Assuming a velocity distribution which is a
parabola having its vertex 12 in from the boundary,
calculate the shear stress at y= 0, 3, 6, 9 and 12 inches.
Fluid’s absolute viscosity is 600 P
600 P=0.6 N⋅sm2=0.01253 lb⋅sft2,600~P=0.6~\frac{N\cdot s}{m^2}=0.01253~\frac{lb\cdot s}{ft^2},600 P=0.6 m2N⋅s=0.01253 ft2lb⋅s,
120−u=a(12−y)2,120-u=a(12-y)^2,120−u=a(12−y)2,
u=0, t=0, ⟹ a=56,u=0,~t=0,\implies a=\frac 56,u=0, t=0,⟹a=65,
u=120−56(12−y)2,u=120-\frac 56(12-y)^2,u=120−65(12−y)2,
dudy=53(12−y),\frac{du}{dy}=\frac 53(12-y),dydu=35(12−y),
y036912dudy20151050η0.2510.1880.12530.06270\def\arraystretch{1.5} \begin{array}{|c|c|c|c|c|c|} \hline y & 0 & 3&6&9&12 \\ \hline \frac{du}{dy} & 20 & 15&10&5&0\\ \hline \eta & 0.251&0.188&0.1253&0.0627 & 0 \\ \hline \end{array}ydyduη0200.2513150.1886100.1253950.06271200
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