• Author / Uploaded
• noahb110

Citation preview

Problem 6.1

[Difficulty: 2]

Given:

Velocity field

Find:

Acceleration of particle and pressure gradient at (1,1)

Solution: NOTE: Units of B are s-1 not ft-1s-1 Basic equations

(2

u ( x , y ) = A⋅ y − x

For this flow

ax = u ⋅

∂ ∂x

u + v⋅

∂ ∂y

) − B⋅ x

2

v ( x , y ) = 2 ⋅ A⋅ x ⋅ y + B⋅ y

(2

u = ⎡⎣A⋅ y − x

(

) − B⋅x⎤⎦ ⋅∂ ⎡⎣A⋅(y2 − x2) − B⋅x⎤⎦ + (2⋅A⋅x⋅y + B⋅y)⋅∂ ⎡⎣A⋅(y2 − x2) − B⋅x⎤⎦

2

∂x

2

ax = ( B + 2 ⋅ A⋅ x ) ⋅ A⋅ x + B⋅ x + A⋅ y ay = u ⋅

∂ ∂x

v + v⋅

∂ ∂y

(2

v = ⎡⎣A⋅ y − x

∂y

)

2

) − B⋅x⎤⎦ ⋅∂ (2⋅A⋅x⋅y + B⋅y) + (2⋅A⋅x⋅y + B⋅y)⋅∂ (2⋅A⋅x⋅y + B⋅y)

2

∂x

∂y

(2

ay = ( B + 2 ⋅ A⋅ x ) ⋅ ( B⋅ y + 2 ⋅ A⋅ x ⋅ y ) − 2 ⋅ A⋅ y ⋅ ⎡⎣B⋅ x + A⋅ x − y ax = ( 1 + 2 ⋅ 1 ⋅ 1 ) ⋅

Hence at (1,1)

ay = ( 1 + 2 ⋅ 1 ⋅ 1 ) ⋅ a =

2

ax + ay

1 s 1 s

(

)s

)⎦

2⎤

2 ft

2

× 1⋅ 1 + 1⋅ 1 + 1⋅ 1 ⋅ × ( 1⋅ 1 + 2⋅ 1⋅ 1⋅ 1) ⋅

2

ft s

ft

ax = 9 ⋅

− 2⋅ 1⋅ 1⋅

1 s

(2

× ⎡⎣1 ⋅ 1 + 1 ⋅ 1 − 1

)⎦ ⋅ s

2 ⎤ ft

⎛ ay ⎞

θ = atan⎜

ay = 7 ⋅

2

s a = 11.4⋅

⎝ ax ⎠

2

s ft

ft 2

θ = 37.9⋅ deg

s

For the pressure gradient lbf

∂ ∂x

p = ρ⋅ g x − ρ⋅ ax = −2 ⋅

slug ft

3

× 9⋅

ft 2

s

2

×

lbf ⋅ s

∂

slug⋅ ft

∂x

2

p = −18⋅

ft

= −0.125 ⋅

ft

psi ft

lbf

∂ ∂y

p = ρ⋅ g y − ρ⋅ ay = 2 ⋅

slug ft

3

× ( −32.2 − 7 ) ⋅

ft 2

s

2

×

lbf ⋅ s

∂

slug⋅ ft

∂y

2

p = −78.4⋅

ft

ft

= −0.544 ⋅

psi ft

Problem 6.2

[Difficulty: 2]

Given:

Velocity field

Find:

Acceleration of particle and pressure gradient at (2,2)

Solution: Basic equations

Given data

For this flow

A = 1⋅

B = 3⋅

s

1 s

x = 2⋅ m

y = 2⋅ m

∂ ∂x

ay = u ⋅

u + v⋅

∂ ∂x

∂y

v + v⋅

ax = ( 1 + 9)

a =

∂

2

1 s

∂ ∂y

kg 3

v ( x , y ) = B⋅ x − A⋅ y

u = ( A ⋅ x + B⋅ y ) ⋅

∂ ∂x

v = ( A ⋅ x + B⋅ y ) ⋅

( A ⋅ x + B⋅ y ) + ( B⋅ x − A ⋅ y ) ⋅

∂ ∂x

∂ ∂y

( B⋅ x − A ⋅ y ) + ( B⋅ x − A ⋅ y ) ⋅

m

× 2⋅ m

ax = 20

2

θ = atan ⎜

ax + ay

ρ = 999 ⋅

m

u ( x , y ) = A⋅ x + B⋅ y

ax = u ⋅

Hence at (2,2)

1

ay = ( 1 + 9)

s

⎛ ay ⎞

a = 28.28

⎝ ax ⎠

∂ ∂y

1 s

(

2

2

)

(

2

2

( A ⋅ x + B⋅ y )

ax = A + B ⋅ x

( B⋅ x − A ⋅ y )

ay = A + B ⋅ y

× 2⋅ m

ay = 20

m

)

m s

θ = 45⋅ deg

s

For the pressure gradient 2

m N⋅s kg × 20⋅ × p = ρ⋅ g x − ρ⋅ ax = −999⋅ 3 2 kg⋅ m ∂x s m ∂

∂ ∂x 2

m N⋅s kg × ( −9.81 − 20) ⋅ × p = −ρ⋅ g y − ρ⋅ ay = 999⋅ 3 2 kg⋅ m ∂y s m ∂

∂ ∂y

p = −20000⋅

Pa

p = −29800⋅

Pa

m

m

= −20.0⋅

kPa

= −29.8⋅

kPa

m

m

Problem 6.3

[Difficulty: 2]

Given:

Velocity field

Find:

Acceleration of particle and pressure gradient at (1,2)

Solution: Basic equations

Given data

A = 1⋅

1

B = 2⋅

s

m 2

x = 1⋅ m

y = 2⋅ m

t = 5⋅ s

ρ = 999 ⋅

s

u ( x , y , t) = −A⋅ x + B⋅ t

kg 3

m v ( x , y , t) = A⋅ y + B⋅ t

The acceleration components and values are axt( x , y , t) =

∂ ∂t

u ( x , y , t) = B

∂ ∂t

m

axt( x , y , t) = 2

2

s

axc( x , y , t) = u ( x , y , t) ⋅

ayt( x , y , t) =

axt( x , y , t) = B

∂ ∂x

u ( x , y , t) + v ( x , y , t) ⋅

v ( x , y , t)

ayc( x , y , t) = u ( x , y , t) ⋅

∂ ∂y

2

u ( x , y , t) axc( x , y , t) = A ⋅ x − A⋅ B⋅ t

axc( x , y , t) = −9

m 2

s

ayt( x , y , t) = B

m

ayt( x , y , t) = 2

2

s ∂ ∂x

v ( x , y , t) + v ( x , y , t) ⋅

ax ( x , y , t) = axt( x , y , t) + axc( x , y , t)

∂ ∂y

2

ayc( x , y , t) = y ⋅ A + B⋅ t⋅ A

v( x , y , t)

ayc( x , y , t) = 12

2

s

2

ax ( x , y , t ) = x ⋅ A − B ⋅ t ⋅ A + B

ax ( x , y , t) = −7

m 2

s ay ( x , y , t) = ayt( x , y , t) + ayc( x , y , t)

m

2

ay ( x , y , t ) = y ⋅ A + B ⋅ t ⋅ A + B

ay ( x , y , t) = 14

m 2

s

For overall acceleration

a( x , y , t ) =

2

ax ( x , y , t ) + ay ( x , y , t )

2

(x⋅A2 − B⋅t⋅A + B) + (y⋅A2 + B⋅t⋅A + B) 2

a( x , y , t ) =

2

a( x , y , t) = 15.7

m 2

s

⎛ ay ( x , y , t ) ⎞

θ = atan⎜

⎝

ax ( x , y , t )

⎠

For the pressure gradient we need

θ = −63.4⋅ deg

−ρ⋅ ax ( x , y , t) = 6.99⋅

kPa m

−ρ⋅ ay ( x , y , t) = −13.99 ⋅

kPa

−ρ⋅ g = −9.80⋅

m

Hence for the pressure gradient 2

m N⋅ s kg × 7⋅ × p = ρ⋅ g x − ρ⋅ ax = 999 ⋅ 3 2 kg⋅ m ∂x s m ∂

∂ ∂x 2

m N⋅ s kg × ( −9.81 − 14) ⋅ × p = −ρ⋅ g y − ρ⋅ ay = 999 ⋅ 3 2 kg⋅ m ∂y s m ∂

∂ ∂y

p = 6990⋅

Pa m

p = −23800 ⋅

= 6.99⋅

Pa m

kPa m

= −23.8⋅

kPa m

kPa m

Problem 6.4

Given:

Velocity field

Find:

Pressure gradient at (1,1) at 1 s

[Difficulty: 2]

Solution: Basic equations

Given data

A = 2⋅

1

B = 1⋅

2

s

1

x = 1⋅ m

2

y = 1⋅ m

t = 1⋅ s

ρ = 1000⋅

s

kg 3

m

u ( x , y , t) = ( −A⋅ x + B⋅ y ) ⋅ t

v ( x , y , t) = ( A⋅ y + B⋅ x ) ⋅ t

The acceleration components and values are axt( x , y , t) =

∂ ∂t

u ( x , y , t) = B⋅ y − A⋅ x

∂ ∂t

m

axt( x , y , t) = −1

2

s

axc( x , y , t) = u ( x , y , t) ⋅

ayt( x , y , t) =

axt( x , y , t) = B⋅ y − A⋅ x

∂ ∂x

u ( x , y , t) + v ( x , y , t) ⋅

v ( x , y , t)

∂ ∂y

(

2

2

u ( x , y , t) axc( x , y , t) = t ⋅ x ⋅ A + B

2

)

axc( x , y , t) = 5

m 2

s

ayt( x , y , t) = A⋅ y + B⋅ x

m

ayt( x , y , t) = 3

2

s

ayc( x , y , t) = u ( x , y , t) ⋅

∂ ∂x

v ( x , y , t) + v ( x , y , t) ⋅

∂ ∂y

v( x , y , t)

2

2

2

)

ayc( x , y , t) = 5

m 2

s 2 2

ax ( x , y , t) = axt( x , y , t) + axc( x , y , t)

(

ayc( x , y , t) = t ⋅ y ⋅ A + B 2 2

ax ( x , y , t ) = x ⋅ A ⋅ t − x ⋅ A + x ⋅ B ⋅ t + y ⋅ B

ax ( x , y , t ) = 4

m 2

s 2 2

ay ( x , y , t) = ayt( x , y , t) + ayc( x , y , t)

2 2

ay ( x , y , t ) = y ⋅ A ⋅ t + y ⋅ A + y ⋅ B ⋅ t + x ⋅ B

ay ( x , y , t ) = 8

m 2

s Hence for the pressure gradient ∂ ∂x

∂ ∂y

p = −ρ⋅ ax = −1000⋅

kg 3

× 4⋅

kg 3

m

2

2

×

s

m

p = −ρ⋅ ay = −1000⋅

m

× 8⋅

m 2

s

N⋅ s

∂

kg⋅ m

∂x

2

×

N⋅ s

∂

kg⋅ m

∂y

p = −4000⋅

Pa

p = −8000⋅

Pa

m

m

= −4 ⋅

kPa

= −8 ⋅

kPa

m

m

Problem 6.5

[Difficulty: 2]

Given:

Velocity field

Find:

Acceleration of particle and pressure gradient at (1,1)

Solution: Basic equations

ax = u ⋅

∂ ∂x

(2

2

(2

2

u ( x , y ) = A⋅ x − y

For this flow

u + v⋅

∂ ∂y

u = ⎡⎣A⋅ x − y

) − 3 ⋅ B⋅ x

v ( x , y ) = −2 ⋅ A⋅ x ⋅ y + 3 ⋅ B⋅ y

) − 3⋅B⋅x⎤⎦ ⋅∂ ⎡⎣A⋅(x2 − y2) − 3⋅B⋅x⎤⎦ + (−2⋅A⋅x⋅y + 3⋅B⋅y)⋅∂ ⎡⎣A⋅(x2 − y2) − 3⋅B⋅x⎤⎦ ∂x

(

∂y

2

ax = ( 2 ⋅ A⋅ x − 3 ⋅ B) ⋅ A⋅ x − 3 ⋅ B⋅ x + A⋅ y

ay = u ⋅

∂ ∂x

v + v⋅

∂ ∂y

(2

v = ⎡⎣A⋅ x − y

)

2

) − 3⋅B⋅x⎤⎦ ⋅∂ (−2⋅A⋅x⋅y + 3⋅B⋅y) + (−2⋅A⋅x⋅y + 3⋅B⋅y)⋅∂ (−2⋅A⋅x⋅y + 3⋅B⋅y)

2

∂x

∂y

(2

ay = ( 3 ⋅ B⋅ y − 2 ⋅ A⋅ x ⋅ y ) ⋅ ( 3 ⋅ B − 2 ⋅ A⋅ x ) − 2 ⋅ A⋅ y ⋅ ⎡⎣A⋅ x − y Hence at (1,1)

ax = ( 2 ⋅ 1 ⋅ 1 − 3 ⋅ 1 ) ⋅

1 s

2

ax + ay

)s

2 ft

2

× 1⋅ 1 − 3⋅ 1⋅ 1 + 1⋅ 1 ⋅

ay = ( 3 ⋅ 1 ⋅ 1 − 2 ⋅ 1 ⋅ 1 ⋅ 1 ) ⋅ a =

(

) − 3⋅B⋅x⎤⎦

2

1

× ( 3⋅ 1 − 2⋅ 1⋅ 1) ⋅

s

ft

ax = 1 ⋅ 1

− 2⋅ 1⋅ 1⋅

s

s

⎛ ay ⎞

2

θ = atan⎜

(2

× ⎡⎣1 ⋅ 1 − 1

⎝ ax ⎠

) − 3⋅1⋅1⎤⎦ ⋅ fts

ay = 7 ⋅

∂x

ft

ft

θ = 81.9⋅ deg

2

s

lbf

p = ρ⋅ g x − ρ⋅ ax = −2 ⋅

slug ft

3

× 1⋅

ft 2

s

2

×

lbf ⋅ s

∂

slug⋅ ft

∂x

2

p = −2 ⋅

ft

ft

= −0.0139⋅

psi ft

lbf

∂ ∂y

p = ρ⋅ g y − ρ⋅ ay = 2 ⋅

slug ft

3

× ( −32.2 − 7 ) ⋅

ft 2

s

2

×

lbf ⋅ s

∂

slug⋅ ft

∂y

2

s

For the pressure gradient ∂

2

s

2

a = 7.1⋅

ft

2

p = −78.4⋅

ft

ft

= −0.544 ⋅

psi ft

Problem 6.6

[Difficulty: 2]

Given:

Velocity field

Find:

Simplest y component of velocity; Acceleration of particle and pressure gradient at (2,1); pressure on x axis

Solution: Basic equations

For this flow

u ( x , y ) = A⋅ x

Hence

v ( x , y ) = −A⋅ y

For acceleration

ax = u ⋅

∂ ∂x

ay = u ⋅

⌠ ⌠ ⎮ ∂ so v ( x, y ) = −⎮ u dy = −⎮ A dy = −A ⋅ y + c u + v =0 ⌡ ∂x ∂y ⎮ ∂x ⌡ is the simplest y component of velocity ∂

u + v⋅

∂ ∂x

∂

u = A ⋅ x⋅

∂y

v + v⋅

∂ ∂y

ax = a =

⎛ 2 ⎞ × 2⋅ m ⎜ ⎝ s⎠ 2

ax + ay

∂ ∂x

v = A⋅ x ⋅

2

Hence at (2,1)

∂

2

( A ⋅ x) + ( −A ⋅ y ) ⋅

∂ ∂x

∂

2

∂y

( −A⋅ y ) + ( −A⋅ y ) ⋅

2

( A ⋅ x) = A ⋅ x

∂ ∂y

ax = A ⋅ x 2

ay = A ⋅ y

( −A⋅ y )

2

⎛ 2 ⎞ × 1⋅ m ⎜ ⎝ s⎠ ⎛ ay ⎞ θ = atan⎜ ⎝ ax ⎠ ay =

ax = 8

m

ay = 4

2

s

a = 8.94

m

θ = 26.6⋅ deg

2

s

2

m N⋅ s kg × 8⋅ × p = ρ⋅ g x − ρ⋅ ax = −1.50⋅ 3 2 kg⋅ m ∂x s m

∂ ∂x

2

m N⋅ s kg × 4⋅ × p = ρ⋅ g y − ρ⋅ ay = −1.50⋅ 3 2 kg⋅ m ∂y s m ∂

∂ ∂z

kg

p = ρ⋅ g z − ρ⋅ az = 1.50 ×

For the pressure on the x axis

1 2 2 p ( x ) = p 0 − ⋅ ρ⋅ A ⋅ x 2

3

× ( −9.81) ⋅

dp =

∂x

p

p ( x ) = 190 ⋅ kPa −

⌠ p − p0 = ⎮ ⌡

x

0

1 2

⋅ 1.5⋅

2

s

m ∂

m

kg 3

m

(

∂y N⋅ s

∂

kg⋅ m

∂y

⌠ ρ⋅ g x − ρ⋅ ax dx = ⎮ ⌡

)

x

0

2

×

∂

2

×

2

s

For the pressure gradient ∂

m

p = −6 ⋅

Pa m

Pa m

p = −14.7⋅

Pa m

(−ρ⋅A2⋅x) dx = − 12 ⋅ρ⋅A2⋅x2

2

⎛ 2 ⎞ × N⋅ s × x 2 ⎜ kg⋅ m ⎝ s⎠

p = −12⋅

p ( x ) = 190 −

3 1000

⋅x

2

(p in kPa, x in m)

Problem 6.7

[Difficulty: 3]

Given:

Velocity field

Find:

Expressions for local, convective and total acceleration; evaluate at several points; evaluate pressure gradient

Solution: A = 2⋅

The given data is

1

ω = 1⋅

s

∂

Check for incompressible flow

∂x ∂

Hence

∂x

u +

u +

1

ρ = 2⋅

s

∂ ∂y ∂ ∂y

kg

u = A⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

3

v = −A⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

m v =0

v = A⋅ sin( 2 ⋅ π⋅ ω⋅ t) − A⋅ sin( 2 ⋅ π⋅ ω⋅ t) = 0

Incompressible flow

The governing equation for acceleration is

The local acceleration is then

∂

x - component

∂t ∂

y - component

∂t

u = 2 ⋅ π⋅ A⋅ ω⋅ x ⋅ cos( 2 ⋅ π⋅ ω⋅ t)

v = −2 ⋅ π⋅ A⋅ ω⋅ y ⋅ cos( 2 ⋅ π⋅ ω⋅ t)

For the present steady, 2D flow, the convective acceleration is

x - component

u⋅

y - component

u⋅

∂ ∂x ∂ ∂x

u + v⋅

v + v⋅

The total acceleration is then

∂

2

∂y ∂

u = A⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ⋅ ( A⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) + ( −A⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) ⋅ 0 = A ⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

2

∂y

2

v = A⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ⋅ 0 + ( −A⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) ⋅ ( −A⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) = A ⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

x - component

y - component

∂ ∂t ∂ ∂t

u + u⋅

v + u⋅

∂ ∂x ∂ ∂x

u + v⋅

v + v⋅

∂ ∂y ∂ ∂y

2

2

u = 2 ⋅ π⋅ A⋅ ω⋅ x ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

2

2

v = −2 ⋅ π⋅ A⋅ ω⋅ y ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

2

Evaluating at point (1,1) at t = 0⋅ s

12.6⋅

Local

m

m

−12.6⋅

and

2

2

s 12.6⋅

Total

m

m

12.6⋅

and

2

12.6⋅

and

2

s t = 1⋅ s

12.6⋅

Local

12.6⋅

Convective

2

0⋅

m

0⋅

and

2

s

∂y Evaluated at (1,1) and time

2

m 2

m

−12.6⋅

and

2

m

Convective

2

s

m

−12.6⋅

and

2

0⋅

m

0⋅

and

2

s

m 2

s

p = −ρ⋅

Du

∂

Dt

∂x

p = −ρ⋅

Dv

∂

Dt

∂x

(

)

2

2

p = −ρ⋅ −2 ⋅ π⋅ A⋅ ω⋅ y ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

t = 0⋅ s

x comp.

t = 0.5⋅ s

x comp.

t = 1⋅ s

2

p = −ρ⋅ 2 ⋅ π⋅ A⋅ ω⋅ x ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)

(

x comp.

−25.1⋅ 25.1⋅

Pa

y comp.

m

Pa

y comp.

m

−25.1⋅

Pa m

m 2

s

Hence, the components of pressure gradient (neglecting gravity) are

∂

m s

The governing equation (assuming inviscid flow) for computing the pressure gradient is

∂x

2

s

s

s

∂

m

2

m

s Total

2

s

s

m

−12.6⋅

0⋅

and

s

s Total

m

m

−12.6⋅

and

2

−12.6⋅

Local

0⋅

s

s t = 0.5⋅ s

Convective

y comp.

25.1⋅

)

Pa m

−25.1⋅ 25.1⋅

2

Pa m

Pa m

Problem 6.8

[Difficulty: 3]

Given:

Velocity field

Find:

Expressions for velocity and acceleration along wall; plot; verify vertical components are zero; plot pressure gradient

Solution: 3

m

q = 2⋅

The given data is

u=

s

h = 1⋅ m

m

ρ = 1000⋅

kg 3

m q⋅ x

2 ⋅ π⎡⎣x + ( y − h ) 2

2⎤

+

⎦

q⋅ x 2 ⋅ π⎡⎣x + ( y + h ) 2

v=

2⎤

⎦

q⋅ ( y − h) 2 ⋅ π⎡⎣x + ( y − h ) 2

2⎤

+

⎦

q⋅ ( y + h) 2 ⋅ π⎡⎣x + ( y + h ) 2

2⎤

⎦

The governing equation for acceleration is

For steady, 2D flow this reduces to (after considerable math!)

x - component

y - component

u=

(

2

π⋅ x + h

∂x

u + v⋅

∂ ∂y

2

u =−

⎡

(2

q ⋅ x⋅ ⎣ x + y

)

2

2

(2

− h ⋅ h − 4⋅ y

2

2⎤

)⎦

2

⎡⎣x2 + ( y + h ) 2⎤⎦ ⋅ ⎡⎣x 2 + ( y − h) 2⎤⎦ ⋅ π2 2 2 ⎡ 2 2 2 2 2⎤ − h ⋅ h + 4⋅ x ⎦ q ⋅ y⋅ ⎣ x + y ∂ ∂ ay = u ⋅ v + v ⋅ v = − 2 2 ∂x ∂y 2 2 2 2 2 π ⋅ ⎡⎣x + ( y + h ) ⎤⎦ ⋅ ⎡⎣x + ( y − h ) ⎤⎦

(

)

2

2

)

2

(

)

y = 0⋅ m

For motion along the wall q⋅ x

ax = u ⋅

∂

v=0

(No normal velocity)

ax = −

(2

q ⋅ x⋅ x − h 2

(2

π ⋅ x +h

)

2

)

2

3

ay = 0

(No normal acceleration)

The governing equation (assuming inviscid flow) for computing the pressure gradient is Hence, the component of pressure gradient (neglecting gravity) along the wall is ∂ ∂x

p = −ρ⋅

Du

∂

Dt

∂x

(2

2

p =

ρ⋅ q ⋅ x ⋅ x − h 2

(2

π ⋅ x +h

)

2

)

2

3

The plots of velocity, acceleration, and pressure gradient are shown below, done in Excel. From the plots it is clear that the fluid experiences an adverse pressure gradient from the origin to x = 1 m, then a negative one promoting fluid acceleration. If flow separates, it will likely be in the region x = 0 to x = h. q = h =

2 1

m 3/s/m m

0.35

∠=

1000

kg/m 3

0.30

0.00000 0.00000 0.01945 0.00973 0.00495 0.00277 0.00168 0.00109 0.00074 0.00053 0.00039

0.00 0.00 -19.45 -9.73 -4.95 -2.77 -1.68 -1.09 -0.74 -0.53 -0.39

u (m/s)

0.00 0.32 0.25 0.19 0.15 0.12 0.10 0.09 0.08 0.07 0.06

dp /dx (Pa/m)

0.25 0.20 0.15 0.10 0.05 0.00 0

1

2

3

4

5

6

7

8

9

10

8

9

10

9

10

x (m)

Acceleration Along Wall Near A Source 0.025 0.020 a (m/s 2)

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

a (m/s2)

0.015 0.010 0.005 0.000 0

1

2

3

4

5

6

7

-0.005 x (m)

Pressure Gradient Along Wall 5 dp /dx (Pa/m)

x (m) u (m/s)

Velocity Along Wall Near A Source

0 0

1

2

3

4

5

-5 -10 -15 -20 -25 x (m)

6

7

8

Problem 6.9

[Difficulty: 2]

Problem 6.10

Given:

Velocity field

Find:

Expression for pressure field; evaluate at (2,2)

[Difficulty: 2]

Solution: Basic equations

Given data

A = 4⋅

1

B = 2⋅

s

1

x = 2⋅ m

s

y = 2⋅ m

u ( x , y ) = A⋅ x + B⋅ y

v ( x , y ) = B⋅ x − A⋅ y

Note that

∂

∂

Then

∂x

∂ ∂y

v( x , y) = 0

ax ( x , y ) = u ( x , y ) ⋅

ay ( x , y ) = u ( x , y ) ⋅

∂ ∂x ∂ ∂x

∂x

u( x , y) + v( x , y) ⋅

v( x , y) + v( x , y) ⋅

∂

The momentum equation becomes

∂x

p = −ρ⋅ ax

v( x , y) −

∂ ∂y ∂ ∂y

∂ ∂y

∂ ∂y

(

x

y

0

0

p ( x , y ) = 80⋅ kPa

(

2

2

)

2

2

−

(

2

2

)

ax ( x , y ) = 40

m 2

s

(

2

ay ( x , y ) = y ⋅ A + B

v( x , y)

2

)

ay ( x , y ) = 40

m 2

s

p = −ρ⋅ ay

⌠ ⌠ p ( x , y ) = p 0 − ρ⋅ ⎮ ax ( x , y ) dx − ρ⋅ ⎮ ay ( x , y ) dy ⌡ ⌡ ρ⋅ A + B ⋅ y

p 0 = 200 ⋅ kPa

3

u( x , y) = 0

ax ( x , y ) = x ⋅ A + B

u( x , y)

Integrating

p( x , y) = p0 −

kg m

For this flow

u( x , y) +

ρ = 1500⋅

2

2

)

ρ⋅ A + B ⋅ x 2

2

and

p = dx⋅

∂ ∂x

p + dy⋅

∂ ∂y

p

Problem 6.11

[Difficulty: 2]

Given:

Velocity field

Find:

Expression for pressure gradient; plot; evaluate pressure at outlet

Solution: Basic equations

Given data

U = 15⋅

m

L = 5⋅ m

s

ρ = 1250⋅

kg 3

m

u ( x ) = U⋅ ⎛⎜ 1 −

Here

p in = 100 ⋅ kPa

⎝

x⎞

u ( 0 ) = 15

L⎠ ρ⋅ u ⋅

The x momentum becomes

dp

The pressure gradient is then

dx

s

u ( L) = 0

d d u = ρ⋅ aa = − p dx dx

ax ( x ) = u ( x ) ⋅

Hence

m

2

= −ρ⋅

∂ ∂x

⋅ ⎛⎜

U

x

⎝L

L

m s

U ⋅ ⎛⎜ 2

ax ( x ) =

u( x)

⎝L

⎠

L

− 1⎞

⎠

x

⌠ p ( x ) = p in − ρ⋅ ⎮ ax ( x ) dx ⌡

Integrating momentum

− 1⎞

x

2

p ( x ) = p in −

U ⋅ ρ⋅ x ⋅ ( x − 2 ⋅ L)

0

2⋅ L

2

2

p ( L) =

Hence

ρ⋅ U 2

+ p in

p ( L) = 241 ⋅ kPa

dp/dx (kPa/m)

60 40 20

0

1

2

3

x (m)

4

5

Problem 6.12

[Difficulty: 2]

Given:

Velocity field

Find:

Expression for acceleration and pressure gradient; plot; evaluate pressure at outlet

Solution: Basic equations

Given data

U = 20⋅

m

L = 2⋅ m

s

u ( x ) = U⋅ e

ρ = 900 ⋅

kg 3

m −

Here

p in = 50⋅ kPa

x L

u ( 0 ) = 20

m

u ( L) = 7.36

s

m s −

The x component of acceleration is then

The x momentum becomes

The pressure gradient is then

ax ( x ) = u ( x ) ⋅

ρ⋅ u ⋅

dp dx

2

∂ ∂x

ax ( x ) = −

u( x)

=

ρ L

−

L

2

⋅U ⋅e

2⋅ x L

⌠ p ( x ) = p in − ρ⋅ ⎮ ax ( x ) dx ⌡ 0

2

Hence

L

d d u = ρ⋅ aa = − p dx dx

x

Integrating momentum

U ⋅e

2⋅ x

p ( L) = p in −

( − 2 − 1)

U ⋅ ρ⋅ e

2

⎛ − 2⋅ x ⎞ 2 ⎜ L U ⋅ ρ⋅ ⎝ e − 1⎠ p ( x ) = p in − 2

p ( L) = 206 ⋅ kPa

dp/dx (kPa/m)

200 150 100 50 0

0.5

1

1.5

2

x (m)

ax (m/s2)

0

0.5

1

− 50 − 100 − 150 − 200

x (m)

1.5

2

Problem 6.13

[Difficulty: 2]

Problem 6.14

[Difficulty: 3]

Given:

Velocity field

Find:

The acceleration at several points; evaluate pressure gradient

Solution: The given data is

3

3

m

q = 2⋅

m

s

K = 1⋅

m

s

m

ρ = 1000⋅

kg

q Vr = − 2 ⋅ π⋅ r

3

m

K Vθ = 2 ⋅ π⋅ r

The governing equations for this 2D flow are

The total acceleration for this steady flow is then 2

2

Vθ Vθ ∂ ar = Vr⋅ Vr + ⋅ Vr − r r ∂θ ∂r

ar = −

θ - component

Vθ Vr⋅ Vθ ∂ ∂ aθ = Vr⋅ Vθ + ⋅ Vθ + r ∂θ r ∂r

aθ = 0

Evaluating at point (1,0)

ar = −0.127

r - component

Evaluating at point (1,π/2)

Evaluating at point (2,0)

From Eq. 6.3, pressure gradient is

∂

ar = −0.127

m

∂ ∂r

s

m

aθ = 0

2

s

m 2

s

Evaluating at point (1,π/2)

Evaluating at point (2,0)

∂ ∂r ∂ ∂r ∂ ∂r

p = 127 ⋅

p = 127 ⋅

∂r

Pa m Pa

p = 15.8⋅

aθ = 0 ∂

p = −ρ⋅ ar

1 ∂ ⋅ p = −ρ⋅ aθ r ∂θ Evaluating at point (1,0)

m Pa m

2 3

4⋅ π ⋅ r

aθ = 0

2

ar = −0.0158

2

q +K

p =

(2

2 3

4⋅ π ⋅ r

1 ∂ ⋅ p =0 r ∂θ 1 ∂ ⋅ p =0 r ∂θ 1 ∂ ⋅ p =0 r ∂θ 1 ∂ ⋅ p =0 r ∂θ

)

2

ρ⋅ q + K

Problem 6.15

[Difficulty: 3]

Problem 6.16

[Difficulty: 3]

Given:

Flow in a pipe with variable area

Find:

Expression for pressure gradient and pressure; Plot them; exit pressure

Solution: Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity Basic equations

Q = V⋅ A

Given data

ρ = 1.75⋅

slug ft

For this 1D flow

2

p i = 35⋅ psi

3

Q = u i⋅ Ai = u ⋅ A

Ai = 15⋅ in

(Ai − Ae)

A = Ai −

L

⋅x

so

2

Ae = 2.5⋅ in

L = 10⋅ ft

Ai u ( x ) = u i⋅ = u i⋅ A

ui = 5⋅

Ai

Ai −

⎡ ( Ai − Ae) ⎤ ⎢ ⋅ x⎥ ⎣ L ⎦

Ai ⎤ Ai ⋅ L ⋅ u i ⋅ ( Ae − Ai) ⎡ ⎥= ax = u ⋅ u + v ⋅ u = u i ⋅ ⋅ ⎢u i⋅ ∂x ∂y ⎡ ( Ai − Ae) ⎤ ∂x ⎢ ⎡ ( Ai − Ae) ⎤ ⎥ ( A ⋅ L + A ⋅ x − A ⋅ x) 3 i e i Ai − ⎢ ⋅ x⎥ ⋅ x⎥ ⎥ ⎢ Ai − ⎢ L ⎣ L ⎦ ⎣ ⎣ ⎦⎦ ∂

For the pressure

∂ ∂x

Ai

∂

2

p = −ρ⋅ ax − ρ⋅ g x = −

2

2

2

∂

(

ρ⋅ Ai ⋅ L ⋅ u i ⋅ Ae − Ai

(Ai⋅ L + Ae⋅ x − Ai⋅ x)

dp =

∂ ∂x

3

⌠ x 2 2 2 ⎮ ⌠ ρ⋅ Ai ⋅ L ⋅ u i ⋅ Ae − Ai ∂ ⎮ ⎮ p − pi = p dx = − dx ⎮ ∂x ⎮ 3 Ai⋅ L + Ae⋅ x − Ai⋅ x ⌡ ⎮ 0 ⌡

p ⋅ dx

(

(

This is a tricky integral, so instead consider the following: x

x

0

0

∂ ∂x

p = −ρ⋅ ax = −ρ⋅ u ⋅

( )

∂

1 ∂ 2 u u = − ⋅ ρ⋅ 2 ∂x ∂x

⌠ ⌠ ρ ρ 2 2 2 ∂ ∂ p − pi = ⎮ p dx = − ⋅ ⎮ u dx = ⋅ u ( x = 0 ) − u ( x ) ⎮ ∂x ⎮ 2 2 ∂x ⌡ ⌡

( )

(

)

)

)

0

Hence

2

) x

and

2

ft s

ρ 2 2 p( x) = pi + ⋅ ⎛ ui − u( x) ⎞ ⎠ 2 ⎝

p( x) = pi +

Hence

⎡ ⎢ ⋅ 1− 2 ⎢ ⎢ ⎣

ρ⋅ u i

2

which we recognise as the Bernoulli equation!

Ai ⎡⎢ ⎤⎥ ⎢ ⎡ ( Ai − Ae) ⎤ ⎥ ⋅ x⎥ ⎥ ⎢ Ai − ⎢ L ⎣ ⎣ ⎦⎦

2⎤

⎥ ⎥ ⎥ ⎦

p ( L) = 29.7 psi

At the exit

Pressure Gradient (psi/ft)

The following plots can be done in Excel

6

4

2

0

2

4

6

8

10

6

8

10

Pressure (psi)

x (ft)

35

30

25

0

2

4

x (ft)

Problem 6.17

[Difficulty: 3]

Given:

Flow in a pipe with variable area

Find:

Expression for pressure gradient and pressure; Plot them

Solution: Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity Basic equations

Given data

Q = V⋅ A

ρ = 1250⋅

kg

2

A0 = 0.25⋅ m

3

a = 1.5⋅ m

L = 5⋅ m

m For this 1D flow

Q = u 0 ⋅ A0 = u ⋅ A

so

A0 u( x) = u0⋅ = A

⎛ − ⎜ A( x ) = A0 ⋅ ⎝ 1 + e

x a

−

−e

x

⎞

2⋅ a

⎠

u 0 = 10⋅

m

u0

⎛ − ⎜ ⎝1 + e

x a

−

−e

x

⎞

2⋅ a

⎠ ⎛ − 2 2⋅ a ⎜ ⋅ ⎝ 2⋅ e u0 ⋅ e −

ax = u ⋅

∂ ∂x

u + v⋅

∂ ∂y

u =

∂ ∂x

u0 ⎡ ⎢ x ⎞ ∂x ⎢ ⎛ x − − − ⎜ ⎢ 2⋅ a a −e −e ⎠ ⎣⎝1 + e

u0

⎛ − ⎜ ⎝1 + e

x a

p = −ρ⋅ ax − ρ⋅ g x = −

⋅

⎛ − 2 2⋅ a ⎜ ⋅ ⎝ 2⋅ e ρ⋅ u 0 ⋅ e −

For the pressure

p 0 = 300 ⋅ kPa

s

⎛ − ⎜ 2 ⋅ a⋅ ⎝ e

x

x a

−

−e

∂

⎞

x 2⋅ a

x 2⋅ a

− 1⎠

⎞ + 1⎠

3

⎤ ⎥= x ⎞⎥ ⎛ − 2⋅ a ⎥ ⎠ ⎦ 2 ⋅ a⋅ ⎜⎝ e

x

x a

−

−e

⎞

x 2⋅ a

x 2⋅ a

− 1⎠

⎞ + 1⎠

3

x

dp =

and

∂ ∂x

⌠ ⎮ − ⎮ x 2 ⌠ ⋅ ⋅ e ρ u ⎮ 0 ∂ p − pi = ⎮ p dx = ⎮ − ⎮ ∂x ⎮ ⌡ ⎛ − 0 ⎮ ⎜ ⎮ 2 ⋅ a⋅ ⎝ e ⌡

p ⋅ dx

⎛ − 2⋅ a ⎜ ⋅ ⎝ 2⋅ e x

x a

−

−e

⎞

x 2⋅ a

− 1⎠

⎞

x 2⋅ a

3

dx

+ 1⎠

0

∂

This is a tricky integral, so instead consider the following:

x

x

0

0

∂x

p = −ρ⋅ ax = −ρ⋅ u ⋅

( )

∂

1 ∂ 2 u u = − ⋅ ρ⋅ 2 ∂x ∂x

⌠ ⌠ ρ ρ 2 2 2 ∂ ∂ p − pi = ⎮ p dx = − ⋅ ⎮ u dx = ⋅ u ( x = 0 ) − u ( x ) ⎮ ∂x ⎮ 2 2 ∂x ⌡ ⌡

Hence

( )

ρ 2 2 p(x) = p0 + ⋅ ⎛ u0 − u(x) ⎞ ⎝ ⎠ 2

p( x) = p0 +

ρ⋅ u 0 2

2

⎡

⋅ ⎢1 −

⎢ ⎢ ⎣

(

)

which we recognise as the Bernoulli equation!

1 ⎡⎢ x ⎢⎛ − − ⎜ ⎢⎣ ⎝ 1 + e a − e

⎤⎥ x ⎞⎥ 2⋅ a ⎥ ⎠⎦

2⎤

⎥ ⎥ ⎥ ⎦

The following plots can be done in Excel 0.26

Area (m2)

0.24

0.22

0.2

0.18

0

1

2

3

x (m)

4

5

Pressure Gradient (kPa/m)

20

0

1

2

3

4

5

− 20

− 40

− 60

x (m)

300

Pressure (kPa)

290

280

270

260

250

0

1

2

3

x (m)

4

5

Problem 6.18

[Difficulty: 3]

Given:

Nozzle geometry

Find:

Acceleration of fluid particle; Plot; Plot pressure gradient; find L such that pressure gradient < 5 MPa/m in absolute value

Solution: The given data is

Di = 0.1⋅ m

Do = 0.02⋅ m D( x ) = Di +

For a linear decrease in diameter

From continuity

Hence

or

Q = V⋅ A = V⋅

V( x ) ⋅

π 4

Do − Di L

π 2 2 ⋅ D = Vi⋅ ⋅ Di 4 4

3

m

⋅x Q = 0.00785

2

⋅ D( x ) = Q

m s

4⋅ Q

V( x ) =

Do − Di ⎞ ⎛ ⋅x π⋅ ⎜ Di + L ⎝ ⎠

Vi Do − Di ⎞ ⎛ ⋅x ⎜1 + L⋅ Di ⎝ ⎠

kg

ρ = 1000⋅

3

π

V( x ) =

m Vi = 1 ⋅ s

L = 0.5⋅ m

2

2

The governing equation for this flow is

or, for steady 1D flow, in the notation of the problem d ax = V⋅ V = dx

Vi Do − Di ⎞ ⎛ ⋅x ⎜1 + L⋅ Di ⎝ ⎠

d ⋅ 2 dx

2

Vi Do − Di ⎞ ⎛ ⋅x ⎜1 + L⋅ Di ⎝ ⎠

This is plotted in the associated Excel workbook From Eq. 6.2a, pressure gradient is ∂ ∂x

p = −ρ⋅ ax

∂ ∂x

2

p =

(

2 ⋅ ρ⋅ Vi ⋅ Do − Di

)

⎡ ( Do − Di) ⎤ Di⋅ L⋅ ⎢1 + ⋅ x⎥ Di⋅ L ⎣ ⎦

5

2

ax ( x ) = −

(

2 ⋅ Vi ⋅ Do − Di

)

⎡ ( Do − Di) ⎤ Di⋅ L⋅ ⎢1 + ⋅ x⎥ Di⋅ L ⎣ ⎦

5

This is also plotted in the associated Excel workbook. Note that the pressure gradient is always negative: separation is unlikely to occur in the nozzle ∂

At the inlet

∂x

p = −3.2⋅

kPa

∂

At the exit

∂x

m

p = −10⋅

To find the length L for which the absolute pressure gradient is no more than 5 MPa/m, we need to solve ∂ ∂x

2

p ≤ 5⋅

MPa m

=

(

2 ⋅ ρ⋅ Vi ⋅ Do − Di

)

⎡ ( Do − Di) ⎤ Di⋅ L⋅ ⎢1 + ⋅ x⎥ Di⋅ L ⎣ ⎦

5

with x = L m (the largest pressure gradient is at the outlet) 2

L≥

Hence

(

2 ⋅ ρ⋅ Vi ⋅ Do − Di

)

5

L ≥ 1⋅ m

⎛ Do ⎞ ∂ Di⋅ ⎜ ⋅ p Di x ∂ ⎝ ⎠ This result is also obtained using Goal Seek in the Excel workbook

From Excel

x (m) a (m/s 2) 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.420 0.440 0.460 0.470 0.480 0.490 0.500

3.20 4.86 7.65 12.6 22.0 41.2 84.2 194 529 843 1408 2495 3411 4761 6806 10000

dp /dx (kPa/m) -3.20 -4.86 -7.65 -12.6 -22.0 -41.2 -84.2 -194 -529 -843 -1408 -2495 -3411 -4761 -6806 -10000

For the length L required for the pressure gradient to be less than 5 MPa/m (abs) use Goal Seek L =

1.00

x (m)

dp /dx (kPa/m)

1.00

-5000

m

MPa m

Acceleration Through A Nozzle 12000

2

a (m/s )

10000 8000 6000 4000 2000 0 0.0

0.1

0.1

0.2

0.2

0.3

0.3

0.4

0.4

0.5

0.5

0.4

0.5

0.5

x (m)

Pressure Gradient Along A Nozzle 0

dp/dx (kPa/m)

0.0

0.1

0.1

0.2

0.2

0.3

-2000 -4000 -6000 -8000 -10000 -12000

x (m)

0.3

0.4

Problem 6.19

[Difficulty: 3]

Given:

Diffuser geometry

Find:

Acceleration of a fluid particle; plot it; plot pressure gradient; find L such that pressure gradient is less than 25 kPa/m

Solution: The given data is

Di = 0.25⋅ m

Do = 0.75⋅ m D( x) = Di +

For a linear increase in diameter

From continuity

Hence

Q = V⋅ A = V⋅

V( x) ⋅

π 4

Do − Di L

π 2 2 ⋅ D = Vi⋅ ⋅ Di 4 4

2

ρ = 1000⋅

4⋅ Q

m s

Vi

V( x) =

or

Do − Di ⎞ ⎛ ⋅x π⋅ ⎜ Di + L ⎝ ⎠

2

Do − Di ⎞ ⎛ ⋅x ⎜1 + L⋅ Di ⎝ ⎠

The governing equation for this flow is

ax = V⋅

or, for steady 1D flow, in the notation of the problem

2

Hence

ax ( x ) = −

(

d V= dx

Vi

d 2 dx

Vi

⋅

Do − Di ⎞ ⎛ ⋅x ⎜1 + L⋅ Di ⎝ ⎠

Do − Di ⎞ ⎛ ⋅x ⎜1 + L⋅ Di ⎝ ⎠

)

⎡ ( Do − Di) ⎤ Di⋅ L⋅ ⎢1 + ⋅ x⎥ Di⋅ L ⎣ ⎦

5

This can be plotted in Excel (see below) From Eq. 6.2a, pressure gradient is

∂ ∂x

3

m

⋅x Q = 0.245

V( x) =

2 ⋅ Vi ⋅ Do − Di

kg

3

π

⋅ D( x) = Q

m Vi = 5⋅ s

L = 1⋅ m

p = −ρ⋅ ax

∂ ∂x

2

p =

(

2 ⋅ ρ⋅ Vi ⋅ Do − Di

)

⎡ ( Do − Di) ⎤ Di⋅ L⋅ ⎢1 + ⋅ x⎥ Di⋅ L ⎣ ⎦

5

2

2

This can also plotted in Excel. Note that the pressure gradient is adverse: separation is likely to occur in the diffuser, and occur near the entrance ∂

At the inlet

∂x

p = 100 ⋅

kPa

∂

At the exit

m

∂x

p = 412 ⋅

Pa m

To find the length L for which the pressure gradient is no more than 25 kPa/m, we need to solve 2

∂ ∂x

p ≤ 25⋅

kPa m

=

(

2 ⋅ ρ⋅ Vi ⋅ Do − Di

)

⎡ ( Do − Di) ⎤ Di⋅ L⋅ ⎢1 + ⋅ x⎥ Di⋅ L ⎣ ⎦

5

with x = 0 m (the largest pressure gradient is at the inlet) 2

L≥

Hence

(

2 ⋅ ρ⋅ Vi ⋅ Do − Di Di⋅

∂ ∂x

)

L ≥ 4⋅ m

p

This result is also obtained using Goal Seek in Excel.

In Excel:

Di Do L Vi

= = = =

0.25 0.75 1 5

m m m m/s

( =

1000

kg/m 3

x (m) a (m/s 2)

dp /dx (kPa/m)

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.40 0.50 0.60

-100 -62.1 -40.2 -26.9 -18.59 -13.17 -9.54 -5.29 -3.125 -1.940

100 62.1 40.2 26.93 18.59 13.17 9.54 5.29 3.125 1.940

0.70 0.80 0.90 1.00

-1.256 -0.842 -0.581 -0.412

1.256 0.842 0.581 0.412

For the length L required for the pressure gradient to be less than 25 kPa/m use Goal Seek L =

4.00

x (m)

dp /dx (kPa/m)

0.0

25.0

m

Acceleration Through a Diffuser 0 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.9

1.0

2

a (m/s )

-20 -40 -60 -80 -100 -120

x (m)

Pressure Gradient Along A Diffuser

dp /dx (kPa/m)

120 100 80 60 40 20 0 0.0

0.1

0.2

0.3

0.4

0.5

x (m)

0.6

0.7

0.8

Problem 6.20

[Difficulty: 4]

Problem 6.21

[Difficulty: 4]

Problem 6.20

Problem 6.22

[Difficulty: 3]

Problem 6.23

[Difficulty: 4]

Given:

Rectangular chip flow

Find:

Velocity field; acceleration; pressure gradient; net force; required flow rate; plot pressure

Solution: Basic equations

→→ ( ∑ V⋅A) = 0

∂ ∂x

CS

The given data is

ρ = 1.23⋅

kg

u +

∂ ∂y

v =0

p atm = 101 ⋅ kPa

3

h = 0.5⋅ mm

b = 40⋅ mm

M length = 0.005 ⋅

m

Assuming a CV that is from the centerline to any point x, and noting that q is inflow per unit area, continuity leads to q ⋅ x ⋅ L = U⋅ h ⋅ L

u ( x ) = U( x ) = q ⋅

or

x h

For acceleration we will need the vertical velocity v; we can use ∂ ∂x Hence

u +

∂ ∂y

v =0

q x du ∂ d = − ⎛⎜ q ⋅ ⎞ = − v =− u =− h dx dx ⎝ h ⎠ ∂y ∂x ∂

or

⌠ v ( y = y ) − v ( y = 0 ) = −⎮ ⎮ ⌡

y

0

But

v( y = 0) = q

For the x acceleration

ax = u ⋅

∂ ∂x

u + v⋅

∂ ∂y

q h

dy = −q ⋅

y h y⎞

so

v ( y ) = q ⋅ ⎛⎜ 1 −

u

x q y ax = q ⋅ ⋅ ⎛⎜ ⎞ + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ( 0 ) h ⎝h⎠ h⎠ ⎝

⎝

h⎠ ax =

q h

2 2

⋅x

kg m

For the y acceleration

ay = u ⋅

∂ ∂x

v + v⋅

∂

x y q ay = q ⋅ ⋅ ( 0 ) + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ⎛⎜ − ⎞ h h⎠ ⎝ h⎠ ⎝

v

∂y

∂

Hence

∂x

ρ⋅

Also

p = −ρ⋅

q h

Dv Dx

Du

ρ⋅

For the pressure gradient we use x and y momentum (Euler equation)

Dx

ax =

q

2

h

⋅ ⎛⎜

y

⎝h

− 1⎞

⎠

⎛ ∂ ∂ ⎞ ∂ u + v ⋅ u = ρ⋅ ax = − p ∂y ⎠ ∂x ⎝ ∂x

= ρ⋅ ⎜ u ⋅

2 2

⋅x

⎛ ∂ ∂ ⎞ ∂ v + v ⋅ v = ρ⋅ ay = − p ∂y ⎠ ∂y ⎝ ∂x

∂

= ρ⋅ ⎜ u ⋅

p = ρ⋅

∂y

q

2

h

⋅ ⎛⎜ 1 −

⎝

y⎞ h⎠

For the pressure distribution, integrating from the outside edge (x = b/2) to any point x x

p ( x = x ) − p ⎛⎜ x =

b⎞

⎝

2⎠

x ⌠ 2 2 2 ⌠ ⎮ q q q 2 2 ⎮ ∂ = p ( x ) − p atm = p dx = ⎮ −ρ⋅ ⋅ x dx = −ρ⋅ ⋅ x + ρ⋅ ⋅b ⎮ ∂x 2 2 2 ⎮ h 2⋅ h 8⋅ h ⎮ ⎮b ⌡b ⌡ 2

2 2

p ( x ) = p atm + ρ⋅

q ⋅b

⎡

⋅ ⎢1 − 4 ⋅ ⎛⎜

⎥ ⎝b⎠ ⎦

8⋅ h ⎣ 2

x⎞

2

2⎤

For the net force we need to integrate this ... actually the gage pressure, as this pressure is opposed on the outer surface by p atm 2 2

pg( x) =

b

ρ⋅ q ⋅ b 8⋅ h

2

⎡

⋅ ⎢1 − 4 ⋅ ⎛⎜

⎣

x⎞

2⎤

⎥ ⎝b⎠ ⎦

b

⌠2 ⌠2 2 2 2 ⎮ ⎮ ρ⋅ q 2⋅ b 2 ⎡ ρ⋅ q ⋅ b ⋅ L ⎛ b x ⎤ 1 b Fnet = 2 ⋅ L⋅ ⎮ p g ( x ) dx = 2 ⋅ L⋅ ⎮ ⋅ ⎢1 − 4 ⋅ ⎛⎜ ⎞ ⎥ dx = ⋅⎜ − ⋅ ⎞ 2 ⎣ 2 ⌡ ⎝b⎠ ⎦ ⎝2 3 2⎠ ⎮ 8⋅ h 4⋅ h 0 ⌡

2 3

Fnet =

ρ⋅ q ⋅ b ⋅ L 12⋅ h

2

0

2 3

The weight of the chip must balance this force

M ⋅ g = M length ⋅ L⋅ g = Fnet =

ρ⋅ q ⋅ b ⋅ L 12⋅ h

2

2 3

or

M length ⋅ g =

ρ⋅ q ⋅ b 12⋅ h 3

m

2

q =

Solving for q for the given mass/length

12⋅ h ⋅ g ⋅ M length ρ⋅ b

q = 0.0432⋅

3

s 2

m

b

The maximum speed

b⎞

b⋅ q Umax = 2⋅ h

Umax = u ⎛⎜ x = = q⋅ h 2⎠ ⎝

2

2 2

The following plot can be done in Excel

pg( x) =

ρ⋅ q ⋅ b 8⋅ h

2

⎡

⋅ ⎢1 − 4 ⋅ ⎛⎜

⎣

x⎞

2⎤

⎥ ⎝b⎠ ⎦

m Umax = 1.73 s

2

2

Pressure (Pa)

1.5

1

0.5

− 0.02

− 0.01

0

0.01

0.02

x (m)

The net force is such that the chip is floating on air due to a Bernoulli effect: the speed is maximum at the edges and zero at the center; pressure has the opposite trend - pressure is minimum (p atm) at the edges and maximum at the center.

Problem 6.24

[Difficulty: 3]

Problem 6.25

[Difficulty: 4]

Given:

Velocity field

Find:

Constant B for incompressible flow; Acceleration of particle at (2,1); acceleration normal to velocity at (2,1)

Solution: Basic equations

For this flow

3

u ( x , y ) = A⋅ x + B⋅ x ⋅ y ∂ ∂x ∂ ∂x

u( x , y) +

u( x , y) +

∂ ∂y ∂ ∂y

2

v( x , y) =

3

∂ ∂x

Hence for ax

ax = u ⋅

u + v⋅

∂ ∂y

(

ay = u ⋅

∂

(2

∂x

v + v⋅

2

∂ ∂y

(2

3

3

0.2

2

2

)

2

)

2

∂y

2

(

2

2

∂x

3

) (A⋅y3 − 3⋅A⋅x2⋅y) + (A⋅y3 − 3⋅A⋅x2⋅y)⋅∂ (A⋅y3 − 3⋅A⋅x2⋅y)

2 ∂

∂x

∂y

2

2

2

ax + ay

1

) (A⋅x3 − 3⋅A⋅x⋅y2) + (A⋅y3 − 3⋅A⋅x2⋅y)⋅∂ (A⋅x3 − 3⋅A⋅x⋅y2)

0.2 ⎞ 2 2 ay = 3 ⋅ ⎛ × 1 ⋅ m × ⎡⎣( 2 ⋅ m) + ( 1 ⋅ m) ⎤⎦ ⎜ 2 ⎝ m ⋅s ⎠ a =

B = −0.6

2 ∂

⎞ × 2⋅ m × ⎡( 2⋅ m) 2 + ( 1 ⋅ m) 2⎤ ⎣ ⎦ ⎜ 2 m ⋅ s ⎝ ⎠

ax = 3 ⋅ ⎛

B = −3 ⋅ A

Hence

v ( x , y ) = A⋅ y − 3 ⋅ A⋅ x ⋅ y

v = A⋅ x − 3 ⋅ A⋅ x ⋅ y ⋅

ay = 3 ⋅ A ⋅ y ⋅ x + y Hence at (2,1)

2

u = A⋅ x − 3 ⋅ A⋅ x ⋅ y ⋅

ax = 3 ⋅ A ⋅ x ⋅ x + y For ay

)=0

2

m ⋅s

3

2

∂y

(2

u ( x , y ) = A⋅ x − 3 ⋅ A⋅ x ⋅ y

∂x

(A⋅x3 + B⋅x⋅y2) + ∂ (A⋅y3 + B⋅x2⋅y) = 0

v ( x , y ) = ( 3 ⋅ A + B) ⋅ x + y

We can write

∂

2

v ( x , y ) = A⋅ y + B⋅ x ⋅ y

2

ax = 6.00⋅

m 2

s 2

ay = 3.00⋅

m 2

s a = 6.71

m 2

s

We need to find the component of acceleration normal to the velocity vector

At (2,1) the velocity vector is at angle

r a

⎛ 1 − 3⋅ 2 ⋅ 1 ⎞ θvel = atan⎜ ⎜ 3 − 3⋅ 2⋅ 12 ⎝2 ⎠

θvel = −79.7⋅ deg

⎛ ay ⎞ θaccel = atan⎜ ⎝ ax ⎠

1 θaccel = atan⎛⎜ ⎞ ⎝2⎠

θaccel = 26.6⋅ deg

∆θ = θaccel − θvel

∆θ = 106 ⋅ deg

3

At (1,2) the acceleration vector is at angle

r V

⎛ A⋅ y 3 − 3⋅ A⋅ x2⋅ y ⎞ v θvel = atan⎛⎜ ⎞ = atan⎜ ⎜ A⋅ x 3 − 3⋅ A⋅ x⋅ y 2 ⎝u⎠ ⎝ ⎠

Hence the angle between the acceleration and velocity vectors is

The component of acceleration normal to the velocity is then

2

an = a⋅ sin( ∆θ) = 6.71⋅

∆θ

m 2

s

⋅ sin( 106 ⋅ deg)

an = 6.45⋅

m 2

s

Problem 6.26

[Difficulty: 4] Part 1/2

Problem 6.26

[Difficulty: 4] Part 2/2

Problem 6.27

[Difficulty: 5]

Given:

Velocity field

Find:

Constant B for incompressible flow; Equation for streamline through (1,2); Acceleration of particle; streamline curvature

Solution: Basic equations

(4

2 2

u ( x, y ) = A ⋅ x − 6⋅ x ⋅ y + y

For this flow

∂ ∂x ∂ ∂x

u ( x, y ) +

u( x , y) +

∂ ∂y

v ( x, y ) =

∂ ∂y

∂ ∂x

)

v ( x, y ) = B⋅ x ⋅ y − x⋅ y

(3

(

)

4

3

)

(

)

⎡⎣A ⋅ x4 − 6⋅ x2⋅ y2 + y 4 ⎤⎦ + ∂ ⎡⎣B⋅ x3⋅ y − x⋅ y 3 ⎤⎦ = 0

(3

∂y

) + A⋅(4⋅x3 − 12⋅x⋅y2) = (4⋅A + B)⋅x⋅(x2 − 3⋅y2) = 0

2

v ( x , y) = B⋅ x − 3⋅ x ⋅ y

B = −4 ⋅ A

Hence

1

B = −8

3

m ⋅s Hence for ax ax = u ⋅

∂ ∂x

u + v⋅

∂ ∂y

(4

)

4 ∂

2 2

u = A⋅ x − 6 ⋅ x ⋅ y + y ⋅ 2

(2

ax = 4 ⋅ A ⋅ x ⋅ x + y

)

2

∂x

(

)

(

)

(

)

⎡⎣A⋅ x 4 − 6⋅ x 2⋅ y2 + y 4 ⎤⎦ + ⎡⎣−4 ⋅ A⋅ x3⋅ y − x⋅ y 3 ⎤⎦ ⋅ ∂ ⎡⎣A⋅ x 4 − 6⋅ x 2⋅ y2 + y 4 ⎤⎦ ∂y

3

For ay ay = u ⋅

∂ ∂x

v + v⋅

∂ ∂y

(4 2

(2

ay = 4 ⋅ A ⋅ y ⋅ x + y For a streamline

dy dx

Let

)

4 ∂

2 2

v = A⋅ x − 6 ⋅ x ⋅ y + y ⋅

u=

=

v u

so

)

2

∂x

(

)

(

)

(

)

⎡⎣−4⋅ A⋅ x 3⋅ y − x ⋅ y3 ⎤⎦ + ⎡⎣−4⋅ A⋅ x 3⋅ y − x ⋅ y3 ⎤⎦ ⋅ ∂ ⎡⎣−4 ⋅ A⋅ x3⋅ y − x⋅ y 3 ⎤⎦ ∂y

3

dy dx

y

du

x

dx

=

=

(3

−4 ⋅ A⋅ x ⋅ y − x ⋅ y

)

3

A⋅ x − 6 ⋅ x ⋅ y + y

4

)

=−

d ⎛⎜

d ⎛⎜

1⎞

(4

2 2

y⎞

(3

4⋅ x ⋅ y − x⋅ y

(x4 − 6⋅x2⋅y2 + y4)

⎝ x ⎠ = 1 ⋅ dy + y⋅ ⎝ x ⎠ = 1 ⋅ dy − y dx

x dx

dx

)

3

x dx

x

2

so

dy dx

= x⋅

du dx

+u

dy

Hence

dx

x⋅

du dx

dx

Separating variables

x

= x⋅

du

(3

4⋅ x ⋅ y − x⋅ y

dx

+u=−

⎡⎢ ⎢ ⎢⎣

=−u+

(x4 − 6⋅x2⋅y2 + y4)

(

)

4

(

2

(

4⋅ 1 − u

(

⎛ 1 − 6⋅ u + u3⎞ ⎜ ⎝u ⎠

)

4⋅ 1 − u

)

2

⎛ 1 − 6⋅ u + u3⎞ ⎜ ⎝u ⎠

)

1 5 3 ln( x ) = − ⋅ ln u − 10⋅ u + 5 ⋅ u + C 5

⋅ du

(u5 − 10⋅u3 + 5⋅u)⋅x5 = c 3 2

)u +

2

(

2

4 2 u ⋅ ( u − 10⋅ u + 5 )

5

=−

4 2 ⎤⎥ u ⋅ u − 10⋅ u + 5 =− 4 2 ⎛ 1 − 6⋅ u + u3⎞ ⎥ u − 6⋅ u + 1 ⎜ ⎥ ⎝u ⎠⎦

4⋅ 1 − u

u − 6⋅ u + 1

=−

)

3

5

3 2

4

y − 10⋅ y ⋅ x + 5 ⋅ y ⋅ x = const

4

y − 10⋅ y ⋅ x + 5 ⋅ y ⋅ x = −38

For the streamline through (1,2)

Note that it would be MUCH easier to use the stream function method here! 2

an = −

To find the radius of curvature we use

2

V

V

R =

or

R

an

r V

We need to find the component of acceleration normal to the velocity vector

(3

)

⎤ ⎥ 4 2 2 4 ⎥ ⎣ x − 6⋅ x ⋅ y + y ⎦

⎡ v θvel = atan⎛⎜ ⎞ = atan⎢− ⎢ ⎝u⎠

At (1,2) the velocity vector is at angle

4⋅ x ⋅ y − x⋅ y

(

4⋅ ( 2 − 8) ⎤ θvel = atan⎡⎢− ⎥ ⎣ 1 − 24 + 16⎦

3

r a

)

∆θ

θvel = −73.7⋅ deg

At (1,2) the acceleration vector is at angle

) ⎤⎥ = atan⎛ y ⎞ ⎥ ⎜ 3 x ⎥ )⎦ ⎝ ⎠

⎡ 2 ⎢ 4⋅ A ⋅ y⋅ x2 + y2 ⎛ ay ⎞ θaccel = atan⎜ = atan⎢ ⎝ ax ⎠ ⎢ 4 ⋅ A2⋅ x ⋅ x 2 + y2 ⎣

( (

3

∆θ = θaccel − θvel

Hence the angle between the acceleration and velocity vectors is

an = a⋅ sin( ∆θ)

The component of acceleration normal to the velocity is then

At (1,2)

(2

2

a =

2

)

2

ax = 4 ⋅ A ⋅ x ⋅ x + y

3

7

2

a = 4472

2

(4

2 2

2

Then

R =

V

an

2

m

an = a⋅ sin( ∆θ)

2

) = −14⋅ ms

4

(3

⎝

m⎞ s

2

2

(2

3

2

ay = 4 ⋅ A ⋅ y ⋅ x + y

)

2

⎠

2

×

) = 48⋅ ms

3

1

an = 3040

m 2

V=

2

s

3040 m

2

u + v = 50⋅

2

⋅

= 4000⋅

m 2

s

s

v = B⋅ x ⋅ y − x ⋅ y R = ⎛⎜ 50⋅

ax + ay

a=

s

s

u = A⋅ x − 6 ⋅ x ⋅ y + y

θaccel = 63.4⋅ deg

∆θ = 137 ⋅ deg

where

⎛ 2 ⎞ = 2000⋅ m ⎜ 3 2 s ⎝ m ⋅s ⎠

7

= 500 ⋅ m × A = 500 ⋅ m ×

2 m

2000 + 4000 ⋅

2 θaccel = atan⎛⎜ ⎞ ⎝1⎠

R = 0.822 m

m s

Problem 6.28

Given:

Velocity field for doublet

Find:

Expression for pressure gradient

[Difficulty: 2]

Solution: Basic equations

For this flow

Hence for r momentum

Λ Vr( r , θ) = − ⋅ cos( θ) 2 r

Λ Vθ( r , θ) = − ⋅ sin( θ) 2 r

Vz = 0

2 ⎛⎜ Vθ Vθ ⎞ ∂ ∂ ρ⋅ g r − p = ρ⋅ ⎜ Vr⋅ Vr + ⋅ V − r ⎠ r ∂θ r ∂r ⎝ ∂r

∂

Ignoring gravity

⎡⎢ ⎢ Λ Λ ∂ ∂ p = −ρ⋅ ⎢⎛ − ⋅ cos( θ) ⎞ ⋅ ⎛ − ⋅ cos( θ) ⎞ + ⎜ ⎜ 2 2 ⎢⎣⎝ r ∂r ⎠ ∂r ⎝ r ⎠ For θ momentum

ρ⋅ g θ −

⎛ − Λ ⋅ sin( θ)⎞ ⎜ 2 ⎝ r ⎠ ⋅ ∂ ⎛ − Λ ⋅ cos( θ) ⎞ − r ∂θ ⎜ r2 ⎝ ⎠

⎛ − Λ ⋅ sin( θ) ⎞ ⎜ 2 ⎝ r ⎠ r

2⎤

⎥ ⎥ ⎥ ⎥⎦

∂ ∂r

2

p =

2⋅ Λ ⋅ ρ 5

r

Vθ Vr⋅ Vθ ⎞ ⎛ ∂ 1 ∂ ∂ ⋅ p = ρ⋅ ⎜ Vr⋅ Vθ + ⋅ Vθ + r ∂θ r ∂θ r ⎠ ⎝ ∂r

Ignoring gravity

⎡ ⎢ Λ ∂ ⎢ Λ ∂ p = −r⋅ ρ⋅ ⎛ − ⋅ cos( θ) ⎞ ⋅ ⎛ − ⋅ sin( θ) ⎞ + ⎜ ⎜ ⎢ 2 2 ∂θ ⎣⎝ r ⎠ ∂r ⎝ r ⎠ The pressure gradient is purely radial

⎛ − Λ ⋅ sin( θ) ⎞ ⎜ 2 ⎝ r ⎠ ⋅ ∂ ⎛ − Λ ⋅ sin( θ)⎞ + r ∂θ ⎜ r2 ⎝ ⎠

⎛ − Λ ⋅ sin( θ)⎞ ⋅ ⎛ − Λ ⋅ cos( θ) ⎞ ⎤ ⎥ ⎜ 2 ⎜ 2 ⎝ r ⎠⎝ r ⎠⎥ ⎥ r ⎦

∂ ∂θ

p =0

Problem 6.29

[Diffculty: 4]

Given:

Velocity field for flow over a cylinder

Find:

Expression for pressure gradient; pressure variation; minimum pressure; plot velocity

Solution: Basic equations

Given data

ρ = 1.23⋅

kg

a = 150 ⋅ mm

3

U = 75⋅

m For this flow

⎤ ⎡ a Vr = U⋅ ⎢⎛⎜ ⎞ − 1⎥ ⋅ cos( θ) r

On the surface r = a

Vr = 0

2

Hence on the surface:

⎣⎝ ⎠

⎦

m s

⎤ ⎡ a Vθ = U⋅ ⎢⎛⎜ ⎞ + 1⎥ ⋅ sin( θ) r 2

⎣⎝ ⎠

⎦

Vθ = 2 ⋅ U⋅ sin( θ)

For r momentum

2 ⎛⎜ ⎛⎜ V 2 ⎞ Vθ Vθ ⎞ θ ∂ ∂ ∂ ρ⋅ ⎜ Vr⋅ Vr + = ρ⋅ ⎜ − ⋅ Vr − =− p r ⎠ r ∂θ ⎝ a ⎠ ∂r ⎝ ∂r

For θ momentum

Vθ Vr⋅ Vθ ⎞ ⎛ ∂ ⎛ Vθ ∂ ⎞ 2 ⋅ U⋅ sin( θ) 1 ∂ ∂ ρ⋅ ⎜ Vr⋅ Vθ + = ρ⋅ ⎜ ⋅ 2 ⋅ U⋅ cos( θ) = − ⋅ p ⋅ Vθ + ⋅ Vθ = ρ⋅ a r ∂θ r ∂θ r ⎠ ⎝ ∂r ⎝ a ∂θ ⎠ 2

∂ ∂θ

p =−

4 ⋅ ρ⋅ U a

∂ ∂r

2

p = ρ⋅

Vθ

2

⋅ sin( θ) ⋅ cos( θ) = −

2 ⋅ ρ⋅ U a

⋅ sin( 2 ⋅ θ)

For the pressure distribution we integrate from θ = 0 to θ = θ, assuming p(0) = p atm (a stagnation point) θ

⌠ ∂ p ( θ) − p atm = ⎮ p dθ = ⎮ ∂θ ⌡ 0

θ

⌠ 2 ⎮ 4 ⋅ ρ⋅ U ⎮ − ⋅ sin( θ) ⋅ cos( θ) dθ a ⎮ ⌡ 0

a

2

= ρ⋅ 4 ⋅ U ⋅ sin( θ)

2

2⌠

θ

2

p ( θ) = −4 ⋅ ρ⋅ U ⎮ sin( θ) ⋅ cos( θ) dθ ⌡

p ( θ) = −2 ⋅ U ⋅ ρ⋅ sin( θ)

2

Minimum p:

0

Pressure (kPa)

0

50

100

p ⎛⎜

π⎞

⎝2⎠

= −13.8⋅ kPa

150

−5

− 10

− 15

x (m)

⎤ ⎡ a Vθ( r) = U⋅ ⎢⎛⎜ ⎞ + 1⎥ r 2

For the velocity as a function of radial position at θ = π/2

Vr = 0

so

V = Vθ

⎣⎝ ⎠

⎦

5

r/a

4

3

2

1 75

100

125

150

V(r) (m/s) The velocity falls off to V = U as directly above the cylinder we have uniform horizontal as the effect of the cylinder decreases m Vθ( 100 ⋅ a) = 75 s

Problem 6.30

[Difficulty: 2]

Given:

Flow in a curved section

Find:

Expression for pressure distribution; plot; V for wall static pressure of 35 kPa

Solution: Basic equation

∂ ∂n

2

p = ρ⋅

V

R

Assumptions: Steady; frictionless; no body force; constant speed along streamline ρ = 999 ⋅

Given data

kg 3

V = 10⋅

m At the inlet section

p = p( y)

m

L = 75⋅ mm

s

∂

hence

∂n 2

p( y) = pc −

Integrating from y = 0 to y = y

ρ⋅ V

R0⋅ L

⋅y

p =−

2

dp dy

R0 = 0.2⋅ m 2

= ρ⋅

V

R

p c = 50⋅ kPa

2 2⋅ y

= ρ⋅ V ⋅

2 2⋅ y

dp = −ρ⋅ V ⋅

L⋅ R 0

p ⎛⎜

p ( 0 ) = 50⋅ kPa

(1)

L⎞

⎝2⎠

L⋅ R 0

⋅ dy

= 40.6⋅ kPa

40

y (mm)

30

20

10

40

42

44

46

48

50

p (kPa)

For a new wall pressure

p wall = 35⋅ kPa

solving Eq 1 for V gives

V =

(

)

4 ⋅ R0 ⋅ p c − p wall ρ⋅ L

V = 12.7

m s

Problem 6.31

[Difficulty: 2]

Problem 6.32

Given:

Velocity field for free vortex flow in elbow

Find:

Similar solution to Example 6.1; find k (above)

[Difficulty: 3]

Solution: Basic equation

∂ ∂r

2

p =

ρ⋅ V

c V = Vθ = r

with

r

Assumptions: 1) Frictionless 2) Incompressible 3) free vortex 2

∂

p =

2

ρ⋅ V ρ⋅ c d = p = 3 r dr r

For this flow

p ≠ p ( θ)

Hence

2⎛ 2 2⎞ ⌠ 2 2 2 ρ⋅ c ⎛ 1 1 ⎞ ρ⋅ c ⋅ ⎝ r2 − r1 ⎠ ⎮ ρ⋅ c dr = ∆p = p 2 − p 1 = ⎮ ⋅⎜ − = 3 2 2 2 2 ⎜ 2 r r1 r2 2⋅ r1 ⋅ r2 ⎮ ⎝ ⎠ ⌡r

so

∂r

r

(1)

1

Next we obtain c in terms of Q ⌠ →→ ⎮ Q = ⎮ V dA = ⌡

r

r

⌠2 ⌠ 2 w⋅ c ⎛ r2 ⎞ dr = w⋅ c⋅ ln⎜ ⎮ V⋅ w dr = ⎮ r ⎮ ⌡r ⎝ r1 ⎠ ⌡r 1 1

Hence

c=

Q

⎛ r2 ⎞

w⋅ ln⎜

⎝ r1 ⎠ ρ⋅ c ⋅ ⎛ r2 − r1 ⎝ 2

Using this in Eq 1

∆p = p 2 − p 1 =

2

2

2 ⋅ r1 ⋅ r2

2

Solving for Q

2

2

2⎞

⎠ =

2 ⋅ r1 ⋅ r2 ⎛ r2 ⎞ Q = w⋅ ln⎜ ⋅ ⋅ ∆p ⎝ r1 ⎠ ρ⋅ ⎛⎝ r2 2 − r12⎞⎠

ρ⋅ Q ⋅ ⎛ r2 − r1 ⎝ 2

2

2⎞

⎠

2

⎛ r2 ⎞ 2 2 2 ⋅ w ⋅ ln⎜ ⋅ r1 ⋅ r2 ⎝ r1 ⎠ 2

2

2

2 ⋅ r1 ⋅ r2 ⎛ r2 ⎞ k = w⋅ ln⎜ ⋅ ⎝ r1 ⎠ ρ⋅ ⎛⎝ r2 2 − r12⎞⎠

Problem 6.33

[Difficulty: 4] Part 1/2

Problem 6.33

[Difficulty: 4] Part 2/2

Problem 6.34

[Difficulty: 2]

Given:

Flow rates in elbow for uniform flow and free vortes

Find:

Plot discrepancy

Solution:

(

)

For Example 6.1 QUniform = V⋅ A = w⋅ r2 − r1 ⋅

⎛ r2 ⎞

For Problem 6.32 Q = w⋅ ln⎜

⎝ r1 ⎠

2

⋅

2 ⋅ r1 ⋅ r2

1

⎛ r2 ⎞ ρ⋅ ln⎜ ⎝ r1 ⎠

⋅ ∆p

or

⎛ r2 ⎞ −1 ⎜ QUniform⋅ ρ ⎝ r1 ⎠ = w⋅ r1 ⋅ ∆p ⎛ r2 ⎞ ln⎜ ⎝ r1 ⎠

2

2 2 ρ⋅ ⎛ r2 − r1 ⎞ ⎝ ⎠

⋅ ∆p

or

Q⋅ ρ w⋅ r1 ⋅ ∆p

=

(1)

⎛ r2 ⎞ ⎛ r2 ⎞ 2 ⎜ r ⋅ ln⎜ ⋅ ⎝ 1 ⎠ ⎝ r1 ⎠ ⎡⎢⎛ r2 ⎞ 2 − ⎢⎜ r ⎣⎝ 1 ⎠

⎤ ⎥ 1 ⎥ ⎦

(2)

It is convenient to plot these as functions of r2/r1 Eq. 1

Eq. 2

Error

1.01 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.40 2.45 2.50

0.100 0.226 0.324 0.401 0.468 0.529 0.586 0.639 0.690 0.738 0.785 0.831 0.875 0.919 0.961 1.003 1.043 1.084 1.123 1.162 1.201 1.239 1.277 1.314 1.351 1.388 1.424 1.460 1.496 1.532 1.567

0.100 0.226 0.324 0.400 0.466 0.526 0.581 0.632 0.680 0.726 0.769 0.811 0.851 0.890 0.928 0.964 1.000 1.034 1.068 1.100 1.132 1.163 1.193 1.223 1.252 1.280 1.308 1.335 1.362 1.388 1.414

0.0% 0.0% 0.1% 0.2% 0.4% 0.6% 0.9% 1.1% 1.4% 1.7% 2.1% 2.4% 2.8% 3.2% 3.6% 4.0% 4.4% 4.8% 5.2% 5.7% 6.1% 6.6% 7.0% 7.5% 8.0% 8.4% 8.9% 9.4% 9.9% 10.3% 10.8%

10.0%

7.5%

Error

r2/r1

5.0%

2.5%

0.0% 1.0

1.2

1.4

1 .6

1 .8 r 2 /r 1

2.0

2.2

2.4

2.6

Problem 6.35

[Difficulty: 4] Part 1/2

Problem 6.35

[Difficulty: 4] Part 2/2

Problem 6.36

[Difficulty: 4]

Given:

x component of velocity field

Find:

y component of velocity field; acceleration at several points; estimate radius of curvature; plot streamlines

Solution: 3

Λ = 2⋅

The given data is

The basic equation (continuity) is

∂ ∂x

u +

m

u=−

s

∂ ∂y

(2

Λ⋅ x − y

)

2

(x2 + y2)

2

v =0

The basic equation for acceleration is

⌠ v = −⎮ ⎮ ⌡

Hence

Integrating (using an integrating factor)

v=−

⌠ ⎮ dy = −⎮ dx ⎮ ⎮ ⌡ du

(2

2⋅ Λ⋅ x⋅ x − 3⋅ y

(x2 + y2)

3

2

) dy

2⋅ Λ⋅ x⋅ y

(x2 + y2)

2

Alternatively, we could check that the given velocities u and v satisfy continuity

u=−

so

∂ ∂x

(2

Λ⋅ x − y

u +

(x2 + y2) ∂ ∂y

v =0

)

2

2

∂ ∂x

u =

(2

2⋅ Λ⋅ x⋅ x − 3⋅ y

(x2 + y2)

3

2

)

v=−

2⋅ Λ⋅ x⋅ y

(x2 + y2)

∂ 2

∂y

v =−

(2

2⋅ Λ⋅ x⋅ x − 3⋅ y

(x2 + y2)

3

2

)

For steady, 2D flow the acceleration components reduce to (after considerable math!): x - component

ax = u ⋅

∂ ∂x

u + v⋅

∂ ∂y

(

u

)

(

⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ x⋅ x2 − 3⋅ y2 ⎥ ⋅⎢ 3 ⎢ 2 2 2⎥ ⎢ 2 2 x +y ⎣ x +y ⎦⎣

ax = ⎢−

y - component

ay = u ⋅

(

∂ ∂x

)

v + v⋅

∂ ∂y

(

(

)

Evaluating at point (0,1)

Evaluating at point (0,2)

Evaluating at point (0,3)

u = 2⋅

(

)

)

m

(

(

)

v = 0⋅

s

u = 0.5⋅

m

v = 0⋅

s

u = 0.222 ⋅

(

)

2 2 2 2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x − y ⎤ 2Λ ⎥ax = − ⋅ ⋅ x ⋅⎢ ⎢ ⎥ 3 3 ⎥ ⎢ x 2 2 2⎥ ⎢ 2 2 2 2 x +y x +y +y ⎦⎣ ⎣ ⎦

⎥ ⎦

(

)

(

)

m s

v = 0⋅

)⎤⎥ + ⎡−

y = 1m

)

(

)

2 2 2 2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ y − x ⎤ 2Λ ⎥ay = − ⋅ ⋅ y ⋅⎢ ⎢ ⎥ 3 3 ⎥ ⎢ x 2 + y2 2⎥ ⎢ 2 2 2 2 x +y x +y ⎣ ⎦⎣ ⎦

⎥ ⎦

m

(

)

ax = 0 ⋅

s

(

m 2

)

s

m

ax = 0 ⋅

s

m 2

ax = 0 ⋅

s

m 2

⎛ 2⋅ m ⎞ ⎜ s ⎝ ⎠ r =

or

m 2

ay = −0.25⋅

m 2

s

ay = −0.0333⋅

s

u aradial = −ay = − r

)

s

s

m

8⋅

(

ay = −8 ⋅

m 2

s

2

The instantaneous radius of curvature is obtained from

For the three points

(

v

⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x2 − y2 ⎥ ⋅⎢ 3 ⎢ 2 2 2⎥ ⎢ 2 2 x +y ⎣ x +y ⎦⎣

ay = ⎢−

)⎤⎥ + ⎡−

r= −

u

2

ay

2

r = 0.5 m

m

2

s

y = 2m

⎛ 0.5⋅ m ⎞ ⎜ s⎠ ⎝ r = 0.25⋅

2

r = 1m

m

2

s

y = 3m

⎛ 0.2222⋅ m ⎞ ⎜ s⎠ ⎝ r = 0.03333 ⋅

m

2

r = 1.5⋅ m

2

s

The radius of curvature in each case is 1/2 of the vertical distance from the origin. The streamlines form circles tangent to the x axis

2⋅ Λ⋅ x⋅ y

−

(x2 + y2) = 2⋅x⋅y = = 2 2 dx u (x2 − y2) Λ⋅ (x − y ) − 2 (x2 + y2) dy

The streamlines are given by

2

v

(2

)

2

−2 ⋅ x ⋅ y ⋅ dx + x − y ⋅ dy = 0

so

This is an inexact integral, so an integrating factor is needed

R=

First we try

F=e

Then the integrating factor is

(

)

⎡d 2 2 d ( −2⋅ x ⋅ y)⎤ = − 2 ⋅⎢ x − y − ⎥ −2 ⋅ x ⋅ y ⎣dx y dy ⎦ 1

⌠ ⎮ 2 ⎮ − dy y ⎮ ⌡

=

1 y

(2

2

) ⋅dy = 0

2

The equation becomes an exact integral

x x −y −2 ⋅ ⋅ dx + 2 y y

So

2 ⌠ x x u = ⎮ −2 ⋅ dx = − + f ( y) ⎮ y y ⌡

ψ=

Comparing solutions

x

and

2

y

+y

(1)

(x2 − y2) dy = − x2 − y + g(x)

⌠ ⎮ u=⎮ ⎮ ⌡

y

2

2

3.50 5.29 4.95 4.64 4.38 4.14 3.95 3.79 3.66 3.57 3.52 3.50

3.75 5.42 5.10 4.82 4.57 4.35 4.17 4.02 3.90 3.82 3.77 3.75

y

2

x + y = ψ⋅ y = const ⋅ y

or

These form circles that are tangential to the x axis, as can be shown in Excel:

x values

The stream function can be evaluated using Eq 1

2.50 2.25 2.00 1.75 1.50 1.25 1.00 0.75 0.50 0.25 0.00

0.10 62.6 50.7 40.1 30.7 22.6 15.7 10.1 5.73 2.60 0.73 0.10

0.25 25.3 20.5 16.3 12.5 9.25 6.50 4.25 2.50 1.25 0.50 0.25

See next page for plot:

0.50 13.0 10.6 8.50 6.63 5.00 3.63 2.50 1.63 1.00 0.63 0.50

0.75 9.08 7.50 6.08 4.83 3.75 2.83 2.08 1.50 1.08 0.83 0.75

1.00 7.25 6.06 5.00 4.06 3.25 2.56 2.00 1.56 1.25 1.06 1.00

1.25 6.25 5.30 4.45 3.70 3.05 2.50 2.05 1.70 1.45 1.30 1.25

1.50 5.67 4.88 4.17 3.54 3.00 2.54 2.17 1.88 1.67 1.54 1.50

1.75 5.32 4.64 4.04 3.50 3.04 2.64 2.32 2.07 1.89 1.79 1.75

2.00 5.13 4.53 4.00 3.53 3.13 2.78 2.50 2.28 2.13 2.03 2.00

2.25 5.03 4.50 4.03 3.61 3.25 2.94 2.69 2.50 2.36 2.28 2.25

y values 2.50 5.00 4.53 4.10 3.73 3.40 3.13 2.90 2.73 2.60 2.53 2.50

2.75 5.02 4.59 4.20 3.86 3.57 3.32 3.11 2.95 2.84 2.77 2.75

3.00 5.08 4.69 4.33 4.02 3.75 3.52 3.33 3.19 3.08 3.02 3.00

3.25 5.17 4.81 4.48 4.19 3.94 3.73 3.56 3.42 3.33 3.27 3.25

4.00 5.56 5.27 5.00 4.77 4.56 4.39 4.25 4.14 4.06 4.02 4.00

4.25 5.72 5.44 5.19 4.97 4.78 4.62 4.49 4.38 4.31 4.26 4.25

4.50 5.89 5.63 5.39 5.18 5.00 4.85 4.72 4.63 4.56 4.51 4.50

4.75 6.07 5.82 5.59 5.39 5.22 5.08 4.96 4.87 4.80 4.76 4.75

5.00 6.25 6.01 5.80 5.61 5.45 5.31 5.20 5.11 5.05 5.01 5.00

Problem 6.37

[Difficulty: 4] Part 1/2

Problem 6.37

[Difficulty: 4 ] Part 2/2

Problem 6.38

Given:

Water at speed 25 ft/s

Find:

Dynamic pressure in in. Hg

[Difficulty: 1]

Solution: Basic equations

p dynamic =

1 2

2

⋅ ρ⋅ V

p = ρHg⋅ g ⋅ ∆h = SGHg⋅ ρ⋅ g ⋅ ∆h

2

Hence

∆h =

∆h =

2

ρ⋅ V

2⋅ SGHg⋅ ρ⋅ g 1 2

× ⎛⎜ 25⋅

⎝

ft ⎞ s

⎠

=

V

2⋅ SG Hg⋅ g

2

×

1 13.6

×

s

2

32.2⋅ ft

×

12⋅ in 1⋅ ft

∆h = 8.56⋅ in

Problem 6.39

Given:

Air speed of 100 km/hr

Find:

Dynamic pressure in mm water

[Difficulty: 1]

Solution: Basic equations

Hence

p dynamic =

∆h =

1

2

⋅ ρair⋅ V

2

p = ρw⋅ g ⋅ ∆h

ρair V2 ⋅ ρw 2 ⋅ g 1.23⋅

kg 3

m

∆h = 999 ⋅

kg 3

m

×

1 2

× ⎛⎜ 100 ⋅

⎝

km ⎞ hr

⎠

2

2

×

2

2

⎛ 1000⋅ m ⎞ × ⎛ 1 ⋅ hr ⎞ × s ⎜ ⎜ 9.81⋅ m ⎝ 1⋅ km ⎠ ⎝ 3600⋅ s ⎠

∆h = 48.4⋅ mm

Problem 6.40

Given:

Air speed

Find:

Plot dynamic pressure in mm Hg

[Difficulty: 2]

Solution: p dynamic =

Basic equations

1 2

2

⋅ ρair⋅ V

kg ρw = 999 ⋅ 3 m

Available data

1

Hence

2

kg ρair = 1.23⋅ 3 m

SG Hg = 13.6

2

⋅ ρair⋅ V = SGHg⋅ ρw⋅ g ⋅ ∆h

V( ∆h) =

Solving for V

p = ρHg⋅ g ⋅ ∆h = SGHg⋅ ρw⋅ g ⋅ ∆h

2 ⋅ SG Hg⋅ ρw⋅ g ⋅ ∆h ρair

250

V (m/s

200

150

100

50

0

50

100

150

Dh (mm)

200

250

Problem 6.41

Given:

Velocity of automobile

Find:

Estimates of aerodynamic force on hand

[Difficulty: 2]

Solution: The basic equation is the Bernoulli equation (in coordinates attached to the vehicle)

p atm +

1 2

2

⋅ ρ⋅ V = p stag

where V is the free stream velocity For air

ρ = 0.00238 ⋅

slug ft

3

We need an estimate of the area of a typical hand. Personal inspection indicates that a good approximation is a square of sides 9 cm and 17 cm A = 9 ⋅ cm × 17⋅ cm

A = 153 ⋅ cm

2

Hence, for p stag on the front side of the hand, and p atm on the rear, by assumption,

(

)

1 2 F = p stag − p atm ⋅ A = ⋅ ρ⋅ V ⋅ A 2 (a)

V = 30⋅ mph 2

ft ⎞ ⎛ 22⋅ ⎜ 1 1 slug s 2 2 F = ⋅ ρ⋅ V ⋅ A = × 0.00238 ⋅ × ⎜ 30⋅ mph⋅ × 153 ⋅ cm × 15⋅ mph ⎠ 3 2 2 ⎝ ft (b)

⎛ 1 ⋅ ft ⎞ ⎜ 12 ⎜ ⎝ 2.54⋅ cm ⎠

2

⎛ 1 ⋅ ft ⎞ ⎜ 12 ⎜ ⎝ 2.54⋅ cm ⎠

2

F = 0.379 ⋅ lbf

V = 60⋅ mph 2

ft ⎞ ⎛ 22⋅ ⎜ 1 1 slug s 2 2 F = ⋅ ρ⋅ V ⋅ A = × 0.00238 ⋅ × ⎜ 60⋅ mph⋅ × 153 ⋅ cm × 15 ⋅ mph 3 2 2 ⎝ ⎠ ft

F = 1.52⋅ lbf

These values pretty much agree with experience. However, they overestimate a bit as the entire front of the hand is not at stagnation pressure - there is flow around the had - so the pressure is less than stagnation over most of the surface.

Problem 6.42

Given:

Air jet hitting wall generating pressures

Find:

Speed of air at two locations

[Difficulty: 2]

Solution: Basic equations

p ρair

2

V

+

p ρair = Rair⋅ T

+ g ⋅ z = const

2

∆p = ρHg⋅ g ⋅ ∆h = SG Hg⋅ ρ⋅ g ⋅ ∆h

Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline Available data

For the air

R = 287 ⋅

J

T = −10 °C

kg⋅ K

kg

ρ = 999 ⋅

p = 200 ⋅ kPa

3

SG Hg = 13.6

m

p ρair = R⋅ T

ρair = 2.65

kg 3

m

Hence, applying Bernoulli between the jet and where it hits the wall directly p atm ρair

Hence

∆h = 25⋅ mm

+

Vj

2

=

2

p wall

p wall =

ρair

p wall = SGHg⋅ ρ⋅ g ⋅ ∆h =

Vj =

hence

ρair⋅ Vj

ρair⋅ Vj

2

(working in gage pressures)

2

2

Vj =

so

2

2 × 13.6 × 999 ⋅

kg 3

m

×

1

2 ⋅ SGHg⋅ ρ⋅ g ⋅ ∆h ρair

3

⋅

m

2.65 kg

× 9.81⋅

m 2

× 25⋅ mm ×

s

1⋅ m

m Vj = 50.1 s

1000⋅ mm

Repeating the analysis for the second point

∆h = 5 ⋅ mm

p atm ρair

+

Vj 2

2

=

p wall ρair

2

Hence

V =

2

+

V

2

V=

2

Vj −

2 ⋅ p wall ρair

=

2

Vj −

2 ⋅ SG Hg⋅ ρ⋅ g ⋅ ∆h ρair

3

⎛ 50.1⋅ m ⎞ − 2 × 13.6 × 999 ⋅ kg × 1 ⋅ m × 9.81⋅ m × 5 ⋅ mm × 1 ⋅ m ⎜ 3 2 s⎠ 2.65 kg 1000⋅ mm ⎝ m s

V = 44.8

m s

Problem 6.43

[Difficulty: 2]

Problem 6.44

[Difficulty: 2]

Given:

Wind tunnel with inlet section

Find:

Dynamic and static pressures on centerline; compare with Speed of air at two locations

Solution: Basic equations

p dyn =

1 2

2

⋅ ρair⋅ U

p 0 = p s + p dyn

p ρair = Rair⋅ T

∆p = ρw⋅ g ⋅ ∆h

p atm = 101⋅ kPa

kg h 0 = −10⋅ mm ρw = 999⋅ 3 m

p s = −1.738 kPa

hs =

Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline Available data

T = −5 °C

m J U = 50R ⋅ = 287⋅ s kg⋅ K

For air

p atm ρair = R⋅ T

ρair = 1.31

p dyn =

1 2

kg 3

m 2

⋅ ρair⋅ U

Also

p 0 = ρw⋅ g ⋅ h 0

and

p 0 = p s + p dyn

p dyn = 1.64⋅ kPa p 0 = −98.0 Pa so

(gage)

p s = p 0 − p dyn

(gage) ∂

Streamlines in the test section are straight so

In the curved section

∂

∂n

p =0

and

2

V p = ρair⋅ R ∂n

so

p w < p centerline

p w = p centerline

ps ρw⋅ g

h s = −177 mm

Problem 6.45

[Difficulty: 2]

Problem 6.46

[Difficulty: 2]

Problem 6.47 4.123

[Difficulty: 4]

Problem 6.48

[Difficulty: 2]

Given:

Flow in pipe/nozzle device

Find:

Gage pressure needed for flow rate; repeat for inverted

Solution: Basic equations

p

Q = V⋅ A

ρ

2

+

V

2

+ g ⋅ z = const

Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline Available data

From continuity

D1 = 1 ⋅ in

D2 = 0.5⋅ in

Q = V1 ⋅ A1 = V2 ⋅ A2

ft V2 = 30⋅ s

z2 = 10⋅ ft

A2

⎛ D2 ⎞ V1 = V2 ⋅ ⎜ ⎝ D1 ⎠

V1 = V2 ⋅ A1

ρ = 1.94⋅

ft

or

2

ft V1 = 7.50 s

Hence, applying Bernoulli between locations 1 and 2 p1 ρ

+

V1

2

2

+0=

p2 ρ

+

V2

2

2

Solving for p 1 (gage)

⎛⎜ V 2 − V 2 ⎞ 1 2 p 1 = ρ⋅ ⎜ + g ⋅ z2 2 ⎝ ⎠

When it is inverted

z2 = −10⋅ ft

⎛⎜ V 2 − V 2 ⎞ 1 2 p 2 = ρ⋅ ⎜ + g ⋅ z2 2 ⎝ ⎠

+ g ⋅ z2 = 0 +

V2 2

slug

2

+ g ⋅ z2working in gage pressures

p 1 = 10.0⋅ psi

p 2 = 1.35⋅ psi

3

Problem 6.49

[Difficulty: 2]

Problem 6.50

Given:

Siphoning of gasoline

Find:

Flow rate

[Difficulty: 2]

Solution: Basic equation

p ρgas

2

+

V

2

+ g ⋅ z = const

Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline Hence, applying Bernoulli between the gas tank free surface and the siphon exit p atm ρgas

=

p atm ρgas

Hence

V=

The flow rate is then

Q = V⋅ A =

2

V

+

2

− g⋅ h

where we assume the tank free surface is slowly changing so V tank