Question

A non-Newtonian drilling fluid (Bingham plastic) is between two flat parallel rectangular plates of area $10\ \mathrm{cm^2}$ each, separated by $1\ \mathrm{cm}$. A force of $300$ dyne is required to initiate motion of the upper plate. A force of $600$ dyne keeps the plate moving at a constant velocity of $10\ \mathrm{cm/s}$. The constitutive law is \[ \tau_{yx} = \mu_p \dot{\gamma} + \tau^o_{yx}. \] Find the Bingham plastic viscosity $\mu_p$ in dyne$\cdot$s/cm$^2$ (rounded to the nearest integer). 

Hint:

For Bingham plastics, determine \(\tau_0\) from the threshold force (\(F_0/A\)), then use a second force–velocity pair to get \(\mu_p=(\tau-\tau_0)/\dot{\gamma}\).

Answer 1

Step 1: Yield stress from start-up force.
Start-up force $F_0=300$ dyne acts over area $A=10\ \mathrm{cm^2}$, so
\[ \tau_0=\frac{F_0}{A}=\frac{300}{10}=30\ \mathrm{dyne/cm^2}. \] Step 2: Shear rate during steady motion.
Gap $h=1\ \mathrm{cm}$, plate speed $V=10\ \mathrm{cm/s}$
\[ \dot{\gamma}=\frac{V}{h}=\frac{10}{1}=10\ \mathrm{s^{-1}}. \] Step 3: Shear stress at steady motion from force $F=600$ dyne.
\[ \tau=\frac{F}{A}=\frac{600}{10}=60\ \mathrm{dyne/cm^2}. \] Step 4: Compute $\mu_p$.
\[ \tau=\mu_p\dot{\gamma}+\tau_0 \Rightarrow \mu_p=\frac{\tau-\tau_0}{\dot{\gamma}}=\frac{60-30}{10}=3\ \mathrm{dyne\cdot s/cm^2}. \]