Unformatted text preview: Problem 6.1 [2] Given: Velocity field Find: Acceleration of particle and pressure gradient at (1,1) Solution:
NOTE: Units of B are s1 not ft1s1
Basic equations (2 ax = u⋅ ∂
∂x u + v⋅ ) − B⋅ x 2 u ( x , y) = A⋅ y − x For this flow v ( x , y) = 2⋅ A⋅ x⋅ y + B⋅ y ( ∂ ) ( ( 2 ax = ( B + 2⋅ A⋅ x) ⋅ A⋅ x + B⋅ x + A⋅ y
ay = u⋅ ∂
∂x v + v⋅ ) ( ) 2
2
2
2
2
2
∂
∂
u = ⎡A⋅ y − x − B⋅ x⎤ ⋅ ⎡A⋅ y − x − B⋅ x⎤ + ( 2⋅ A⋅ x⋅ y + B⋅ y) ⋅ ⎡A⋅ y − x − B⋅ x⎤
⎣
⎦⎣
⎦
⎣
⎦
∂y
∂x
∂y ( ∂ ) 2 ) 2
2
∂
∂
v = ⎡A⋅ y − x − B⋅ x⎤ ⋅ ( 2⋅ A⋅ x⋅ y + B⋅ y) + ( 2⋅ A⋅ x⋅ y + B⋅ y) ⋅ ( 2⋅ A⋅ x⋅ y + B⋅ y)
⎣
⎦
∂y
∂x
∂y ( ) 2
2
ay = ( B + 2⋅ A⋅ x) ⋅ ( B⋅ y + 2⋅ A⋅ x⋅ y) − 2⋅ A⋅ y⋅ ⎡B⋅ x + A⋅ x − y ⎤
⎣
⎦ ( ) 1
2
2 ft
ax = ( 1 + 2⋅ 1⋅ 1) ⋅ × 1⋅ 1 + 1⋅ 1 + 1⋅ 1 ⋅
s
s Hence at (1,1) ( ft
ax = 9⋅
2
s
ft
ay = 7⋅
2
s ) 1
ft
1
2
2 ft
ay = ( 1 + 2⋅ 1⋅ 1) ⋅ × ( 1⋅ 1 + 2⋅ 1⋅ 1⋅ 1) ⋅ − 2⋅ 1⋅ 1⋅ × ⎡1⋅ 1 + 1⋅ 1 − 1 ⎤ ⋅
⎣
⎦
s
s
s
s
a= 2 ax + ay ⎛ ay ⎞
⎟
⎝ ax ⎠ 2 θ = atan ⎜ a = 11.4⋅ ft θ = 37.9⋅ deg 2 s For the pressure gradient
lbf ∂
∂x p = ρ⋅ gx − 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⋅ gy − 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 [2] Given: Velocity field Find: Acceleration of particle and pressure gradient at (0.7,2) Solution:
Basic equations u ( x , y) = A⋅ x − B⋅ y For this flow ax = u⋅ ay = u⋅ ∂
∂x
∂
∂x u + v⋅ v + v⋅ ∂
∂y
∂
∂y v ( x , y) = −A⋅ y u = ( A⋅ x − B⋅ y) ⋅ v = ( A⋅ x − B⋅ y) ⋅ ∂
∂x ( A⋅ x − B⋅ y) + ( −A⋅ y) ⋅ ∂
∂x ( − A ⋅ y) + ( − A ⋅ y) ⋅ ∂
∂y ∂
∂y ( A⋅ x − B⋅ y) 2 ax = A ⋅ x
2 ( − A ⋅ y) ay = A ⋅ y 2 1⎞
ax = ⎛ ⎟ × 0.7⋅ m
⎜
⎝ s⎠ Hence at (0.7,2) m
ax = 0.7
2
s 2 1⎞
ay = ⎛ ⎟ × 2⋅ m
⎜
⎝ s⎠
a= 2 ax + ay 2 m
ay = 2
2
s ⎛ ay ⎞
⎟
⎝ ax ⎠ θ = atan ⎜ a = 2.12 m θ = 70.7⋅ deg 2 s For the pressure gradient
2 kg
m N⋅ s
× 0.7⋅ ×
p = ρ⋅ gx − ρ⋅ ax = −1000⋅
3
2 kg⋅ m
∂x
m
s ∂ ∂
∂x
2 kg
m N⋅ s
× ( −9.81 − 2) ⋅ ×
p = ρ⋅ gy − ρ⋅ ay = 1000⋅
3
2 kg⋅ m
∂y
m
s ∂ ∂
∂y p = −700⋅ Pa
kPa
= −0.7⋅
m
m p = −11800⋅ Pa
kPa
= −11.8⋅
m
m Problem 6.3 [2] Problem 6.4 [2] Problem 6.5 [2] Given: Velocity field Find: Acceleration of particle and pressure gradient at (1,1) Solution:
Basic equations (2 2 u ( x , y) = A⋅ x − y For this flow ax = u⋅ ∂
∂x u + v⋅ ∂
∂y ) − 3⋅ B⋅ x v ( x , y) = −2⋅ A⋅ x⋅ y + 3⋅ B⋅ y (2 u = ⎡A⋅ x − y
⎣ 2 ) − 3⋅ B⋅ x⎤ ⋅ ∂
⎦ ∂x + ( −2⋅ A⋅ x⋅ y + 3⋅ B⋅ y) ⋅ ( 2 2 ax = ( 2⋅ A⋅ x − 3⋅ B) ⋅ A⋅ x − 3⋅ B⋅ x + A⋅ y
ay = u⋅ ∂
∂x v + v⋅ ∂
∂y (2 ( ) 2
2
∂⎡
⎣A⋅ x − y − 3⋅ B⋅ x⎤
⎦
∂y ) ) − 3⋅ B⋅ x⎤ ⋅ ∂
⎦ v = ⎡A⋅ x − y
⎣ ⎡A⋅ ( x2 − y2) − 3⋅ B⋅ x⎤ ...
⎣
⎦ 2 ∂x ( −2⋅ A⋅ x⋅ y + 3⋅ B⋅ y) + ( −2⋅ A⋅ x⋅ y + 3⋅ B⋅ y) ⋅ (2 ay = ( 3⋅ B⋅ y − 2⋅ A⋅ x⋅ y) ⋅ ( 3⋅ B − 2⋅ A⋅ x) − 2⋅ A⋅ y⋅ ⎡A⋅ x − y
⎣ ( ) − 3⋅ B⋅ x⎤
⎦ ∂
∂y ( −2⋅ A⋅ x⋅ y + 3⋅ B⋅ y) 2 ) 1
2
2 ft
ax = ( 2⋅ 1⋅ 1 − 3⋅ 1) ⋅ × 1⋅ 1 − 3⋅ 1⋅ 1 + 1⋅ 1 ⋅
s
s Hence at (1,1) ( ft
ax = 1⋅
2
s
ft
ay = 7⋅
2
s ) 1
ft
1
ft
2
2
ay = ( 3⋅ 1⋅ 1 − 2⋅ 1⋅ 1⋅ 1) ⋅ × ( 3⋅ 1 − 2⋅ 1⋅ 1) ⋅ − 2⋅ 1⋅ 1⋅ × ⎡1⋅ 1 − 1 − 3⋅ 1⋅ 1⎤ ⋅
⎣
⎦
s
s
s
s
a= 2 ax + ay ⎛ ay ⎞
⎟
⎝ ax ⎠ 2 θ = atan ⎜ a = 7.1⋅ ft θ = 81.9⋅ deg 2 s For the pressure gradient
lbf ∂
∂x p = ρ⋅ gx − ρ⋅ ax = −2⋅ slug
ft 3 × 1⋅ ft
2 s 2 × lbf ⋅ s
slug⋅ ft ∂
∂x 2 p = −2⋅ ft = −0.0139⋅ ft psi
ft lbf ∂
∂y p = ρ⋅ gy − 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 6.6 [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 Check for incompressible flow 1
s ω = 1⋅
∂
∂x
∂ Hence ∂x u+ u+ 1
s ∂
∂y
∂
∂y ρ = 2⋅ kg
3 u = A⋅ x⋅ sin ( 2⋅ π⋅ ω⋅ t) 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⋅ ∂
∂y
∂
∂y The total acceleration is then 2 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 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 Local 12.6⋅ m and 2 m −12.6⋅ s
Total 12.6⋅ Local m and 2 and 2 12.6⋅ and 2 12.6⋅ s
t = 1⋅ s Local 12.6⋅ 12.6⋅ m
2 s 2 m Convective 2 0⋅ m and 2 0⋅ s m
2 s m
2 s m and 2 −12.6⋅ s
Total 0⋅ s s m −12.6⋅ and 2 m −12.6⋅ s
Total m s m −12.6⋅ 0⋅ s s
t = 0.5⋅ s Convective 2 m Convective 2 s m and 2 −12.6⋅ s 0⋅ m and 2 0⋅ s m
2 s m
2 s The governing equation (assuming inviscid flow) for computing the pressure gradient is (6.1) Hence, the components of pressure gradient (neglecting gravity) are
∂
∂x
∂
∂y
Evaluated at (1,1) and time p = − ρ⋅ Du
Dt ∂ p = − ρ⋅ Dv
Dt ∂ ∂x ∂x ( 2 p = −ρ⋅ 2⋅ π⋅ A⋅ ω⋅ x⋅ cos ( 2⋅ π⋅ ω⋅ t) + A ⋅ x⋅ sin ( 2⋅ π⋅ ω⋅ t) ( ) 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. −25.1⋅ t = 0.5⋅ s x comp. 25.1⋅ t = 1⋅ s x comp. −25.1⋅ Pa
m Pa
m
Pa
m Pa
m y comp. 25.1⋅ y comp. −25.1⋅ y comp. 25.1⋅ Pa
m Pa
m ) 2 Problem 6.7 [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⋅
ay = u⋅ ∂
∂x
∂
∂x ∂x ∂
∂y ∂
∂y u = A⋅ x⋅ v = A⋅ x⋅ ∂
∂x
∂
∂x 2 Hence at (2,1) 2 ax + ay v =0 so ( A⋅ x) + ( −A⋅ y) ⋅ ∂ 2 ∂y ( − A ⋅ y) + ( − A ⋅ y) ⋅ 2 ( A⋅ x) = A ⋅ x ∂
∂y ax = A ⋅ x
2 ( − A ⋅ y) ay = A ⋅ y 2 2⎞
ax = ⎛ ⎟ × 2⋅ m
⎜
⎝ s⎠
a= ∂y ⌠
⌠
⎮∂
⎮
v ( x , y) = −⎮
u dy = −⎮ A dy = −A⋅ y + c
⌡
∂x
⌡ is the simplest y component of velocity u + v⋅ v + v⋅ u+ ∂ 2⎞
ay = ⎛ ⎟ × 1⋅ m
⎜
⎝ s⎠ m
ax = 8
2
s ⎛ ay ⎞
⎟
⎝ ax ⎠ 2 θ = atan ⎜ m
ay = 4
2
s
a = 8.94 m θ = 26.6⋅ deg 2 s For the pressure gradient
2 kg
m N⋅ s
× 8⋅ ×
p = ρ⋅ gx − ρ⋅ ax = −1.50⋅
3
2 kg⋅ m
∂x
m
s ∂ 2 ∂ ∂ ∂x kg
m N⋅ s
× 4⋅ ×
p = ρ⋅ gy − ρ⋅ ay = −1.50⋅
3
2 kg⋅ m
∂y
m
s ∂ ∂
∂z p = ρ⋅ gz − ρ⋅ az = 1.50 × For the pressure on the x axis
1
22
p ( x) = p0 − ⋅ ρ⋅ A ⋅ x
2 kg
3 × ( −9.81) ⋅ m
dp = ∂
∂x p ( x) = 190⋅ kPa − 2 s
x p m ∂y
2 × N⋅ s
kg⋅ m ∂
∂y
x ( p = −12⋅ p = −6⋅ Pa
m Pa
m p = −14.7⋅ Pa
m ) ⌠
⌠
1
2
22
p − p0 = ⎮ ρ⋅ gx − ρ⋅ ax dx = ⎮ −ρ⋅ A ⋅ x dx = − ⋅ ρ⋅ A ⋅ x
⌡0
2
⌡0
1
2 ⋅ 1.5⋅ ( ) 2 2 2⎞
N⋅ s
2
×⎛ ⎟ ×
×x
⎜
3 ⎝ s⎠
kg⋅ m
m
kg p ( x) = 190 − 3
1000 2 ⋅x (p in kPa, x in m) Problem 6.8 [3] Given: Velocity field Find: Expressions for velocity and acceleration along wall; plot; verify vertical components are zero; plot pressure gradient Solution: q = 2⋅ The given data is u= m
s 3 h = 1⋅ m m kg ρ = 1000⋅ 3 m
q⋅ x 2⋅ π⎡x + ( y − h) ⎦
⎣
2 2⎤ + q⋅ x q⋅ ( y − h) v= 2⎤ 2⋅ π⎡x + ( y + h) ⎦
⎣
2 2⎤ 2⋅ π⎡x + ( y − h) ⎦
⎣
2 + q⋅ ( y + h)
2
2
2⋅ π⎡x + ( y + h) ⎤
⎣
⎦ The governing equation for acceleration is For steady, 2D flow this reduces to (after considerable math!) x  component y  component ∂
∂x u + v⋅ u=− ∂y v + v⋅ ∂
∂y )2 − h2⋅ (h2 − 4⋅ y2)⎤
⎦ 2 q ⋅ x⋅ ⎣ x + y 2 ) 2 π⋅ x + h 2 ⎡x2 + ( y + h) 2⎤ ⋅ ⎡x2 + ( y − h) 2⎤ ⋅ π2
⎣
⎦⎣
⎦
2 v =− (2 ⎡ 2 q ⋅ y⋅ ⎣ x + y )2 − h 2⋅ ( h 2 + 4⋅ x2)⎤
⎦
2 22
2
2
2
π ⋅ ⎡x + ( y + h ) ⎤ ⋅ ⎡x + ( y − h ) ⎤
⎣
⎦⎣
⎦ (2 2 q⋅ x (2 ay = u⋅ ∂x (2 ⎡ 2 ∂ 2 y = 0⋅ m For motion along the wall u= ax = u⋅ ∂ v=0 (No normal velocity) ax = − 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 (6.1) Hence, the component of pressure gradient (neglecting gravity) along the wall is ∂
∂x p = − ρ⋅ Du ∂ Dt ∂x 2 p= (2 2 (2 ) 2 ρ ⋅ q ⋅ x⋅ x − h ) 2 π ⋅ x +h 3 The plots of velocity, acceleration, and pressure gradient are shown in the associated Excel workbook. 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. The velocity, acceleration and pressure gradient are given by q=
h= 2
1 ρ= 1000 m3/s/m
m
kg/m3 x (m) u (m/s) a (m/s2) dp /dx (Pa/m)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0 0.00
0.32
0.25
0.19
0.15
0.12
0.10
0.09
0.08
0.07
0.06 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 Velocity Along Wall Near A Source
0.35 u (m/s) 0.30
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 a (m/s2) 0.020
0.015
0.010
0.005
0.000
0.005 0 1 2 3 4 5 6 7 x (m) Pressure Gradient Along Wall dp /dx (Pa/m) 5
0
5 0 1 2 3 4 5 10
15
20
25
x (m) 6 7 8 Problem 6.9 [2] Problem 6.10 [2] Problem 6.11 [2] Problem 6.12 [2] Problem 6.13 [3] Given: Velocity field Find: The acceleration at several points; evaluate pressure gradient Solution:
The given data is q = 2⋅ m
s 3 K = 1⋅ m m
s 3 m ρ = 1000⋅ kg Vr = − 3 m The governing equations for this 2D flow are The total acceleration for this steady flow is then
2 Vθ ∂
∂
⋅V
ar = Vr⋅ Vr +
r ∂θ r
∂r ar = − Vθ ∂
∂
⋅V
aθ = Vr⋅ Vθ +
r ∂θ θ
∂r aθ = Evaluating at point (1,0) m
ar = −0.101
2
s m
aθ = 0.0507
2
s Evaluating at point (1,π/2) m
ar = −0.101
2
s m
aθ = 0.0507
2
s Evaluating at point (2,0) m
ar = −0.0127
2
s m
aθ = 0.00633
2
s ∂ ∂ r  component θ  component From Eq. 6.3, pressure gradient is ∂r p = −ρ⋅ ar ∂r q 23 4⋅ π ⋅ r
q⋅ K 23 4⋅ π ⋅ r 2 p= ρ⋅ q 23 4⋅ π ⋅ r 1∂
⋅ p = −ρ⋅ aθ
r ∂θ
Evaluating at point (1,0) Evaluating at point (1,π/2) Evaluating at point (2,0) 1∂
ρ ⋅ q⋅ K
⋅ p=−
23
r ∂θ
4⋅ π ⋅ r ∂
∂r
∂
∂r
∂
∂r p = 101⋅ Pa
m 1∂
Pa
⋅ p = −50.5⋅
r ∂θ
m p = 101⋅ Pa
m 1∂
Pa
⋅ p = −50.5⋅
r ∂θ
m Pa
m 1∂
Pa
⋅ p = −6.33⋅
r ∂θ
m p = 12.7⋅ q
2⋅ π⋅ r Vθ = K
2⋅ π⋅ r Problem 6.14 [3] Problem 6.15 [4] 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 Q = V⋅ A For this 1D flow Q = ui⋅ Ai = u⋅ A A = Ai − (Ai − Ae) ⋅ x Ai
u ( x) = ui⋅
= ui⋅
A so L Ai ⎡ (Ai − Ae) ⎤
⋅ x⎥
⎣L
⎦ Ai − ⎢
22 2 ( )
) Ai
⎤ A ⋅ L ⋅ ui ⋅ Ae − Ai
∂⎡
⎥= i
⋅ ⎢ui⋅
ax = u⋅ u + v⋅ u = ui⋅
3
∂x
∂y
⎡ Ai − Ae ⎤ ∂x ⎢
⎡ Ai − Ae ⎤ ⎥
Ai⋅ L + Ae⋅ x − Ai⋅ x
⋅ x⎥
⋅ x⎥ ⎥
Ai − ⎢
⎢ Ai − ⎢ L
⎣L
⎦⎣
⎣
⎦⎦
∂ For the pressure ∂
∂x ∂ ( Ai 22 p = −ρ⋅ ax − ρ⋅ gx = − 2 ) ( ( ) ) ρ⋅ Ai ⋅ L ⋅ ui ⋅ Ae − Ai (Ai⋅ L + Ae⋅ x − Ai⋅ x) 3
x and dp = ∂
∂x x
⌠
222
⎮
⌠
ρ⋅ Ai ⋅ L ⋅ ui ⋅ Ae − Ai
∂
⎮−
⎮
dx
p − pi =
p dx =
⎮
⎮ ∂x
3
Ai⋅ L + Ae⋅ x − Ai⋅ x
⌡0
⎮
⌡0 p ⋅ dx ( ( This is a tricky integral, so instead consider the following:
x Hence ( x ∂
∂x p = −ρ⋅ ax = −ρ⋅ u⋅ ) ) ∂ () 1∂
2
u = − ⋅ ρ⋅
u
2 ∂x
∂x ⌠
⌠
ρ
ρ
2
2
2
∂
∂
p − pi = ⎮
p dx = − ⋅ ⎮
u dx = ⋅ u ( x = 0) − u ( x)
⎮ ∂x
2 ⎮ ∂x
2
⌡0
⌡0 () ρ
2
2
p ( x) = pi + ⋅ ⎛ ui − u ( x) ⎞
⎠
2⎝
2
ρ⋅ ui ⎡ ⎡
p ( x) = pi +
⋅ ⎢1 − ⎢
2⎢⎢ ⎢
⎣ ( ) which we recognise as the Bernoulli equation! ⎤
⎥
⎡ (Ai − Ae) ⎤ ⎥
⋅ x⎥ ⎥
⎢ Ai − ⎢ L
⎣
⎣
⎦⎦
Ai 2⎤ ⎥
⎥
⎥
⎦ The following plots can be done in Excel Pressure Gradient (kPa/m) 30 20 10 0 0.5 1 1.5 2 1.5 2 x (m) Pressure (kPa) 250
248
246
244
242
240 0 0.5 1 x (m) Problem 6.16 [4] 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 Q = V⋅ A x
x⎞
⎛
−
−
⎜
a
2⋅ a ⎟
A ( x) = A0⋅ ⎝ 1 + e
−e
⎠ For this 1D flow Q = u0⋅ A0 = u⋅ A so A0
u ( x) = u0⋅
=
A
⎛ x
x⎞
−
−
⎜
a
2⋅ a ⎟
⎝1 + e − e
⎠ ax = u⋅ For the pressure u0 ∂
∂x ∂
∂x u + v⋅ ∂
∂y u= u0 ⋅ ∂⎡
⎢ p = −ρ⋅ ax − ρ⋅ gx = − − ρ⋅ u0 ⋅ e x
2⋅ a − x
2⋅ a ⎞
⎛ −x
⎜
⎟
2⋅ a
⋅ ⎝ 2⋅ e
− 1⎠ ⎞
⎛ −x − x
⎜a
⎟
2⋅ a
2⋅ a ⋅ ⎝ e
−e
+ 1⎠ 3 x and dp = ∂
∂x p ⋅ dx ⌠
x⎛
x
⎞
⎮
−
−
x
⎟
2
2⋅ a ⎜
2⋅ a
⎮
⌠
⋅ ⎝ 2⋅ e
− 1⎠
ρ⋅ u0 ⋅ e
∂
dx
p − pi = ⎮
p dx = ⎮ −
⎮ ∂x
3
⎮
x
x
⎞
⎛−
⌡0
⎮
⎜ a − 2⋅ a
⎟
⎮
2⋅ a ⋅ ⎝ e
−e
+ 1⎠
⌡0 This is a tricky integral, so instead consider the following: ∂
∂x ⎞
⎛ −x
⎜
⎟
2⋅ a
⋅ ⎝ 2⋅ e
− 1⎠ ⎤ u0 ⋅ e
⎥=
x ⎞⎥
3
−
⎞
⎛ −x − x
2⋅ a ⎟ ⎥
⎜a
⎟
2⋅ a
−e
⎠⎦
2⋅ a ⋅ ⎝ e
−e
+ 1⎠ u0 x
x ⎞ ∂x ⎢ ⎛
x
⎛
−
−
−
⎜
a
2⋅ a ⎟
⎢⎜1 + e a
⎝1 + e − e
⎠ ⎣⎝ 2 2 p = −ρ⋅ ax = −ρ⋅ u⋅ ∂ () 1∂
2
u = − ⋅ ρ⋅
u
2 ∂x
∂x x x ⌠
⌠
ρ
ρ
2
2
2
∂
∂
p − pi = ⎮
p dx = − ⋅ ⎮
u dx = ⋅ u ( x = 0) − u ( x)
⎮ ∂x
2 ⎮ ∂x
2
⌡0
⌡0 Hence () ρ
2
2
p ( x) = p0 + ⋅ ⎛ u0 − u ( x) ⎞
⎠
2⎝
2 ) which we recognise as the Bernoulli equation! ρ⋅ u0 ⎡
⎡
p ( x) = p0 +
⋅ ⎢1 −
2 ⎢ ⎢⎛ ⎢
⎣ ( ⎤
x
x ⎞⎥
−
−
⎢⎜
a
2⋅ a ⎟ ⎥
⎣⎝1 + e − e
⎠⎦
1 2⎤ ⎥
⎥
⎥
⎦ The following plots can be done in Excel Pressure Gradient (kPa/m) 0.1 0 2 4 6 8 10 6 8 10 6 8 10 − 0.1
− 0.2
− 0.3
− 0.4 x (m) Pressure (kPa) 200 199.9 199.8 199.7 0 2 4 x (m) Area (m2) 0.1 0.09 0.08 0.07 0 2 4 x (m) Problem 6.17 [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 D ( x) = Di + For a linear decrease in diameter From continuity Q = V⋅ A = V⋅ Hence V ( x) ⋅ or Do = 0.02⋅ m L = 0.5⋅ m
Do − Di
L π2
π2
⋅ D = Vi⋅ ⋅ Di
4
4 π
2
⋅ D ( x) = Q
4 kg ρ = 1000⋅ 3 m 3 Q = 0.00785 m
s 4⋅ Q V ( x) = Do − Di ⎞
⎛
⋅ x⎟
π⋅ ⎜ D i +
L
⎝
⎠ ⎛ Do − Di ⎞
⋅ x⎟
⎜1 +
L⋅ D i
⎝
⎠ m
s ⋅x Vi V ( x) = Vi = 1⋅ 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⋅ D i
⎝
⎠ d
⋅
2 dx ⎛ Do − Di ⎞
⋅ x⎟
⎜1 +
L⋅ D i
⎝
⎠ (
)
5
⎡ (Do − Di) ⎤
⋅x
D ⋅ L⋅ 1 +
2 Vi
2 ax ( x) = − 2⋅ V i ⋅ D o − D i i ⎢
⎣ ⎥
⎦ Di⋅ L 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) ⎤
⋅ x⎥
Di⋅ L⋅ ⎢1 +
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 m ∂
∂x p = −10⋅ To find the length L for which the absolute pressure gradient is no more than 5 MPa/m, we need to solve
2 MPa
p ≤ 5⋅
=
m
∂x ( ) 2⋅ ρ⋅ Vi ⋅ Do − Di ∂ ⎡ (Do − Di) ⎤
⋅ x⎥
Di⋅ L⋅ ⎢1 +
Di⋅ L
⎣
⎦ 5 with x = L m (the largest pressure gradient is at the outlet)
2 Hence L≥ ( ) 2⋅ ρ⋅ Vi ⋅ Do − Di ⎛ Do ⎞ Di⋅ ⎜ ⎝ Di 5 ∂
⎟⋅ p
⎠ ∂x This result is also obtained using Goal Seek in the Excel workbook L ≥ 1⋅ m MPa
m The acceleration and pressure gradient are given by
Di = 0.1 m Do =
L=
Vi = 0.02
0.5
1 m
m
m/s ρ= 1000 kg/m3 x (m) a (m/s2) dp /dx (kPa/m)
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 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 Acceleration Through A Nozzle
12000 a (m/s2) 10000
8000
6000
4000
2000
0
0.0 0.1 0.1 0.2 0.2 0.3
x (m) 0.3 0.4 0.4 0.5 0.5 0.4 0.5 0.5 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 6.18 [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 Q = V⋅ A = V⋅ Hence V ( x) ⋅ L = 1⋅ m
Do − Di
L m
s ρ = 1000⋅ m
s 4⋅ Q
Do − Di ⎞
⎛
⋅ x⎟
π⋅ ⎜ D i +
L
⎝
⎠ 2 Vi V ( x) = or ⎛ Do − Di ⎞
⋅ x⎟
⎜1 +
L⋅ D i
⎝
⎠ The governing equation for this flow is d
V=
dx Vi d
2 dx Vi ⋅ ⎛ Do − Di ⎞
⋅ x⎟
⎜1 +
L⋅ D i
⎝
⎠ ⎛ Do − Di ⎞
⋅ x⎟
⎜1 +
L⋅ D i
⎝
⎠ 2 (
)
5
⎡ (Do − Di) ⎤
⋅x
D ⋅ L⋅ 1 +
2 Hence ax ( x) = − 2⋅ V i ⋅ D o − D i i ⎢
⎣ Di⋅ L ⎥
⎦ This is plotted in the associated Excel workbook
From Eq. 6.2a, pressure gradient is ∂
∂x 3 m ⋅x Q = 0.245 V ( x) = or, for steady 1D flow, in the notation of the problem ax = V⋅ kg 3 π2
π2
⋅ D = Vi⋅ ⋅ Di
4
4 π
2
⋅ D ( x) = Q
4 Vi = 5⋅ p = −ρ⋅ ax ∂
∂x 2 p= ( ) 2⋅ ρ⋅ Vi ⋅ Do − Di ⎡ (Do − Di) ⎤
⋅ x⎥
Di⋅ L⋅ ⎢1 +
Di⋅ L
⎣
⎦ 5 This is also plotted in the associated Excel workbook. Note that the pressure gradient is adverse: separation is likely to
occur in the diffuser, and occur near the entrance 2 At the inlet ∂
∂x kPa p = 100⋅ 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
∂
∂x 2 p ≤ 25⋅ kPa
m = ( ) 2⋅ ρ⋅ Vi ⋅ Do − Di ⎡ (Do − Di) ⎤
⋅ x⎥
Di⋅ L⋅ ⎢1 +
Di⋅ L
⎣
⎦ 5 with x = 0 m (the largest pressure gradient is at the inlet)
2 Hence L≥ ( ) 2⋅ ρ⋅ Vi ⋅ Do − Di
Di⋅ ∂
∂x p This result is also obtained using Goal Seek in the Excel workbook L ≥ 4⋅ m a The acceleration and pressure gradient are given by
Di = 0.25 m Do =
L=
Vi = 0.75
1
5 m
m
m/s ρ= 1000 kg/m3 x (m) a (m/s2) dp /dx (kPa/m)
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00 100
62.1
40.2
26.9
18.59
13.17
9.54
5.29
3.125
1.940
1.256
0.842
0.581
0.412 100
62.1
40.2
26.93
18.59
13.17
9.54
5.29
3.125
1.940
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 6.19 [4] Problem 6.20 [4] Problem 6.20 Problem 6.21 [5] Problem 6.22 [4] Part 1/2 Problem 6.19 cont'd Problem 6.22 [4] Part 2/2 Problem 6.23 [5] Problem 6.24 [2] Problem 6.25 Given: Velocity field for doublet Find: [2] Expression for pressure gradient Solution:
Basic equations For this flow Vr ( r , θ) = − Λ
2 ⋅ cos ( θ) r
Hence for r momentum Vθ ( r , θ) = − Λ
2 ⋅ sin ( θ) Vz = 0 r 2⎞
⎛
Vθ ∂
Vθ ⎟
⎜∂
⋅ V−
ρ⋅ gr − p = ρ⋅ ⎜ Vr⋅ Vr +
r ∂θ r
r⎟
∂r
⎝ ∂r
⎠ ∂ Ignoring gravity ⎡
⎢
⎢Λ
Λ
∂
∂
p = −ρ⋅ ⎢⎛ − ⋅ cos ( θ)⎞ ⋅ ⎛ − ⋅ cos ( θ)⎞ +
⎜2
⎟ ∂r ⎜ 2
⎟
∂r
⎣⎝ r
⎠ ⎝r
⎠
For θ momentum ρ⋅ gθ − ⎛ − Λ ⋅ sin ( θ)⎞
⎜2
⎟
⎝r
⎠ ⋅ ∂ ⎛ − Λ ⋅ cos ( θ)⎞ −
⎜
⎟
r
∂θ r2
⎝
⎠ 2
⎛ − Λ ⋅ sin ( θ)⎞ ⎤
⎜2
⎟⎥
⎝r
⎠⎥
⎥
r
⎦ ∂
∂r 2 p= 2⋅ Λ ⋅ ρ Vθ ∂
Vr⋅ Vθ ⎞
⎛∂
1∂
⋅ p = ρ⋅ ⎜ Vr⋅ Vθ +
⋅ Vθ +
⎟
r ∂θ
r⎠
r ∂θ
⎝ ∂r Ignoring gravity ⎡
⎢
∂
⎢⎛ Λ
∂⎛ Λ
p = −r⋅ ρ⋅ ⎜ − ⋅ cos ( θ)⎞ ⋅ ⎜ − ⋅ sin ( θ)⎞ +
⎟ ∂r 2
⎟
⎢2
∂θ
⎣⎝ 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 5 r p=0 Problem 6.26 [2] Problem 6.27 [2] Problem 6.28 [3] Given: Velocity field for free vortex flow in elbow Find: Similar solution to Example 6.1; find k (above) Solution:
Basic equation ∂
∂r p= ρ V
r 2 c
V = Vθ =
r with Assumptions: 1) Frictionless 2) Incompressible 3) free vortex
For this flow p ≠ p ( θ) Hence ⌠
⎮
Δp = p2 − p1 = ⎮
⎮
⌡r r2 1 2 ∂ so ∂r p= 2 ρ⋅ c
ρ⋅ V
d
=
p=
3
r
dr
r 2
2
2
2
ρ⋅ c ⋅ ⎛ r2 − r1 ⎞
ρ⋅ c ⎛ 1
1⎞
⎝
⎠
dr =
⋅⎜
−
⎟=
3
2
2
22
2 ⎜r
r
r2 ⎟
2⋅ r1 ⋅ r2
⎝1
⎠
2 ρ⋅ c (1) Next we obtain c in terms of Q
r r 2
⌠ →→ ⌠2
⌠ w⋅ c
⎛ r2 ⎞
⎮
⎮
dr = w⋅ c⋅ ln ⎜ ⎟
Q = ⎮ V dA = ⎮ V⋅ w dr =
⎮
r
⌡r
⌡
⎝ r1 ⎠
1
⌡r
1 Hence c= Q ⎛ r2 ⎞
⎟
⎝ r1 ⎠ w⋅ ln ⎜ Using this in Eq 1 Δp = p2 − p1 = 2
2
2
ρ⋅ c ⋅ ⎛ r2 − r1 ⎞
⎝
⎠
2 2⋅ r1 ⋅ r2 2 Solving for Q 2 2 = 2⋅ r1 ⋅ r2
⎛ r2 ⎞
Q = w⋅ ln ⎜ ⎟ ⋅
⋅ Δp
2
2
⎝ r1 ⎠ ρ⋅ ⎛ r2 − r1 ⎞
⎝
⎠ 2
2
2
ρ⋅ Q ⋅ ⎛ r2 − r1 ⎞
⎝
⎠
2 ⎛ r2 ⎞ 2 2
2⋅ w ⋅ ln ⎜ ⎟ ⋅ r1 ⋅ r2
⎝ r1 ⎠
2 2 2 2⋅ r1 ⋅ r2
⎛ r2 ⎞
k = w⋅ ln ⎜ ⎟ ⋅
2
2
⎝ r1 ⎠ ρ⋅ ⎛ r2 − r1 ⎞
⎝
⎠ Problem 6.29 From Example 6.1: or From Problem 6.28: Eq. 1 Eq. 2 Instead of plotting as a function of inner radius we plot as a function of r 2/r1
Eq. 1
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 Eq. 2
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 Error
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
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 5.0% 2.5% 0.0%
1.0 1.2 1.4 1.6 1.8
r2/r1 2.0 2.2 2.4 2.6 Problem 6.30 [3] Part 1/2 Problem 6.30 [3] Part 2/2 Problem 6.31 [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 2 3 u ( x , y) = A⋅ x + B⋅ x⋅ y
∂
∂x
∂
∂x u ( x , y) + u ( x , y) + ∂
∂y
∂
∂y v ( x , y) = ∂
∂x ( A⋅ x3 + B⋅ x⋅ y2) + ∂ ( A⋅ y3 + B⋅ x2⋅ y) = 0
∂y (2 3 2 Hence for ax ax = u⋅ u + v⋅ 2 ∂ ay = u⋅ ∂
∂x 3 ∂y ( ( ∂
∂y 2 3 ( (2 v + v⋅ 2 3 ) (A⋅ x3 − 3⋅ A⋅ x⋅ y2) + (A⋅ y3 − 3⋅ A⋅ x2⋅ y)⋅ ∂ (A⋅ x3 − 3⋅ A⋅ x⋅ y2)
∂x ) (A⋅ y3 − 3⋅ A⋅ x2⋅ y) + (A⋅ y3 − 3⋅ A⋅ x2⋅ y)⋅ ∂ (A⋅ y3 − 3⋅ A⋅ x2⋅ y) 2∂ v = A⋅ x − 3⋅ A⋅ x⋅ y ⋅ ) 2 ∂x ∂y 2 2 2
0.2 ⎞
2
2
ay = 3⋅ ⎛
× 1⋅ m × ⎡( 2⋅ m) + ( 1⋅ m) ⎤
⎣
⎦
⎜2⎟
⎝ m ⋅s ⎠
2 ∂y )2 2 ax + ay 2 2∂ 2
0.2 ⎞
2
2
ax = 3⋅ ⎛
× 2⋅ m × ⎡( 2⋅ m) + ( 1⋅ m) ⎤
⎣
⎦
⎜2⎟
⎝ m ⋅s ⎠ a= B = −0.6 2 ay = 3⋅ A ⋅ y⋅ x + y
Hence at (2,1) B = − 3⋅ A Hence v ( x , y) = A⋅ y − 3⋅ A⋅ x ⋅ y u = A⋅ x − 3⋅ A⋅ x⋅ y ⋅ ax = 3⋅ A ⋅ x⋅ x + y
For ay )=0 1
2 m ⋅s u ( x , y) = A⋅ x − 3⋅ A⋅ x⋅ y
∂x 2 v ( x , y) = ( 3⋅ A + B) ⋅ x + y We can write ∂ 2 v ( x , y) = A⋅ y + B⋅ x ⋅ y 2 m
ax = 6.00⋅
2
s
m
ay = 3.00⋅
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
V ⎛ A⋅ y3 − 3⋅ A⋅ x2⋅ y ⎞
⎛ v⎞
⎟
θvel = atan ⎜ ⎟ = atan ⎜
⎜ A⋅ x3 − 3⋅ A⋅ x⋅ y2 ⎟
⎝ u⎠
⎝
⎠ r
a ⎛ 1 − 3⋅ 2 ⋅ 1 ⎞
⎟
θvel = atan ⎜
⎜ 23 − 3⋅ 2⋅ 12 ⎟
⎝
⎠ θvel = −79.7⋅ deg ⎛ ay ⎞
θaccel = atan ⎜ ⎟
⎝ ax ⎠ ⎛ 1⎞
θaccel = atan ⎜ ⎟
⎝ 2⎠ θaccel = 26.6⋅ deg Hence the angle between the acceleration and velocity vectors is Δθ = θaccel − θvel Δθ = 106⋅ deg The component of acceleration normal to the velocity is then m
an = a⋅ sin ( Δθ) = 6.71⋅ ⋅ sin ( 106⋅ deg)
2
s m
an = 6.45⋅
2
s 3 At (1,2) the acceleration vector is at
angle 2 Δθ Problem 6.32 [4] Part 1/2 Problem 6.32 [4] Part 2/2 Problem 6.33 [4] Part 1/2 Problem 6.33 [4] Part 2/2 Problem 6.34 [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 m
s Λ = 2⋅ The given data is The governing equation (continuity) is ∂
∂x ∂ u+ ∂y Hence v=− (2 ) 2 Λ⋅ x − y (x2 + y2)2 v =0 (2 ⌠
⎮
du
dy = −⎮
dx
⎮
⎮
⌡ ⌠
v = −⎮
⎮
⌡ Integrating (using an integrating factor) u=− 2 2⋅ Λ⋅ x⋅ x − 3⋅ y ( x2 + y2) 3 ) dy 2⋅ Λ⋅ x⋅ y (x2 + y2)2 Alternatively, we could check that the given velocities u and v satisfy continuity u=− (2 ) 2 Λ⋅ x − y (x2 + y2)2 so ∂
∂x
∂
∂x (2 u+ (x2 + y2)3
∂
∂y v=0 The governing equation for acceleration is For steady, 2D flow this reduces to (after considerable math!)
x  component ax = u⋅ ∂
∂x u + v⋅ ∂
∂y ) 2 2⋅ Λ⋅ x⋅ x − 3⋅ y u= u v=− 2⋅ Λ⋅ x⋅ y ∂ (x2 + y2)2 ∂y v=− (2 (x2 + y2)3 ) 2 2⋅ Λ⋅ x⋅ x − 3⋅ y ⎡ Λ⋅ (x2 − y2)⎤ ⎡ 2⋅ Λ⋅ x⋅ (x2 − 3⋅ y2)⎤ ⎡ 2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ (3⋅ x2 − y2)⎤
⎥⋅⎢
⎥ + ⎢−
⎥
⎥⋅⎢
⎢ ( 2 2)2 ⎥ ⎢ ( 2 2)3 ⎥ ⎢ ( 2 2)2⎥ ⎢ ( 2 2)3 ⎥
x +y
x +y
⎣ x +y ⎦ ⎣
⎦ ⎣ x +y ⎦ ⎣
⎦ ax = ⎢− y  component ay = u⋅ ∂
∂x v + v⋅ ∂
∂y ⎡ Λ⋅ (x2 − y2)⎤ ⎡ 2⋅ Λ⋅ y⋅ (3⋅ x2 − y2)⎤ ⎡ 2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ (3⋅ y2 − x2)⎤
⎥⋅⎢
⎥ + ⎢−
⎥
⎥⋅⎢
⎢ ( 2 2)2 ⎥ ⎢ ( 2 2)3 ⎥ ⎢ ( 2 2)2⎥ ⎢ ( 2 2)3 ⎥
x +y
x +y
⎣ x +y ⎦ ⎣
⎦ ⎣ x +y ⎦ ⎣
⎦
m
s u = 2⋅ Evaluating at point (0,2) u = 0.5⋅ Evaluating at point (0,3) u = 0.222⋅ (x2 + y2)3 2 ay = − 2⋅ Λ ⋅ y (x2 + y2)3 v = 0⋅ m
ax = 0⋅
2
s m
ay = −8⋅
2
s v = 0⋅ m
s m
ax = 0⋅
2
s m
ay = −0.25⋅
2
s v = 0⋅ m
s m
ax = 0⋅
2
s m
ay = −0.0333⋅
2
s m
s m
s m
s 2 2 u
The instantaneous radius of curvature is obtained from aradial = −ay = −
r For the three points 2⋅ Λ ⋅ x v ay = ⎢− Evaluating at point (0,1) 2 ax = − ⎛ 2⋅ m ⎞
⎜
⎟
s⎠
r=⎝ y = 1m 8⋅ or r=− u
ay 2 r = 0.5 m m 2 s ⎛ 0.5⋅ m ⎞
⎜
⎟
s⎠
r=⎝ y = 2m 0.25⋅ 2 r = 1m m 2 s ⎛ 0.2222⋅ m ⎞
⎜
⎟
s⎠
r=⎝ y = 3m 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 = 2⋅ x⋅ y
2
2
(x2 − y2)
Λ⋅ ( x − y )
−
(x2 + y2)2 The streamlines are given by dy
v
=
=
dx
u so −2⋅ x⋅ y⋅ dx + x − y ⋅ dy = 0 (2 ) 2 This is an inexact integral, so an integrating factor is needed First we try Then the integrating factor is R= ( ) 1
2
⎤
⎡d 2 2 d
⋅ ⎢ x − y − ( −2⋅ x⋅ y)⎥ = −
−2⋅ x⋅ y ⎣dx
y
dy
⎦
⌠
⎮ − 2 dy
y
⎮
⌡ F=e 1 = 2 y (2 ) 2 The equation becomes an exact integral x
x −y
−2⋅ ⋅ dx +
⋅ dy = 0
2
y
y So 2
⌠
x
x
u = ⎮ −2⋅ dx = − + f ( y)
⎮
y
y
⌡ and 2 Comparing solutions ψ= x
+y
y or (x2 − y2) dy = − x2 − y + g (x) ⌠
⎮
u=⎮
⎮
⌡
2 2 y 2 x + y = ψ⋅ y = const⋅ y These form circles that are tangential to the x axis, as shown in the associated Excel workbook y x values This function is computed and plotted below 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 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 2.75
5.00 5.02
4.53 4.59
4.10 4.20
3.73 3.86
3.40 3.57
3.13 3.32
2.90 3.11
2.73 2.95
2.60 2.84
2.53 2.77
2.50 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 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 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 6.35 [4] Part 1/2 Problem 6.35 [4] Part 2/2 Problem 6.36 [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 22 ∂
∂x
∂
∂x u ( x , y) + u ( x , y) + ∂
∂y
∂
∂y v ( x , y) = ) (3 4 u ( x , y) = A⋅ x − 6⋅ x ⋅ y + y For this flow ( ) ) 3 v ( x , y) = B⋅ x ⋅ y − x⋅ y ( ) 4
22
4
3
3
∂⎡
∂
⎣A⋅ x − 6⋅ x ⋅ y + y ⎤ + ⎡B⋅ x ⋅ y − x⋅ y ⎤ = 0
⎦
⎣
⎦
∂x
∂y (3 ) + 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 B = −8 1
3 m ⋅s
Hence for ax
ax = u⋅ ∂
∂x u + v⋅ ∂
∂y (4 ) 4∂ 22 u = A⋅ x − 6⋅ x ⋅ y + y ⋅
2 (2 ∂x ⎡A⋅ (x4 − 6⋅ x2⋅ y2 + y4)⎤ + ⎡−4⋅ A⋅ (x3⋅ y − x⋅ y3)⎤ ⋅ ∂ ⎡A⋅ (x4 − 6⋅ x2⋅ y2 + y4)⎤
⎣
⎦⎣
⎦⎣
⎦
∂y )3 2 ax = 4⋅ A ⋅ x⋅ x + y
For ay
ay = u⋅ ∂
∂x v + v⋅ ∂
∂y (4
2 (2 Let dy
v
=
dx
u u= y
x so ) 2 ay = 4⋅ A ⋅ y⋅ x + y
For a streamline ) 4∂ 22 v = A⋅ x − 6⋅ x ⋅ y + y ⋅ ∂x ⎡−4⋅ A⋅ (x3⋅ y − x⋅ y3)⎤ + ⎡−4⋅ A⋅ (x3⋅ y − x⋅ y3)⎤ ⋅ ∂ ⎡−4⋅ A⋅ (x3⋅ y − x⋅ y3)⎤
⎣
⎦⎣
⎦⎣
⎦
∂y 3 dy
=
dx (3 (4 ) 3 −4⋅ A⋅ x ⋅ y − x⋅ y
22 ) 4 A⋅ x − 6⋅ x ⋅ y + y =− (3 ) 3 4⋅ x ⋅ y − x⋅ y (x4 − 6⋅ x2⋅ y2 + y4) 1⎞
⎛ y⎞
d⎛ ⎟
d⎜ ⎟
⎜
du
1 dy
x
1 dy
y
x
so
= ⎝ ⎠= ⋅
+ y⋅ ⎝ ⎠ = ⋅
−
dx
dx
dx
x dx
x dx x2 dy
du
= x⋅
+u
dx
dx (3 ) ( 3 ) ( 2 ) 2 dy
du
4⋅ x ⋅ y − x⋅ y
4⋅ 1 − u
4⋅ 1 − u
= x⋅
+u = −
=−
u+
4
22
4
dx
dx
⎛ 1 − 6⋅ u + u3⎞
⎛ 1 − 6⋅ u + u3⎞
x − 6⋅ x ⋅ y + y
⎜
⎟
⎜
⎟
u
⎝
⎠
⎝u
⎠ ( Hence ( ) (4 ) ⋅ du 1
5
3
ln ( x) = − ⋅ ln u − 10⋅ u + 5⋅ u + C
5 ⎡
⎤
du
4⋅ 1 − u
⎢
⎥ = − u⋅ u − 10⋅ u + 5
= − u+
⎢ ⎛1
4
2
dx
3⎞ ⎥
u − 6⋅ u + 1
⎢ ⎜ u − 6⋅ u + u ⎟ ⎥
⎣⎝
⎠⎦ x⋅ dx
=−
x Separating variables ) 4 2 u⋅ (u 4 ( 2 u − 6⋅ u + 1
− 10⋅ u 2 + 5) (u5 − 10⋅ u3 + 5⋅ u)⋅ x5 = c
5 32 2 5 ) 32 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 2 V
an = −
R To find the radius of curvature we use V
an R= or We need to find the component of acceleration normal to the velocity vector (3 ) r
V ⎡ 4⋅ x ⋅ y − x⋅ y
v⎞
θvel = atan ⎛ ⎟ = atan ⎢−
⎜
⎢ x4 − 6⋅ x2⋅ y2 + y4
⎝ u⎠
⎣ ( At (1,2) the velocity vector is at angle 3 ⎤
⎥
)⎥
⎦ r
a
Δθ 4⋅ ( 2 − 8) ⎤
θvel = atan ⎡−
⎢
⎥
⎣ 1 − 24 + 16⎦ θvel = −73.7⋅ deg 3⎤
⎡2
⎢ 4⋅ A ⋅ y⋅ (x2 + y2) ⎥
⎛ ay ⎞
⎛ y⎞
θaccel = atan ⎜ ⎟ = atan ⎢
⎥ = atan ⎜ ⎟
3
ax
⎝ x⎠
⎝⎠
⎢ 4⋅ A2⋅ x⋅ (x2 + y2) ⎥
⎣
⎦ At (1,2) the acceleration vector is at
angle 2⎞
θaccel = atan ⎛ ⎟
⎜
⎝ 1⎠ θaccel = 63.4⋅ deg Hence the angle between the acceleration and velocity vectors is Δθ = θaccel − θvel Δθ = 137⋅ deg The component of acceleration normal to the velocity is then an = a⋅ sin ( Δθ) a= 2 (2 2 2 At (1,2) )3 = 500⋅ m7 × A2 = 500⋅ m7 × ⎛
⎜ (2 2 ax = 4⋅ A ⋅ x⋅ x + y
ay = 4⋅ A ⋅ y⋅ x + y
a= 2 a = 4472 V 2 an m an = a⋅ sin ( Δθ) 2 s ) = −14⋅ m (3 4 u = A⋅ x − 6⋅ x ⋅ y + y R= 2 2⎞
m
= 4000⋅
3⎟
2
s
⎝ m ⋅s ⎠ s Then 2 2 )3 = 1000⋅ m7 × A2 = 1000⋅ m7 × ⎛
⎜ 2m
2 22 2 ax + ay 2⎞
m
= 2000⋅
3⎟
2
s
⎝ m ⋅s ⎠ 2000 + 4000 ⋅ (4 where s R = ⎛ 50⋅
⎜ ⎝ m⎞ 2 ⎟× s⎠ ) = 48⋅ m 3 v = B⋅ x ⋅ y − x⋅ y s m
an = 3040
2
s
V= 2 2 1s
⋅
3040 m 2 u + v = 50⋅ R = 0.822 m m
s Problem 6.37 Given: Water at speed 10 ft/s Find: [1] Dynamic pressure in in. Hg Solution:
1
2
⋅ ρ⋅ V
2 Basic equation pdynamic = Hence Δh = ρ⋅ V
V
=
2⋅ SGHg⋅ ρ⋅ g
2⋅ SGHg⋅ g Δh = 1 ⎛ ft ⎞
1
s
12⋅ in
× ⎜ 10⋅ ⎟ ×
×
×
2⎝
s⎠
13.6 32.2⋅ ft
1⋅ ft 2 p = ρHg⋅ g⋅ Δh = SGHg⋅ ρ⋅ g⋅ Δh
2 2 2 Δh = 1.37⋅ in Problem 6.38 [1] Problem 6.39 Given: Velocity of automobile Find: [1] Estimates of aerodynamic force on hand Solution:
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 2 A = 153 cm The governing equation is the Bernoulli equation (in coordinates attached to the vehicle)
patm + 1
2
⋅ ρ⋅ V = pstag
2 where V is the free stream velocity
Hence, for pstag on the front side of the hand, and patm on the rear, by assumption, ( ) 1
2
F = pstag − patm ⋅ A = ⋅ ρ⋅ V ⋅ A
2 (a) V = 30⋅ mph
2 ft ⎞
⎛
⎛ 1 ⋅ ft ⎞
22⋅ ⎟
⎜
⎜ 12 ⎟
1
1
slug
s
2
2
F = ⋅ ρ⋅ V ⋅ A = × 0.00238⋅
× ⎜ 30⋅ mph⋅
⎟ × 153⋅ cm × ⎜
⎟
3
2
2
15⋅ mph ⎠
⎝
⎝ 2.54⋅ cm ⎠
ft
(b) 2 F = 0.379 lbf V = 60⋅ mph
2 ft ⎞
⎛
⎛ 1 ⋅ ft ⎞
22⋅ ⎟
⎜
⎜ 12 ⎟
1
1
slug
s
2
2
F = ⋅ ρ⋅ V ⋅ A = × 0.00238⋅
× ⎜ 60⋅ mph⋅
⎟ × 153⋅ cm × ⎜
⎟
3
2
2
15⋅ mph ⎠
⎝
⎝ 2.54⋅ cm ⎠
ft 2 F = 1.52 lbf Problem 6.40 Given: Air jet hitting wall generating pressures Find: [2] Speed of air at two locations Solution:
Basic equation 2 p V
+ g⋅ z = const
2 ρair + ρair = p
Rair⋅ T Δp = ρHg⋅ g⋅ Δh = SGHg⋅ ρ⋅ g⋅ Δh Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the jet and where it hits the wall directly
patm
ρair
For air 2 + Vj
2 ρair = 14.7⋅ 2 pwall = pwall = ρair
2 lbf
2 × 144⋅ in
1⋅ ft in × 2 ρair⋅ Vj
2 (working in gage pressures) lbm⋅ R
1⋅ slug
1
×
×
53.33⋅ ft⋅ lbf 32.2⋅ lbm ( 50 + 460) ⋅ R
2 Hence pwall = SGHg⋅ ρ⋅ g⋅ Δh = Hence Vj = 2 × 13.6 × 1.94⋅ ρair⋅ Vj
2 slug
ft 3 × Vj = so 1 ρair = 2.42 × 10 ft 2⋅ SGHg⋅ ρ⋅ g⋅ Δh
ρair 3 ft
ft
1ft
× 32.2⋅ × 0.15⋅ in ×
− 3 slug
2
12⋅ in
2.42 × 10
s
⋅ Vj = 93.7 ft
s Repeating the analysis for the second point
patm
ρair 2 + Vj
2 = pwall
ρair
2 Hence V= 2 + V
2 2 V= − 3 slug
3 Vj − 2⋅ pwall
ρair = 2 Vj − 2⋅ SGHg⋅ ρ⋅ g⋅ Δh
ρair 3 1
ft
ft
1ft
⎛ 93.7⋅ ft ⎞ − 2 × 13.6 × 1.94⋅ slug ×
⋅
× 32.2⋅ × 0.1⋅ in ×
⎜
⎟
3
− 3 slug
2
s⎠
12⋅ in
⎝
ft
2.42 × 10
s V = 54.1 ft
s Problem 6.41 [2] Problem 6.42 [2] Problem 6.43 [2] Problem 6.44 [2] Problem 6.45
4.123 [4] Problem 6.46 [2] Problem 6.47 [2] Problem 6.48 Given: Siphoning of gasoline Find: [2] Flow rate Solution:
Basic equation p
ρgas 2 + V
+ g⋅ z = const
2 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
patm
ρgas = patm
ρgas Hence V= The flow rate is then Q = V⋅ A = 2 + V
− g⋅ h
2 where we assume the tank free surface is slowly changing so Vtank <<,
and h is the difference in levels 2⋅ g ⋅ h
2 π⋅ D
⋅ 2⋅ g⋅ h
4
2 Q= π
1⋅ ft
2
× ( 1⋅ in) ×
×
2
4
144⋅ in 2 × 32.2 ft
2 s × 1
⋅ ft
2 3 Q = 0.0309 ft
s Q = 13.9 gal
min Problem 6.49 Given: Ruptured pipe Find: [2] Pressure in tank Solution:
Basic equation p
ρben 2 + V
+ g⋅ z = const
2 Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the pipe and the rise height of the benzene
ppipe
ρben = patm
ρben + g⋅ h where we assume Vpipe <<, and h is the rise height Hence ppipe = ρben⋅ g⋅ h = SGben⋅ ρ⋅ g⋅ h From Table A.2 SGben = 0.879 Hence pben = 0.879 × 1.94⋅ slug
ft 3 × 32.2⋅ where ppipe is now the gage pressure ft
2 s 2 × 25⋅ ft × lbf⋅ s
slugft
⋅ pben = 1373 lbf
ft 2 pben = 9.53 psi (gage) Problem 6.50 Given: Ruptured Coke can Find: [2] Pressure in can Solution:
Basic equation p
ρCoke 2 + V
+ g⋅ z = const
2 Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the coke can and the rise height of the coke
pcan
ρCoke = patm
ρCoke + g⋅ h where we assume VCoke <<, and h is the rise height Hence pCoke = ρCoke⋅ g⋅ h = SGCoke⋅ ρ⋅ g⋅ h From a web search SGDietCoke = 1 Hence pDiet = 1 × 1.94⋅ slug
ft Hence 3 SGRegularCoke = 1.11 × 32.2⋅ pRegular = 1.11 × 1.94⋅ ft
2 2 × 20⋅ in × s
slug
ft 3 where ppipe is now the gage pressure × 32.2⋅ ft
2 s 1⋅ ft
lbf⋅ s
×
⋅
12⋅ in slugft pDiet = 104⋅ lbf
ft 2 × 20⋅ in × 1⋅ ft
lbf⋅ s
×
⋅
12⋅ in slugft pDiet = 0.723⋅ psi 2 pRegular = 116⋅ lbf
ft 2 (gage) pRegular = 0.803⋅ psi (gage) Problem 6.51 Given: Flow rate through siphon Find: [2] Maximum height h to avoid cavitation Solution:
2 Basic equation pV
+
+ g⋅ z = const
ρ
2 Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
From continuity V= 3 Q
4⋅ Q
=
2
A
π⋅ D V= 2 4
ft
1 ⎞ ⎛ 12⋅ in ⎞
× 0.7⋅
×⎛
⎜
⎟ ×⎜
⎟
s ⎝ 2⋅ in ⎠
π
⎝ 1⋅ ft ⎠ 2 V = 32.1 ft
s Hence, applying Bernoulli between the free surface and point A
patm
ρ = 2 pA + g⋅ h + ρ V
2 where we assume VSurface <<
2 Hence V
pA = patm − ρ⋅ g⋅ h − ρ⋅
2 pv = 0.363⋅ psi From the steam tables, at 70oF the vapor pressure is
This is the lowest permissible value of pA
2 Hence Hence V
pA = pv = patm − ρ⋅ g⋅ h − ρ⋅
2
h = ( 14.7 − 0.363) ⋅ lbf
2 in ×⎛
⎜ 12⋅ in ⎞ 2 ⎟× ⎝ 1⋅ ft ⎠ or
3 h=
2 patm − pv
ρ⋅ g − V 2 2⋅ g
2 2 ft
s
slug⋅ ft 1 ⎛
ft
s
×
×
− × ⎜ 32.18 ⎞ ×
⎟
32.2⋅ ft
1.94 slug 32.2⋅ ft lbf ⋅ s2 2 ⎝
s⎠
1 ⋅ h = 17.0 ft Problem 6.52 [2] h1 = H= (h2) Given: Flow through tankpipe system Find: Velocity in pipe; Rate of discharge Solution:
2 Basic equation pV
+
+ g⋅ z = const
ρ
2 Δp = ρ⋅ g⋅ Δh Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the free surface and the manometer location
patm
ρ 2 = p
V
− g⋅ H +
ρ
2 where we assume VSurface <<, and H = 4 m
2 Hence V
p = patm + ρ⋅ g⋅ H − ρ⋅
2 For the manometer p − patm = SGHg⋅ ρ⋅ g⋅ h2 − ρ⋅ g⋅ h1 Combining equations ρ ⋅ g ⋅ H − ρ⋅ Hence V= Note that we have water on one side and mercury on
the other of the manometer 2 V
= SGHg⋅ ρ⋅ g⋅ h2 − ρ⋅ g⋅ h1
2 2 × 9.81⋅ m
2 or V= ( × ( 4 − 13.6 × 0.15 + 0.75) ⋅ m V = 7.29 s
The flow rate is Q = V⋅ π⋅ D
4 ) 2⋅ g⋅ H − SGHg⋅ h2 + h2 2 Q= π
m
2
× 7.29⋅ × ( 0.05⋅ m)
4
s m
s
3 Q = 0.0143 m
s Problem 6.53 [2] Problem 6.54 [2] Problem 6.55 Given: Air flow over a wing Find: [2] Air speed relative to wing at a point Solution:
2 Basic equation pV
+
+ g⋅ z = const
ρ
2 p = ρ⋅ R ⋅ T Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the upstream point (1) and the point on the wing (2)
p1
ρ 2 + Hence V2 = For air ρ= V1
2 2 p2 V2
=
+
2
ρ
2 V1 + 2⋅ V= ( p1 − p2 )
ρ
kg⋅ K
1
3N
×
×
2 286.9⋅ N ⋅ m ( 4 + 273) ⋅ K p
R⋅ T ρ = ( 75 + 101) × 10 ⋅ ρ = 2.21 m 2 Then where we ignore gravity effects m 3 ⎛ 60⋅ m ⎞ + 2 × m × ( 75 − 3) × 103⋅ N × kg⋅ m
⎜
⎟
2
2
2.21⋅ kg
⎝ s⎠
m
N ⋅s NOTE: At this speed, significant density changes will occur, so this result is not very realistic kg V = 262 m
s 3 Problem 6.56 [2] Problem 6.57 Given: Flow through fire nozzle Find: [2] Maximum flow rate Solution:
2 Basic equation pV
+
+ g⋅ z = const
ρ
2 Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ 2 + 2 V1 p2 V2
=
+
2
ρ 2 where we ignore gravity effects
2 But we have Q = V1⋅ A1 = V1⋅ π⋅ D
π⋅ d
= V2⋅ A2 =
4
4 4 2
2d
V2 − V2 ⋅ ⎛ ⎞ =
⎜⎟
⎝ D⎠ Hence V2 = ( 2⋅ p1 − p2 ⎡ ⎛
ρ⋅ ⎢1 − ⎜ ( 2⋅ p2 − p1
ρ 2 ) ) 4
d⎞ ⎤ 3 V2 = 2× 2 ⎟⎥
⎝ D⎠ ⎦ ⎣ Then d
V 1 = V 2⋅ ⎛ ⎞
⎜⎟
D⎠
⎝ so 2 ft
lbf ⎛ 12⋅ in ⎞
× ( 100 − 0) ⋅
×⎜
⎟×
2 ⎝ 1⋅ ft ⎠
1.94⋅ slug
in 2 π⋅ d
Q = V2⋅
4 1 ⎛ 1⎞
1−⎜ ⎟
⎝ 3⎠ 1
Q=
× 124⋅ × ⎛ ⋅ ft⎞
⎜
⎟
4
s ⎝ 12 ⎠
π ft 2 3 × slugft
⋅ V2 = 124⋅ 2 lbf⋅ s 3 ft
Q = 0.676⋅
s Q = 304⋅ ft
s gal
min Problem 6.58 [2] Problem 6.59 [2] Problem 6.60 Given: Velocity field for plane doublet Find: [3] Pressure distribution along x axis; plot distribution Solution:
3 The given data is m
s Vr = − From Table 6.1 ρ = 1000⋅ Λ Λ = 3⋅ Vθ = − 2 kg p0 = 100⋅ kPa 3 m ⋅ cos ( θ) r Λ
2 ⋅ sin ( θ) r where Vr and Vθ are the velocity components in cylindrical coordinates (r,θ). For points along the x axis, r = x, θ = 0, Vr = u and
Vθ = v = 0
u=− Λ v=0 2 x The governing equation is the Bernoulli equation
p12
+ ⋅ V + g⋅ z = const
ρ2
so (neglecting gravity) V= where p 12
+ ⋅ u = const
ρ2 Apply this to point arbitrary point (x,0) on the x axis and at infinity
At
At point (x,0) x→
u=− u→ 0 p → p0 Λ
2 x
Hence the Bernoulli equation becomes
p0
ρ 2 = p
Λ
+
4
ρ
2⋅ x or The plot of pressure is shown in the associated Excel workbook p ( x) = p0 − ρ⋅ Λ 2 4 2⋅ x 2 2 u +v The given data is
Λ= 3 m3/s ρ=
p0 = 1.5
100 kg/m3
kPa x (m) p (Pa)
99.892
99.948
99.972
99.984
99.990
99.993
99.995
99.997
99.998
99.998
99.999
99.999
99.999
99.999
99.999
100.000 Pressure Distribution Along x axis
100.0 p (kPa) 0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0 100.0 99.9 99.9 99.8
0.0 0.2 0.4 0.6 0.8 1.0
x (m) 1.2 1.4 1.6 1.8 2.0 Problem 6.61 [3] Given: Velocity field Find: Pressure distribution along wall; plot distribution; net force on wall Solution:
The given data is q = 2⋅ u= m
s 3 h = 1⋅ m m kg ρ = 1000⋅ 3 m
q⋅ x + 2⎤ 2⋅ π⎡x + ( y − h) ⎦
⎣
2 q⋅ x v= 2⎤ 2⋅ π⎡x + ( y + h) ⎦
⎣
2 q ⋅ (y − h )
2⎤ 2⋅ π⎡x + ( y − h ) ⎦
⎣
2 + q ⋅ (y + h )
2
2
2⋅ π⎡x + ( y + h ) ⎤
⎣
⎦ The governing equation is the Bernoulli equation
p12
+ ⋅ V + g⋅ z = const
ρ2 2 V= where u +v 2 Apply this to point arbitrary point (x,0) on the wall and at infinity (neglecting gravity)
At
At point (x,0) x →0
u= ( u→ 0
q⋅ x
2 V→ 0 v=0 ) 2 π⋅ x + h v→ 0 V= q⋅ x (2 patm Hence the Bernoulli equation becomes = ρ p 1⎡
q⋅ x
⎤
+ ⋅⎢
2⎥
ρ 2 π x2
⎣ ⋅ +h ⎦ ( ) 2 π⋅ x + h
2 ) ρ
q⋅ x
⎤
p ( x) = − ⋅ ⎡
⎢
2
2⎥
2 π⋅ x + h
⎣
⎦ ( or (with pressure expressed as gage pressure) 2 ) (Alternatively, the pressure distribution could have been obtained from Problem 6.8, where the momentum equation
was used to find the pressure gradient ∂
∂x 2 p= (2 2 (2 ) along the wall. Integration of this with respect to x 2 ρ ⋅ q ⋅ x⋅ x − h ) 2 π ⋅ x +h 3 leads to the same result for p(x))
The plot of pressure is shown in the associated Excel workbook. From the plot it is clear that the wall experiences a
negative gage pressure on the upper surface (and zero gage pressure on the lower), so the net force on the wall is upwards,
towards the source
10⋅ h The force per width on the wall is given by 10⋅ h ⌠
F=⎮
pupper − plower dx
⌡− 10⋅ h ( ) 2⌠
2
ρ⋅ q ⎮
x
⋅
dx
F=−
2⎮
2
2
2
2⋅ π ⎮
x +h
⌡− 10⋅ h ( ) The integral is ⌠
⎮
⎮
⎮
⎮
⌡ 2 x (x2 + h2) 2 dx → x⎞
atan ⎛ ⎟
⎜
⎝ h⎠
2⋅ h x − 2 2 2⋅ h + 2⋅ x 2 so F=− ⎛ 10 + atan ( 10)⎞
⎟
⎠
2⋅ π ⋅ h ⎝ 101
ρ⋅ q 2 ⋅ ⎜− 2 2
⎛ m2 ⎞
1
⎛ 10 + atan ( 10)⎞ × N⋅ s
F=−
× 1000⋅
× ⎜ 2⋅ ⎟ ×
× ⎜−
⎟
2
3 ⎝ s⎠
1⋅ m ⎝ 101
⎠ kg⋅ m
2⋅ π
m 1 kg F = −278 N
m The given data is
m3/s/m
m
ρ = 1000 kg/m3 q=
h= 2
1 x (m) p (Pa)
0.00
50.66
32.42
18.24
11.22
7.49
5.33
3.97
3.07
2.44
1.99 Pressure Distribution Along Wall
0
0 1 2 3 4 5 10
p (Pa) 0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0 20
30
40
50
60
x (m) 6 7 8 9 10 Problem 6.62 [3] Rx Given: Flow through fire nozzle Find: Maximum flow rate Solution:
2 pV
+
+ g⋅ z = const
ρ
2 Basic equation Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ 2 + 2 V1 p2 V2
=
+
2
ρ 2 where we ignore gravity effects
2 Q = V1⋅ A1 = V1⋅ But we have π⋅ D
π⋅ d
= V2⋅
4
4 ( 2⋅ p2 − p1
ρ 4 2
2d
V2 − V2 ⋅ ⎛ ⎞ =
⎜⎟
⎝ D⎠ V2 = Hence ( 2⋅ p1 − p2 ⎡ 2 d
V1 = V2⋅ ⎛ ⎞
⎜⎟
D⎠
⎝ so ) 2 ) 4
⎛d⎞ ⎤
⎟⎥
⎝D⎠ ⎦ ρ⋅ ⎢1 − ⎜ ⎣ 3 V2 = 2× m
3N
× ( 700 − 0) × 10 ⋅
×
2
1000⋅ kg
m
2 1 ⎛ 25 ⎞
⎟
⎝ 75 ⎠ π⋅ d
Q = V2⋅
4 From x momentum Rx + p1⋅ A1 = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 Hence Rx = −p1⋅ ( π
4 ) × 37.6⋅ ( m
s × × ( 0.025⋅ m) ) kg⋅ m 2 3 Q = 0.0185⋅ m
s Q = 18.5⋅ L
s using gage pressures 2
2
2
⎡
π⋅ D
π⋅ D
d⎤
+ ρ⋅ Q⋅ V2 − V1 = −p1⋅
+ ρ⋅ Q⋅ V2⋅ ⎢1 − ⎛ ⎞ ⎥
⎜⎟
4
4
⎣ ⎝D⎠ ⎦ ( ) 3
3
2
π
kg
m
m⎡
25 ⎞ ⎤ N⋅ s
3N
2
× ⋅ ( 0.075⋅ m) + 1000⋅
× 0.0185⋅
× 37.6⋅ × ⎢1 − ⎛ ⎟ ⎥ ×
⎜
24
3
s
s ⎣ ⎝ 75 ⎠ ⎦ kg⋅ m Rx = −700 × 10 ⋅ m m
s V2 = 37.6 2 N⋅ s 1−⎜ Then Q= 4 Rx = −2423 N m This is the force of the nozzle on the fluid; hence the force of the fluid on the nozzle is 2400 N to the right; the nozzle is in tension Problem 6.63 [3] Problem 6.64 [3] Problem 6.65 [3] Given: Flow through reducing elbow Find: Mass flow rate in terms of Δp, T1 and D1 and D2 Solution:
2 Basic equations: pV
+
+ g⋅ z = const
ρ
2 Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Ignore elevation change 6) p2 = patm
3 Available data: From contnuity Q = 20⋅ gpm V1 = Q = 0.0446 Q V1 = 3.63 ⎛ π⋅ D
⎜
⎟
⎝4⎠ 2⎞ ft
s D = 1.5⋅ in d = 0.5⋅ in slug
ft ft
s V2 = p1 Hence, applying Bernoulli between the inlet (1) and exit (2) ρ
ρ⎛ 2
2
⋅ V − V1 ⎞
⎠
2⎝ 2 Q ⎛ π⋅ d
⎜
⎟
⎝4⎠
2 + V2 = 32.7 2⎞ V1
2 = p2
ρ 3 ft
s 2 + V2
2 or, in gage pressures p1g = From xmomentum Rx + p1g⋅ A1 = u1⋅ −mrate + u2⋅ mrate = −mrate⋅ V1 = −ρ⋅ Q⋅ V1 ( ρ = 1.94⋅ p1g = 7.11 psi ) ( ) because u1 = V1 u2 = 0 2 π⋅ D
Rx = −p1g⋅
− ρ⋅ Q ⋅ V 1
4
The force on the supply pipe is then Rx = −12.9 lbf
K x = −R x Kx = 12.9 lbf on the pipe to the right Problem 6.66 [2] Given: Flow nozzle Find: Mass flow rate in terms of Δp, T1 and D1 and D2 Solution:
2 Basic equation pV
+
+ g⋅ z = const
ρ
2 Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ
But we have 2 + 2 V1 p2 V2
=
+
2
ρ 2 Q = V1⋅ A1 = V1⋅ where we ignore gravity effects π⋅ D 1 2 4 = V 2⋅ π⋅ D 2 2 ⎛ D2 ⎞
V 1 = V 2⋅ ⎜
⎟
⎝ D1 ⎠ so 4 2 Note that we assume the flow at D2 is at the same pressure as the entire section 2; this will be true if there is turbulent mixing
Hence 2 2 ⎛ D2 ⎞ 4 V2 − V2 ⋅ ⎜
V2 = ⎟=
⎝ D1 ⎠ ( 2⋅ p1 − p2 ( 2⋅ p2 − p1
ρ ) ) ⎡ ⎛ D ⎞ 4⎤
2⎥
⎢
ρ⋅ 1 − ⎜
⎟⎥
⎢
⎣ ⎝ D1 ⎠ ⎦
2 Then the mass flow rate is mflow = ρ⋅ V2⋅ A2 = ρ⋅ π⋅ D 2
4 ⋅ ( 2⋅ p1 − p2 For a flow nozzle p = ρ⋅ R ⋅ T mflow = k⋅ Δp mflow = where 2 4⎤ ⎡ ⎛D ⎞
2⎥
⎢
ρ⋅ 1 − ⎜
⎟
⎢
D1 ⎥
⎣ ⎝ ⎠⎦
2 Using ) π⋅ D 2 2⋅ 2 ⋅ π⋅ D 2 = 2⋅ 2 ⋅ Δp⋅ ρ ⎡ ⎛ D ⎞ 4⎤
2⎥
⎢
⎢1 − ⎜ D ⎟ ⎥
⎣ ⎝ 1⎠ ⎦ Δp⋅ p1 ⎡ ⎛ D ⎞ 4⎤
2⎥
⎢
R⋅ T1⋅ 1 − ⎜
⎟⎥
⎢
⎣ ⎝ D1 ⎠ ⎦
k= π⋅ D 2 2 2⋅ 2 ⋅ p1 ⎡ ⎛ D ⎞ 4⎤
2⎥
⎢
R ⋅ T 1⋅ 1 − ⎜
⎟⎥
⎢
⎣ ⎝ D1 ⎠ ⎦ We can expect the actual flow will be less because there is actually significant loss in the device. Also the flow will experience a vena co
that the minimum diameter is actually smaller than D2. We will discuss this device in Chapter 8. Problem 6.67 Given: Flow through branching blood vessel Find: [4] Blood pressure in each branch; force at branch Solution:
2 Basic equation ∑Q=0 pV
+
+ g⋅ z = const
ρ
2 Q = V⋅ A Δp = ρ⋅ g⋅ Δh CV Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
For Q3 we have ∑ Q = −Q 1 + Q 2 + Q 3 = 0 Q3 = Q1 − Q2 so Q3 = 1.5⋅ L
min CV We will need each velocity
V1 = Similarly V2 = Q1
A1 4⋅ Q1 = 3 4
L
0.001⋅ m
1⋅ min ⎛ 1 ⎞
V1 =
× 4⋅
×
×
×⎜
⎟
1⋅ L
π
min
60⋅ s ⎝ 0.01⋅ m ⎠ 2 π⋅ D 1 4⋅ Q2 V2 = 0.943 2 π⋅ D 2 m V3 = s 4⋅ Q3
2 π⋅ D 3 2 V1 = 0.849 V3 = 5.09 m
s m
s Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ 2 + V1
2 = p2
ρ 2 V2 + where we ignore gravity effects 2 ρ
2
2
p2 = p1 + ⋅ ⎛ V1 − V2 ⎞
⎝
⎠
2
p1 = SGHg⋅ ρ⋅ g⋅ h1
p1 = 13.6 × 1000⋅ where h1 = 100 mm Hg kg
3 m × 9.81⋅ m
2 s 2 × 0.1⋅ m × N⋅ s
kg⋅ m p1 = 13.3⋅ kPa p2 = 13300⋅ Hence N
2 + m ( ) 2 3 p2
SGHg⋅ ρ⋅ g 2 1
kg
N⋅ s
2
2 ⎛m
⋅ 1000⋅
× 0.849 − 0.943 ⋅ ⎜ ⎞ ×
⎟
3
2
⎝ s ⎠ kg⋅ m
m In mm Hg h2 = Similarly for exit (3) 2 1m
1
s
N
kg⋅ m
×
⋅
×
× 13200⋅
×
2
2
13.6 1000 kg 9.81⋅ m
m
s ⋅N ρ
2
2
p3 = p1 + ⋅ ⎛ V1 − V3 ⎞
⎠
2⎝
p3 = 13300⋅ N
2 m
h3 = In mm Hg p3
SGHg⋅ ρ⋅ g h2 = p2 = 13.2⋅ kPa + ( ) 2 3 h2 = 98.9⋅ mm 2 2 1
kg
N⋅ s
2
2 ⎛m
⋅ 1000⋅
× 0.849 − 5.09 ⋅ ⎜ ⎞ ×
⎟
3
2
⎝ s ⎠ kg⋅ m
m
h3 = p3 = 706⋅ Pa 1m
1
s
N
kg⋅ m
×
⋅
×
× 706⋅
×
2
2
13.6 1000 kg 9.81⋅ m
m
s ⋅N h3 = 5.29⋅ mm Note that all pressures are gage. ( ) ( Rx + p3⋅ A3⋅ cos ( 60⋅ deg) − p2⋅ A2⋅ cos ( 45⋅ deg) = u3⋅ ρ⋅ Q3 + u2⋅ ρ⋅ Q2 For x momentum ) ( Rx = p2⋅ A2⋅ cos ( 45⋅ deg) − p3⋅ A3⋅ cos ( 60⋅ deg) + ρ⋅ Q2⋅ V2⋅ cos ( 45⋅ deg) − Q3⋅ V3⋅ cos ( 60⋅ deg)
Rx = 13200⋅ N
2 2 × m 2 π⋅ ( 0.0075⋅ m)
N
π⋅ ( 0.0025⋅ m)
× cos ( 45⋅ deg) − 706⋅
×
× cos ( 60⋅ deg) ...
2
4
4
m
−3 3 2 L
m
L
m
10 ⋅ m
1⋅ min
N⋅ s
kg
+ 1000⋅ ⋅ ⎛ 2.5⋅
⋅ 0.943⋅ ⋅ cos ( 45⋅ deg) − 1.5⋅
⋅ 5.09⋅ ⋅ cos ( 60⋅ deg)⎞ ×
×
×
⎜
⎟
3⎝
1⋅ L
kg × m
min
s
min
s
60⋅ s
⎠
m ( ) ( Ry − p3⋅ A3⋅ sin ( 60⋅ deg) − p2⋅ A2⋅ sin ( 45⋅ deg) = v3⋅ ρ⋅ Q3 + v2⋅ ρ⋅ Q2 For y momentum ( N
2 m 2 × Rx = 0.375 N ) Ry = p2⋅ A2⋅ sin ( 45⋅ deg) + p3⋅ A3⋅ sin ( 60⋅ deg) + ρ⋅ Q2⋅ V2⋅ sin ( 45⋅ deg) + Q3⋅ V3⋅ sin ( 60⋅ deg)
Ry = 13200⋅ ) ) 2 π⋅ ( 0.0075⋅ m)
N
π⋅ ( 0.0025⋅ m)
× sin ( 45⋅ deg) + 706⋅
×
⋅ sin ( 60⋅ deg) ...
2
4
4
m
−3 3 2 L
m
L
m
kg
⎞ 10 ⋅ m × 1⋅ min × N⋅ s
+ 1000⋅ ⋅ ⎛ 2.5⋅
⋅ 0.943⋅ ⋅ sin ( 45⋅ deg) + 1.5⋅
⋅ 5.09⋅ ⋅ sin ( 60⋅ deg)⎟ ×
⎜
3⎝
1⋅ L
kg × m
min
s
min
s
60⋅ s
⎠
m Ry = 0.553 N Problem 6.68 [3] Problem 6.69 [3] Part 1/2 Problem 6.69 [3] Part 2/2 Problem 6.70 [4] H V
CS W
y x Ry Given: Flow through kitchen faucet Find: Area variation with height; force to hold plate as function of height Solution:
2 pV
+
+ g⋅ z = const
ρ
2 Basic equation Q = V⋅ A Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the faucet (1) and any height y
2 V1
2 2 + g⋅ H = where we assume the water is at patm 2 V ( y) = Hence V
+ g⋅ y
2 V1 + 2⋅ g⋅ ( H − y)
V1 = 0.815 The problem doesn't require a plot, but it looks like m
s V ( 0⋅ m) = 3.08 m
s 5 V (m/s) 4
3
2
1
0 5 10 15 20 25 30 35 40 45 y (cm)
The speed increases as y decreases because the fluid particles "trade" potential energy for kinetic, just as a falling solid particle does!
2 But we have
Hence Q = V1⋅ A1 = V1⋅
A= V1⋅ A1
V π⋅ D
= V⋅ A
4
A ( y) = 2 π⋅ D 1 ⋅ V 1
2 4⋅ V1 + 2⋅ g⋅ ( H − y) 45 y (cm) The problem doesn't require a plot, but it looks like 2 A ( H) = 1.23 cm 30
15 2 A ( 0) = 0.325 cm 0 0.5 1 1.5 A (cm2)
The area decreases as the speed increases. If the stream falls far enough the flow will change to turbulent.
For the CV above ( ) Ry − W = uin⋅ −ρ⋅ Vin⋅ Ain = −V ⋅ ( −ρ⋅ Q)
2 2 Ry = W + ρ⋅ V ⋅ A = W + ρ⋅ Q⋅ V1 + 2⋅ g⋅ ( H − y)
2 Hence Ry increases in the same way as V as the height y varies; the maximum force is when y = Hymax = W + ρ⋅ Q⋅ V1 + 2⋅ g⋅ H
R Problem 6.71 [4] An old magic trick uses an empty thread spool and a playing card. The playing card is
placed against the bottom of the spool. Contrary to intuition, when one blows downward
through the central hole in the spool, the card is not blown away. Instead it is ‘‘sucked’’
up against the spool. Explain. OpenEnded Problem Statement: An old magic trick uses an empty thread spool and a
playing card. The playing card is placed against the bottom of the spool. Contrary to
intuition, when one blows downward through the central hole in the spool, the card is not
blown away. Instead it is ‘‘sucked’’ up against the spool. Explain.
Discussion: The secret to this “parlor trick” lies in the velocity distribution, and hence
the pressure distribution, that exists between the spool and the playing cards.
Neglect viscous effects for the purposes of discussion. Consider the space between the
end of the spool and the playing card as a pair of parallel disks. Air from the hole in the
spool enters the annular space surrounding the hole, and then flows radially outward
between the parallel disks. For a given flow rate of air the edge of the hole is the crosssection of minimum flow area and therefore the location of maximum air speed.
After entering the space between the parallel disks, air flows radially outward. The flow
area becomes larger as the radius increases. Thus the air slows and its pressure increases.
The largest flow area, slowest air speed, and highest pressure between the disks occur at
the outer periphery of the spool where the air is discharged from an annular area.
The air leaving the annular space between the disk and card must be at atmospheric
pressure. This is the location of the highest pressure in the space between the parallel
disks. Therefore pressure at smaller radii between the disks must be lower, and hence the
pressure between the disks is subatmospheric. Pressure above the card is less than
atmospheric pressure; pressure beneath the card is atmospheric. Each portion of the card
experiences a pressure difference acting upward. This causes a net pressure force to act
upward on the whole card. The upward pressure force acting on the card tends to keep it
from blowing off the spool when air is introduced through the central hole in the spool.
Viscous effects are present in the narrow space between the disk and card. However, they
only reduce the pressure rise as the air flows outward, they do not dominate the flow
behavior. Problem 6.72 [4] Part 1/2 Problem 6.72 [4] Part 2/2 Problem 6.73 [4] CS Given: Air jet striking disk Find: Manometer deflection; Force to hold disk; Force assuming p0 on entire disk; plot pressure distribution Solution:
Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction
2 pV
+
+ g⋅ z = constant
ρ
2 Δp = SG⋅ ρ⋅ g⋅ Δh Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying Bernoulli between jet exit and stagnation point
patm
ρair
But from hydrostatics + 2
p0
V
=
+0
2
ρair p0 − patm = SG⋅ ρ⋅ g⋅ Δh
Δh = 0.002377⋅ For x momentum 1
2
p0 − patm = ⋅ ρair⋅ V
2
1
2
2
⋅ρ ⋅V
ρair⋅ V
2 air
Δh =
=
2⋅ SG⋅ ρ⋅ g
SG⋅ ρ⋅ g so
2 3 2 ft
1
ft
s
× ⎛ 225⋅ ⎞ ×
×
×
⎜
⎟
3
s⎠
2⋅ 1.75 1.94⋅ slug 32.2⋅ ft
⎝
ft slug 2 π⋅ d
Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅
4 ( ) Δh = 0.55⋅ ft 2 2 ⎛ 0.4 ⋅ ft⎞
2 π⋅ ⎜
⎟
2
slug ⎛
ft
12 ⎠
lbf ⋅ s
Rx = −0.002377⋅
× ⎜ 225⋅ ⎞ × ⎝
×
⎟
3
slug⋅ ft
s⎠
4
⎝
ft
The force of the jet on the plate is then F = −Rx The stagnation pressure is F = 0.105⋅ lbf 1
2
p0 = patm + ⋅ ρair⋅ V
2 The force on the plate, assuming stagnation pressure on the front face, is
1
2 π⋅ D
F = p0 − p ⋅ A = ⋅ ρair⋅ V ⋅
4
2 ( ) Rx = −0.105⋅ lbf 2 Δh = 6.60⋅ in 2 F= 2 2 π
slug ⎛
ft
⎛ 7.5 ⋅ ft⎞ × lbf ⋅ s
× 0.002377⋅
× ⎜ 225⋅ ⎞ × ⎜
⎟
⎟
3
8
s⎠
⎝
⎝ 12 ⎠ slug⋅ ft
ft F = 18.5 lbf Obviously this is a huge overestimate!
For the pressure distribution on the disk, we use Bernoulli between the disk outside edge any radius r for radial flow
patm 1
12
p
2
+ ⋅ vedge =
+ ⋅v
2
ρair
ρair 2
We need to obtain the speed v as a function of radius. If we assume the flow remains constant thickness h, then
2 Q = v⋅ 2⋅ π⋅ r⋅ h = V⋅ 2 π⋅ d
4 v ( r) = V⋅ d
8⋅ h⋅ r We need an estimate for h. As an approximation, we assume that h = d (this assumption will change the scale of p(r) but not the basic shap
v ( r) = V⋅ Hence d
8⋅ r
22 ρair⋅ V ⋅ d
4
1⎞
1
2
2
p ( r) = patm + ⋅ ρair⋅ ⎛ vedge − v ( r) ⎞ = patm +
⋅⎛
⎜ 2 − 2⎟
⎝
⎠
128
2
⎝D r ⎠
22
ρair⋅ V ⋅ d
4
1⎞
p ( r) =
⋅⎛
⎜ 2 − 2⎟
128
⎝D r ⎠ Using this in Bernoulli Expressed as a gage pressure p (psi) 0 0.25 0.5 0.75 1 1.25 1.5 1.75 − 0.1
− 0.2
− 0.3 r (in) 2 2.25 2.5 2.75 3 3.25 3.5 3.75 Problem 6.74 [4] Part 1/2 Problem 6.74 [4] Part 2/2 Problem 6.75 [4] Problem 6.76 [4] Problem 6.77 [4] Given: Water flow out of tube Find: Pressure indicated by gage; force to hold body in place Solution:
Basic equations: Bernoulli, and momentum flux in x direction
2 pV
+
+ g⋅ z = constant
ρ
2 Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying Bernoulli between jet exit and stagnation point
p1
ρ 2 + p1 = V1
2 2 2 p2 V2
V2
=
+
=
2
2
ρ where we work in gage pressure ρ⎛ 2
2
⋅ V2 − V1 ⎞
⎝
⎠
2
2
A1
D
V2 = V1⋅
= V1⋅
2
2
A2
D −d But from continuity Q = V1⋅ A1 = V2⋅ A2 ⎞
ft ⎛
2
⎟
V2 = 20⋅ ⋅ ⎜
s ⎜ 22 − 1.52 ⎟
2 ⎝ p1 = Hence The x mometum is V2 = 45.7 ⎠ ( 1
slug
2
2
× 1.94⋅
× 45.7 − 20
3
2
ft )⋅ ⎛ ft ⎞
⎜⎟
⎝s⎠ ( 2 ) 2 lbf
2 in p1 = 1638 ft ( 2 lbf F = 14.1 lbf 2 p1 = 11.4 psi 2 (gage) ) using gage pressures π⋅ ( 2⋅ in)
π⋅ ⎡( 2⋅ in) − ( 1.5⋅ in)
π⋅ ( 2⋅ in)
slug ⎡⎛
ft ⎞
ft
+ 1.94⋅
× ⎢⎜ 20⋅ ⎟ ×
− ⎛ 45.7⋅ ⎞ × ⎣
⎜
⎟
3 ⎣⎝
4
4
4
s⎠
s⎠
⎝
ft
2 × lbf⋅ s
slugft
⋅ −F + p1 ⋅ A1 − p2 ⋅ A2 = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2
F = p1 ⋅ A1 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2 ⎞
⎝
⎠ F = 11.4⋅ ft
s 2 × where D = 2 in and d = 1.5 in 2 in the direction shown 2 2 2⎤ ⎤ 2 2 ⎦ ⎥ × ⎛ 1⋅ ft ⎞ × lbf⋅ s
⎜
⎟
⋅
⎦ ⎝ 12⋅ in ⎠ slugft Problem 6.78 [4] Part 1/2 Problem 6.78 [4] Part 2/2 Problem 6.79 [4] Part 1/2 Problem 6.79 [4] Part 2/2 Problem 6.80 [5] Describe the pressure distribution on the exterior of a multistory building in a steady
wind. Identify the locations of the maximum and minimum pressures on the outside of
the building. Discuss the effect of these pressures on infiltration of outside air into the
building. OpenEnded Problem Statement: Describe the pressure distribution on the exterior of a
multistory building in a steady wind. Identify the locations of the maximum and
minimum pressures on the outside of the building. Discuss the effect of these pressures
on infiltration of outside air into the building.
Discussion: A multistory building acts as a bluffbody obstruction in a thick
atmospheric boundary layer. The boundarylayer velocity profile causes the air speed
near the top of the building to be highest and that toward the ground to be lower.
Obstruction of air flow by the building causes regions of stagnation pressure on upwind
surfaces. The stagnation pressure is highest where the air speed is highest. Therefore the
maximum surface pressure occurs near the roof on the upwind side of the building.
Minimum pressure on the upwind surface of the building occurs near the ground where
the air speed is lowest.
The minimum pressure on the entire building will likely be in the lowspeed, lowpressure wake region on the downwind side of the building.
Static pressure inside the building will tend to be an average of all the surface pressures
that act on the outside of the building. It is never possible to seal all openings completely.
Therefore air will tend to infiltrate into the building in regions where the outside surface
pressure is above the interior pressure, and will tend to pass out of the building in regions
where the outside surface pressure is below the interior pressure. Thus generally air will
tend to move through the building from the upper floors toward the lower floors, and
from the upwind side to the downwind side. Problem 6.81 [5] Imagine a garden hose with a stream of water flowing out through a nozzle. Explain why
the end of the hose may be unstable when held a half meter or so from the nozzle end. OpenEnded Problem Statement: Imagine a garden hose with a stream of water
flowing out through a nozzle. Explain why the end of the hose may be unstable when
held a half meter or so from the nozzle end.
Discussion: Water flowing out of the nozzle tends to exert a thrust force on the end of the
hose. The thrust force is aligned with the flow from the nozzle and is directed toward the
hose.
Any misalignment of the hose will lead to a tendency for the thrust force to bend the hose
further. This will quickly become unstable, with the result that the free end of the hose
will “flail” about, spraying water from the nozzle in all directions.
This instability phenomenon can be demonstrated easily in the backyard. However, it will
tend to do least damage when the person demonstrating it is wearing a bathing suit! Problem 6.82 [5] An aspirator provides suction by using a stream of water flowing through a venturi.
Analyze the shape and dimensions of such a device. Comment on any limitations on its
use. OpenEnded Problem Statement: An aspirator provides suction by using a stream of
water flowing through a venturi. Analyze the shape and dimensions of such a device.
Comment on any limitations on its use.
Discussion: The basic shape of the aspirator channel should be a converging nozzle
section to reduce pressure followed by a diverging diffuser section to promote pressure
recovery. The basic shape is that of a venturi flow meter.
If the diffuser exhausts to atmosphere, the exit pressure will be atmospheric. The pressure
rise in the diffuser will cause the pressure at the diffuser inlet (venturi throat) to be below
atmospheric.
A small tube can be brought in from the side of the throat to aspirate another liquid or gas
into the throat as a result of the reduced pressure there.
The following comments can be made about limitations on the aspirator:
1. It is desirable to minimize the area of the aspirator tube compared to the flow area
of the venturi throat. This minimizes the disturbance of the main flow through the
venturi and promotes the best possible pressure recovery in the diffuser.
2. It is desirable to avoid cavitation in the throat of the venturi. Cavitation alters the
effective shape of the flow channel and destroys the pressure recovery in the
diffuser. To avoid cavitation, the reduced pressure must always be above the
vapor pressure of the driver liquid.
3. It is desirable to limit the flow rate of gas into the venturi throat. A large amount
of gas can alter the flow pattern and adversely affect pressure recovery in the
diffuser.
The best combination of specific dimensions could be determined experimentally by a
systematic study of aspirator performance. A good starting point probably would be to
use dimensions similar to those of a commercially available venturi flow meter. Problem 6.83 [5] Problem 6.84 [2] Carefully sketch the energy grade lines (EGL) and hydraulic grade lines (HGL) for the
system shown in Fig. 6.6 if the pipe is horizontal (i.e., the outlet is at the base of the
reservoir), and a water turbine (extracting energy) is located at (a) point , or (b) at point
. In Chapter 8 we will investigate the effects of friction on internal flows. Can you
anticipate and sketch the effect of friction on the EGL and HGL for cases (a) and (b)? (a) Note that the effect of friction would be that the EGL would tend to drop:
suddenly at the contraction, gradually in the large pipe, more steeply in the
small pipe. The HGL would then “hang” below the HGL in a manner similar
to that shown. EGL Turbine HGL (b) Note that the effect of friction would be that the EGL would tend to drop:
suddenly at the contraction, gradually in the large pipe, more steeply in the
small pipe. The HGL would then “hang” below the HGL in a manner similar
to that shown. EGL Turbine HGL Problem 6.85 [2] Carefully sketch the energy grade lines (EGL) and hydraulic grade lines (HGL) for the
system shown in Fig. 6.6 if a pump (adding energy to the fluid) is located at (a) point ,
or (b) at point , such that flow is into the reservoir. In Chapter 8 we will investigate the
effects of friction on internal flows. Can you anticipate and sketch the effect of friction
on the EGL and HGL for cases (a) and (b)? (a) Note that the effect of friction would be that the EGL would tend to drop from
right to left: steeply in the small pipe, gradually in the large pipe, and
suddenly at the expansion. The HGL would then “hang” below the HGL in a
manner similar to that shown. EGL Flow Pump HGL (b) Note that the effect of friction would be that the EGL would tend to drop from
right to left: steeply in the small pipe, gradually in the large pipe, and
suddenly at the expansion. The HGL would then “hang” below the HGL in a
manner similar to that shown. EGL Flow HGL
Pump Problem *6.86 Given: Unsteady water flow out of tube Find: [2] Pressure in the tank Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying unsteady Bernoulli between reservoir and tube exit
2 2
2⌠
2
p
V
dV ⌠
V
∂
+⎮
+ g⋅ h =
V ds =
+
⋅ ⎮ 1 ds
⎮ ∂t
ρ
2
2
dt ⌡1
⌡1 ⎛ V2 Hence p = ρ⋅ ⎜ Hence p = 1.94⋅ ⎝2 − g⋅ h + where we work in gage pressure dV ⎞
⋅ L⎟
dt ⎠ 2
⎞ ft 2 lbf⋅ s2
slug ⎛ 6
×⎜
− 32.2 × 4.5 + 7.5 × 35 ⎟ ⋅ ⎛ ⎞ ×
⎜⎟
3 ⎝2
⋅
⎠ ⎝ s ⎠ slugft
ft p = 263⋅ lbf
ft 2 p = 1.83⋅ psi (gage) Problem *6.87 Given: Unsteady water flow out of tube Find: [2] Initial acceleration Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying unsteady Bernoulli between reservoir and tube exit
2 2
⌠
p
⎮ ∂ V ds = dV ⋅ ⌠ 1 ds = a ⋅ L
⎮
+ g⋅ h =
x
⎮ ∂t
ρ
dt ⌡1
⌡1 where we work in gage pressure Hence 1p
⎞
ax = ⋅ ⎛ + g⋅ h⎟
⎜
L ⎝ρ
⎠ Hence ⎡ lbf ⎛ 12⋅ in ⎞
⎤
1
ft
slugft
⋅
ft
ax =
× ⎢3⋅
×⎜
×
+ 32.2⋅ × 4.5⋅ ft⎥
⎟×
2
1.94⋅ slug 2⋅ lbf
⎥
35⋅ ft ⎢ in2 ⎝ 1⋅ ft ⎠
s
s
⎣
⎦
2 3 ft
ax = 10.5⋅
2
s Note that we obtain the same result if we treat the water in the pipe as a single body at rest with gage pressure p + ρgh at the left end! Problem *6.88 [5] Problem *6.89 [4] Problem *6.90 [4] Given: Unsteady water flow out of tube Find: Differential equation for velocity; Integrate; Plot v versus time Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying unsteady Bernoulli between reservoir and tube exit
2 2
2⌠
2
2
V
p
dV ⌠
dV
V
V
∂
+⎮
+ g⋅ h =
V ds =
+
⋅ ⎮ 1 ds =
+
⋅L
⎮ ∂t
ρ
2
2
dt ⌡1
2
dt
⌡1 where we work in gage pressure 2 dV V
1p
⎞
+
= ⋅ ⎛ + g⋅ h⎟
⎜
dt
2⋅ L
L ⎝ρ
⎠
L⋅ dV
= dt
2
p
V
+ g⋅ h −
ρ
2 Hence
Separating variables is the differential equation for the flow Integrating and using limits V(0) = 0 and V(t) = V ⎛p
⎞
⎜
+ g⋅ h ⎟
ρ
p
⎞
2⋅ ⎛ + g⋅ h⎟ ⋅ tanh⎜
⋅ t⎟
⎜
⎜ 2⋅ L2 ⎟
⎝ρ
⎠
⎝
⎠ V ( t) =
25 V (ft/s) 20
15
10
5
0 1 2 3 t (s) This graph is suitable for plotting in Excel
For large times V= 2⋅ ⎛
⎜ p ⎝ρ ⎞
+ g⋅ h⎟
⎠ V = 22.6 ft
s 4 5 Problem *6.91 [5] Part 1/2 Problem *6.91 4.44 [5] Part 2/2 Problem *6.92 [5] Problem *6.93 [2] Problem *6.94 [2] Given: Stream function Find: If the flow is irrotational; Pressure difference between points (1,4) and (2,1) Solution:
u= Basic equations: Incompressibility because ψ exists ∂
∂y v=− ψ ∂
∂x ψ ∂ Irrotationality ∂x v− ∂
∂y 2 ψ ( x , y) = A⋅ x ⋅ y
u ( x , y) = ∂
∂y v ( x , y) = − Hence ∂ ∂
∂x v ( x , y) − ∂x ψ ( x , y) = ∂
∂y ψ ( x , y) = − ∂
∂y (A⋅ x2⋅ y)
∂
∂x u ( x , y) = A⋅ x (A⋅ x2⋅ y) 2 v ( x , y) = −2⋅ A⋅ x⋅ y
∂ u ( x , y) → −2⋅ A⋅ y ∂x v− ∂
∂y u≠0 so flow is NOT IRROTATIONAL Since flow is rotational, we must be on same streamline to be able to use Bernoulli
At point (1,4) ψ ( 1 , 4) = 4 A ψ ( 2 , 1) = 4 A and at point (2,1) Hence these points are on same streamline so Bernoulli can be used. The velocity at a point is
2 Hence at (1,4) V1 = ⎡ 2.5 × ( 1⋅ m) 2⎤ + ⎛ −2 × 2.5 × 1⋅ m × 4⋅ m⎞
⎢
⎥⎜
⎟
m⋅ s
⎣ m⋅ s
⎦⎝
⎠ Hence at (2,1) V2 = ⎡ 2.5 × ( 2⋅ m) 2⎤ + ⎛ −2 × 2.5 × 2⋅ m × 1⋅ m⎞
⎢
⎥⎜
⎟
m⋅ s
⎣ m⋅ s
⎦⎝
⎠ 2 Using Bernoulli p1
ρ + Δp = 2 2 × 1200⋅ kg
3 m ( m
s V2 = 14.1 m
s 2 Δp = ) × 14.1 − 20.2 ⋅ ⎛
⎜
2 2 m⎞ 2 ⎟× ⎝s⎠ 2 u ( x , y) + v ( x , y) V1 = 20.2 p2 1
1
2
2
⋅ V1 =
+ ⋅ V2
2
2
ρ
1 V ( x , y) = ρ⎛ 2
2
⋅ V2 − V1 ⎞
⎝
⎠
2 2 N⋅ s
kg⋅ m Δp = −126⋅ kPa 2 u=0 Problem *6.95 [2] Problem *6.96 [3] Given: Data from Table 6.2 Find: Stream function and velocity potential for a source in a corner; plot; velocity along one plane Solution:
From Table 6.2, for a source at the origin ψ ( r , θ) = q
⋅θ
2⋅ π ϕ ( r , θ) = − q
⋅ ln ( r)
2⋅ π Expressed in Cartesian coordinates ψ ( x , y) = q
y⎞
⋅ atan ⎛ ⎟
⎜
2⋅ π
⎝ x⎠ ϕ ( x , y) = − q
2
2
⋅ ln x + y
4⋅ π ( ) To build flow in a corner, we need image sources at three locations so that there is symmetry about both axes. We need
sources at (h,h), (h, h), ( h,h), and ( h, h)
Hence the composite stream function and velocity potential are
ψ ( x , y) = q⎛
y − h⎞
⎛ y + h ⎞ + atan ⎛ y + h ⎞ + atan ⎛ y − h ⎞ ⎞
⋅ ⎜ atan ⎛
⎜
⎟ + atan ⎜
⎟
⎜
⎟
⎜
⎟⎟
2⋅ π ⎝
x − h⎠
⎝
⎝ x − h⎠
⎝ x + h⎠
⎝ x + h ⎠⎠ ϕ ( x , y) = − q
q⎡
2
2
2
2
2
2
2
2
⋅ ln ⎡⎡( x − h) + ( y − h) ⎤ ⋅ ⎡( x − h) + ( y + h) ⎤⎤ −
⋅ ⎣( x + h) + ( y + h) ⎤ ⋅ ⎡( x + h) + ( y − h) ⎤
⎣⎣
⎦⎣
⎦⎦
⎦⎣
⎦
4⋅ π
4⋅ π By a similar reasoning the horizontal velocity is given by
u= q⋅ ( x − h)
2⎤ 2⋅ π⎡( x − h) + ( y − h) ⎦
⎣
2 + q⋅ ( x − h)
2⎤ 2⋅ π⎡( x − h) + ( y + h) ⎦
⎣
2 + q⋅ ( x + h)
2⋅ π⎡( x + h) + ( y + h) ⎦
⎣
2 2⎤ + q⋅ ( x + h)
2
2
2⋅ π⎡( x + h) + ( y + h) ⎤
⎣
⎦ Along the horizontal wall (y = 0)
u= or q⋅ ( x − h)
2⋅ π⎡( x − h) + h ⎦
⎣ u ( x) = 2 2⎤ + q⋅ ( x − h)
2⋅ π⎡( x − h) + h ⎦
⎣
2 x−h
x+h
q⎡
⎤
⋅
+
π ⎢ ( x − h) 2 + h2 ( x + h) 2 + h2⎥
⎣
⎦ 2⎤ + q⋅ ( x + h)
2⋅ π⎡( x + h) + h ⎦
⎣
2 2⎤ + q⋅ ( x + h)
2⋅ π⎡( x + h) + h ⎤
⎣
⎦
2 2 Stream Function y #NAME? Stream Function #NAME? Velocity Potential Note that the plot is
from x = 0 to 5 and y = 0 to 5 Velocity Potential x y x Problem *6.97 [3] Given: Velocity field of irrotational and incompressible flow Find: Stream function and velocity potential; plot Solution:
q⋅x The governing equations are u= Hence for the stream function ⌠
q⎛
y−h⎞
⎮
⎛ y + h ⎞ ⎞ + f ( x)
ψ = ⎮ u ( x , y) dy =
⋅ ⎜ atan⎛
⎜
⎟ + atan⎜
⎟⎟
2⋅ π ⎝
x⎠
⎝
⎝ x ⎠⎠
⌡ 2⎤ 2⋅ π⎡x + ( y − h ) ⎦
⎣
∂
∂y v=− ψ ∂
∂x ψ 2⎤ 2⋅ π⎡x + ( y + h ) ⎦
⎣
2 u=− ∂
∂x ϕ v= q ⋅ (y − h ) u= 2 + q⋅x The velocity field is 2⋅ π⎡x + ( y − h ) ⎦
⎣ v=− 2 ∂
∂y ϕ ⌠
q⎛
y−h⎞
⎮
⎛ y + h ⎞ ⎞ + g( y)
ψ = −⎮ v ( x , y) dx =
⋅ ⎜ atan⎛
⎜
⎟ + atan⎜
⎟⎟
2⋅ π ⎝
⎝x⎠
⎝ x ⎠⎠
⌡
q⎛
y − h⎞
⎛ y + h ⎞⎞
⋅ ⎜ atan ⎛
⎜
⎟ + atan ⎜
⎟⎟
2⋅ π ⎝
x⎠
⎝
⎝ x ⎠⎠ The simplest expression for ψ is ψ ( x , y) = For the stream function ϕ=− ⌠
q
2
2
2
2
⎮
u ( x , y) dx = −
⋅ ln ⎡⎡x + ( y − h) ⎤ ⋅ ⎡x + ( y + h) ⎤⎤ + f ( y)
⎣⎣
⎦⎣
⎦⎦
⎮
4⋅ π
⌡ ϕ=− ⌠
q
2
2
2
2
⎮
v ( x , y) dy = −
⋅ ln ⎡⎡x + ( y − h) ⎤ ⋅ ⎡x + ( y + h) ⎤⎤ + g ( x)
⎣⎣
⎦⎣
⎦⎦
⎮
4⋅ π
⌡ The simplest expression for φ is ϕ ( x , y) = − q
4⋅ π 2
2
2
2
⋅ ln ⎡⎡x + ( y − h) ⎤ ⋅ ⎡x + ( y + h) ⎤⎤
⎣⎣
⎦⎣
⎦⎦ 2⎤ + q ⋅ (y + h )
2
2
2⋅ π⎡x + ( y + h ) ⎤
⎣
⎦ Stream Function y x Velocity Potential
#NAME? Stream Function #NAME? Velocity Potential Note that the plot is
from x = 2.5 to 2.5 and y = 0 to 5 y x Problem *6.98 [3] Given: Data from Table 6.2 Find: Stream function and velocity potential for a vortex in a corner; plot; velocity along one plane Solution:
From Table 6.2, for a vortex at the origin ϕ ( r , θ) = K
⋅θ
2⋅ π ψ ( r , θ) = − K
⋅ ln ( r )
2⋅ π Expressed in Cartesian coordinates ϕ ( x , y) = q
y
⋅ atan⎛ ⎞
⎜⎟
2⋅ π
⎝ x⎠ ψ ( x , y) = − q
2
2
⋅ ln x + y
4⋅ π ( ) To build flow in a corner, we need image vortices at three locations so that there is symmetry about both axes. We need
vortices at (h,h), (h, h), ( h,h), and ( h, h). Note that some of them must have strengths of  K!
Hence the composite velocity potential and stream function are
ϕ ( x , y) = K⎛
y−h⎞
⎛ y + h ⎞ + atan⎛ y + h ⎞ − atan⎛ y − h ⎞ ⎞
⋅ ⎜ atan⎛
⎜
⎟ − atan⎜
⎟
⎜
⎟
⎜
⎟⎟
2⋅ π ⎝
⎝x−h⎠
⎝x−h⎠
⎝x+h⎠
⎝ x + h ⎠⎠ ⎡ (x − h ) + (y − h ) (x + h ) + (y + h ) ⎤
K
⎥
⋅ ln ⎢
⋅
4⋅ π ⎢ ( x − h ) 2 + ( y + h ) 2 ( x + h ) 2 + ( y − h ) 2⎥
2 ψ ( x , y) = − 2 2 2 ⎣ ⎦ By a similar reasoning the horizontal velocity is given by
u=− K ⋅ ( y − h)
2⎤ 2⋅ π⎡( x − h) + ( y − h) ⎦
⎣
2 + K ⋅ ( y + h)
2⋅ π⎡( x − h) + ( y + h) ⎦
⎣
2 2⎤ − K ⋅ ( y + h)
2⎤ 2⋅ π⎡( x + h) + ( y + h) ⎦
⎣
2 + Along the horizontal wall (y = 0)
u= or K⋅ h
2⋅ π⎡( x − h) + h ⎦
⎣ u ( x) = 2 2⎤ + K⋅ h
2⋅ π⎡( x − h) + h ⎦
⎣
2 2⎤ 1
1
K⋅ h ⎡
⎤
⋅
−
π ⎢ ( x − h) 2 + h2 ( x + h) 2 + h2⎥
⎣
⎦ − K⋅ h
2⋅ π⎡( x + h) + h ⎦
⎣
2 2⎤ − K⋅ h
2⋅ π⎡( x + h) + h ⎤
⎣
⎦
2 2 K ⋅ ( y − h)
2
2
2⋅ π⎡( x + h) + ( y − h) ⎤
⎣
⎦ y
Stream Function x Velocity Potential
#NAME? Stream Function y #NAME? #NAME?
Note that the plot is
from x = 5 to 5 and y = 5 to 5 Velocity Potential x Problem *6.99 [NOTE: Typographical Error  Wrong Function!] [2] Problem *6.100 [2] Given: Stream function Find: Velocity field; Show flow is irrotational; Velocity potential Solution:
Basic equations: Incompressibility because ψ exists
∂ Irrotationality ∂x 5 v− ∂ ∂
∂y v=− ψ ∂
∂x u=− ψ ∂
∂x φ v=− ∂
∂y φ u=0 ∂y 32 u= 4 ψ ( x , y) = x − 10⋅ x ⋅ y + 5⋅ x⋅ y
u ( x , y) = ∂
∂y v ( x , y) = − ∂
∂x
Hence v ( x , y) − u=− v=− ∂
∂x
∂
∂y 3 ∂
∂x 3 22 ψ ( x , y) 4 u ( x , y) → 20⋅ x⋅ y − 20⋅ x ⋅ y
4 ψ ( x , y) v ( x , y) → 30⋅ x ⋅ y − 5⋅ x − 5⋅ y u ( x , y) → 0 Hence flow is IRROTATIONAL ∂
∂y φ so ⌠
4
23
⎮
φ ( x , y) = −⎮ u ( x , y) dx + f ( y) = 5⋅ x ⋅ y − 10⋅ x ⋅ y + f ( y)
⌡ φ so ⌠
4
23
5
⎮
φ ( x , y) = −⎮ v ( x , y) dy + g ( x) = 5⋅ x ⋅ y − 10⋅ x ⋅ y + y + g ( x)
⌡ Comparing, the simplest velocity potential is then 4 23 5 φ ( x , y) = 5⋅ x ⋅ y − 10⋅ x ⋅ y + y Problem *6.101 [2] Problem *6.102 Given: Velocity potential Find: [2] Show flow is incompressible; Stream function Solution:
u= Basic equations: Irrotationality because φ exists
∂ Incompressibility
6 ∂x u+ 42 ∂
∂y ∂
∂y v=− ψ ∂ u=− ∂x ψ ∂
∂x φ v=− ∂
∂y φ v =0 24 6 φ ( x , y) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
u ( x , y) = − v ( x , y) = − Hence Hence ∂
∂x u ( x , y) + u= ∂
∂y v=− ψ ∂
∂x ψ ∂
∂x
∂
∂y 32 5 4 4 23 5 φ ( x , y) u ( x , y) → 60⋅ x ⋅ y − 6⋅ x − 30⋅ x⋅ y φ ( x , y) v ( x , y) → 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6⋅ y v ( x , y) → 0 Hence flow is INCOMPRESSIBLE ∂
∂y so ⌠
33
5
5
⎮
ψ ( x , y) = ⎮ u ( x , y) dy + f ( x) = 20⋅ x ⋅ y − 6⋅ x ⋅ y − 6⋅ x⋅ y + f ( x)
⌡ so ⌠
33
5
5
⎮
ψ ( x , y) = −⎮ v ( x , y) dx + g ( y) = 20⋅ x ⋅ y − 6⋅ x ⋅ y − 6⋅ x⋅ y + g ( y)
⌡ Comparing, the simplest stream function is then 33 5 5 ψ ( x , y) = 20⋅ x ⋅ y − 6⋅ x ⋅ y − 6⋅ x⋅ y Problem *6.103 [4] Given: Complex function Find: Show it leads to velocity potential and stream function of irrotational incompressible flow;
Show that df/dz leads to u and v Solution:
u= Basic equations: Irrotationality because φ exists
∂ Incompressibility ∂x 6 f ( z) = z = ( x + i⋅ y)
Expanding 6 u+ ∂
∂y v=0 ∂
∂y v=− ψ ∂ Irrotationality ∂x ∂
∂x
∂ v− u=− ψ ∂y ∂
∂x v=− φ ∂
∂y φ u=0 6 42 24 ( 6 5 5 ) 33 f ( z) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y + i⋅ 6⋅ x⋅ y + 6⋅ x ⋅ y − 20⋅ x ⋅ y We are thus to check the following
6 42 24 6 5 φ ( x , y) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
u ( x , y) = − v ( x , y) = − ∂ 32 ∂ ∂y v ( x , y) = − 4 4 23 5 v ( x , y) → 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6⋅ y 5 ψ ( x , y) ∂ 33 u ( x , y) → 60⋅ x ⋅ y − 6⋅ x − 30⋅ x⋅ y φ ( x , y) ∂x ∂ 5 φ ( x , y) ∂y
An alternative derivation of u and v is
u ( x , y) = 5 ψ ( x , y) = 6⋅ x⋅ y + 6⋅ x ⋅ y − 20⋅ x ⋅ y 32 4 u ( x , y) → 6⋅ x − 60⋅ x ⋅ y + 30⋅ x⋅ y
23 ψ ( x , y) 4 5 v ( x , y) → 60⋅ x ⋅ y − 30⋅ x ⋅ y − 6⋅ y ∂x
Note that the values of u and v are of opposite sign using ψ and φ!different which is the same result using φ! To
resolve this we could either let f = φ+iψ; altenatively we could use a different definition of φ that many authors use:
u= Hence Hence ∂
∂x
∂
∂x ∂
∂x v= φ v ( x , y) − u ( x , y) + ∂
∂y
∂
∂y ∂
∂y φ u ( x , y) → 0 Hence flow is IRROTATIONAL v ( x , y) → 0 Hence flow is INCOMPRESSIBLE ( 6) = 6⋅ z5 = 6⋅ (x + i⋅ y)5 = (6⋅ x5 − 60⋅ x3⋅ y2 + 30⋅ x⋅ y4) + i⋅ (30⋅ x4⋅ y + 6⋅ y5 − 60⋅ x2⋅ y3) Next we find df
dz
=
dz
dz Hence we see df
= u − i⋅ v
dz Hence the results are verified; These interesting results are explained in Problem 6.104! u = Re ⎛
⎜ df ⎞ ⎟
⎝ dz ⎠ and v = −Im ⎛
⎜ df ⎞ ⎟
⎝ dz ⎠ Problem *6.104 [4] Given: Complex function Find: Show it leads to velocity potential and stream function of irrotational incompressible flow;
Show that df/dz leads to u and v Solution:
Basic equations: First consider Hence u=
∂
∂x
∂ 2
2 ∂ ∂y f= ∂x
Combining ∂ 2
2 ∂x v=− ψ
∂ ∂
∂x u=− ψ d
d
d
f = 1⋅ f = f
dz
dz
∂x dz f= f+ z⋅ ∂ ⎛∂ ⎜ ⎞ f⎟ = ∂x ⎝ ∂x ⎠
∂ 2
2 ∂y f= 2 d 2 ∂
∂x φ (1) v=− ∂
∂y φ
∂ and also d ⎛d ⎞
d
⎜ f ⎟ = 2f
dz ⎝ dz ⎠
dz ∂y 2 f− dz 2 d 2 ∂ and f= 2
2 ∂y f =0 ∂ d
d
d
f = i⋅ f = i⋅ f
dz
dz
∂y dz f= z⋅ ∂ ⎛∂ ⎜ ⎞ f ⎟ = i⋅ ∂y ⎝ ∂y ⎠ (2) d⎛d ⎞
d
⎜ i⋅ f ⎟ = − 2 f
dz ⎝ dz ⎠
dz
2 Any differentiable function f(z) automatically satisfies the Laplace
Equation; so do its real and imaginary parts! dz We demonstrate derivation of velocities u and v
From Eq 1 d
d
∂
∂
∂
f = ( φ + i ⋅ ψ) = ( φ + i ⋅ ψ) = φ + i ⋅ ψ = − u − i ⋅ v
dz
dz
∂x
∂x
∂x From Eq 2 1∂
d
d
∂
∂
f = ( φ + i ⋅ ψ) = ⋅ ( φ + i ⋅ ψ) = − i ⋅ φ + ψ = i ⋅ v + u
i ∂y
dz
dz
∂y
∂y
∂ There appears to be an incompatibilty here,
but many authors define φ as u= Alternatively, we can use out φ but set ∂ f = −φ + i ⋅ ψ ∂x φ v= ∂y φ or in other words, as the negative
of our definition Then
From Eq 1 d
d
∂
∂
∂
f = ( φ + i ⋅ ψ) = ( φ + i ⋅ ψ) = φ + i ⋅ ψ = u − i ⋅ v
dz
dz
∂x
∂x
∂x From Eq 2 1∂
d
d
∂
∂
f = ( φ + i ⋅ ψ) = ⋅ ( φ + i ⋅ ψ) = − i ⋅ φ + ψ = − i ⋅ v + u
i ∂y
dz
dz
∂y
∂y Hence we have demonstrated that df
= u − i⋅ v
dz if we set u= ∂
∂x φ v= ∂
∂y φ Problem *6.105 [2] Problem *6.106 [3] Problem *6.107 [2] Part 1/2 Problem *6.107 [2] Part 2/2 Problem *6.108 [3] Problem *6.109 [3] Part 1/2 Problem *6.109 [3] Part 2/2 Problem *6.110 [2] Problem *6.111 [3] Consider flow around a circular cylinder with freestream velocity from right to left and a
counterclockwise free vortex. Show that the lift force on the cylinder can be expressed as
FL = −ρUΓ, as illustrated in Example 6.12.
OpenEnded Problem Statement: Consider flow around a circular cylinder with
freestream velocity from right to left and a counterclockwise free vortex. Show that the
lift force on the cylinder can be expressed as FL = −ρUΓ, as illustrated in Example 6.12.
Discussion: The only change in this flow from the flow of Example 6.12 is that the
directions of the freestream velocity and the vortex are changed. This changes the sign of
the freestream velocity from U to −U and the sign of the vortex strength from K to −K.
Consequently the signs of both terms in the equation for lift are changed. Therefore the
direction of the lift force remains unchanged.
The analysis of Example 6.12 shows that only the term involving the vortex strength
contributes to the lift force. Therefore the expression for lift obtained with the changed
freestream velocity and vortex strength is identical to that derived in Example 6.12. Thus
the general solution of Example 6.12 holds for any orientation of the freestream and
vortex velocities. For the present case, FL = −ρUΓ, as shown for the general case in
Example 6.12. Problem *6.112 [3] Problem *6.113 [3] Problem *6.114 [3] Problem *6.115 [3] Part 1/2 Problem *6.115 [3] Part 2/2 Problem *6.116 [4] Problem *6.117 [4] Part 1/2 Problem *6.117 [4] Part 2/2 Problem *6.118 [3] Part 1/2 Problem *6.118 [3] Part 2/2 ...
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This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .
 Spring '11
 TUZLA

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