Question

A drilling fluid with time-dependent rheology is used for the rotary drilling of a reservoir. The following equation describes the dependence of shear stress (\( \tau \)) on shear rate (\( \dot{\gamma} \)):
\[ \tau + \frac{\mu_0}{\alpha} \frac{d\tau}{dt} = \mu_0 \dot{\gamma} \] where \( \mu_0 \) and \( \alpha \) are constants.
If the rotation of the drill pipe is stopped at time \( t = 0 \), then the relaxation behavior of the fluid stress with time is:

• A. \( \tau \propto e^{-\frac{\mu_0 t}{\alpha}} \)
• B. \( \tau \propto \frac{\mu_0 t}{\alpha} \)
• C. \( \tau \propto e^{-\frac{t}{\mu_0}} \)
• D. \( \tau \propto \frac{\alpha t}{\mu_0} \)

Hint:

In problems involving time-dependent rheology and first-order differential equations, it's useful to recognize the exponential decay solutions for relaxation processes after a perturbation is stopped.

Answer 1

The Correct Option is C

We are given the equation that describes the shear stress \( \tau \) in relation to the shear rate \( \dot{\gamma} \) and time: \[ \tau + \frac{\mu_0}{\alpha} \frac{d\tau}{dt} = \mu_0 \dot{\gamma} \] At the moment when the rotation of the drill pipe stops at time \( t = 0 \), the shear rate \( \dot{\gamma} \) becomes zero, and the equation reduces to: \[ \tau + \frac{\mu_0}{\alpha} \frac{d\tau}{dt} = 0 \] This is a first-order linear differential equation. Solving it, we find the relaxation behavior of the shear stress \( \tau \) with respect to time \( t \). The solution to this equation is of the form: \[ \tau(t) = \tau_0 e^{-\frac{t}{\mu_0}} \] Therefore, the relaxation behavior of the fluid stress is: \[ \tau \propto e^{-\frac{t}{\mu_0}} \] Thus, the correct answer is (C).