Energy Balance Equation for Diffusion of Specimens Across a Membrane

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The discussion focuses on deriving the energy balance equation for the diffusion of two specimens, A and B, across a membrane under steady-state conditions. Specimen A diffuses from a concentration cA at temperature TA to the other side with temperature TB, while specimen B diffuses in the opposite direction. The heat flux is calculated using the specific enthalpy of each specimen at their respective temperatures. The expressions for heat flux are given as mA*(hA(TA)-hA(TB)) and mB*(hB(TB)-hB(TA)). The participants seek confirmation on the correctness of this energy balance equation.
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Imagine a wall (membrane). Concentration of a specimen A on one side of the wall is cA and it has the temperature TA. The concentration of a specimen B on another side of the wall is cB and it has the temperature TB. Specimen A nad B diffuse against each other (assume steady state). Can somebody write the energy balance equation for this case?
 
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The specimen A diffuses from the side of the membrane with the concentration cA (side 1) to the other side (side 2) and specimen B diffuses in the opposite direction. Because the temperature at the side 1 is TA and the temperature at the side 2 is TB is heat flux in both directions:
mA*(hA(TA)-hA(TB))
mB*(hB(TB)-hB(TA))

mA, mB - diffusion flux of specimen A/B
hA(TA) - specific enthalpy of specimen A at TA
hA(TB) - specific enthalpy of specimen B at TB
 
Is that correct?
 
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