Two slabs of thicknesses d,2d;and thermal conductivities k and 2k,are joined face to face in the longitudinal way.Under steady state condition find rate of heat transfer per unit face area and the temperature at the interface

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a) Thermal Resistance of the first slab is R1=dkA  (A:face area) Thermal Resistance of the second slab is R2=d2kA  (A:face area)As these slaba are connected in series so equivalent thermal resistance isReqv=R1+R2Reqv=dkA+d2kAReqv=3d2kALet the temperature of one face of the slab =T1 and the temperature of the other side of the slab is T2heat current i=T1 - T2Reqv=2kA(T1 - T2)3drate transfer per unit area is =iA=2k(T1 - T2)3db)Let temperature of the interface is T0heat current through first slab=T1 - ToR1T1 - ToR1=T1 - T2ReqvT1 - TodkA=T1 - T23d2kAT1 - T0=23(T1 - T2)3T1-3T0=2T1-2T23T0=T1+2T2T0=T1+2T23Regards

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