Question

The two‐dimensional velocity field for an incompressible Newtonian fluid is described by the relationship V = ( 12 x y 2 − 6 x 3 ) ˆ i + ( 18 x 2 y − 4 y 3 ) ˆ j V=(12xy2−6x3)iˆ+(18x2y−4y3)jˆ where the velocity has units of m / s m/s when x x and y y are in meters. Determine the stresses σ x x σxx, σ y y σyy, and τ x y τxy at the point x = 0. 5 m x=0. 5 m, y = 1. 0 m y=1. 0 m if pressure at this point is 6 kPa 6 kPa and the fluid is glycerin at 20 ° C 20°C. Show these stresses on a sketch

Answer 1

To determine the stresses at the given point (x = 0.5 m, y = 1.0 m) in the fluid described by the velocity field V = (12xy^2 - 6x^3)i + (18x^2y - 4y^3)j, we can use the equations of fluid mechanics.

The stresses in a fluid are related to the velocity field through the Navier-Stokes equations. However, in this case, we are only interested in determining the stresses at a specific point and not analyzing the fluid flow. Therefore, we can use the simplified equation for the stress components:

σxx = -p + 2μ(∂V/∂x)
σyy = -p + 2μ(∂V/∂y)
τxy = μ(∂V/∂x + ∂V/∂y)

Where:
- σxx and σyy are the normal stresses in the x and y directions, respectively.
- τxy is the shear stress in the xy plane.
- p is the pressure at the point.
- μ is the dynamic viscosity of the fluid.

Given:
x = 0.5 m
y = 1.0 m
p = 6 kPa = 6,000 Pa
μ (viscosity of glycerin at 20°C) = 1.49 kg/(m·s)

Let's calculate the stresses at the given point:

1. Partial derivative ∂V/∂x:
∂V/∂x = (12y^2 - 18x^2)i + (36xy - 0)j
= (12y^2 - 18x^2)i + 36xyj

2. Partial derivative ∂V/∂y:
∂V/∂y = (24xy)i + (18x^2 - 12y^2)j

3. Substituting the given values into the stress equations:
σxx = -p + 2μ(∂V/∂x)
= -6,000 + 2(1.49)((12(1^2) - 18(0.5^2))(1) + 36(0.5)(1))
= -6,000 + 2(1.49)(12 - 4.5 + 18)
= -6,000 + 2(1.49)(25.5)
= -6,000 + 75.39
= -5,924.61 Pa

σyy = -p + 2μ(∂V/∂y)
= -6,000 + 2(1.49)((24(0.5)(1)) + (18(0.5^2) - 12(1^2)))
= -6,000 + 2(1.49)(12 + 4.5 - 12)
= -6,000 + 2(1.49)(4.5)
= -6,000 + 13.41
= -5,986.59 Pa

τxy = μ(∂V/∂x + ∂V/∂y)
= 1.49((12y^2 - 18x^2) + (24xy + 18x^2 - 12y^2))
= 1.49((12(1^2) - 18(0.5^2)) + (24(0.5)(1) + 18(0.5^2) - 12(1