Step 1: Rework the same problem using SI units instead of cgs units, as an independent check:
$D_i = 4.276 \times 0.0254 = 0.108610$ m
Step 2: Convert the flow rate to cubic metres per second:
$1$ gallon $= 3785.4$ cm$^3 = 0.0037854$ m$^3$
$Q = 600 \times 0.0037854 = 2.27124$ m$^3$/min
$Q = 2.27124 / 60 = 0.037854$ m$^3$/s
Step 3: Compute the cross-sectional area in square metres:
Radius $= 0.108610/2 = 0.054305$ m
$A = \pi (0.054305)^2 = \pi \times 0.00294906 = 0.0092647$ m$^2$
Step 4: Compute the average velocity in metres per second:
$V = Q/A = 0.037854 / 0.0092647 = 4.0858$ m/s
Step 5: Note that 4.0858 m/s equals 408.58 cm/s, matching the cgs calculation exactly, which confirms the unit conversion is consistent:
$4.0858 \times 100 = 408.58$ cm/s
Step 6: Compute the power law coefficient once again:
$\frac{3n+1}{4n} = \frac{3(0.67)+1}{4(0.67)} = \frac{3.01}{2.68} = 1.12313$
Step 7: Compute 8V/Di using the SI values, keeping the units of V and Di consistent so the ratio still comes out in s^-1:
$8V = 8 \times 4.0858 = 32.6864$ m/s
$8V/D_i = 32.6864 / 0.108610 = 300.95$ s$^{-1}$
Step 8: Multiply by the power law coefficient to obtain the wall shear rate:
$\dot{\gamma}_w = 1.12313 \times 300.95 = 338.0$ s$^{-1}$, which matches the cgs unit calculation exactly.
Final Answer:
\[ \boxed{\dot{\gamma}_w = 338.0 \text{ s}^{-1}} \]