SOLUTIONS MANUAL FOR Fundamental Mechanics of Fluids Fourth Edition by: I.G. Currie ,All Chapters

SOLUTIONS MANUAL FOR
Fundamental
Mechanics of Fluids
Fourth Edition


by
I.G. Currie

, 1

BASIC CONSERVATION LAWS

, BASIC CONSERVATION LAWS


Problem 1.1




Inflow through x = constant:  u y z

Outflow through x +  x = constant:  u y z + ( u  y  z) x +
x

Net inflow through x = constant surfaces: − ( u ) x y  z +
x

Net inflow through y = constant surfaces: − ( v ) x  y  z +
y

Net inflow through z = constant surfaces: − ( w)  x y  z +
z
But the rate at which the mass is accumulating inside the control volume is:

(  x y  z)
t
Then the equation of mass conservation becomes:
  
t  x y  z = −x (u) + y (v) + z
 (w)  x  y  z +
 

Taking the limits as the quantities  x,  y and  z become vanishingly small, we get:


   
+ (u) + (v) + (w) = 0
t x y z




Page 1-1

, BASIC CONSERVATION LAWS


Problem 1.2




Inflow through R = constant:  u R R  z

Outflow through R +  R = constant:  u R R  z + ( u RR  z) R +
R

Net inflow through R = constant surfaces: − ( Ru ) R  z +
R R

Net inflow through  = constant surfaces: − ( u ) R  z +



Net inflow through z = constant surfaces: − ( uz ) R R  z +
z
But the rate at which the mass is accumulating inside the control volume is:

( R  R  z)
t
Then the equation of mass conservation becomes:
     
R  R   z = −  (  Ru R ) + (  u  ) + R (  u z )   R   z +
t  R  z 

Taking the limits as the quantities  R,  and  z become vanishingly small, we get:

 1  1  
+ (  Ru R ) + ( u  ) + ( u z ) = 0
t R R R  z




Page 1-2