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. 1.2-2(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. 1.2-2 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

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 1.2-1 and 1.2-2

1.3 Differential mass-balance equation According to Eq. 1.2-4 [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]

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. 1.3-4 becomes Example 1.3-2 Normal stress due to creeping flow around a sphere

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. 1.4-1. 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 :

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]

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

[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

Example 1.4-1 Force exerted by an impinging jet

Example 1.4-2 Back trust of a jet

Example 1.4-3 Thrust on a pipe bend

Example 1.4-4 Friction force on a pipe wall

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

Example 1.4-5

Example 1.4-6 Laminar flow over a flat plate

1.5 Differential momentum-balance equation 1.5.1 Derivation Eq. 1.4-3 represented the integral form of momentum-balance equation [1.5-1] The surface integrals in Eq. 1.5-1 can be converted into their corresponding volume integrals. From Eq. [A.4-4], we have Eq. 1.5-1 can be rewritten as follows:

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

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. 1.5-9) 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

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

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:

Therefore, for forced convection Continuity: Motion: Where

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.

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

M: a+1 =0 => a = -1 L: -3a+b+c+1=0 => c = -2 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.

1.5.3 Boundary Conditions

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

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

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

Example 1.5-5 Adjacent flow of two immiscible fluids