Question

A pressure drop in a 8cm diameter pipe is 75kpa. at a distance of 15 m.the shear stress at pipe wall in kpa is

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Answer 1

Answer:

0.2kpa

Explanation:

hope it will helps you

Answer 2

Answer:

The shear stress, $\tau$, in the wall of the pipe calculated is $0.066kPa$.

Explanation:

The pressure drop in the pipe, Δp = $75kPa$

The distance of the pipe, L = $15m$

The diameter of the pipe, d = $8cm$ = $0.08m$

The shear stress in the wall of the pipe, $\tau$ =?

As we know,

• The shear stress, $\tau$ = $\frac{\Delta p}{L}\times\frac{r}{2}$

Here, r = radius of the pipe = $\frac{d}{2}$ = $\frac{0.08}{2}$ = $0.04m$

After putting the values of the pressure drop, radius, and length in the equation, we get:

• The shear stress, $\tau$ = $\frac{75}{15}\times\frac{0.04}{2}$
• The shear stress, $\tau$ = $0.066kPa$

Hence, the shear stress at the pipe wall calculated is $0.066kPa$.