Download Tutorial 5

MECH 221 FLUID MECHANICS
(Fall 06/07)
Tutorial 5
1
Outline
1.
Equations of motion for inviscid flow
1.
2.
2.
Conservation of mass
Conservation of momentum
Bernoulli Equation
1.
2.
3.
Bernoulli equation for steady flow
Static, dynamic, stagnation and total pressure
Example
2
1. Equations of Motion for Inviscid Flow

Conservation of Mass

Conservation of Momentum
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1.1. Conservation of Mass

Mass in fluid flows must conserve. The total mass in
V(t) is given by:
m  
V (t )

 dV
Therefore, the conservation of mass requires that
dm/dt = 0.
d 

dm / dt 

dV

dt  V (t )

 dV 
 
dV   

V ( t ) t
dt

S

 
dV    v  ds
V ( t ) t
S
where the Leibniz rule was invoked.
4
1.1. Conservation of Mass

Hence:

V ( t ) t dV   S  v  ds  0
This is the Integral Form of mass conservation
equation.
5
1.1. Conservation of Mass

As V(t)→0, the integrand is independent of V(t)
and therefore,

   ( v)  0
t
This is the Differential Form of mass
conservation and also called as continuity
equation.
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1.2. Conservation of Momentum

The Newton’s second law,
dM
F
dt
is Lagrangian in a description of momentum
conservation. For motion of fluid particles that
have no rotation, the flow is termed irrotational.
An irrotational flow does not subject to shear force,
i.e., pressure force only. Because the shear force is
only caused by fluid viscosity, the irrotational flow
is also called as “inviscid” flow
7
1.2. Conservation of Momentum

For fluid subjecting to earth gravitational acceleration,
the net force on fluids in the control volume V enclosed
by a control surface S is:
F    pd s  
S
Pressure
force
V(t)
ρgdV
Body
force
where s is out-normal to S from V and the divergence
theorem is applied for the second equality.

This force applied on the fluid body will leads to the
acceleration which is described as the rate of change in
momentum.
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1.2. Conservation of Momentum
d 

dM / dt 

v
dV

dt  V ( t )

dV 

 
( v )dV  ( v ) 
V ( t ) t
dt  S


 
( v )dV   vv  ds
V ( t ) t
S
where the Leibniz rule was invoked.
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1.2. Conservation of Momentum

Hence:

V ( t ) t ( v )dV   S vv  ds   S pds  V ( t )  gdV
This is the Integral Form of momentum
conservation equation.
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1.2. Conservation of Momentum

As V→0, the integrands are independent of V.
Therefore,
This is the Differential Form of momentum
conservation equation for inviscid flows.
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1.2. Conservation of Momentum

By invoking the continuity equation,

   ( v )  0
t

The momentum equation can take the following
alternative form:

 ( v )   ( v   )v  p  g
t
which is commonly referred to as Euler’s equation
of motion.
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2.1. Bernoulli Equation for Steady Flows

1.
2.
From differential form of the momentum
conservation equation
g=-gVz

 ( v )   ( v   )v  p  g
t
By vector identity,
1
( v  ) v  ( v  v)  v  (  v)
2

Therefore, we get,

1
p  gz   ( v)   (v  v)  v  (  v)
t
2
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2.1. Bernoulli Equation for Steady Flows

Assumption,

Steady flow;


v and t are independent
Irrotational flow;

Vxv=0

1
p  gz   ( v)   (v  v)  v  (  v)
t
2
=0 (Steady flow)
=0 (irrotational flow)
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2.1. Bernoulli Equation for Steady Flows

Finally, we can get,
1
p  gz   (v  v)  0
2
Or
p
1
2
 gz  (v )  0

2
where v=magnitude of velocity vector v,
i.e. v=√(u2+v2+w2)
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2.1. Bernoulli Equation for Steady Flows

Since, f  dr  df for dr in any direction, we have:

For anywhere of irrotational fluids


v2

 gz  constant

2
dp
For anywhere of incompressible fluids
v2

 gz  constant
 2
p
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2.1. Bernoulli Equation for Steady Flows

Bernoulli Equation in different form:
1.
Energy density:
p
2.
v 2
2
 gz  constant
[N
m
2
or J
m3
]
Total head (H):
p
v2

zH
g 2 g
[m ]
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2.2. Static, Dynamic, Stagnation and Total Pressure

Consider the Bernoulli equation,
p

v 2
2
 gz  constant
(for incompressible fluid)
The static pressure ps is defined as the pressure
associated with the gravitational force when the
fluid is not in motion. If the atmospheric pressure
is used as the reference for a gage pressure at
z=0.
18
2.2. Static, Dynamic, Stagnation and Total Pressure


Then we have ps   gz as also from chapter 2.
The dynamic pressure pd is then the pressure
deviates from the static pressure, i.e., p = pd+ps.
The substitution of p = pd+ps. into the Bernoulli
equation gives
pd 
v 2
2
 constant
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2.2. Static, Dynamic, Stagnation and Total Pressure

The maximum dynamic pressure occurs at the
stagnation point where v=0 and this maximum
pressure is called as the stagnation pressure p0.
Hence,
pd 

v
2
2
 po
The total pressure pT is then the sum of the
stagnation pressure and the static pressure, i.e.,
pT= p0 - ρgz. For z = -h, the static pressure is ρgh
and the total pressure is p0 + ρgh.
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2.3.1. Example (1)

Determine the flowrate through the pipe.
21
2.3.1. Example (1)

Procedure:


Choose the reference point
From the Bernoulli equation




From the balance of static pressure




P, V, Z all are unknowns
For same horizontal level, Z1=Z2
V = V(P1, P2)
P = ρgh
Δh is given, ρm, ρwater are known
V = V(Δh, ρm, ρwater)
Q = AV = πD2V/4
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2.3.1. Example (1)

From the Bernoulli equation,
p1
 water g
2

2
v1
p2
v
 z1 
 2  z2
2g
 water g 2 g
Since,
at stagnation point, v 2  0
at the same horizontia l level, z1  z 2
p1
2
v1
p2


 water g 2 g  water g
v1  2
( p2  p1 )
 water
23
2.3.1. Example (1)

From the balance of static pressure,
p1   water gl   m gh  p2   water g (h  l )
p2  p1  (  water   m ) gh
Therefore,
v1  2
( p2  p1 )
 water
 2
Given  m  900 kg
v1  2
3
v1  2.215 m
 water
,  m  1000 kg
m
(  water   m ) gh
 water
(  water   m ) gh
, h  2.5m
m3
(1000  900)(9.81)( 2.5)
 2
1000
s
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2.3.1. Example (1)

Volume flow rate (Q),
Given D  0.08m
Q  A1v1 
Q
D 2
4
 (0.08) 2
4
v1
3
m
(2.215)  0.0111
s
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2.3.2. Example (2)

A conical plug is used to regulate the air flow
from the pipe. The air leaves the edge of the
cone with a uniform thickness of 0.02m. If
viscous effects are negligible and the flowrate
is 0.05m3/s, determine the pressure within the
pipe.
26
2.3.2. Example (2)

Procedure:


Choose the reference point
From the Bernoulli equation



P, V, Z all are unknowns
For same horizontal level, Z1=Z2
Flowrate conservation

Q=AV
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2.3.2. Example (2)

From the Bernoulli equation,
2
2
p1 v1
p 2 v2

 z1 

 z2
g 2 g
g 2 g
Since,
at the same horizontia l level, z1  z 2
2
2
p1 v1
p 2 v2



g 2 g g 2 g
p1  p2 

2
(v2  v1 )
2
2
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2.3.2. Example (2)

From flowrate conservation,
Q  A1v1  A2 v2
3
m
Given Q  0.5
, D  0.23m, t  0.02m, r  0.2m
s
D 2  0.232
A1 

 0.0415m 2
4
4
A2  2rt  2 (0.2)(0.02)  0.0251m 2
Therefore,
v1  0.5 / 0.0415  12.034 m
s
v2  0.5 / 0.0251  19.894 m
s
29
2.3.2. Example (2)

Sub. into the Bernoulli equation,
p1  p2 

(v2  v1 )
2
2
2
v1  12.034 m , v2  19.894 m
s
s
For standard air@1 atm, 25C,   1.184 kg
m3
Set p 2 becomes reference point, p 2  0
1.184
p1  0 
(19.894 2  12.034 2 )
2
p1  148.565 N 2
m
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The End
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