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# Chapter 4 Continuity, Energy, and Momentum Equations

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(1)

## Continuity, Energy, and Momentum

Equations

(2)

Contents

4.1 Conservation of Matter in Homogeneous Fluids 4.2 The General Energy Equation

4.3 Linear Momentum Equation for Finite Control Volumes

4.4 The Moment of Momentum Equation for Finite Control Volumes Objectives

- Apply finite control volume to get integral form of continuity, energy, and momentum equations

- Compare integral and point form equations

- Derive the simplified equations for continuity, energy, and momentum equations

(3)

Infinitesimal elements and control volumes

 Each of the observational laws of mass, heat, and momentum transport may be formulated in the Eulerian sense of focusing attention on a fixed point in space.

 There are two basic method of arriving the Eulerian equation.

• Material method (Particle approach)

• Control volume method:

 Finite control volume

 Differential control volume

• If fluid is considered as a continuum, end result of either method is identical.

(4)

 Material method (Particle approach)

• Describe flow characteristics at a fixed point (x, y, z) by observing the motion of a material particle of a infinitesimal mass

• Laws of conservation of mass, momentum, and energy can be stated in the differential form, applicable at a point.

• Newton's 2nd law

• Material-particle approach is used to develop a stress-strain relationship in Ch. 5.

## dF  = dma 

(5)

 Control volume method

① Finite control volume – arbitrary control volume

② Differential (infinitesimal) control volume – parallelepiped control volume

[Re] Control volume

- fixed volume which consists of the same fluid particles and whose bounding surface moves with the fluid

(6)

 Finite control volume method

• Frequently used for 1D analysis (Ch. 4)

• Gross descriptions of flow

• Analytical formulation is easier than differential control volume method

• Integral form of equations for conservation of mass, momentum, and energy

• Continuity equation: conservation of mass

CV

dV

CS

q dA

0

t

ρ ρ

∂ + ⋅ =

∫ ∂  ∫

(7)
(8)

 Differential control volume method

• Concerned with a fixed differential control volume (=∆x∆y∆z) of fluid

• 2D or 3D analysis (Ch. 6)

( ) ( )

d d

F mq x y zq

dt dt ρ

∆ =  ∆  = ∆ ∆ ∆ 

• ∆x, ∆y, ∆z become vanishingly small

• Point form of equations for conservation of mass, momentum, and energy

(9)
(10)
(11)

4.1.1 Finite control volume method-arbitrary control volume

• Consider an arbitrary control volume

• Although control volume remains fixed, mass of fluid originally enclosed (regions A+B) occupies the volume within the dashed line (regions B+C).

• Since mass m is conserved:

( )

mA t +

( )

mB t =

( )

mB t dt+ +

( )

mC t dt+

( ) m

B t dt

( ) m

B t

( ) m

A t

( ) m

C t dt

dt dt

+

− −

+

=

(4.1)

(4.2)

(12)

• LHS of Eq. (4.2) = time rate of change of mass in the original control volume in the limit

( )

B t dt

( )

B t

( )

B

( )

CV

m m m

dt t t

ρ

dV

+ − ≈ ∂ = ∂

∂ ∂

(4.3)

where dV = volume element

• RHS of Eq. (4.2)

= net flux of matter through the control surface

= flux in – flux out

1 2

n n

q dA q dA

ρ ρ

= ∫ − ∫

(13)

where = component of velocity vector normal to the surface of

CV = q

cos

φ

( )

n 1 n 2

CV

dV

CS

q dA

CS

q dA

t ρ ρ ρ

∴ ∂ = −

∂ ∫ ∫ ∫

(4.4)

※ Flux (= mass/time) is due to velocity of the flow.

• Vector form is

( )

CV dV CS q dA

t

ρ ρ

∂ = − ⋅

∫  ∫

(4.5)

q

n
(14)

where = vector differential area pointing in the outward direction over an enclosed control surface

dA

## cos q dA q dA φ

∴ ⋅   =  

positive for an outflow from cv negative for inflow into cv,

, 90

90 180

φ φ

 ≤

=    ≤ ≤

If fluid continues to occupy the entire control volume at subsequent times

→ time independent

LHS:

( )

CV dV CV dV

t t

ρ ρ

∂ ∂

(4.5a)
(15)

CV dV CS q dA 0 t

ρ ρ

∂ + ⋅ =

 ∫

Eq. (4.4) becomes

(4.6)

→ General form of continuity equation→ Integral form [Re] Differential form

Use Gauss divergence theorem

V A i

i

F dV FdA x

∂ =

∫ ∂ ∫

(16)

Transform surface integral of Eq. (4.6) into volume integral

CS

ρ q dA ⋅ =

CV

∇⋅ ( ) ρ q dV

∫   ∫ 

Then, Eq. (4.6) becomes

( )

0

CV q dV

t

ρ ρ

∂ + ∇⋅  =

 ∂ 

 

Eq. (4.6a) holds for any volume only if the integrand vanishes at every point.

(4.6b) (4.6a)

( )

q 0

t

ρ ρ

∂ + ∇⋅ =

→ Differential (point) form

(17)

## Simplified form of continuity equation

 Steady flow of a compressible fluid

CV

dv

0

t ρ

∂ =

∫ ∂

Therefore, Eq. (4.6) becomes

CS

ρ

q dA⋅ = 0

(4.7)

const. 0, d 0

t dt

ρ ρ

ρ

= → = =

(18)

CSq dA⋅ = 0

(4.8)

Therefore, Eq. (4.6) becomes

(19)

c c 0 uc D

t x x

∂∂ + ∂∂  − ∂∂  =

c uc c

t x x D x

∂ ∂ ∂  ∂ 

→ ∂ + ∂ = ∂  ∂  [Cf] Non-homogeneous fluid mixture

• Conservation of mass equations for the individual species

= conservation of mass equation + mass flux equation due to advection and diffusion

c q

0

t x

∂ + ∂ =

∂ ∂

q uc D c x

= − ∂

(20)

4.1.2 Stream - tube control volume analysis for steady flow

• Steady flow: There is no flow across the longitudinal boundary of the stream tube.

• Eq. (4.7) becomes

(4.9)

1 1 1 2 2 2

q dA q dA q dA

ρ ⋅ = − ρ + ρ =

∫  

## const.

ρ qdA =

Flux = 0

(21)

where dQ = volume rate of flow If density = const.

1 1 2 2

q dA = q dA = dQ

(4.10)

 For flow in conduit with variable density V qdA

=

A → average velocity

→ average density dQ

Q

ρ

=

∫ ρ

1V A1 1 2V A2 2

ρ = ρ (4.11)

(22)

 For a branching conduit

0 ρ q dA ⋅ =

∫  

1 2 3

1 1 1 2 2 2 3 3 3

### 0

A

ρ q dA

A

ρ q dA

A

ρ q dA

− ∫ + ∫ + ∫ =

1

V A

1 1 2

V A

2 2 3

V A

3 3

ρ = ρ + ρ

(4.12)
(23)

 At the centroid of the control volume,

 rate of mass flux across the surface perpendicular to x is

 Equation of Continuity

Use Infinitesimal (differential) control volume method

, u v w , ,

ρ

(24)

net mass flux across the surface perpendicular to y flux in

( )

2

## u dx dydz

u x

ρ ρ

=

 ∂ 

 − 

 

flux out

( )

2

u dx dydz

u x

ρ ρ

=

 ∂ 

 + 

 

net flux = flux in flux out

( ) u

dxdydz x

∂ ρ

= − ∂

( )

v

dydxdz y

ρ

= − ∂

net mass flux across the surface perpendicular to z

( )

w

dzdxdy z

ρ

= − ∂

∂ Time rate of change of mass inside the c.v.

( dxdydz )

t ρ

= ∂

(25)

(A1) Time rate of change of mass inside = sum of three net rates

( dxdydz ) ( ) u ( ) v ( ) w

## dxdydz

t x y z

ρ ρ ρ ρ

∂  ∂ ∂ ∂ 

= −  + + 

∂  ∂ ∂ ∂ 

By taking limit dV = dxdydz

( ) u ( ) v ( ) w q div ( ) q

t x y z

ρ ρ ρ

ρ ∂ ∂ ∂ ρ ρ

− ∂ = + + = ∇ ⋅ =

∂ ∂ ∂ ∂

 

0 t q

ρ ρ

∂ + ∇ ⋅ =

→ point (differential) form of Continuity Equation (the same as Eq. 4.6b)

(26)

[Re]

( ) u ( ) v ( ) w q div ( ) q

x y z

ρ ρ ρ ρ ρ

∂ ∂ ∂

+ + = ∇ ⋅ =

∂ ∂ ∂

 

By the way,

q q q

ρ ρ ρ

∇⋅

= ⋅∇ + ∇⋅

 

Thus, (A1) becomes

0

q q

t

ρ ρ ρ

∂ + ⋅∇ + ∇⋅ =

  (A2)

(27)

1) For incompressible fluid 0 ( const.)

d

dt

ρ = ρ =

d

0

t q dt

ρ ρ ρ

→ ∂ + ⋅∇ = =

Therefore Eq. (A2) becomes

0 0

q q

ρ ∇⋅ =  → ∇⋅  =

u v w 0

x y z

∂ + ∂ + ∂ =

∂ ∂ ∂

In scalar form,

(A3)

(A4)

→ Continuity Eq. for 3D incompressible fluid

(28)

For 2D incompressible fluid,

u v

0

x y

∂ + ∂ =

∂ ∂

2) For steady flow, t 0

ρ

∂ =

Thus, (A1) becomes

### 0

q q q

ρ ρ ρ

∇ ⋅   = ⋅∇ + ∇⋅ = 

(4.13)
(29)

4.2.1 The 1st law of thermodynamics

 The 1st law of thermodynamics:

The difference between the heat added to a system of masses and the work done by the system depends only on the initial and final states of the system (→ change in energy).

→ Conservation of energy

Q W

E

(30)

where δQ = heat added to the system from surroundings

δW = work done by the system on its surroundings δΕ = increase in energy of the system

Q W dE

δ − δ =

(4.14) Q W

E

(31)

[Re]

• property of a system: position, velocity, pressure, temperature, mass, volume

• state of a system: condition as identified through properties of the system

Consider time rate of change

Q W dE

dt dt dt

δ

δ

= (4.15)
(32)

 Energy

Consider e = energy per unit mass = E/mass

eu = internal energy associated with fluid temperature = u ep = potential energy per unit mass = gh

where h = local elevation of the fluid eq= kinetic energy per unit mass =

Work

Wpressure = work of normal stresses acting on the system boundary Wshear = work of tangential stresses done at the system boundary

on adjacent external fluid in motion

Wshaft = shaft work done on a rotating element in the system

2

2

q

(33)

enthalpy

u p

+ ρ =

2

u p q

2

e = e + e + e = u + gh + q

(4.16)

• Internal energy

= activity of the molecules comprising the substance

= force existing between the molecules

~ depend on temperature and change in phase

(34)

(4.15) 4.2.2 General energy equation

Q W dE

dt dt dt δ − δ =

Consider work done

pressure shaft shear

W W W

W

dt dt dt dt

δ δ δ

δ = + +

(4.15a)

pressure

W dt

δ

= net rate at which work of pressure is done by the fluid on the surroundings

= work fluxout – work fluxin

( )

CS

p q dA ⋅

=  ∫  

(35)
(36)

p = pressure acting on the surroundings = F/A = F/ L2 positive for outflow into CV

negative for inflow

q dA

 

⋅ =  

 

3

/ q dA  ⋅  = = Q L t

( ) F L

2 3

/ /

p q dA FL t E t

= L t = =

⋅ 

Thus, (4.15a) becomes

( )

shaft shear

CS

W W

W p q dA

dt dt dt

δ δ

δ = + +

(4.15b)
(37)

= total rate change of stored energy

= net rate of energy flux through C.V.

Now, consider energy change term dE

dt

+ time rate of change inside C.V.

( ) ( )

CS

e q dA

CV

e dV

ρ ∂ t ρ

= ⋅ +

∫   ∂ ∫

(4.15c)

( )

/ ; /

## e = E mass ρ q dA

= mass time

( )

/

e ρ q dA

= E t

(38)

Substituting (4.15b) and (4.15c) into Eq. (4.15) yields

( )

shaft shear

CS

Q W W

p q dA

dt dt dt

δ

δ

δ

( ) ( )

CS

e q dA

CV

e dV

ρ ∂ t ρ

= ⋅ +

∫   ∂ ∫

shaft shear

Q W W

dt dt dt

δ − δ − δ

( ) ( )

CS CV

p e q dA e dV

ρ t ρ

ρ

 +  ∂

=  ∫      ⋅  + ∂ ∫

(4.16)

(4.17)

(39)

Assume potential energy ep = gh (due to gravitational field of the earth)

Then

Then, Eq. (4.17) becomes

2

2 e = u + gh + q

( )

2

2

shaft shear

CS CV

W W

Q

dt dt dt

p q

u gh q dA e dV

t

δ δ

δ

ρ ρ

ρ

− −

  ∂

=  ∫   + + +  

+ ∂ ∫

(4.17)
(40)

 Application: generalized apparatus

At boundaries normal to flow lines → no shear

shear

### 0

→ W =

(4.18)

( )

2

2

shaft

CS CV

Q W p q

e dV

u gh q dA

dt dt t

δ δ ρ ρ

ρ

  ∂

− =  ∫   + + +  

+ ∂ ∫

(4.19)

(4.20)

( )

2

2

shaft

CS

Q W p q

u gh q dA

dt dt

δ δ ρ

ρ

 

− =  + + +  ⋅

 

∫  

(41)
(42)

 Effect of friction

• This effect is accounted for implicitly.

• This results in a degradation of mechanical energy into heat which may be transferred away (Q, heat transfer), or may cause a temperature change → modification of internal energy.

• Thus, Eq. (4.20) can be applied to both viscous fluids and non-viscous fluids (ideal frictionless processes).

(43)

4.2.3 1 D Steady flow equations

For flow through conduits, properties are uniform normal to the flow direction.

Integrated form of Eq. (4.20) = ② - ①

2 2

2 2

shaft

Q W p V p V

Q Q

u gh u gh

dt dt

δ δ ρ ρ

ρ ρ

   

− =  + + +  −  + + + 

   

( )

2

2

shaft

CS

Q W

dt dt

p q

u gh q dA

δ δ

ρ ρ

− =

 

+ + + ⋅

 

 

∫  

(44)

2

where average kinetic energy per unit mass

2 V =

( )

1 mass flow rate into CV

1:

## Section ∫ ρ q dA  ⋅  = − ρ Q =

( )

2 mass flow from CV

2:

Section ∫ ρ q dA  ⋅  = ρ Q =

ρ Q

Divide by (mass/time)

2 2

2 2

shaft

heat transfer W p V p V

u gh u gh

mass mass

ρ ρ

   

− =  + + +  −  + + + 

   

M = ρ Q dt

(45)

 Energy Equation for 1-D steady flow: Eq. (4.21)

• use average values for p, γ, h, u, and V at each flow section

• use Ke(energy correction coeff.) to account for non-uniform velocity distribution over flow cross section

Divide by

g

2 2

2 2

shaft

heat transfer W u p V u p V

h h

weight weight g

γ

g g

γ

g

   

− =  + + +  −  + + + 

   

(4.21)

(46)

---- kinetic energy/time

2 2

2 2

K

e

ρ V Q = ρ q dQ

∫ = 1 2 mV t

2

2

2

2 1

2

e

q dQ K

V Q ρ

= ∫ ρ ≥

2 2

2 1

2 2

shaft

e e

heat transfer W p V p V u u

h K h K

weight weight

γ

g

γ

g g

    −

− =  + +  −  + +  +

   

(4.22)

(4.23) Ke = 2, for laminar flow (parabolic velocity distribution)

1.06, for turbulent flow (smooth pipe)

(47)

For a fluid of uniform density γ

(4.24)

→ unit: m (energy per unit weight) For viscous fluid;

1 2

2 1

L

## heat transfer u u

weight g H

− + − =

→ loss of mechanical energy

~ irreversible in liquid

1 2

2 2

1 1 2 2 2 1

1 2

2 2

shaft

e e

p V p V W heat transfer u u

h K h K

g g weight weight g

γ γ

+ + = + + + − + −

(48)

where H1 , H2 = weight flow rate average values of total head

where ∆HM = shaft work transmitted from the system to the outside Then, Eq. (4.24) becomes

1 2 1 2

2 2

1 1 2 2

1 2

2 2

e e M L

p V p V

h K h K H H

g g

γ

+ + =

γ

+ + + ∆ + ∆ (4.24a)

1 2 M L1 2

H = H + ∆ H + ∆ H

(4.24b)
(49)

 Bernoulli Equation Assume

① ideal fluid → friction losses are negligible

② no shaft work →

③ no heat transfer and internal energy is constant →

M 0

H =

1 2 0

HL

∆ =

1 2

2 2

1 1 2 2

1 2

2 2

e e

p V p V

h K h K

g g

γ + + = γ + +

1 2

H = H

(4.25)

(50)

If Ke1 = Ke2 = 1, then Eq. (4.25) reduces to

2 2

1 1 2 2

1 2

2 2

p V p V

H h h

g g

γ γ

= + + = + + (4.26)

~ total head along a conduct is constant

(51)

1) Energy (total head) line (E.L) ~ above datum

= above datum

H

p h γ

 + 

 

 

For flow through a pipe with a constant diameter

2 2

1 2

1 2

2 2

V V

V V

g g

= → =

(52)
(53)

1) If the fluid is real (viscous fluid) and if no energy is being added, then the energy line may never be horizontal or slope upward in the direction of flow.

2) Vertical drop in energy line represents the head loss or energy dissipation.

(54)

4.3.1 Momentum Principle

 The momentum equation can be derived from Newton's 2nd law of motion

= boundary (surface) forces: normal to boundary - pressure, tangential to boundary - shear, body forces - force due to gravitational or magnetic fields,

(4.27)

( )

dq d mq dM

F ma m

dt dt dt

= = = =

 

  

M  =

= mq 

linear momentum vector

F

=

external force

F

p

Fs

Fb

(55)

where = body force per unit mass

p s b

F F F dM + + = dt

   

( ) ,

b b

F  = ∫

CV

f ρ dv f

b

(4.28)

(56)

4.3.2 The general linear momentum equation

Consider change of momentum

= total rate of change of momentum

= net momentum flux across the CV boundaries + time rate of increase of momentum within CV

where = momentum flux =

= vector unit area pointing outward over the control surface dM

dt

( )

CS

q q dA

CV

q dV

ρ ∂ t ρ

= ⋅ +

∫    ∂ ∫ 

(4.29)

( )

q 

ρ

q dA velocity

×

mass per time

dA

(57)
(58)

Substitute (4.29) into (4.28)

( )

p s b CS CV

F F F q q dA q dV

ρ ∂ t ρ

+ + = ⋅ +

∫ ∂ ∫

      

(4.30)

For steady flow and negligible body forces

( )

p s

F

+ F

=  ∫

CS

q

 

ρ q dA ⋅

(4.31)

 Eq. (4.30)

• It is applicable to both ideal fluid systems and viscous fluid systems involving friction and energy dissipation.

• It is applicable to both compressible fluid and incompressible fluid.

(59)

• Combined effects of friction, energy loss, and heat transfer appear implicitly in the magnitude of the external forces, with corresponding effects on the local flow velocities.

• Knowledge of the internal conditions is not necessary.

• We can consider only external conditions.

(60)

4.3.3 Inertial control volume for a generalized apparatus

• Three components of the forces

( )

. : px sx bx CS CV

## x dir F F F u q dA u dV

ρ ∂ t ρ

− + + = ⋅ +

∫ ∂ ∫

    

( )

. : py sy by

CS CV

y dir F F F v q dA v dV

ρ ∂ t ρ

− + + = ⋅ +

∫ ∂ ∫

    

( )

. : pz sz bz

CS CV

z dir F F F w q dA w dV

ρ ∂ t ρ

− + + = ⋅ +

∫ ∂ ∫

    

(4.32)
(61)

 For flow through generalized apparatus

2 1

. : px sx bx

x dir F F F u dQ u dQ CV u dV

ρ ρ

t

ρ

− + + = − +

∫ ∫

  

CV q dV 0

t

ρ

∂ =

(62)

• Velocity and density are constant normal to the flow direction.

( ) (

2

)

1

.: px sx bx x x x

x dirF +F +F =

F = V

ρ

QV

ρ

Q

1Q1 2 Q2 Q (4.12)

ρ = ρ = ρ

(

2 1

) ( )

2 2 2 1 1 1 x x xout xin

x x V V V V

V

ρ

Q V

ρ

Q Q

ρ

Q

ρ

= − = =

( ) (

2

)

1

.: y

V

y

Q V

y

Q

y − dir ∑ F = ρ − ρ

( ) (

2

)

1

.: z z z

zdir

F = V

ρ

QV

ρ

Q where V = average velocity in flow direction
(63)

If velocity varies over the cross section, then introduce momentum flux coefficient

( )

m

( )

q ρ q dA ⋅ = K V ρ VA

q ρ dQ = K V Q

m

ρ

m

K q dQ

V Q

ρ

=

ρ

• Non-uniform velocity profile

(64)

where

V = magnitude of average velocity over cross section = Q/A

= average velocity vector

Km= momentum flux coefficient ≥ 1

= 1.33 for laminar flow (pipe flow)

1.03-1.04 for turbulent flow (smooth pipe) V

( ) (

2

)

1

x m x m x

F = K V

ρ

QK V

ρ

Q

(

m y

) (

2 m y

)

1

y K V Q K V Q

F =

ρ

ρ

( ) (

2

)

1

z m z m z

F = K V Q

ρ

K V Q

ρ

(65)

[Cf] Energy correction coefficient

2

2 2

e

q dQ K

VQ ρ

=

ρ

(66)

A large tank mounted on rollers is filled with water to a depth of 16 ft above a discharge port. At time t = 0, the fast-acting valve on the

discharge nozzle is opened.

Determine depth h, discharge rate Q, and force F necessary to keep the tank stationary at t = 50 sec .

[Example 4-4] Continuity, energy, and linear momentum with unsteady flow

(67)

Continuity, energy, and linear momentum equations

CV dV CS q dA 0

t

ρ ρ

∂ + ⋅ =

 ∫

(4.6)

( )

2

2

shaft shear

CS CV

W W

Q

dt dt dt

p q

u gh q dA e dV

t

δ δ

δ

ρ ρ

ρ

− −

  ∂

=  ∫   + + +  

+ ∂ ∫

(4.17)

( )

p s b

CS CV

F F F q q dA q dV

ρ ∂ t ρ

+ + = ⋅ +

∫ ∂ ∫

      

(4.30)

(68)

i) Use integral form of continuity equation, Eq. (4.6)

ii) Energy equation, Eq. (4.17)

~ no shaft work

~ heat transfer and temperature changes due to friction are negligible (because no inflow across the Section 1)

1 2

n n

CV dV q dA q dA

t

ρ ρ ρ

∂ = −

∫ ∫ ∫

1 , n 1 0

dV = A dh

ρ

q dA =

1 2 2

0

A hdh V A

ρ

t

ρ

∴ = −

1 2 2

A dh V A

dt = − (A)

(69)

e = energy per unit mass = Q

dt

δ

Wshaft

dt

δ

Wshear

dt

δ

( )

2

2

CS CV

p q

e dV

u gh q dA

ρ t ρ

ρ

  ∂

=  ∫   + + +    ⋅  + ∂ ∫

I II

2

2 u + gh + q

I = 2

( )

2

CS

p q

u gh ρ q dA

ρ

 

+ + + ⋅

 

 

2 2

2 2 1 1

2 1

2 2

p q p q

V A V A

u gh ρ u gh ρ

ρ ρ

   

=  + + +  −  + + + 

   

(70)

II =

( )

2

2 2 1

2

2 0

p q

V A V

u gh

ρ

ρ

 

=  + + +  ≈

 

A dh1 2

CV 2

e dV u gh q

t

ρ

t

∂ ∂

= + +

CV

ρ

dV

 

nearly constant in the tank except near the nozzle

( )

1 0

A h u gh dh

ρ

t

= +

( )

2

2 2 1 0

2

0 2

p q h

V A A dh

u gh u gh

ρ ρ

t

ρ

  ∂

∴ =  + + +  + ∂

+
(71)

Assume

ρ =

const. ,

p

2

= p

atm

=

0 ,

h

2

=

0 (datum)

2 2

2 2 2 2 1 1

0 2

V dh dh

uV A V A uA A gh

dt dt

= + + +

(B)

Substitute (A) into (B)

2 2

0 = uV A 22 2 2

(

2 2

)

2

V V A u V A

+ + − + gh

(

V A2 2

)

2 2

2 2 2 2

2

V V A ghV A

∴ =

2 2

V = gh (C)

1 2 2

A dh V A dt = −

(72)

Substitute (C) into (A)

2 2 1 dh

A gh A

= − dt

2 1

dh A

2

A g dt h = −

Integrate

0 0 0

1

2 2 1

0 2

1

2 2

h t h h

h h h

dh A

g dt h dh

A h h

  

= −   =    

∫ ∫ ∫

1 2 2 2 0

1

2 2

A g

h h t

A

 

=  − 

 

(73)

( ) 2 0.1 2 32.2

16 20 2

h =  − t

 

( 4 0.0201 t )

2

= −

( )

2

At

## t = 50sec , h = 4 − 0.0201 50 × = 8.98 ft ( )( )

2

2 2 32.2 8.98 24.05 fps

V = gh = =

( ) ( )

2 2

24.05 0.1 2.405 cfs

Q = VA = =

(74)

iii) Momentum equation, Eq. (4.30)

II = Time rate of change of momentum inside CV is negligible if tank area is large compared to the nozzle area

p s

F  + F 

F

b

+ 

CS

q ( q dA )

CV

q dV

ρ ∂ t ρ

= ⋅ +

∫    ∂ ∫ 

I II

( )

A1

( )

A2 .

( )

n n 2 n n 1

CS

q ρ q dA ⋅ = q ρ q dA − q ρ q dA

 

∫ ∫

 = V V A

2

ρ

2 2

2 2 2 2 2

F

px

V V A ρ V ρ Q

∴ = =

(

24.05 1.94

)( )(

2.405

)

112 lb

F

px

= =

I =

(75)

→ The vector sum of all the external moments acting on a fluid mass

equals the time rate of change of the moment of momentum (angular

momentum) vector of the fluid mass.

Example: rotary lawn sprinklers, ceiling fans, wind turbines

(

r F×

) (

r M×

)

4.4.1 The Moment of momentum principle for inertial reference systems Apply Newton's 2nd law to rotating fluid masses

(76)
(77)

where = position vector of a mass in an arbitrary curvilinear motion

= linear momentum

( )

T r F d r M

= ×

 

= dt

×

(4.35)

r 

M

(78)

[Re] Derivation of (4.35)

Take the vector cross product of

By the way,

Eq. (4.27): dM F = dt

 

r 

r F r dM

× = × dt

 

 

I

( )

d dr dM

r M M r

dt × = dt × + × dt

 

 

 

(79)

where = angular momentum (moment of momentum)

dr

0

dr

I M q m q q

dt dt

 

= × = × =  = 

 

      

(

∴ ×q q = q q  sin 0 = 0

)

( )

dM d r r M

dt dt

 

∴   ×   = ×

  

( )

r F d r M

∴∴

×

= dt

×

r  × M 

(80)

[Re] Torque

• translational motion → Force – linear acceleration

• rotational motion →

Torque – angular acceleration [Re] Vector Product

Magnitude = = area of parallelogram

direction = perpendicular to plane of and → right-handed triple

T

= × r

F

V  = × a  b 

a sin

V  =  × b  γ

a 

b

fig_05_05

(81)

• External moments arise from external forces

where = external torque

( )

b × = − a  a  × b

( ) ka  × b  = k a (  × b  )

( ) ( ) ( )

a  × b  + c  = a  × b  + a  × b 

Tb

Ts

Tp

( r F

p

) ( r F

s

) ( r F

b

) d ( r M )

×  + ×  + ×  = dt × 

   

( )

p s b

T T T d r M

+ + = dt ×

     (4.36)

, ,

p s b

T    T T

(82)

4.4.2 The general moment of momentum equation

(4.37)

( )

CS CV

dM q q dA q dV

dt = ρ ⋅ + ∂ t ρ

∫ ∂ ∫

    

( )

CS

( ) ( )

CV

( )

d r M r q q dA r q dV

dt ρ ∂ t ρ

∴ × = × ⋅ + ×

∫ ∂ ∫

 

     

( ) ( ) ( )

CS CV

T T T r q q dA r q dV

p + s + b = × ρ ⋅ + ∂ t × ρ

∫ ∂ ∫

        

angle between

q

yz and

r

yz

( ) ( )

## . : sin cos

yz yz

2

yz yz

r q

x − dir × = r q    π − α    = rq α

 

(4.29):

(83)

( ) ( ) ( )

. : py sy by cos zx cos zx

CS CV

y dir T T T rq q dA rq dV

ρ t ρ

α ∂ α

− + + = ⋅ +

∫ ∂ ∫

    

( ) ( ) ( )

. : pz sz bz cos xy cos xy

CS CV

z dir T T T rq q dA rq dV

ρ t ρ

α ∂ α

− + + = ⋅ +

∫ ∂ ∫

    

(4.38)

( ) ( ) ( )

. : px sx bx cos yz cos yz

CS CV

x dir T T T rq q dA rq dV

ρ t ρ

α ∂ α

− + + = ⋅ +

∫ ∂ ∫

    

fig_05_05

(84)

4-12. Derive the expression for the total force per unit width exerted by the sluice gate on the fluid in terms of vertical distances shown in Fig. 4-20.

Homework Assignment # 4 Due: 1 week from today

4-11. Derive the equation for the volume rate of flow per unit width for the sluice gate shown in Fig. 4-20 in terms of the geometric variable b, y1, and CC. Assume the pressure in hydrostatic at y1 and ccb and the velocity is constant over the depth at each of these sections.

(85)

4-14. Consider the flow of an incompressible fluid through the Venturi meter shown in Fig. 4-22. Assuming uniform flow at sections (1) and (2) neglecting all losses, find the pressure difference between these sections as a function of the flow rate Q, the diameters of the sections, and the density of the fluid, P. Note that for a given configuration, Q is a function of only the pressure drop and fluid density.

(86)

through a horizontal pipe as shown in Fig. 4-23. The rates of inflow and outflow are the same, and the water surface in the tank remains a

distance h above the discharge pipe centerline. All velocities in the tank are negligible compared to those in the pipe. The head loss between the tank and the pipe exit is HL (a) Find the discharge Q in terms of h, A, and HL. (b) What is the horizontal force, FX required to keep the tank

from moving? (c) If the supply line has an area A’, what is the vertical force exerted on the water in the tank by the vertical jet?

(87)

4-28. Derive the one-dimensional continuity equation for the unsteady, non-uniform flow of an incompressible liquid in a horizontal open channel as shown in Fig. 4-29. The channel has a rectangular cross section of a constant width, b. Both the depth, y0 and the mean velocity, V are

functions of x and t.

참조

관련 문서

• Useful for investigating the shear forces and bending moments throughout the entire length of a beam. • Helpful when constructing shear-force and

– Useful for investigating the shear forces and bending moments throughout the entire length of a beam. – Helpful when constructing shear-force and