Ch. 4 Continuity Equation

Chapter 4 Continuity Equation and Reynolds Transport Theorem

4.1 Control Volume

4.2 The Continuity Equation for One-Dimensional

Steady Flow

4.3 The Continuity Equation for Two-Dimensional

Steady Flow

4.4 The Reynolds Transport Theorem

Objectives:

• apply the concept of the control volume to derive equations for the conservation of mass for steady one- and two-dimensional flows

• derive the Reynolds Transport Theorem for three-dimensional flow

• show that continuity equation can recovered by simplification of the Reynolds Transport

Theorem

4-1

Ch. 4 Continuity Equation

4.1 Control Volume

▪ Physical system surroundings

~ is defined as a particular collection of matter or a region of space chosen for study

~ is identified as being separated from everything external to the system by closed boundary

• The boundary of a system: fixed vs. movable boundary

•Two types of system: closed system (control mass) ~ consists of a fixed mass, no mass can cross its boundary

open system (control volume) ~ mass and energy can cross the boundary of a control volume

4-2

Ch. 4 Continuity Equation

→ A system-based analysis of fluid flow leads to the Lagrangian equations of motion in which particles of fluid are tracked.

A fluid system is mobile and very deformable.

A large number of engineering problems involve mass flow in and out of a system.

→ This suggests the need to define a convenient object for analysis. → control volume

• Control volume

~ a volume which is fixed in space and through whose boundary matter, mass, momentum, energy can flow

~ The boundary of control volume is a control surface.

~ The control volume can be any size (finite or infinitesimal), any space.

~ The control volume can be fixed in size and shape.

→ This approach is consistent with the Eulerian view of fluid motion, in which attention is focused on particular points in the space filled by him fluid rather than on the fluid particles.

4-3

Ch. 4 Continuity Equation

4.2 The Continuity Equation for One-Dimensional Steady Flow

• Principle of conservation of mass

The application of principle of conservation of mass to a steady flow in a streamtube results in the continuity equation.

• Continuity equation

~ describes the continuity of flow from section to section of the streamtube

• One-dimensional steady flow

No flow

Fig. 4.1

Consider the element of a finite streamtube

- no net velocity normal to a streamline

- no fluid can leave or enter the stream tube except at the ends

4-4

Ch. 4 Continuity Equation

Now, define the control volume as marked by the control surface that bounds the region

between sections 1 and 2.

→ To be consistent with the assumption of one-dimensional steady flow, the velocities at

sections 1 and 2 are assumed to be uniform.

→ The control volume comprises volumes I and R.

→ The control volume is fixed in space, but in dt the system moves downstream.

From the conservation of system mass

()()mmIRtROtt mm (1)

m For steady flow, the fluid properties at points in space are not functions of time,  0 t

→ ()()mmR tRtt  (2)

Substituting (2) into (1) yields

()mmIt ( Ot ) t (3) Inflow Outflow

4-5

Ch. 4 Continuity Equation

Express inflow and outflow in terms of the mass of fluid moving across the control surfce in time dt

()mAdsIt 11 1

()mAdsOt t 22 2 (4)

Substituting (4) into (3) yields

11A ds 1  2 A 2 ds 2

Dividing by dt gives

mAdsAds 11 1 2 2 2 (4.1)

→ Continuity equation

In steady flow, the mass flow rate, m passing all sections of a stream tube is constant.

mAV   = constant (kg/sec)

dAV()0  (4.2.a)

→ dAVdAVdVA() () ()0 (5)

4-6

Ch. 4 Continuity Equation

Dividing (a) by  AV results in

ddAdV  0 (4.2.b)  AV

→ 1-D steady compressible fluid flow

For incompressible fluid flow; constant density

 → d  0 t

From Eq. (4.2a)

dAV()0

dAV()0 (6)

Set Q =volume flowrate (m3/s, cms)

Then (6) becomes

QAVconst.  AVAV11  2 2 (4.5)

For 2-D flow, flowrate is usually quoted per unit distance normal to the plane of the flow, b

→ q  flowrate per unit distance normal to the plane of flow msm3  

4-7

Ch. 4 Continuity Equation

QAV qhV  bb

hV11 hV 2 2 (4.7)

[re] For unsteady flow

masstt  mass t+ inflow outflow

()mmmmR tt ()()() Rt It Ott 

Divide by dt

()mm () Rt t Rt()(mm ) dt It Ot t

Define

mm() () m ( vol ) Rt t Rt tdtt

Then

() vol ()(mm ) t It Ot t

4-8

Ch. 4 Continuity Equation

• Non-uniform velocity distribution through flow cross section

Eq. (4.5) can be applied. However, velocity in Eq. (4.5) should be the mean velocity.

Q V  A

QdQvdA AA

1  VvdA A A

•The product AV remains constant along a streamline in a fluid of constant density.

→ As the cross-sectional area of stream tube increases, the velocity must decrease.

→ Streamlines widely spaced indicate regions of low velocity, streamlines closely spaced

indicate regions of high velocity.

A11VAVAA 2 2: 1 2 VV 1 2

4-9

Ch. 4 Continuity Equation

[IP 4.3] p. 113

The velocity in a cylindrical pipe of radius R is represented by an axisymmetric parabolic distribution. What is V in terms of maximum velocity, vc ?

[Sol] dr

dA=2rdr r R

vc

r 2 vvc 1 2 ← equation of parabola R

2 Qr11R  V v dA  vc 12   r dr A 220  AA R R

32422R 222vvvvR rrrRR     ccccrdr → Laminar flow 222220   RRRRR 24 0  2 4  2

[Cf] Turbulent flow

→ logarithmic velocity distribution

4-10

Ch. 4 Continuity Equation

4.3 The Continuity Equation for Two-Dimensional Steady Flow

(1) Finite control volume

Fig. 4.3

Consider a general control volume, and apply conservation of mass

()()mmIRtROtt mm (a)

For steady flow: ()()mmR tRtt 

Then (a) becomes

()mmIt ( Ot ) t (b)

i) Mass in O moving out through control surface

()mdsdA  (cos) Ot t C.. S out

mass vol area  1  ds dA cos

4-11

Ch. 4 Continuity Equation

• Displacement along a streamline;

ds vdt (c)

Substituting (c) into (b) gives

()m  (cos) v dAdt (d) Ot t CSout..

By the way, vcos = normal velocity component normal to C.S. at dA   Set n = outward unit normal vector at dA  n 1  ← scalar or dot product (e) vvnvn cos

Substitute (e) into (d)   ()mdtvndAdtvdA Ot t C.. S out C .. S out

  where dA ndA=directed area element

[Cf] tangential component of velocity does not contribute to flow through the C.S.

→ Circulation

ii) Mass flow into I

()mdsdA  (cos) 90 cos  0 It CSin..   (cos)vdAdtdtvndA  ( ) CSin.. CSin ..

4-12

Ch. 4 Continuity Equation

  dt v  ndA  dt v  dA  CSin..  CSin .. 

For steady flow, mass in = mass out   dt v dA dt v  dA CSout.. CSin .. 

Divide by dt     vdA vdA CSin.. CSout ..   vdA vdA 0 (f) C.. S out C .. S in

Combine c.s. in and c.s. out

  vdA vndA  0 (4.9) CS.. CS ..

where  integral around the control surface in the counterclockwise direction CS..

→ Continuity equation for 2-D steady flow of compressible fluid

[Cf] For unsteady flow

 (..)mass inside c v = mass flowrate in – mass flowrate out t

4-13

Ch. 4 Continuity Equation

(2) Infinitesimal control volume

udx udx u  u  x 2 x 2

Apply (4.9) to control volume ABCD

    v ndA v  ndA v  ndA v  ndA 0 (f) AB BC CD DA

By the way, to first-order accuracy  dx  AB x 2

  dy v dy vndA v  dx AB  yy22

  vdy vn v y 2

  dx u dx vndA u  dy BC xx22

  dy v dy vndA v  dx (g) CD  yy22

  dx u dx vndA u  dy DA xx22

4-14

Ch. 4 Continuity Equation

Substitute (g) to (f), and expand products, and then retain only terms of lowest order (largest order of magnitude)

vdy dy v() dy 2 vdx dx  v dx  dx yyyy22 4

vdy dy v() dy 2 vdx dx  v dx  dx yyyy22 4

udx dx u() dx 2 udy dy  u dy  dy x 22xxx 4

udx dx() dx 2 udy dy  u dy  dy 0 x 22xxx 4

vu    dxdy v dxdy dxdy u dxdy 0 yy  xx

vu   vu0 yy  xx

 ()uv ()0 (4.10) xy

→ Continuity equation for 2-D steady flow of compressible fluid

• Continuity equation of incompressible flow for both steady and unsteady flow (   const.)

uv 0 (4.11) xy

4-15

Ch. 4 Continuity Equation

[Cf] Continuity equation for unsteady 3-D flow of compressible fluid

   ()uvw () ( )0  tx  y  z

For steady 3-D flow of incompressible fluid

uvw  0 xyz

• Continuity equation for polar coordinates

D C

A

B

Apply (4.9) to control volume ABCD         v ndA V  ndA  V  ndA  V  ndA 0 AB BC  CD  DA

  dvdt  VndA vt  dr AB 22    dr vr dr VndA vr () r  drd BC rr22   dvdt  VndA vt  dr CD 22

4-16

Ch. 4 Continuity Equation

   dr v dr VndA v r rd DA r rr22

Substituting these terms yields

vd d v() d 2 v drtt dr  v dr  dr tt22 4

vd d v() d 2 v drtt dr  v dr  dr tt22 4

vdr vdr vrd vdrd  rr rd   drd  rr rr22

22 dr  dr  dr vrr   dr v vrr rd v drd  rd   drd  rrrrrr222  2

2 vdrrr dr v dr vrr rd rd   v rd   rd  0 rrrr22 2

vv   trd dr v d dr rdrd  v rdrd trrr 

vv11  1 v drdrr() dr22 d   v () dr d   () dr 3 d   0 rrrrrr22 2

Divide by drd

vv   v11   v 1 trvvrvdrvdrdr   r   r 0 trrrrr  r22  r  rr 2

vv   rtrv r v v 0 rrrr  t

4-17

Ch. 4 Continuity Equation

Divide by r

vvv   rrtvv 0 rrrrrrt  

()vv ()v  rrt 0 (4.12) rrr

For incompressible fluid

vv v rrt 0 (4.13) rrr

4-18

Ch. 4 Continuity Equation

[IP 4.4] p. 117

A mixture of ethanol and gasoline, called "gasohol," is created by pumping the two liquids into the "wye" pipe junction. Find Qeth and Veth

3 mix  691.1 kg m No flow in and out Vmix 1.08 m s

33 Qlsgas 30 / 30 10 m /s

3 gas  680.3 kg m  n

3 eth  788.6 kg m

[Sol]

 A (0.2)22 0.031m ; A  0.0079m2 ; A  0.031m2 1 4 2 3

3 V1 30 10 / 0.031  0.97 m/s (4.4)    vndA vndA  vndA 0 (4.9) 123  vndA680.3 0.97 0.031  20.4 kg/s 1  vndA788.6 V22 0.0079  6.23 V 2  vndA691.1  1.08 0.031  23.1 kg/s 3   vndA20.4 6.23 V2  23.1  0 cs.

V2  0.43 m/s

33 QVAeth 22 (0.43)(0.0079)  3.4 10 m /s 3.4 l /s 4-19

Ch. 4 Continuity Equation

4.4 The Reynolds Transport Theorem Osborne Reynolds (1842-1912); • Reynolds Transport Theorem (RTT) English engineer

→ a general relationship that converts the laws such as mass conservation and Newton’s 2nd law from the system to the control volume

→ Most principles of fluid mechanics are adopted from solid mechanics, where the physical laws dealing with the time rates of change of extensive properties are expressed for systems.

→ Thesre is a need to relate the changes in a control volume to the changes in a system.

• Two types of properties

Extensive properties ( E ): total system mass, momentum, energy

Intensive properties (i ): mass, momentum, energy per unit mass

E i system mass, m 1   system momentum, mv v

 2  system energy, mv ()v 2

E i dm i dvol (4.14) system  system

4-20

Ch. 4 Continuity Equation

▪ Derivation of RTT

Consider time rate of change of a system property

EtdtEEE t()() R 0 tdt  EE R  I t (a)  ()E0 tdt dt i v dA (b.1) c.. s out  ()EItdt i v  dA (b.2) csin..

()ERtdt  idvol (b.3) R tdt

()ERt idvol (b.4) R t

Substitute (b) into (a) and divide by dt

EE 1  tdt ti dvol i dvol dt dt RRtdt    t       i v dA i v dA csout..  csin ..

4-21

Ch. 4 Continuity Equation

dE    idvol ivdA (4.15) dt t cv..   cs ..

dE ① = time rate of change of E in the system dt

 ② idvol = time rate change within the control volume → unsteady term t cv..   ③ ivdA  = fluxes of E across the control surface cs..

▪ Application of RTT to conservation of mass

For application of RTT to the conservation of mass,

dm in Eq. (4.15), E  m , i 1 and  0 because mass is conserved. dt

           dvol vdA    vdA   vdA  (4.16) t cv..   cs ..  csout ..  csin .. 

Unsteady flow: mass within the control volume may change if the density changes

For flow of uniform density or steady flow, (4.16) becomes   vdA vdA 0 ~ same as Eq. (4.9) csout..  csin ..

4-22

Ch. 4 Continuity Equation

For one-dimensional flow  vdA22 VA 2 c.. s out  vdA11 VA 1 csin..

111AVAV  2 22 (4.1)

• In Ch. 5 & 6, RTT is used to derive the work-energy, impulse-momentum, and moment of momentum principles.

4-23

Ch. 4 Continuity Equation

Homework Assignment # 4

Due: 1 week from today

Prob. 4.9

Prob. 4.12

Prob. 4.14

Prob. 4.20

Prob. 4.31

4-24