22. The control surface A can be considered to consist of three regions:
Ain: the surface of the the region where the fluid enters the control volume;
Aout: the surface of the region where the fluid leaves
Awall; the fluid is in contact with a wall
A = Ain + Aout + Awall
As shown in Fig (b) the outward mass flow
rate through dA is rv．ndA, since n is points outward,
thus the inward mass flow rate through dA is - rv．ndA
For a homogeneous fluid of a pure material
undergoing no chemical reactions, the conservation
law for mass is
(b)
Fig Volume flow rate through a
differential surface element dA: (a) three-
dimensional view; (b) two-dimensional view
Rate of mass accumulation (term 1)
= Rate of mass in (term 2)
- Rate of mass out (term 3)
The mass of the fluid in the differential volume element dΩ is dM= rdΩ

23. The mass of the fluid in the differential volume element dΩ is dM= rdΩ
The total mass in the control volume is M =
The rate of mass change in the control volume is
The net inward mass flow rate into Ω is
That is
[1.2-4]
Examples and 1.2-2

24. 1.3 Differential mass-balance equation
According to Eq
[1.2-4]
[1.3-1’]
According to Gauss divergence theorem [A.4-1]
[1.3-2]
Substituting Eq.[1.3-1’] and [1.3-2] into [1.2-4], we obtained
[1.3-3]
The integrand must be zero everywhere since the equation must hold for
any arbitrary Ω. Therefore, we have
[1.3-4]

25. Equation [1.4-4] is the differential mass-balance equation, that is, the equation
of continuity. For an incompressible fluid, the density r is constant and Eq
becomes
Example Normal stress due to creeping flow around a sphere

26. 1.4 Overall momentum-balance equation
Consider an arbitrary stationary control volume Ω bounded by
surface A through a moving fluid is flowing, as shown in Fig
The control surface A can be considered to consist of three parts:
A = Ain + Aout + Awall
Consider conservation law for momentum
of the control volume:
Rate of
momentum
accumulation in
out
Sum of forces
acting on
system = - +
[1.4-2]
This equation is consistent with Newton’s second law of motion
The mass contained in a differential volume element dΩ in the control volume is
rdΩ and its momentum is dP = rvdΩ. The momentum of the fluid in Ω is :

27. The rate of momentum change in Ω is :
The inward mass flow rate through dA : -r(v．n)dA
The inward momentum flow rate through dA : -rv(v．n)dA
The net inward momentum flow rate into Ω : or
Since v = 0 at the wall, the equation can be expressed as
Define v =｜v．n｜
[1.4-b]

28. The pressure force acting on dA is dFp = -pndA, The pressure force acting
on the entire control surface A is
[1.4-c]
The viscous force exerted on the fluid by the surrounding over dA is
dFv = -t‧ndA, the viscous force exerted on the fluid by the surroundings over A is
[1.4-d]
The body force acting on the differential volume element dΩ is dFb = fbdΩ,
the body force acting on the entire control volume is
[1.4-e]
Substituting [1.4-a] through [1.4-e] into [1.4-2]
[1.4-3] or

29. [1.4-8]
Where P = momentum of fluid in control volume
m = mass flow rate at inlet or outlet
Fp = pressure force acting on control volume by surrounding
Fv = viscous force acting on control volume by surrounding
Fb = body force acting on control volume
The correction factor a
a = 4/3 for laminar flow in round pipes
~ 1 for turbulent flow

30. Example 1.4-1 Force exerted by an impinging jet

31. Example 1.4-2 Back trust of a jet

32. Example 1.4-3 Thrust on a pipe bend

33. Example 1.4-4 Friction force on a pipe wall

34. Example 1.4-5Creeping flow around a sphere: Stoke’s law

35. Example 1.4-5

36. Example 1.4-6 Laminar flow over a flat plate

37. 1.5 Differential momentum-balance equation
1.5.1 Derivation
Eq represented the integral form of momentum-balance equation
[1.5-1]
The surface integrals in Eq can be converted into their corresponding
volume integrals. From Eq. [A.4-4], we have
Eq can be rewritten as follows:

38. The integrand must be zero everywhere since the equation must hold for any
arbitrary region Ω, therefore ∵
and
It can be further simplified with the help of the equation of continuity as following:
(for constant r and m)
[1.5-8]
If the gravity force is the only body force involved, that is
[1.5-9]
The equation is called Navier-Stokes equation

39. The equation of motion is often expressed in terms of the Stokes derivative D/Dt,
Which is defined as follows:
The equation of motion (Eq ) becomes
(1) (2) (3) (4)
Where
term (1): the inertia force per unit volume, namely, the mass per unit volume
times acceleration
(2):pressure force, (3) viscous force, and (4) gravity force per unit volume

40. 1.5.2 Dimensionless form
The equation of motion and the continuity equation can be expressed in
the dimensionless form.
In forced convection, a characteristic velocity V, length L, and time L/V are
used to express the equation qualitatively. By substituting V for v, (L/V)-1 for
D/Dt, L-1 for ▽, and L-2 for ▽2.
inertia force: pressure force:
viscous force: gravity force: rg
Define:
dimensionless velocity dimensionless pressure
dimensionless time dimensionless coordinates
Dimensionless operator

41. Substituting Eqs. [1. 5-12] through [1. 5-17] into Eqs. [1
Substituting Eqs. [1.5-12] through [1.5-17] into Eqs.[1.5-18] and [1.5-19], we have
Continuity:
Motion:
Multiplying Eq.[1.5-21] by L/V and Eq.[1.5-22] by L/rV2, we have
Continuity:
Motion:

42. Therefore, for forced convection
Continuity:
Motion:
Where

43. Given a physical problem in which the dependent parameter is a function of n-1
independent parameters. We may express the relationship among the variables in
functional form as
where q1 is the dependent parameter, and q2, …, qn are n-1 independent parameters.
Mathematically we can express the functional relationship in the equivalent form
Where g is an unspecified function, different from f.
The Buckingham Pi theorem states that: Given a relation among n parameters
of the form
then the n parameters may be grouped into n-m independent dimensionless
ratios, or n parameters, expressed in a functional form by or
The number m is usually, but not always, equal to the minimum number of
independent dimensions required to specify the dimensions of all the parameters
q1, q2, …and qn.

44. Example
The drag force F, on a sphere depends on the relative velocity, V,
the sphere diameter, D, the fluid density, r, and the fluid viscosity, m, Obtain a
set of dimensionless groups that can be used to correlate experimental data.
Given: F=f(r, V, D, m) for a smooth sphere
Find: An appropriate set of dimensionless groups
Solution:
Step 1: Find the parameters: F, r, V, D, and m n=5
Step 2: Select primary dimensions: M, L, T or F, L, T r=3
Step 3: Find the dimension of the parameters selected in step 1
F r V D m
ML/T2 M/L3 L/T L M/(LT)
Step 4: Select repeating parameters with the number equal to
the primary dimensions r, V, D m=3
Step 5: n-m=2. Two dimensionless groups will result.
Step 6: Setting up dimensional equations Π1= raVbDcF, and Π2= rdVeDfm
Step 7: Summing exponents

45. M: a+1 =0 => a = -1
L: -3a+b+c+1=0 => c = Therefore
T:-b-2=0 => b =-2
Similarly
Step 8: Check using F, L, t dimensions
and
The functional relationship is Π1=f(Π2), or
As noted before. The form of the function, f, must be determined experimentally.

46. 1.5.3 Boundary Conditions

52. Example 1.5-1 Tangential annual flow (Incompressible, Newtonian, laminar)
Flow is in q direction only, vr=vz=gq=0
Flow is axisymmetric, no pressure variation in the q direction. Neglect end effects, 
The equation of continuity in cylinder coordinates is
Which can be reduced to
The velocity distribution vq(r), the q component of the equation of motion is
The boundary conditions are vq =0 at r = r1, vq = r2w2 at r = r2

53. Example 1.5-2 Laminar flow through a vertical tube, find the velocity, volume flow rate,
and shear stress (Steady-state, incompressible, Newtonian)
Flow is in z direction only, vr=vq=0, vz is independent of q
Subjected to B.Cs
Find vav

54. Example 1.5-3 Flow of a rising film
Continuity Eq.
Eq. of motion
B.Cs

55. Example 1.5-5 Adjacent flow of two immiscible fluids