{\,\mathrm{\,R\,}}_{\,\mathrm{\,Al\,}}}``
`` =\frac{l}{{K}_{\,\mathrm{\,Al\,}}A}+\frac{{\displaystyle \frac{l}{{K}_{\,\mathrm{\,Cu\,}}A}}\times {\displaystyle \frac{l}{{K}_{\,\mathrm{\,Al\,}}l}}}{{\displaystyle \frac{l}{{K}_{\,\mathrm{\,Cu\,}}A}}+{\displaystyle \frac{l}{{K}_{\mathit{Al}}\,\mathrm{\,A\,}}}}``
`` =\frac{\mathit{l}}{\mathit{A}\mathit{·}{\mathit{K}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{l}}{\mathit{A}\left({\mathit{K}}_{\mathit{Al}}\mathit{+}{\mathit{K}}_{\mathit{Cu}}\right)}``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{1}{200}+\frac{1}{200+400}\right)``
`` =\frac{l}{\,\mathrm{\,A\,}}\left(\frac{1}{200}+\frac{1}{600}\right)``
`` =\frac{4}{600}.\frac{l}{\,\mathrm{\,A\,}}``
Rate of flow of heat is given by
`` q=\frac{{\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}}{{\,\mathrm{\,R\,}}_{\,\mathrm{\,net\,}}}``
`` =\frac{\left(100-0\right)}{4l}600\,\mathrm{\,A\,}``
`` =\frac{100\times 600}{4}\times \frac{1}{200}``
`` =75\,\mathrm{\,W\,}``
For arrangement

#30-c
Thermal condcutivities of aluminium and copper are 200 W m^-1°C^-1 and 400 W m^-1°C^-1 respectively.
Figure
Ans : ,

`` \frac{\mathit{1}}{{\mathit{R}}_{\mathit{net}}}\mathit{=}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Cu}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}``
`` =\frac{{K}_{\,\mathrm{\,Al\,}}A}{l}+\frac{{K}_{\,\mathrm{\,Cu\,}}A}{l}+\frac{{K}_{\,\mathrm{\,Al\,}}A}{l}``
`` \frac{1}{{R}_{\,\mathrm{\,net\,}}}=\left({K}_{\,\mathrm{\,Al\,}}+{K}_{\,\mathrm{\,Cu\,}}+{K}_{\,\mathrm{\,Al\,}}\right)\frac{\mathit{A}}{\mathit{l}}``
`` \frac{1}{{R}_{\,\mathrm{\,net\,}}}=\left(200+400+200\right)\times \frac{1}{200}``
`` \Rightarrow {R}_{\,\mathrm{\,net\,}}=\frac{200}{800}``
`` =\frac{1}{4}``
`` ``
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}\,\mathrm{\,flow\,}=\frac{\Delta T}{{R}_{\,\mathrm{\,net\,}}}``
`` =\frac{100}{{\displaystyle \frac{1}{4}}}``
`` =400\,\mathrm{\,W\,}``
Page No 100: