Thermal Analysis & Rheology ENERGY STORAGE AND DISSIPATION IN VISCOELASTIC MATERIALS Performing an energy balance on a viscoelastic material For liquids the energy stored, Es(t), is zero. undergoing shear can be far from straightforward, a point which is not always recognized in the rheological literature. It For solids the energy dissipated is zero, and the energy is difficult to recommend a text where the matter is dealt with stored per unit volume is given by: clearly and fully. The most complete analysis is that of σγ γ Tschoegl [1], from which the following is mainly taken. * (4) (V =≡ During a rheological experiment, a body is supplied with mechanical energy. A full energy balance would consider the For viscoelastic materials part of the mechanical energy is resulting in changes in kinetic, surface, potential, thermal and stored and part is dissipated, by definition. The energy all other forms of energy within the sample. But rheology is dissipated under steady state conditions is given by concerned only with the deformation and rate of deformation equations 1 to 3, the stored component is obtained from the of materials, and the contributions from most of these other transient response, i.e. the growth in shear rate to steady forms of energy are considered to be negligible. In other state, or its decay after removal of the stress. For low words the sample is considered to have constant volume and stresses, where the condition of linear viscoelasticity holds, surface area. The change in kinetic energy arising from its the analysis is relatively straightforward, but in practice at acceleration from rest is ignored, and it is considered to be in higher stresses it is difficult to separate the elastic (stored constant thermal equilibrium with its surroundings. The energy) response from that due to thixotropic and other time heating effect of the thermal energy generated is also dependent effects. ignored. The analysis of the energy stored and dissipated during In these circumstances it is only necessary to consider the oscillation is more complicated than is sometimes recognized. energy stored elastically, as potential energy, or dissipated The principle difficulty is that a liquid will continue to thermally. In the limiting cases of solids and liquids the dissipate energy continuously during oscillation, but the analysis is not difficult. If the energy supplied to the material total energy stored by a solid during a cycle is zero. In the E(t) is considered as a function of time, then for a liquid, the case of the solid, there are two points in the cycle at which energy dissipated per unit time, per unit volume, Ed(t), is the strain is zero, at which points the stored potential energy given by: is also zero. Integrating over the complete cycle, or even half cycle, would not be useful. G( W Gγ (1) G σ GW GW The first thought on how to overcome this might be to consider the energy stored at the point of maximum strain, in where is σ the shear stress and γ is the shear strain. Since η σ γ .γ γ other words integrate over a quarter cycle, and this is what is the viscosity, , is defined as /., where = d /dt, this sometimes done. But it turns out that this type of analysis is expression could equally well be been written: flawed, because the various mechanisms for storing energy do not act coherently. A better way of proceeding is to G(G W =ηγ& (2) consider the average amount of energy stored during a cycle. GW It is possible then to make a comparison with the total energy dissipated per cycle. This is the approach taken by Tschoegl. or After much analysis, it is found that: σ G(G W = (3) η γ GW ω = ′ ω (5) (*VDY RH-065 and so: ωπγω= ′′ (6) ωπγω= ′′ (15) (*G F\F (*GF\F so For solids, combining equations 4 and 8 gives: ω γω (GF\F = πδ (7) ω = *W VLQ (16) ω WDQ (V (DYV Equations 5 and 6 can be checked for liquids and solids for The total energy stored in a quarter cycle is then: sinusoidal oscillation: π (8) γ ω γ = γ ω * (17) VLQ W = ∫ VLQ ωWGW so γ = γ ω ω (9) π & FRV W γ ω * WWVLQ ω =− (18) and ω γγωω& = (10) π FRV W which, since sin = 0, For a liquid, combining with equation 2 gives: γπ = * (19) ω G( =ηγ ω ω (11) FRV W GW The average energy stored over a quarter cycle, which is The energy dissipated per cycle is then: equal to the average energy stored over a full cycle, is then: π γπ π ω = * (20) (12) (WGWGF\F ωηγωω = ∫ FRV ω ω π so ω = ηγ ω∫ ω (13) FRVWGW γ ω = ′ ω (21) (*VDY π ω ω =+ηγ ω WWVLQ (14) [1] Tschoegl, N.W., The Phenomenological Theory of ω Linear Viscoleastic Behaviour, chapter 9, pages 443 - 483, Springer-Verlag, Berlin, 1989. but sin 4π = 0, and for liquids η = G″ / ω, For more information or to place an order, contact: TA Instruments, Inc., 109 Lukens Drive, New Castle, DE 19720, Telephone: (302) 427-4000, Fax: (302) 427-4001 TA Instruments S.A.R.L., Paris, France, Telephone: 33-01-30489460, Fax: 33-01-30489451 TA Instruments N.V./S.A., Gent, Belgium, Telephone: 32-9-220-79-89, Fax: 32-9-220-83-21 TA Instruments GmbH, Alzenau, Germany, Telephone: 49-6023-30044, Fax: 49-6023-30823 TA Instruments, Ltd., Leatherhead, England, Telephone: 44-1-372-360363, Fax: 44-1-372-360135 TA Instruments Japan K.K., Tokyo, Japan, Telephone: 813-3450-0981, Fax: 813-3450-1322 Internet: http://www.tainst.com e-mail: [email protected] Thermal Analysis & Rheology RH-065 A SUBSIDIARY OF WATERS CORPORATION.