An oil of relative density `0.9` and viscosity `0.12 kg//ms` flows through a `2.5 cm` diameter pipe with a pressure drop of `38.4 kN//m^(2)` in a length of `30 m`. Determine
Determine the shear stress at the pipe wall

Options

A. `8xx10^(-6)N//m^(2)`
B. `3.9xx10^(-6)N//m^(2)`
C. `2.3xx10^(-6)N//m^(2)`
D. `10.6xx10^(-6)N//m^(2)`

Answer

A

Explanation

Using Poiseuille's equation we get
`Q=((p_(1)-p_(2))piD^(4))/(128etaL)`
here `p_(1)-p_(2)=38.4x10^(3)N//m^(2)`
`D=2.5cm=2.5xx10^(-2)m`
`L=30m,h=0.12N//m^(2)`
`:. Q=((38.4xx10^(3))(3.14)(2.5xx12^(-2))^(4))/(128(0.12)(30))=1.0xx10^(-4)m^(3)//s`
b. We know that `tau=(dp)/(dx) r/2`
At the wall of the pipe ie `r=R`
`tau_("max")=(dp)/(dx) R/2=((p_(1)-p_(2))/L)D/4`
`=((38.4xx10^(-3))/30)(2.5xx10^(-2))/4=8xx10^(-6)N//m^(2)`
c. We know that `P=(128etaQ^(2)L)/(piD^(4)) =(p_(1)-p_(2))Q`
`:. (38.4xx10^(3))(1xx10^(-4))` or `P=3.84W`