(a) Whether the flow satisfy conservation of mass.

Question

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Answer 1

Question

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y

−2ρx2y+y3∂y=∂p∂p=−2ρx2y+y3∂y

Integrate the equation on both sides

∫∂p=−2ρ∫ x 2 y+ y 3 ∂yp=−2ρx2 y 2 2+ y 4 4+fx+constant ... (V)

Differentiate the Equation with respect to x.

∂p∂x=∂∂x−2ρ x 2 y 2 2 + y 4 4 +fx+constant=−2ρ× 2x y 2 2 +0+ ∂ f x ∂x+0=−2ρ× x y 2 +0+ ∂ f x ∂x=−2ρxy2+∂ f x ∂x

Substitute −2ρxy2+∂fx∂x for ∂p∂x in Equation (IV).

−2ρxy2+∂ f x ∂x=−2ρxy2+x3∂ f x ∂x=−2ρxy2+x3+2ρxy2∂ f x ∂x=−2ρxy2−2ρx3+2ρxy2∂ f x ∂x=−2ρx3

∂fx=−2ρx3∂x

Integrating the above equation.

∫∂ f x =−2ρ∫ x 3 ∂xfx=−2ρ x 4 4

Substitute −2ρx44 for fx in Equation (V).

p=−2ρx2 y 2 2+ y 4 4+−2ρ x 4 4+constant=−2ρx2 y 2 2+ y 4 4+ x 4 4+constant=−2ρ×142x2y2+y4+x4+constant=−ρ22x2y2+y4+x4+constant ... (VI).

Substitute 0 for x, 0 for y, pa for p in Equation (VI).

pa=−ρ22×0+0+0+constantpa=constant

Substitute pa for constant in Equation (VI).

p=−ρ22x2y2+y4+x4+pa

Conclusion:

The pressure field for px,y is −ρ22x2y2+x4+y4+pa ..

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Answer 2

Question

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y

−2ρx2y+y3∂y=∂p∂p=−2ρx2y+y3∂y

Integrate the equation on both sides

∫∂p=−2ρ∫ x 2 y+ y 3 ∂yp=−2ρx2 y 2 2+ y 4 4+fx+constant ... (V)

Differentiate the Equation with respect to x.

∂p∂x=∂∂x−2ρ x 2 y 2 2 + y 4 4 +fx+constant=−2ρ× 2x y 2 2 +0+ ∂ f x ∂x+0=−2ρ× x y 2 +0+ ∂ f x ∂x=−2ρxy2+∂ f x ∂x

Substitute −2ρxy2+∂fx∂x for ∂p∂x in Equation (IV).

−2ρxy2+∂ f x ∂x=−2ρxy2+x3∂ f x ∂x=−2ρxy2+x3+2ρxy2∂ f x ∂x=−2ρxy2−2ρx3+2ρxy2∂ f x ∂x=−2ρx3

∂fx=−2ρx3∂x

Integrating the above equation.

∫∂ f x =−2ρ∫ x 3 ∂xfx=−2ρ x 4 4

Substitute −2ρx44 for fx in Equation (V).

p=−2ρx2 y 2 2+ y 4 4+−2ρ x 4 4+constant=−2ρx2 y 2 2+ y 4 4+ x 4 4+constant=−2ρ×142x2y2+y4+x4+constant=−ρ22x2y2+y4+x4+constant ... (VI).

Substitute 0 for x, 0 for y, pa for p in Equation (VI).

pa=−ρ22×0+0+0+constantpa=constant

Substitute pa for constant in Equation (VI).

p=−ρ22x2y2+y4+x4+pa

Conclusion:

The pressure field for px,y is −ρ22x2y2+x4+y4+pa ..

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Answer 3

Question

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y

−2ρx2y+y3∂y=∂p∂p=−2ρx2y+y3∂y

Integrate the equation on both sides

∫∂p=−2ρ∫ x 2 y+ y 3 ∂yp=−2ρx2 y 2 2+ y 4 4+fx+constant ... (V)

Differentiate the Equation with respect to x.

∂p∂x=∂∂x−2ρ x 2 y 2 2 + y 4 4 +fx+constant=−2ρ× 2x y 2 2 +0+ ∂ f x ∂x+0=−2ρ× x y 2 +0+ ∂ f x ∂x=−2ρxy2+∂ f x ∂x

Substitute −2ρxy2+∂fx∂x for ∂p∂x in Equation (IV).

−2ρxy2+∂ f x ∂x=−2ρxy2+x3∂ f x ∂x=−2ρxy2+x3+2ρxy2∂ f x ∂x=−2ρxy2−2ρx3+2ρxy2∂ f x ∂x=−2ρx3

∂fx=−2ρx3∂x

Integrating the above equation.

∫∂ f x =−2ρ∫ x 3 ∂xfx=−2ρ x 4 4

Substitute −2ρx44 for fx in Equation (V).

p=−2ρx2 y 2 2+ y 4 4+−2ρ x 4 4+constant=−2ρx2 y 2 2+ y 4 4+ x 4 4+constant=−2ρ×142x2y2+y4+x4+constant=−ρ22x2y2+y4+x4+constant ... (VI).

Substitute 0 for x, 0 for y, pa for p in Equation (VI).

pa=−ρ22×0+0+0+constantpa=constant

Substitute pa for constant in Equation (VI).

p=−ρ22x2y2+y4+x4+pa

Conclusion:

The pressure field for px,y is −ρ22x2y2+x4+y4+pa ..

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Answer 4

Question

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y

−2ρx2y+y3∂y=∂p∂p=−2ρx2y+y3∂y

Integrate the equation on both sides

∫∂p=−2ρ∫ x 2 y+ y 3 ∂yp=−2ρx2 y 2 2+ y 4 4+fx+constant ... (V)

Differentiate the Equation with respect to x.

∂p∂x=∂∂x−2ρ x 2 y 2 2 + y 4 4 +fx+constant=−2ρ× 2x y 2 2 +0+ ∂ f x ∂x+0=−2ρ× x y 2 +0+ ∂ f x ∂x=−2ρxy2+∂ f x ∂x

Substitute −2ρxy2+∂fx∂x for ∂p∂x in Equation (IV).

−2ρxy2+∂ f x ∂x=−2ρxy2+x3∂ f x ∂x=−2ρxy2+x3+2ρxy2∂ f x ∂x=−2ρxy2−2ρx3+2ρxy2∂ f x ∂x=−2ρx3

∂fx=−2ρx3∂x

Integrating the above equation.

∫∂ f x =−2ρ∫ x 3 ∂xfx=−2ρ x 4 4

Substitute −2ρx44 for fx in Equation (V).

p=−2ρx2 y 2 2+ y 4 4+−2ρ x 4 4+constant=−2ρx2 y 2 2+ y 4 4+ x 4 4+constant=−2ρ×142x2y2+y4+x4+constant=−ρ22x2y2+y4+x4+constant ... (VI).

Substitute 0 for x, 0 for y, pa for p in Equation (VI).

pa=−ρ22×0+0+0+constantpa=constant

Substitute pa for constant in Equation (VI).

p=−ρ22x2y2+y4+x4+pa

Conclusion:

The pressure field for px,y is −ρ22x2y2+x4+y4+pa ..

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Answer 5

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y

−2ρx2y+y3∂y=∂p∂p=−2ρx2y+y3∂y

Integrate the equation on both sides

∫∂p=−2ρ∫ x 2 y+ y 3 ∂yp=−2ρx2 y 2 2+ y 4 4+fx+constant ... (V)

Differentiate the Equation with respect to x.

∂p∂x=∂∂x−2ρ x 2 y 2 2 + y 4 4 +fx+constant=−2ρ× 2x y 2 2 +0+ ∂ f x ∂x+0=−2ρ× x y 2 +0+ ∂ f x ∂x=−2ρxy2+∂ f x ∂x

Substitute −2ρxy2+∂fx∂x for ∂p∂x in Equation (IV).

−2ρxy2+∂ f x ∂x=−2ρxy2+x3∂ f x ∂x=−2ρxy2+x3+2ρxy2∂ f x ∂x=−2ρxy2−2ρx3+2ρxy2∂ f x ∂x=−2ρx3

∂fx=−2ρx3∂x

Integrating the above equation.

∫∂ f x =−2ρ∫ x 3 ∂xfx=−2ρ x 4 4

Substitute −2ρx44 for fx in Equation (V).

p=−2ρx2 y 2 2+ y 4 4+−2ρ x 4 4+constant=−2ρx2 y 2 2+ y 4 4+ x 4 4+constant=−2ρ×142x2y2+y4+x4+constant=−ρ22x2y2+y4+x4+constant ... (VI).

Substitute 0 for x, 0 for y, pa for p in Equation (VI).

pa=−ρ22×0+0+0+constantpa=constant

Substitute pa for constant in Equation (VI).

p=−ρ22x2y2+y4+x4+pa

Conclusion:

The pressure field for px,y is −ρ22x2y2+x4+y4+pa ..

Answer 6

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

Answer 7

Chapter 4, Problem 4.29P

To determine

(a)

Whether the flow satisfy conservation of mass.

Answer 8

Expert Solution

Answer to Problem 4.29P

Yes, the flow satisfy conservation of mass

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The velocity in x -direction is −2xy, the velocity in y direction is y2−x2.

Write the expression for dot product of the gradient and the velocity vector.

∇⋅V=∂u∂x+∂v∂y+∂w∂z ... (I)

Here, velocity component along x direction is u, velocity component along y direction is v, the velocity component along z direction is w.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w in Equation (I).

∇⋅V=∂∂x−2xy+∂∂yy2−x2+∂∂z0=∂∂x−2xy+∂∂yy2−x2=−2y+2y=0

Conclusion:

Since the dot product of the gradient and velocity vector is zero therefore the field satisfies conservation of mass.

Answer 13

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y

−2ρx2y+y3∂y=∂p∂p=−2ρx2y+y3∂y

Integrate the equation on both sides

∫∂p=−2ρ∫ x 2 y+ y 3 ∂yp=−2ρx2 y 2 2+ y 4 4+fx+constant ... (V)

Differentiate the Equation with respect to x.

∂p∂x=∂∂x−2ρ x 2 y 2 2 + y 4 4 +fx+constant=−2ρ× 2x y 2 2 +0+ ∂ f x ∂x+0=−2ρ× x y 2 +0+ ∂ f x ∂x=−2ρxy2+∂ f x ∂x

Substitute −2ρxy2+∂fx∂x for ∂p∂x in Equation (IV).

−2ρxy2+∂ f x ∂x=−2ρxy2+x3∂ f x ∂x=−2ρxy2+x3+2ρxy2∂ f x ∂x=−2ρxy2−2ρx3+2ρxy2∂ f x ∂x=−2ρx3

∂fx=−2ρx3∂x

Integrating the above equation.

∫∂ f x =−2ρ∫ x 3 ∂xfx=−2ρ x 4 4

Substitute −2ρx44 for fx in Equation (V).

p=−2ρx2 y 2 2+ y 4 4+−2ρ x 4 4+constant=−2ρx2 y 2 2+ y 4 4+ x 4 4+constant=−2ρ×142x2y2+y4+x4+constant=−ρ22x2y2+y4+x4+constant ... (VI).

Substitute 0 for x, 0 for y, pa for p in Equation (VI).

pa=−ρ22×0+0+0+constantpa=constant

Substitute pa for constant in Equation (VI).

p=−ρ22x2y2+y4+x4+pa

Conclusion:

The pressure field for px,y is −ρ22x2y2+x4+y4+pa ..

Answer 14

To determine

(b)

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa.

Answer 15

Expert Solution

Answer to Problem 4.29P

The pressure field p(x,y), if the pressure at the point x=0,y=0 is equal to pa is px,y=−ρ22x2y2+x4+y4+pa.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Information:

The pressure at the point x=0,y=0 is pa.

Write the expression for incompressible Navier stoke Equation using x relation.

ρu∂u∂x+v∂u∂y+w∂u∂z=ρgx−∂p∂x+μ∂2u∂x2+∂2u∂y2+∂2u∂z2 ... (II)

Here the density of the fluid is ρ The acceleration due to gravity in the x direction is gx, Here, pressure change in x direction is ∂p∂x, viscosity of the fluid is μ.

Write the expression for incompressible Navier stoke Equation using y relation.

ρu∂v∂x+v∂v∂y+w∂v∂z=ρgy−∂p∂y+μ∂2v∂x2+∂2v∂y2+∂2v∂z2 ... (III)

Here the density of the fluid is ρ the acceleration due to gravity in the y direction is gy, Here, pressure change in y direction is ∂p∂y.

Calculation:

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gx in Equation (II).

ρ −2xy ∂ −2xy ∂x + y 2 − x 2 ∂ −2xy ∂y + 0 ∂ −2xy ∂z =ρ×0−∂p∂x+μ ∂ 2 −2xy ∂ x 2 + ∂ 2 −2xy ∂ y 2 + ∂ 2 −2xy ∂ z 2 ρ −2xy+ −2y+ y 2 − x 2 −2x+0=0−∂p∂x+μ0+0+0ρ4xy2+ −2x y 2 +2 x 3 =−∂p∂xρ2xy2+2x3=−∂p∂x

2ρxy2+x3=−∂p∂x∂p∂x=−2ρxy2+x3 ... (IV)

Substitute −2xy for u, y2−x2 for v, 0 for w, 0 for gy in Equation (III)

ρ −2xy ∂ y 2 − x 2 ∂x + y 2 − x 2 ∂ y 2 − x 2 ∂y +0× ∂ y 2 − x 2 ∂z =ρ×0−∂p∂y+μ ∂ 2 y 2 − x 2 ∂ x 2 + ∂ 2 y 2 − x 2 ∂ y 2 + ∂ 2 y 2 − x 2 ∂ z 2 ρ −2xy+ −2x+ y 2 − x 2 −2y+0=0−∂p∂y+μ −2+2+0ρ4xy2+ 2 y 3 −2 x 2 y=−∂p∂y