Question

In polycrystalline Ni, Nabarro–Herring diffusion creep was found to be the rate controlling creep mechanism at a certain temperature. At that temperature, if the steady state strain rate is $10^{-8$ s$^{-1}$ at a stress of 10 MPa, the steady state strain rate of $10^{-9}$ s$^{-1}$ will be obtained at a stress value of ............. MPa (in integer). Assume that the same creep mechanism is rate controlling during the creep deformation.}

Hint:

For diffusion creep (Nabarro–Herring or Coble), the stress exponent $n$ is one. This means strain rate is directly proportional to applied stress.

Answer 1

Step 1: Recall stress dependence of Nabarro–Herring creep.
For diffusion creep, steady-state creep rate is proportional to stress to the power one: \[ \dot{\varepsilon} \propto \sigma \] Step 2: Write ratio relation.
\[ \frac{\dot{\varepsilon}_1}{\dot{\varepsilon}_2} = \frac{\sigma_1}{\sigma_2} \] Step 3: Substitute given values.
\[ \frac{10^{-8}}{10^{-9}} = \frac{10}{\sigma_2} \] \[ 10 = \frac{10}{\sigma_2} \] \[ \sigma_2 = 1 \, MPa \] Wait, let's check carefully: Given: At $\sigma_1 = 10$ MPa, $\dot{\varepsilon}_1 = 10^{-8}$ s$^{-1}$. We want $\dot{\varepsilon}_2 = 10^{-9}$ s$^{-1}$. So: \[ \frac{10^{-9}}{10^{-8}} = \frac{\sigma_2}{10} \] \[ 0.1 = \frac{\sigma_2}{10} \] \[ \sigma_2 = 1 \, MPa \] Final Answer: \[ \boxed{1 \, MPa} \]