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A fluid of specific gravity 1.05 flows through a pipe of diameter 120mm. The viscosity of oil is 12 poise and the velocity of flow along the centre line of the pipe is 3.2 m/s.

Find:
1) Pressure gradient in flow direction.
2) Shear stress at the pipe wall.
3) Reynold's number.

Answer :

To solve this problem, we'll handle each part individually. We're dealing with fluid mechanics, specifically looking at flow in a pipe. The properties given are:

  • Specific gravity of the fluid is 1.05.
  • Diameter of the pipe is 120 mm (= 0.12 m).
  • Viscosity of the fluid is 12 poise which we convert to SI units: 12 poise = 1.2 Pa·s.
  • Velocity of the flow along the center line is 3.2 m/s.

1) Pressure Gradient in Flow Direction

For a Newtonian fluid flowing through a pipe, the pressure gradient [tex]\frac{dP}{dx}[/tex] can be determined by the relation:

[tex]\frac{dP}{dx} = \frac{4 \cdot \mu \cdot u}{D^2}[/tex]

where:

  • [tex]\mu[/tex] is the dynamic viscosity of the fluid (1.2 Pa·s).
  • [tex]u[/tex] is the mean velocity of the fluid (3.2 m/s).
  • [tex]D[/tex] is the diameter of the pipe (0.12 m).

Plugging in the values:

[tex]\frac{dP}{dx} = \frac{4 \times 1.2 \times 3.2}{(0.12)^2} = \frac{15.36}{0.0144} = 1066.67 \text{ Pa/m}[/tex]

So the pressure gradient is 1066.67 Pa/m.

2) Shear Stress at the Pipe Wall

The shear stress [tex]\tau[/tex] at the pipe wall is given by:

[tex]\tau = \mu \cdot \frac{du}{dy} \bigg|_{wall} = \mu \cdot \frac{8 \cdot u}{D}[/tex]

Substituting for viscosity [tex]\mu[/tex] (1.2 Pa·s), velocity [tex]u[/tex] (3.2 m/s), and diameter [tex]D[/tex] (0.12 m):

[tex]\tau = 1.2 \cdot \frac{8 \cdot 3.2}{0.12} = 1.2 \cdot \frac{25.6}{0.12} = 1.2 \cdot 213.33 = 256 \text{ Pa}[/tex]

The shear stress at the pipe wall is 256 Pa.

3) Reynold's Number

The Reynold's number [tex]ext{Re}[/tex] helps determine the flow regime and is calculated using:

[tex]ext{Re} = \frac{\rho \cdot u \cdot D}{\mu}[/tex]

Where the fluid density [tex]\rho[/tex] can be calculated from its specific gravity. If the specific gravity is 1.05, then:

[tex]\rho = 1.05 \times 1000 \text{ kg/m}^3 = 1050 \text{ kg/m}^3[/tex]

Now substitute for [tex]\rho[/tex] (1050 kg/m³), [tex]u[/tex] (3.2 m/s), [tex]D[/tex] (0.12 m), and [tex]\mu[/tex] (1.2 Pa·s):

[tex]ext{Re} = \frac{1050 \times 3.2 \times 0.12}{1.2} = \frac{403.2}{1.2} = 336[/tex]

The Reynold's number for the flow is 336, indicating that the flow is likely laminar.

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