Question

In the frame shown in figure made up of metallic rods of identical size, the ends `A` and `F` are maintained at `T_(1)` and `T_(2)` respectively. The rate of heat flow from `A` to `C` is `omega`. If `C` and `D` are instead maintained at `T_(1)` & `T_(2)` respectively find, find the total rate of heat flow.

Answer 1

equivalent thermal lesitance b//w `A` & `F`

`=3R` where `R=(l)/(KA)`
`l=` Length of Rod
`K=` Thermal conductivity
`A=` Area of cross section
Heat current through `AC=` Heat current through `AD=omega`
`implies` Heat current through `AB=2omega`
So `(dH)/(dt)=(DeltaT)/(Req)=(T_(1)T_(2))/((3l)/(KA))=omega`
where `C` & `D` are maintained at `T_(1)` & `T_(2)` then equivalent thermal resistance `b//w C`
`R_(eq)=R//2`
`implies i_(2)=(DeltaT)/(R_(eq))=(DeltaT)/(R//2)=(2DeltaT)/(R )`
`i_(2)=(2DeltaT)/((l)/(KA))=2KA(2DeltaT)/(l)`
`implies (KADeltaT)/(3l)=2omega`
`implies(KADeltaT)/(l)=6omega`
`implies i_(2)=6omegaxx2=12omega`