Question

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

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

Hint:

For viscoelastic or time-dependent fluids, stress relaxation follows an exponential decay of the form \(\tau = \tau_0 e^{-t/\lambda}\), where \(\lambda\) is the relaxation time.

Answer 1

The Correct Option is A

Step 1: Governing equation after stopping rotation.
When the drill pipe rotation stops (\(\dot{\gamma} = 0\)), the governing equation becomes: \[ \tau + \frac{\mu_0}{\alpha} \frac{d\tau}{dt} = 0 \] Step 2: Rearrangement.
\[ \frac{\mu_0}{\alpha} \frac{d\tau}{dt} = -\tau \] \[ \frac{d\tau}{dt} = -\frac{\alpha}{\mu_0}\tau \] Step 3: Solve differential equation.
This is a first-order linear ODE. The solution is exponential decay: \[ \tau(t) = \tau(0) e^{-\frac{\alpha}{\mu_0}t} \] Step 4: Compare with given options.
The exponent is \(-\alpha t / \mu_0\). This matches Option (C). Correct Final Answer: \[ \boxed{\text{(C) } \tau \propto e^{-\alpha t / \mu_0}} \]