calculate the momentum deficit correction factor.

CONSERVATION OF MASS 

Density and Volume are NOT constant! In general the density is NOT uniform!

( , , , )m dV x y z t dxdydz   

Control Mass (Closed Systems) A control mass system does not exchange mass with its surroundings.

Mass is fixed/constant.

0mmm 12  0 dm dt



( , , , )x y z t  

If the density can be assumed uniform,

( )t m V     

0 dV d

V dt dt

   

Control Volume (Open Systems) A control volume system does exchange mass with its surroundings.

m CV

in out inlets outlets

dm m m

dt    

MASS‐FLOW RATE

: ( cos ) ( / )Mass Flow Rate m v dA v dA kg s      

iv 

idA 

edA 

ev 

 z

r 0: 0 ( 180 )

( ) : ( cos ) 0

i i

e e

Inlet m v dA

Exit Outlet m v dA

 

 

  

 

 





SPECIAL CASES

If the flow is normal (=0, 180o): ( , )m v r z dA  Ad 

r z

z)v(r,v 

If the flow is normal AND 1‐dimensional  with Uniform density:

m vA Ad 

r z

z)v(r,v v = v(z)Convenient to define the average speed: 1

avgv v vdAA   

Helpful for uniform density flows since m vA 

IN‐CLASS EXAMPLES

The axial velocity profile of steady flow of H2O  of 989.5kg/m3 through a cylindrical pipe is  given by

1/

( ) 1 ; 0 n

mx r

v r v r R R

      

  where R is the radius of the pipe and n is an integer, 6≤n≤10. Develop an expression for the mass‐flow rate.

Then calculate the value if R=0.1cm, vmx=0.2m/s, n=7.

MASS‐FLOW RATE EXAMPLE

MASS CONSERVATION EXAMPLE Study Example 3.12 

Study Examples 3.6, 3.7, 3.8 

(See Table 3.1, p193.)

0 20 40 60 80 100 120 140 0

1

2

3

4

5

(sec)t

( ) ( ) L t ft

30 0.048 1 exp

9 t

ft s

                     

-kt ρAimL(t)= 1-e

k

MASS CONSERVATION EXAMPLE CV

in out inlets outlets

dm m m

dt    

H

(t)

A

 em = kL

 im

Asymptote at 30/9ft=3.333ft

If H=2ft it will fill up in ~20s.

If H=4ft it will never fill up.

dx

dy

dz

z

y

x

Cartesian             coordinates

MASS CONSERVATION (REVISITED) CV

i e dm

m m dt

    ( , , , )CVm x y z t dV and

i e i i i e e e d

dV m m v A v A dt

           For 1‐D normal flow

[ ( )]e x xm v d v dydz  

Outlet right‐x: ( )

[ ( )]e x x vxm v d v dydz v dx dydzx x

   

    

      



Applying the same approach for all y‐ and z‐faces and since

d dV dV dxdydz

dt t t  

  

    

( )( ) ( )yx z x x y y z z

vv v dxdydz v dydz v dydz dxdydz v dxdz v dxdz dxdydz v dydx v dydx dxdydz

t x y z  

       

            

( ) ( ) ( ) 0x y zv v vt x y z 

      

       

Continuity PDEquation (Assumption: no  mass source/sinks, , v continuous  functions)

i xm v dydz

Inlet left‐x: i xm v dydz

ˆˆ ˆ x y zv v i v j v k   

 v

Define the vector gradient operator: ˆˆ ˆi j k x y z   

      

 ( ) 0v

t 

 

  

  

Example: An incompressible velocity vector field is given by 2 2 ˆˆ ˆ( ) where a,b are constants.yv a x y i v j bk   



What must the form for the velocity component, vy be to satisfy mass conservation?