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MECH 221 FLUID MECHANICS(Fall 06/07)Tutorial 5

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Outline

1. Equations of motion for inviscid flow1. Conservation of mass2. Conservation of momentum

2. Bernoulli Equation1. Bernoulli equation for steady flow2. Static, dynamic, stagnation and total pressure3. Example

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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:

Therefore, the conservation of mass requires thatdm/dt = 0.

where the Leibniz rule was invoked.

)(

/tVdV

dt

ddtdm

S

dt

dVdVt)t(V

)(tVdVm

S)t(V

ddVt

sv

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1.1. Conservation of Mass

Hence:

This is the Integral Form of mass conservation equation.

0ddVt S)t(V

sv

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1.1. Conservation of Mass

As V(t)→0, the integrand is independent of V(t) and therefore,

This is the Differential Form of mass conservation and also called as continuity equation.

0 )(

vt

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1.2. Conservation of Momentum

The Newton’s second law,

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

dt

dMF

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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:

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.

V(t)S

dVρpd gsF

Pressure force

Body force

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1.2. Conservation of Momentum

where the Leibniz rule was invoked.

)t(V dV

dt

ddt/d vM

S

dt

dV)(dV)(

t)t(Vvv

S)t(V

ddV)(t

svvv

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1.2. Conservation of Momentum

Hence:

This is the Integral Form of momentum conservation equation.

dVpdddV)(t S )t(VS)t(V

gss vvv

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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,

The momentum equation can take the following alternative form:

which is commonly referred to as Euler’s equation of motion.

0 )

v(

t

g

p)()(t

vvv

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2.1. Bernoulli Equation for Steady Flows

From differential form of the momentum conservation equation

1. g=-gVz2. By vector identity,

Therefore, we get,

)v(vv)(v2

1)(

vt

zgp

)()(2

1 )( vvvvvv

g

p)()(t

vvv

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2.1. Bernoulli Equation for Steady Flows

Assumption, Steady flow;

v and t are independent Irrotational flow;

Vxv=0

)v(vv)(v

2

1)(

vt

zgp

=0 (Steady flow) =0 (irrotational flow)

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2.1. Bernoulli Equation for Steady Flows

Finally, we can get,

Or

where v=magnitude of velocity vector v,i.e. v=√(u2+v2+w2)

0v)(v2

1 zgp

0)(2

1 2

vzgp

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2.1. Bernoulli Equation for Steady Flows

Since, for dr in any direction, we have:

For anywhere of irrotational fluids

For anywhere of incompressible fluids

dfdf r

constantz 2

g2

vp

constantz 2

g2

vdp

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2.1. Bernoulli Equation for Steady Flows

Bernoulli Equation in different form:

1. Energy density:

2. Total head (H):

][m H z 2

]m

Jor m

N[ constant z 2

2

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2

g

v

g

p

gv

p

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2.2. Static, Dynamic, Stagnation and Total Pressure

Consider the Bernoulli equation,

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.

constantz 2

g2

vp

(for incompressible fluid)

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Then we have 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

zgps

constant 2

2

vpd

2.2. Static, Dynamic, Stagnation and Total Pressure

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The maximum dynamic pressure occurs at the stagnation point where v=0 and this maximum pressure is called as the stagnation pressure p0. Hence,

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.

od p2

vp

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2.2. Static, Dynamic, Stagnation and Total Pressure

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2.3.1. Example (1)

Determine the flowrate through the pipe.

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2.3.1. Example (1)

Procedure: Choose the reference point From the Bernoulli equation

P, V, Z all are unknowns For same horizontal level, Z1=Z2

V = V(P1, P2)

From the balance of static pressure 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,

water

waterwater

waterwater

ppv

g

p

g

v

g

p

g

v

g

p

g

v

g

p

)(2

2

zz level, lhorizontia same at the

0 vpoint, stagnationat

Since,

z 2

z 2

121

22

11

21

2

2

222

1

211

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2.3.1. Example (1)

From the balance of static pressure,

smv

ghv

mkg

mkg

ghppv

ghpp

lhgpghglp

water

mwater

mm

water

mwater

water

mwater

watermwater

215.2

1000

)5.2)(81.9)(9001000(2

)(2

2.5mh ,1000 ,900Given

)(2

)(2

Therefore,

)(

)(

1

1

33

121

12

21

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2.3.1. Example (1)

Volume flow rate (Q),

smQ

vD

vAQ

D

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1

2

11

0111.0)215.2(4

)08.0(

4

0.08mGiven

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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.

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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

2

zz level, lhorizontia same at the

Since,

z 2

z 2

21

2221

222

211

21

2

222

1

211

vvpp

g

v

g

p

g

v

g

p

g

v

g

p

g

v

g

p

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2.3.2. Example (2)

From flowrate conservation,

smv

smv

mrtA

mD

A

mrmtmDsmQ

vAvAQ

894.190251.0/5.0

034.120415.0/5.0

Therefore,

0251.0)02.0)(2.0(22

0415.04

23.0

4

2.0,02.0 ,23.0 ,5.0Given

2

1

22

222

1

3

2211

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2.3.2. Example (2)

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221

22

3

21

21

2221

565.148

)034.12894.19(2

184.10

0p point, reference becomes pSet

184.1 C,25 atm, air@1 standardFor

894.19 ,034.12

)(2

mNp

p

mkg

smvs

mv

vvpp

Sub. into the Bernoulli equation,

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The End