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ME309 Exam Formula Sheet
Fluid Mechanics (Purdue University)
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ME 309 Exam Formula Sheet
p = ρRT
∂u
τ yx = µ x
∂y
dp
= −ρg
dz
dN
D
d
ηρ dV = ∫ ηρ dV + ∫ η ( ρ urel ⋅ dA ) or
∫
dt
Dt Vsystem
dt CV
CS
z
g
 
∂
ηρ dV + ∫ ηρV ⋅ dA
∫
∂t CV
CS
=
system
 
∂
d
ρ dV + ∫ ρV ⋅ dA = 0
ρ dV + ∫ ( ρ urel ⋅ dA ) = 0 or
∫
∫
∂t
dt CV
CS
CV
CS
  


∂ 

d
∫ u xyz ρ dV + CS∫ u xyz ( ρurel ⋅ dA ) = FS + FB − CV∫ a xyz / XYZ ρ dV or ∂t CV∫ Vxyz ρ dV + CS∫ Vxyz ρVxyz ⋅ dA = FS + FB − CV∫ arf ρ dV
dt CV
d
2
2
p
∫ eρ dV + CS∫ h + 12 V + gz ρurel ⋅ dA = Q into CV + Won CV where e = u + 12 V + gz and h = u + ρ
dt CV
(
)(
)
z
g
p V2
+
+ z = constant (for inviscid flow)
ρ g 2g
1/ 2
2
⎡⎛ x ∂R ⎞2 ⎛ x ∂R ⎞ 2
⎛ x ∂R
⎞ ⎤
u R = ± ⎢⎜ 1
u1 ⎟ + ⎜ 2
u 2 ⎟ + ⋅⋅⋅ + ⎜ n
un ⎟ ⎥
R ∂x1 ⎠ ⎝ R ∂x 2 ⎠
⎝ R ∂x n ⎠ ⎦⎥
⎣⎢⎝
⎛ p
⎞ ⎛ p
⎞
V2
V2
+α
+ z⎟ = ⎜
+α
+ z ⎟ − H L ,1→2 + H S ,1→2
⎜
g
2
g
g
2
g
ρ
ρ
⎝
⎠2 ⎝
⎠1
z
g
⎛p
⎞ ⎛p
⎞
V2
V2
+ gz ⎟ − ⎜ + α
+ gz ⎟ = hL,1→2 − hS ,1→2
⎜ +α
2
2
⎝ρ
⎠1 ⎝ ρ
⎠2
where
H L ,1→ 2 = ∑ K i
i
W
Δp
Vi 2
⎛L ⎞
⎛L⎞
; K≡ 1
; K major = f ⎜ ⎟ ; K = f ⎜ e ⎟ ; H S ,1→2 = S ,1→2 ; h = gH
2

mg
ρV
2g
⎝D⎠
⎝D⎠
2
∂V ⎤
∂p
∂z
⎡ ∂V
+V
= − − ρg
⎥
∂s ⎦
∂s
∂s
⎣ ∂t
ρ⎢
ρ
n
V2
∂p
∂z
= − + ρg
R
∂n
∂n
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z
g
lOMoARcPSD|21735616
ME 309 Exam Formula Sheet
g = 9.81 m/s2 = 32.2 ft/s2
⎫⎪
⎬ Water
µ = 1.00 × 10 kg/ (m ⋅s) ⎭⎪
ρ = 1000 kg/m 3
−3
⎪⎫
⎬ Air (STP)
µ = 1.79 × 10 kg/ (m ⋅s) ⎪⎭
ρ = 1.23 kg/m 3
−5
Properties of air (treated as a perfect gas)
N⋅ m
N⋅m
k = 1.4
R = 287 kg
c p = 1005 kg
⋅K
⋅K
R = 53.3
ft ⋅lbf
lbm ⋅°R
c p = 187
cv = 718
ft ⋅lbf
lbm ⋅°R
cv = 133
N⋅ m
kg⋅K
ft ⋅lbf
lbm ⋅°R
Specific gravity, SG = ρ/ρH2O (at 4 oC)
Kinematic viscosity, ν ≡ µ/ρ
1 Pa = N/m2; 1 atm = 101 kPa = 14.7 psia
1 ft = 0.305 m; g = 9.81 m/s2 = 32.2 ft/s2
1 lbm = 0.454 kg; 1 slug = 32.2 lbm
1 lbf = 32.2 lbm⋅ft/s2 = 1 slug⋅ft/s2
1 ft3 = 7.48 gal; 1 m3 = 103 L; 1000 L = 1 m3
1 hp = 550 ft⋅lbf/s = 746 W
1 Btu = 778 ft⋅lbf = 1.06 kJ
1 rpm = 0.1047 rad//s
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ME 309 Exam Formula Sheet
Continuity equation rectangular coordinates (x, y, z)
∂ρ ∂(ρu x ) ∂ (ρu y ) ∂(ρu z )
=0
+
+
+
∂z
∂y
∂t
∂x
cylindrical coordinates (r, θ, z)
∂ρ 1 ∂ (ρru r ) 1 ∂ (ρuθ ) ∂ (ρu z )
+
+
+
=0
r ∂θ
∂t r ∂r
∂z
Stress tensor components for a Newtonian fluid rectangular coordinates (x, y, z)
cylindrical coordinates (r, θ, z)
⎡ ∂u x 2
⎤
− ( ∇ ⋅ u )⎥
⎣ ∂x 3
⎦
⎡ ∂u
⎤
2
σ yy = − p + µ ⎢ 2 y − ( ∇ ⋅ u )⎥
⎣ ∂y 3
⎦
⎡ ∂u 2
⎤
σ zz = − p + µ ⎢ 2 z − ( ∇ ⋅ u )⎥
∂
z
3
⎣
⎦
⎡ ∂u ∂u ⎤
σ xy = σ yx = µ ⎢ x + y ⎥
∂x ⎦
⎣ ∂y
⎡ ∂u ∂u ⎤
σ xz = σ zx = µ ⎢ x + z ⎥
∂x ⎦
⎣ ∂z
⎡ ∂u
∂u ⎤
σ yz = σ zy = µ ⎢ y + z ⎥
∂y ⎦
⎣ ∂z
⎡ ∂ur 2
⎤
− ( ∇ ⋅ u )⎥
⎣ ∂r 3
⎦
⎡ ⎛ 1 ∂uθ ur ⎞ 2
⎤
+ ⎟ − ( ∇ ⋅ u )⎥
σ θθ = − p + µ ⎢ 2 ⎜
r ⎠ 3
⎣ ⎝ r ∂θ
⎦
⎡ ∂u 2
⎤
σ zz = − p + µ ⎢ 2 z − ( ∇ ⋅ u )⎥
3
∂
z
⎣
⎦
σ xx = − p + µ ⎢ 2
∇⋅u =
∂u x ∂u y ∂u z
+
+
∂x
∂y
∂z
σ rr = − p + µ ⎢ 2
⎡ ∂ ⎛ uθ ⎞ 1 ∂ur ⎤
⎜ ⎟+
⎥
⎣ ∂r ⎝ r ⎠ r ∂θ ⎦
σ rθ = σ θ r = µ ⎢ r
⎡ ∂u
1 ∂u ⎤
⎡ ∂u
∂u ⎤
z
σ θ z = σ zθ = µ ⎢ θ +
⎥
⎣ ∂z r ∂θ ⎦
σ zr = σ rz = µ ⎢ z + r ⎥
∂z ⎦
⎣ ∂r
∇⋅u =
1 ∂ (ru r ) 1 ∂uθ ∂u z
+
+
∂z
r ∂r
r ∂θ
Navier-­‐Stokes equations for a Newtonian fluid with constant density (ρ) and dynamic viscosity (µ) rectangular coordinates (x, y, z):
⎡∂ 2u
∂u x
∂u x
∂u x ⎞
∂ 2u x ∂ 2u x ⎤
⎛ ∂u x
∂p
⎟⎟ = −
+ ux
+uy
+
+ uz
+ µ ⎢ 2x +
⎥ + ρf x
∂x
∂y
∂x
∂z ⎠
∂z 2 ⎦
∂y 2
⎝ ∂t
⎣ ∂x
ρ ⎜⎜
∂u y
∂u y
∂u y
⎛ ∂u y
ρ ⎜⎜
+ uz
+uy
+ ux
∂z
∂y
∂x
⎝ ∂t
∂u z
∂u z
∂u z
⎛ ∂u z
+ ux
+uy
+ uz
∂x
∂y
∂z
⎝ ∂t
ρ ⎜⎜
⎡∂ 2u y ∂ 2u y ∂ 2u y ⎤
⎞
∂p
⎟=−
+ µ⎢ 2 +
+
⎥ + ρf y
⎟
∂y
∂y 2
∂z 2 ⎥⎦
⎢⎣ ∂x
⎠
⎡∂ 2u
∂ 2u z ∂ 2u z ⎤
⎞
∂p
⎟⎟ = −
+
+ µ ⎢ 2z +
⎥ + ρf z
∂z
∂z 2 ⎦
∂y 2
⎠
⎣ ∂x
cylindrical coordinates (r, θ, z):
⎛ ∂ur
⎡ ∂ ⎛1 ∂
∂u u ∂u u 2
∂u ⎞
∂p
1 ∂ 2u ∂ 2u
2 ∂u ⎤
+ ur r + θ r − θ + u z r ⎟ = − + µ ⎢ ⎜
( rur ) ⎞⎟ + 2 2r + 2r − 2 θ ⎥ + ρ f r
r ∂θ
r
r ∂θ ⎦
∂r
∂z ⎠
∂r
∂z
⎠ r ∂θ
⎣ ∂r ⎝ r ∂r
⎝ ∂t
ρ⎜
∂u u ∂u u u
∂u
⎛ ∂uθ
+ ur θ + θ θ + r θ + u z θ
r ∂θ
r
∂r
∂z
⎝ ∂t
ρ⎜
2
2
⎡ ∂ ⎛1 ∂
1 ∂p
⎞
⎞ 1 ∂ uθ ∂ uθ 2 ∂ur ⎤
ru
=
−
+
+
+
+
µ
(
)
θ ⎟
⎢ ⎜
⎥ + ρ fθ
⎟
2
2
r ∂θ
∂z 2 r 2 ∂θ ⎦
⎠ r ∂θ
⎠
⎣ ∂r ⎝ r ∂r
⎡ 1 ∂ ⎛ ∂u z ⎞ 1 ∂ 2u z ∂ 2u z ⎤
∂u u ∂u z
∂u ⎞
∂p
⎛ ∂u z
+ ur z + θ
+ uz z ⎟ = − + µ ⎢
+ 2 ⎥ + ρ fz
⎜r
⎟+ 2
2
∂r
∂z ⎠
∂z
∂z ⎦
r ∂θ
⎝ ∂t
⎣ r ∂r ⎝ ∂r ⎠ r ∂θ
ρ⎜
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ME 309 Exam Formula Sheet
Math Formulas d
(cos x ) = − sin x
dx
d
( tan x ) = sec2 x
dx
d
(csc x ) = − csc x cot x
dx
d
( ln x ) = 1x
dx
d
(sin x ) = cos x
dx
d
(sec x ) = sec x tan x
dx
d
(cot x ) = − csc2 x
dx
d
⎡exp ( ax ) ⎤⎦ = aexp ( ax )
dx ⎣
( )
( )
tan α + tan β
tan (α + β ) =
1− tan α tan β
∫ sin x dx = x − sin 2x
∫ cos x dx = x + sin 2x
x
dx
=
tan
x
−
x
tan
∫
∫ sin x dx = − ( 2 + sin x ) cos x
∫ cos x dx = ( 2 + cos x ) sin x ∫ tan x dx = tan x + ln cos x
2
1
2
1
2
4
2
1
3
3
2
1
3
2
1
3
2
(
cos α + β = cos α cos β − sin α sin β
1
2
3
⎛θ⎞ 1
sin 2 ⎜ ⎟ = 1− cosθ
⎝ 2⎠ 2
sin α + β = sin α cos β + cos α sin β
1
4
2
)
⎛θ⎞ 1
cos2 ⎜ ⎟ = 1+ cosθ
⎝ 2⎠ 2
(
)
Miscellaneous vector operations (In the table below: Ν is a scalar and n is a vector.) rectangular coordinates (x, y, z)
cylindrical coordinates (r, θ, z)
∇N =
∂N
∂N
∂N
ê +
ê +
ê
∂x x ∂ y y ∂z z
∇⋅n =
∇N =
∂nx ∂ny ∂nz
+
+
∂x ∂ y ∂z
∇⋅n =
⎛ ∂n ∂n ⎞
∇ × n = ⎜ z − y ⎟ ê x +
⎝ ∂ y ∂z ⎠
∇2 N =
1 ∂N
∂N
∂N
ê +
ê +
ê
∂r r r ∂θ θ ∂z z
∂n ∂n
1 ∂
( rn ) + 1 θ + ∂zz
r ∂r r r ∂θ
⎛ 1 ∂nz ∂nθ ⎞
∇×n=⎜
−
ê +
∂z ⎟⎠ r
⎝ r ∂θ
⎛ ∂nx ∂nz ⎞
⎜⎝ ∂z − ∂x ⎟⎠ ê y +
⎛ ∂nr ∂nz ⎞
⎜⎝ ∂z − ∂r ⎟⎠ êθ +
⎛ ∂ny ∂nx ⎞
⎜ ∂x − ∂ y ⎟ ê z
⎠
⎝
∂n ⎞
1⎛ ∂
( rn ) − ∂θr ⎟⎠ ê z
r ⎜⎝ ∂r θ
∂2 N ∂2 N ∂2 N
+
+
∂x 2 ∂ y 2 ∂z 2
∇2 N =
Lagrangian (aka material, substantial) Acceleration rectangular coordinates (x, y, z)
Dux ∂ux
∂u
∂u
∂u
=
+ ux x + u y x + uz x
Dt
∂t
∂x
∂y
∂z
Du y ∂u y
∂u
∂u
∂u
=
+ ux y + u y y + uz y
Dt
∂t
∂x
∂y
∂z
Duz ∂uz
∂u
∂u
∂u
=
+ ux z + u y z + uz z
Dt
∂t
∂x
∂y
∂z
1 ∂ ⎛ ∂N ⎞ 1 ∂2 N ∂2 N
r
+
+
r ∂r ⎜⎝ ∂r ⎟⎠ r 2 ∂θ 2 ∂z 2
cylindrical coordinates (r, θ, z)
Dur ∂ur
∂ur uθ ∂ur uθ2
∂u
=
+ ur
+
− + uz r
Dt
∂t
∂r
r ∂θ r
∂z
Duθ ∂uθ
∂uθ uθ ∂uθ ur uθ
∂u
=
+ ur
+
+
+ uz θ
Dt
∂t
∂r
r ∂θ
r
∂z
Du z ∂u z
∂u z uθ ∂u z
∂u z
=
+ ur
+
+ uz
Dt
∂t
∂r
r ∂θ
∂z
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ME 309 Exam Formula Sheet
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ME 309 Exam Formula Sheet
Average Roughness of Commercial Pipes Material (new)
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Asphalted cast iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
ft
0.003-0.03
0.001-0.01
0.0006-0.003
0.00085
0.0005
0.0004
0.00015
0.000005
0.0 (smooth)
mm
0.9-9.0
0.3-3.0
0.18-0.9
0.26
0.15
0.12
0.045
0.0015
0.0 (smooth)
Table of Minor Loss Coefficients Component
a.
Elbows
Regular 90o, flanged
Regular 90o, threaded
Long radius 90o, flanged
Long radius 90o, threaded
Long radius 45o, flanged
Regular 45o, threaded
b.
c.
d.
K
Component
0.3
1.5
0.2
0.7
0.2
0.4
180o return bends
180o return bends, flanged
180o return bends, threaded
0.2
1.5
Tees
Line flow, flanged
Line flow, threaded
Branch flow, flanged
Branch flow, threaded
0.2
0.9
1.0
2.0
Union, threaded
0.06
e.
f.
g.
h.
K
Valves
Globe, fully open
Angle, fully open
Gate, fully open
Gate, ¼ closed
Gate, ½ closed
Gate, ¾ closed
Swing check, forward flow
Swing check, backward flow
Ball valve, fully open
Ball valve, 1/3 closed
Ball valve, 2/3 closed
10
2
0.15
0.26
2.1
17
2
∞
0.05
5.5
210
Entrances
Re-entrant
Sharp-edged
Slightly rounded
Well rounded
0.8
0.5
0.2
0.04
Exits
Re-entrant, sharp-edged,
slightly rounded, well-rounded
1
Sudden Contraction/Expansion:
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ME 309 Exam Formula Sheet
Pumps Ψ=
gH
ω 2 D2
NS =
Π=
W
ρω 3 D5
Φ=
Q
ω D3
η=
ρQgH
W
1
1
ω ( rpm ) Q ( gpm )
Φ2
ωQ 2
or N sd ≡
=
3
3
3
4
Ψ 4 ( gH ) 4
⎡⎣ H ( ft )⎤⎦
Nsd = 2733 [rpm⋅(gpm)1/2/(ft)3/4] Ns
⎛ p V2 ⎞
pv
NPSH = ⎜ +
⎟ −
ρ
g
2g
ρ
g
⎝
⎠S
δ
δD = δ* = ∫ (1 −
0
CD =
1
u
)dy
U
FD
2
ρ
2 V A
CL =
δ
δM = θ = ∫
0
1
FL
2
ρ
2 V A
Parameter
u
u
(1 − )dy
U
U
τw
d
dU
=
(U 2 θ) + δ* U
ρ dx
dx
Laminar (Re < 500,000)
99% thickness, δ
δ
x
displacement thickness, δD or δ
*
=
5.0
1
Re x2
5
δ D 1.72
=
x Re x
δ M 0.664
=
x
Re x
τw
0.664
Cf =
=
1 ρU 2
Re x
2
1.328
D
=
CD =
1 ρU 2 LW
Re L
2
2
1
2
friction coefficient, Cf
1
2
drag coefficient, CD
δ 0.382
=
x Re x
δ D 0.0478
=
x
Re x
δ M 0.0371
=
x
Re x
τw
0.0594
Cf =
=
1 ρU 2
Re x
2
0.0742
D
=
CD =
1 ρU 2 LW
Re L
2
1
1
momentum thickness, δM or Θ
Turbulent (Re > 500,000)
1
2
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1
5
1
5
1
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lOMoARcPSD|21735616
ME 309 Exam Formula Sheet
Ideal gas relations p = ρ RT
k=
du = cv dT
cp
dh = c p dT
1
k
R
cv =
R
k −1
k −1
⎛T ⎞
⎛ν ⎞
⎛T ⎞
⎛P ⎞
s2 − s1 = cv ln ⎜ 2 ⎟ + R ln ⎜ 2 ⎟ = c p ln ⎜ 2 ⎟ − R ln ⎜ 2 ⎟
⎝ T1 ⎠
⎝ ν1 ⎠
⎝ T1 ⎠
⎝ P1 ⎠
c p = cv + R
cv
c = kRT
cp =
sinα=1/Ma
Adiabatic relations for a perfect gas T0 = T +
−1
T ⎛ c ⎞ ⎛ k −1
⎞
Ma 2 ⎟
= ⎜ ⎟ = ⎜1 +
T0 ⎝ c0 ⎠ ⎝
2
⎠
2
V2
2c p
Isentropic relations for a perfect gas k
p2 ⎛ T2 ⎞
=⎜ ⎟
p1 ⎝ T1 ⎠
k
ρ2 ⎛ T2 ⎞
=⎜ ⎟
ρ1 ⎝ T1 ⎠
k −1
1
p2 ⎛ ρ2 ⎞
=⎜ ⎟
p1 ⎝ ρ1 ⎠
k −1
k
Po ⎛ k − 1
⎞ k −1
Ma 2 ⎟
= ⎜1 +
P ⎝
2
⎠
k+1
ρ o ⎛ k −1 2 ⎞
= ⎜1 +
Ma ⎟
ρ ⎝
2
⎠
1
k −1
2( k−1)
⎛
k −1
2⎞
1 ⎜ 1+ 2 Ma ⎟
A
=
⎜
⎟
A* Ma ⎜ 1+ k − 1 ⎟
⎝
⎠
2
k+1
m choked
M
0.00
0.20
0.40
0.60
0.80
1.00
1.50
2.00
2.50
3.00
⎛ k − 1 ⎞ 2(1−k )
k *
= ⎜ 1+
p0
A
RT0
2 ⎟⎠
⎝
T/To
1.0000
0.9921
0.9690
0.9328
0.8865
0.8333
0.6897
0.5556
0.4444
0.3571
P/Po
1.0000
0.9725
0.8956
0.7840
0.6560
0.5283
0.2724
0.1278
0.05853
0.02722
A/A*
ρ/ρo
1.0000
0.9803
0.9243
0.8405
0.7400
0.6339
0.3950
0.2301
0.1317
0.07623
∞
2.964
1.590
1.188
1.038
1.000
1.160
1.688
2.637
4.235
Conditions across a normal shock wave Ma 22 =
( k − 1) Ma12 + 2
2kMa12 − ( k − 1)
2
( k + 1) Ma1
ρ2 V1
=
=
ρ1 V2 2 + ( k − 1) Ma12
p2
k −1
2k
=
Ma12 −
p1 k + 1
k +1
2kMa12 − ( k − 1)
T2 ⎡
= 2 + ( k − 1) Ma12 ⎤
2
⎦
T1 ⎣
⎡⎣( k + 1) Ma1 ⎤⎦
2
p02 ρ02 A1* ⎡ ( k + 1) Ma1 ⎤
=
= * =⎢
⎥
p01 ρ01 A2 ⎢⎣ 2 + ( k − 1) Ma12 ⎥⎦
k
( k −1)
⎡
⎤
k +1
⎢
⎥
2
⎢⎣ 2kMa1 − ( k − 1) ⎥⎦
1
( k −1)
T02
=1
T01
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