Dear Chhatrasal,

Let me give a try based on my understanding of homogenization.

If the material A & B are subjected to same temperature difference (i.e. if they are parallel), then effective Cp for homogenized material should be average of the two (A&B). However, in case of sandwich structure, the two materials are in series. Hence, the effective Cp cannot be average of the two. Try the following equation.

Cp(eff) = N / D

N = N1 + N2

N1 = rhoA * CpA*tA*tA/kA

N2 = rhoB * CpB*tB*tB/kB

D = rho(eff)*(tA+tB)*(tA/kA+tB/kB)

where, A and B are materials, eff - effective or homogenized material

rho = density

Cp = Specific heat per unit mass

k = thermal conductivity

rho(eff)=(rhoA*tA+rhoB*tB)/(tA+tB)

The above equation should justify your results.

Regards,

Chandrashekhar

forgot to mention that

t = thickness....

Hello,

I would disagree with the above answer. Whatever the microstructure of your material is, the specific heat capacity cannot decrease when assembling two materials (say A and B) to make a composite material. To demonstrate this, one may proceed as follows. The total mass of the composite material is:

m=ma+mb

Also, enthalpy being an extensive quantity, the total enthalpy of the composite material is:

H=HA+HB

For each material (A, B or composite), the corresponding specific enthalpy is:

ha=HA/ma

hb=HB/mb

h=H/m=ma/m ha + mb/m hb

For a thermodynamic system, the specific heat capacity c (at constant pressure) is defined as the infinitesimal variation of specific enthalpy caused by an infinitesimal variation of temperature:

c=dh/dT

=ma/m dha/dT+mb/m dhb/dT

=ma/m ca +mb/m cb

where ca and cb are the specific heat capacities of material A and B, respectively. The above relation shows that c is necessarily comprised between ca and cb, whatever the composite microstructure is.

One possible explanation of your experimental observation is that, in comparison with the pristine materials, the respective properties (including heat capacity) of A and/or B are changed during the elaboration of the composite material.

The equation given by Charles is a special case in which kA=kB and tA=tB. If we substitute these values, Cp(eff) will match with c (by Charles). It should be noted that the temperature difference across A and B is assumed to be constant (since kA=kB and tA=tB) which may not be the case for sandwich material.

To calculate effective specific heat, we need to consider layer-wise variation in specific heat, density, thickness and thermal conductivity. This is to say that the temperature difference across layers will be different depending upon layer thickness and thermal conductivity. Due to different temperature difference across the layers (in case of heat flow along the thickness direction) and different specific heat, effective specific heat of the sandwich material will be a function of Cp, k, t and rho.

Regards,

Chandrashekhar

Okay, what do you feel, how about the interface between two materials, when they are atomically not joined. In effective Cp, those interfaces might have to be matter, when heat is flowing in parallel/perpendicular (surrounding) direction. Generally, thermal conductivity of layered materials are quantified as, (kA+kB)/2 - interfacial resistance (R) originated through the junction. In such terminology, is it possible effective Cp is influenced by the interfaces. I am not much sure but suspect. Any opinion in this regard.

Dear Chhatrasal,

If the interface are not joined atomically, it can be considered as delamination or interface debonding (depending upon the case). It will affect the effective thermal conductivity. You can refer my paper in which I have studied effect of fiber cracks and interface debonding in CFRP on effective thermal conductivity.

Thermal conductivity of layered material are quantified as

a. if the heat is flowing along the layered material (in-plane): (kA+kB)/2 (for tA=tB)

b. if the heat is flowing along the thickness of the layered material (out-of-plane): 1/((1/kA)+(1/kB)) (for tA=tB)

Since the effective thermal conductivity changes with introduction of damage, effective specific heat will also change, in case of heat flowing along the thickness of the layered material (out-of-plane).

Regards,

Chandrashekhar

Dear all,

In the above analysis, Ta is different of Tb, which means the sandwich structure is not in thermal equilibrium. It is therefore impossible to define the effective temperature of the sandwich structure. Since heat capacity is obtained from the derivation of enthalpy with respect to temperature, it is also impossible to define an effective heat capacity for a structure which is out of equilibrium (The consequence being that heat capacity cannot depend on thermal conductivity. If there was a such dependence, heat capacity would depend on the heating rate). For your measurements, you need to make sure thermal equilibrium is reached otherwise your results do not make any sense.

Regards.

I agree with Charles.. Thanks. We can not define effective Cp terminology, as it depends on heat flow. But meanwhile, in DSC thermal equilibrium always be there, as these instrument are calibrated and controlled. Therefore, Cp should be intrinsic material property, not to be the sum.

Needless to say that the system (as discussed earlier) will be in thermal equilibrium. We need to appreciate that thermal gradient can exist along the thickness of sandwich material ( especially in case of heat flow along thickness direction). Therefore, despite having different temperature difference across individual layer, system will be in thermal equilibrium.