Solutions Manual
Fundamental Mechanics of Fluids 4th Edition
By
I. G. Currie
( All Chapters Included - 100% Verified Solutions )
1
, 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:
x y z ( u ) ( v) ( w) x y z
t 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
2
,BASIC CONSERVATION LAWS
Problem 1.2
Inflow through R = constant: u R R z
Outflow through R R = constant: u R R z ( u R R z ) R
R
Net inflow through R = constant surfaces: ( R u R ) R z
R
Net inflow through = constant surfaces: ( u ) R z
Net inflow through z = constant surfaces: ( u z ) 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
( R u R ) ( u ) ( u z ) 0
t R R R z
Page 1-2
3
, BASIC CONSERVATION LAWS
Problem 1.3
Inflow through r = constant: u r r 2 sin
Outflow through r r = constant: u r r 2 sin
( r 2u r sin ) r
r
Net inflow through r = constant surfaces: ( r 2u r )sin r
r
Net inflow through = constant surfaces: ( u sin ) r r
Net inflow through = constant surfaces: ( u ) r r
But the rate at which the mass is accumulating inside the control volume is:
( r 2 sin r )
t
Then the equation of mass conservation becomes:
2
r sin r ( r 2u r ) sin r ( u sin ) r ( u ) r
t r
Taking the limits as the quantities r, and become vanishingly small, we get:
1 1 1
2 ( r 2u r ) ( u sin ) ( u ) 0
t r r r sin r sin
Page 1-3
4
Fundamental Mechanics of Fluids 4th Edition
By
I. G. Currie
( All Chapters Included - 100% Verified Solutions )
1
, 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:
x y z ( u ) ( v) ( w) x y z
t 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
2
,BASIC CONSERVATION LAWS
Problem 1.2
Inflow through R = constant: u R R z
Outflow through R R = constant: u R R z ( u R R z ) R
R
Net inflow through R = constant surfaces: ( R u R ) R z
R
Net inflow through = constant surfaces: ( u ) R z
Net inflow through z = constant surfaces: ( u z ) 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
( R u R ) ( u ) ( u z ) 0
t R R R z
Page 1-2
3
, BASIC CONSERVATION LAWS
Problem 1.3
Inflow through r = constant: u r r 2 sin
Outflow through r r = constant: u r r 2 sin
( r 2u r sin ) r
r
Net inflow through r = constant surfaces: ( r 2u r )sin r
r
Net inflow through = constant surfaces: ( u sin ) r r
Net inflow through = constant surfaces: ( u ) r r
But the rate at which the mass is accumulating inside the control volume is:
( r 2 sin r )
t
Then the equation of mass conservation becomes:
2
r sin r ( r 2u r ) sin r ( u sin ) r ( u ) r
t r
Taking the limits as the quantities r, and become vanishingly small, we get:
1 1 1
2 ( r 2u r ) ( u sin ) ( u ) 0
t r r r sin r sin
Page 1-3
4