Oldroyd-B model

Example

The Oldroyd-B model is a constitutive law for an incompressible viscoelastic fluid, such as a dilute polymer solution. It has a relaxation time $\lambda_1 > 0$, a retardation time $\lambda_2$ with $0 \le \lambda_2 < \lambda_1$, and a total viscosity $\eta_0 > 0$. The extra stress $\tau$ satisfies $$\tau + \lambda_1 \overset{\nabla}{\tau} = 2 \eta_0 \big( D + \lambda_2 \overset{\nabla}{D} \big),$$ where $\overset{\nabla}{\ }$ is the upper-convected time derivative and $D$ is the strain rate tensor.

Equivalently, $\tau = 2 \eta_s D + \tau_p$, where the solvent viscosity is $\eta_s = \eta_0 \lambda_2 / \lambda_1$ and the polymer stress $\tau_p$ obeys the upper-convected Maxwell model $\tau_p + \lambda_1 \overset{\nabla}{\tau_p} = 2 \eta_p D$ with $\eta_p = \eta_0 - \eta_s$.

Check in steady simple shear $u = (\dot\gamma y, 0, 0)$: the solvent part gives no normal stress, and the polymer part gives $(\tau_p)_{12} = \eta_p \dot\gamma$ and $(\tau_p)_{11} - (\tau_p)_{22} = 2 \lambda_1 \eta_p \dot\gamma^2$. So the shear stress is $\eta_0 \dot\gamma$, and the first normal stress difference $2 \lambda_1 \eta_p \dot\gamma^2$ is nonzero. A Newtonian fluid has no normal stress difference in this flow, so the model is a non-Newtonian fluid.