Q1 Q2 Passage of an electric current through a long conducting rod of radius r, and thermal conductivity k, results in uniform volumetric heating at a rate of d. The conduct- ing rod is wrapped in an electrically nonconducting faced, 28 kg/m³), and a 12-mm layer of gypsum (vermi- cladding material of outer radius r, and thermal conduc- tivity ke, and convection cooling is provided by an adjoining fluid. Consider a composite wall that includes an 8-mm-thick hardwood siding, 40-mm by 130-mm hardwood studs on 0.65-m centers with glass fiber insulation (paper culite) wall board. Wood siding Stud 130 mm Insulation Conducting rod, ġ, k, Wall board 40 mm What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide T, h PROPERTIES: Table A-3 (T = 300K): Hardwood siding, ka = 0.094 W/m-K; Hardwood, kg =0.16 W/m K; Gypsum, kç=0.17 W/m-K; Insulation (glass fiber paper faced, 28 kg/m³), Cladding, ke kp = 0.038 W/m-K. For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express ap- propriate boundary conditions for the solution of these equations.

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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yo l::V i %V) l.
Q1
Q2
Passage of an electric current through a long conducting Consider a composite wall that includes an 8-mm-thick
rod of radius r, and thermal conductivity k, results in
uniform volumetric heating at a rate of ġ. The conduct-
ing rod is wrapped in an electrically nonconducting
cladding material of outer radius r, and thermal conduc-
tivity ke, and convection cooling is provided by an
adjoining fluid.
hardwood siding, 40-mm by 130-mm hardwood studs
on 0.65-m centers with glass fiber insulation (paper
faced, 28 kg/m'), and a 12-mm layer of gypsum (vermi-
culite) wall board.
Wood siding
Stud
130 mm
Insulation
Conducting
rod, ġ, k,
Wall board
40 mm
is the
that is 2.5 m high by 6.5 m wide
resistance associated
a wall
T, h
PROPERTIES: Table A-3 (T 300K): Hardwood siding, ka = 0.094 W/m-K; Hardwood,
kg = 0.16 W/m-K; Gypsum, kc = 0.17 W/m-K; Insulation (glass fiber paper faced, 28 kg/m'),
Cladding, ke
kp = 0.038 W/m-K.
For steady-state conditions, write appropriate forms of
the heat equations for the rod and cladding. Express ap-
propriate boundary conditions for the solution of these
equations.
Transcribed Image Text:yo l::V i %V) l. Q1 Q2 Passage of an electric current through a long conducting Consider a composite wall that includes an 8-mm-thick rod of radius r, and thermal conductivity k, results in uniform volumetric heating at a rate of ġ. The conduct- ing rod is wrapped in an electrically nonconducting cladding material of outer radius r, and thermal conduc- tivity ke, and convection cooling is provided by an adjoining fluid. hardwood siding, 40-mm by 130-mm hardwood studs on 0.65-m centers with glass fiber insulation (paper faced, 28 kg/m'), and a 12-mm layer of gypsum (vermi- culite) wall board. Wood siding Stud 130 mm Insulation Conducting rod, ġ, k, Wall board 40 mm is the that is 2.5 m high by 6.5 m wide resistance associated a wall T, h PROPERTIES: Table A-3 (T 300K): Hardwood siding, ka = 0.094 W/m-K; Hardwood, kg = 0.16 W/m-K; Gypsum, kc = 0.17 W/m-K; Insulation (glass fiber paper faced, 28 kg/m'), Cladding, ke kp = 0.038 W/m-K. For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express ap- propriate boundary conditions for the solution of these equations.
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