The expression for stream function. The plot some streamlines of the flow.

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Answer 1

Question

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 2

Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 3

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Answer 2

Question

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 2

Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 3

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 2

Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 3

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 2

Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 3

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 2

Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 3

Answer 6

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 2

Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 3

Answer 7

Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Answer 8

Expert Solution & Answer

Answer to Problem 63EP

The expression for the stream function is $ax2y−bxy22+C$ .

The following figure represents the streamlines of the flow.

Fluid Mechanics: Fundamentals and Applications, Chapter 9, Problem 63EP , additional homework tip 1

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=ax2−bxy$ , where $a=0.45 (ft⋅s)−1$ , $b=0.75 (ft⋅s)−1$ and the velocity along the $x$ axis is $0$ . The stream function along the $x$ axis is $0$ . The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .

Write the expression for the velocity along $x$ direction.

$v=−∂ψ∂x$ ....... (I)

Here, the stream function along $x$ direction is $∂ψ$ , the change in distance along $x$ direction is $∂x$ and the velocity along $x$ direction is $v$ .

Write the expression for the velocity along $x$ direction.

$u=∂ψ∂y$ ...... (II)

Here, the stream function along $y$ direction is $∂ψ$ , the change in distance along $y$ direction is $∂y$ and the velocity along $y$ direction is $u$ .

Write the expression for quadric stream function.

$y=ax2±a2x4−2bxψbx$ ....... (III)

Here, the stream function is $ψ$ , the distance along $x$ direction is $x$ , the distance along $y$ direction is $y$ and the constants are $a$ and $b$ .

Calculation:

Substitute $ax2−bxy$ for $u$ in Equation (II).

$ax2−bxy=∂ψ∂y∂ψ∂y=ax2−bxy∂ψ=(ax2−bxy)⋅∂y$ ...... (IV)

Integrate Equation (IV) with respect to $y$ .

$∂ψ=(ax2−bxy)⋅∂y∫∂ψ=∫(a x 2−bxy)⋅∂y+f(x)ψ=ax2y−bxy1+12+f(x)ψ=ax2y−bxy22+f(x)$ ....... (V)

Here, the constant is $f(x)$ .

Substitute $ax2y−bxy22+f(x)$ for $ψ$ and $0$ for $v$ in Equation (I).

$0=−∂(ax2y− bx y 2 2+f(x))∂x−∂(ax2y− bx y 2 2+f(x))∂x=0$ ....... (VI)

Differentiate Equation (V) with respect to $x$ .

$−∂(ax2y− bx y 2 2+f(x))∂x=0−2ax2−1y+bx1−1y22−∂f(x)∂x=0−2ax1y+by22−∂f(x)∂x=0$ ....... (VII)

Substitute $0$ for $y$ in Equation (VII).

$−2ax1(0)+b(0)22−∂f(x)∂x=0−∂f(x)∂x=0∂f(x)∂x=0$ ....... (VIII)

Substitute $0$ for $y$ and $C$ for $ψ$ in Equation (V).

$ax2(0)−bx(0)22+f(x)=C0−0+f(x)=Cf(x)=C$ ....... (IX)

Here, the constant is $C$ .

Substitute $C$ for $f(x)$ in Equation (V).

$ψ=ax2y−bxy22+C$ ....... (X)

Substitute $C$ for $f(x)$ in Equation (X).

$ψ=ax2y−bxy22+0ψ=ax2y−bxy22$

Substitute $0.45 (ft⋅s)−1$ for $a$ and $0.75 (ft⋅s)−1$ for $b$ in Equation (III).

$y=(0.45 ( ft⋅s ) −1)x2± ( 0.45 ( ft⋅s ) −1 )2x4−2( 0.75 ( ft⋅s) −1 )xψ(0.75 ( ft⋅s ) −1)xy=0.45 ( ft⋅s)−1×x2±0.2025 ( ft⋅s ) −1×x4−1.50 ( ft⋅s ) −1×xψ(0.75 ( ft⋅s ) −1)x$ ..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

$1=0.45 ( ft⋅s)−1×(1)2±0.2025 ( ft⋅s ) −1× ( 1 )4−1.50 ( ft⋅s ) −1×(1)ψ(0.75 ( ft⋅s ) −1)(1)±0.2025 ( ft⋅s)−1×(1)4−1.50 ( ft⋅s)−1×(1)ψ=0.75 (ft⋅s)−1−0.45 (ft⋅s)−10.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2$ .. (XII)

Solve Equation (XII) by taking positive sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=±(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.08ψ=0.05333$

Solve Equation (XII) by taking negative sign.

$0.2025 (ft⋅s)−1×(1)4−1.50 (ft⋅s)−1×(1)ψ=−(0.35 ( ft⋅s)−1)2−1.50 (ft⋅s)−1×(1)ψ=−0.325ψ=0.216666$

The following table shows that the value of stream function with respect to value of $x$ and $y$ .

$x$	$y$	$−ψ$	$+ψ$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$ .