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The pressure drop needed to force water through a horizontal 1-in.-diameter pipe is 0.55 psi for every 8-ft length of pipe. (a) Determine the shear stress on the pipe wall. Determine the shear stress at distances (b) 0.3 and (c) 0.5 in. away from the pipe wall.

Respuesta :

Answer:

(a). The shear stress on the pipe wall is 0.2062 lb/ft²

(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²

(c).  The shear stress at the distance 0.5 in away from the pipe wall is zero.

Explanation:

Given that,

Diameter = 1 in

Pressure = 0.55 psi

Length = 8 ft

We need to calculate the radius of the pipe

Using formula of radius

[tex]r=\dfrac{D}{2}[/tex]

Put the value into the formula

[tex]r=\dfrac{1}{2}[/tex]

[tex]r=0.5\ in[/tex]

(a). We need to calculate the shear stress on the pipe wall

Using formula of shear stress

[tex]\dfrac{\Delta p}{L}=\dfrac{2\tau}{r}[/tex]

[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]

Put the value into the formula

[tex]\tau=\dfrac{0.55\times144\times0.5}{2\times8\times12}[/tex]

[tex]\tau=0.2062\ lb/ft^2[/tex]

(b). We need to calculate the shear stress at the distance 0.3 in

Using formula of shear stress

[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]

Put the value into the formula

[tex]\tau=\dfrac{0.55\times144\times0.3}{2\times8\times12}[/tex]

[tex]\tau=0.12375\ lb/ft^2[/tex]

(c). We need to calculate the shear stress at the distance 0.5 in away from the pipe wall

r = 0.5-0.5 = 0

Using formula of shear stress

[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]

Put the value into the formula

[tex]\tau=\dfrac{0.55\times144\times0}{2\times8\times12}[/tex]

[tex]\tau=0[/tex]

Hence, (a). The shear stress on the pipe wall is 0.2062 lb/ft²

(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²

(c).  The shear stress at the distance 0.5 in away from the pipe wall is zero.