 The continuum approach is usually not valid when the gas pressure is very small (few millitorr like in a vacuum), or the aperture size is small (like in an orifice)
 Mathematically, for the continuum approach based model to hold good, where is the mean free path of the gas molecule and is the characteristic length of the system. Alternatively, Knudsen defined as .
Fluid Statics

 Concerned with forces in a stationary fluid
Recall. In a stationary fluid, shear stress . However, fluid can sustain the normal stress.

 Fluid pressure: It is a normal force per unit area of the fluid element, acting inward onto the element. Consider a fluid element, ABC:


(Fig. 3b)

Pressure acts inward on all three faces of the element.

Consider another fluid element:



Pressure also acts inward on the differential element dA
 Pascal's TheoremPressure at any point within a static fluid is the same in all directions. Let us prove it. Consider a triangular fluid element ABC in a 2D(xz) scenario. Pressure acts on all 3faces of the element inward and normal to the surface.


  
(Fig. 3d)

(Fig. 3e)
 

Force balance:



Taking limit

The basic equation of fluid statics 
 First, let us calculate force on a fluid element because of pressure variation. Consider an elemental fluid volume bounded by (ABCD)(A'B'C'D') surfaces.


(Fig. 3f)

 Pressure at is
 Volume of the fluid under consideration is
 Force acting on the face AA'C'C:
(+ ve direction) Force acting on the face (ve direction) (Neglecting the higher order terms or assuming linear variation over ) Differential force, (acting along –ve xdirection) Similarly,

 
 Therefore, we have expression for differential pressure–force
or,
 If there is no pressure variation or pressure is uniform around an element, the net force is zero,
or
The basic equation of fluid statics (continued)

2Now, let us calculate pressure–gradient in a static fluid.

 Identify all forces: pressure forces, and external body force which is gravity in the present case.Consider a differential fluid elemental volume, .
 Force Balance (in vector form )
(pressure force + gravitational force) = 0

or,
or, (form previous lecture)
or,
This is the basic (vector) equation for fluid statics. 
The equation in the scalar form:
: x direction
: y direction
: z direction 
Where, 
 If gravity acts in the negative zdirection

or
where
Notes: 
 Pressure gradient in a static fluid is zero if there is no gravity. Alternatively, the net force acting on a fluid volume is zero because pressure–force balances the force due to gravity. Alternatively, pressurevariation occurs in a static fluid because of gravity only.
 If the fluid is water, static pressure is often referred as ‘hydrostatic' pressure.
If a body is submerged in a fluid, which is in contact with (or open to) atmosphere, the atmospheric pressure acts uniformly on the body. For most of the engineering applications one is interested in calculating pressure due to the fluid only, ignoring the atmospheric pressure. In such case, the pressure is specified as a gauge pressure (above the atmosphere pressure): . Therefore, the gauge pressure may be negative if the pressure in the fluid is sub or below atmospheric. Absolute pressure is always positive
 Pressure in the atmosphere may vary, of course, over a relatively longer altitude, because the density of air is small (1 kg/m^{3} in comparison to 1000 kg/m^{3} for water).
Consider an atmosphere consisting of ideal gases. Static pressure gradient




And,
 
Substituting, 
(Fig. 4a)

or,
 
If or an isothermal atmosphere

And,

or,

The pressure varies (decreases) exponentially with altitude, if temperature variation is considered negligible.
 Engineering calculations based on the fluid static equation


 Pressure variation in an incompressible fluid





Integrating,


(Fig. 4b)

Therefore, pressure decreases lineally with altitude, or increases with depth in the fluid 

(Fig. 4c)

or, 

(Fig. 4d)

(, where is the depth of the fluid measured from the top ) 
(2) Pressure variation in twoimmiscible liquids (one is heavier than the other) 

(Fig. 5a)


Consider two liquids of densities . Depth (measured form the top) of the topfluid is and that of the bottomfluid is
Put the coordinate axis at the bottom (left). The governing equation for the static fluid is
or
On integration ,
Similarly,
or
One can get similar expression by putting the coordinate axis at the top of the surface, with zaxis pointing vertically downward (right):
or,
On integration
Similarly ,
or ,
Note: Both expressions are same!
(3) Barometer (Instrument to measure atmospheric pressure) 



On integration ,

But, at the same level and

Therefore, . is nothing but the vapor pressure of Hg in the space over the liquid Hg in the tube. Generally, (very small at Room temperature). Hence,. One can experimentally measure to show that 

There are two problems in using waterbarometer 
 Tube is long!
 Vapor pressure of water is significantly larger than that of Hg, i.e., and should be considered in the calculation!!
4) Monometer (To measure pressure of the fluid in a device or container) 
On integration between C and D 

But, 
Therefore, 
Therefore, 
However, 
( NOTE at sea  level)
In this lecture we discuss steady state/unsteady state condition for velocity fields.

Steady state 
The steady state condition for a flowfield implies that the velocity field and any property associated with the flow field remain unchanged with time.In other words, local derivative of the velocity is zero. 
Mathematically, 0. The concentration and temperature fields, if associated will also be under steady–state:
As explained below, may or may not be zero.

Consider the flow of a fluid through a convergent nozzle. If the velocity field is steady, or the flow–field is under steadystate conditions, . However, an observer moving with the flow field will experience ‘acceleration' as he moves from the larger diametersection to the smaller diametersection of the nozzle. This is because the velocity increases as the diameter of the nozzle decreases. Therefore, velocity remains unchanged with time anywhere in the flow field. However, it has a spatial variation, or


(Fig. 10a)

, , (between 12) 
The last expression is tantamount to saying that the velocity of a labeled or marked materialparticle is not constant, as it moves from the larger section to the smaller section of the nozzle.
Example 1: At a certain time, the concentration of measured at a location near a powerplant (A) is 200 ppm (parts per million). The concentration at ‘B', 1 km downstream, is measured to be 10 ppm. At this instance, the average wind velocity is 1 km/hr. The wind is blowing in the direction of the higher (A) to lower concentration (B). If a balloonsensor floating with the wind records a decrease of 50 ppm per hour, as it flows past ‘B', determine the rate of increase of the concentration measured by a stationary observer at ‘B'.



(Fig. 10b)






(Assume that the concentration gradient between two locations is linear)
 
 
(Note:
Mathematical analysis of fluid motion and associated properties 
There are two ways to carry out the analysis:
 Control mass: In such an analysis, a fixed mass of fluid element in the flowfield is identified and conservation equations for properties such as momentum, energy or concentration are written. The identified mass moves around in the flowfield. Its property may change from one location to another; however, the property must correspond to the same contents of the identified fluid element. In general, such approach is mathematically or experimentally difficult to apply.
 Control volume: This approach is popular and widely applied in the analysis. An arbitrary fixed volume located at a certain place in the flowfield is identified and the conservation equations are written. The property under consideration or analysis may change with time.
 If is a general property associated with fluid, then and may be considered to describe change in the property, of the fluid, following the control mass (CM) and control volume (CV)approach, respectively.
 The surface which bounds CV is called control surface (CS). In the CV approach, the coordinate axis is first fixed. A ‘CV’ is then marked in the flowfield. Choice of location and shape of CV are important for mathematical formulation.
Reynolds Transport Theorem 
We exclude the derivation of the theorem from the present course; however discuss the significance and the application of the same. The theorem relates the rate of change of a property associated with a CV to the material or particle rate of change of that property. Consider a CV bound by the CS through which a fluid flows, described by the flow field, relative to coordinates . As per the definition, CV is fixed in the flow field:

(Fig. 11a)

N is a property associated with flow field in CV. N could be either mass or momentum or energy or concentration of a species dispersed in the fluid. The theorem states:

where is the specific property, N/M, where M = mass of fluid in volume .
or,
Note that if the property is mass, .
If the property is momentum, .
If the property is kinetic energy, .
Let us give physical interpretation to the above three forms of the theorem.
 The first term is the rate of change in the property contained in the CV, which is fixed in the flow field. This term also describes if the property is under steady state or not, or if there is any accumulation of the property, N in the CV.
 The 2^{nd} term represents the material rate of change of the property and may be seen as the rate of generation of the property. Note that if because mass of the fluid particles initially marked in CV cannot change, nor it can be generated or destroyed, as the particles move in the flow field.
 The 3^{rd} term represents the net rate of change in the property because of the flow of the fluid in and out of the CV through CS. is the rate of volumetric flux in and out of dA, a differential area on CS. is the rate of mass flux. is the rate of the propertyflux in and out of the differential area is the rate of the total property over the entire CV.
Take a simple CV in the flow field, shown by the dotted line. 

(Fig. 11b)

It is easier to show that the last term 

 
 
In such case, the fluid enters and leaves the CS only through and . The remaining surface of CS is impervious to the mass.


Conservation of mass 
N = property =M; Therefore, Apply Reynold’s Transport Theorem: 

(Fig. 12a)


 (Particlemass cannot be created or destroyed) 
The 1st term of the righthandside is zero: particlemass cannot be created or destroyed. 
Therefore, : conservation of mass equation 
 Note that of a fluid in a CV may change with time, for example, in a tank filled with air and water. If water is drained out, of the mixture (air + water) will vary with time. Message: Be careful while evaluating the integral, whether under evaluation refers to the density of a pure fluid or fluidsmixture.
The RHS of the above equation is the integral (over the entire crosssection) of the differential volumetric flow rate: 




 
Or Ans.
Apply the conservation law again over larger to show that,
Ans.
Further, (for the 2^{nd} section) (for the 1^{st} section)
or,
Revisit,
or, (volumetric flow rate)
or, , average velocity in the section
Compare to to show that .
Differential form of massconservation 
In the previous lecture, we considered the conservation of total mass over a CV, by applying the general Reynolds Transport theorem. Now, we obtain the differential form of the equation, also known as the continuity equation. Consider CV of



(Fig. 13a)


 CV is bound by six CSs.
 fluid density

Consider CS marked 1:
Rate of massin:
Differential form of massconservation 
In the previous lecture, we considered the conservation of total mass over a CV, by applying the general Reynolds Transport theorem. Now, we obtain the differential form of the equation, also known as the continuity equation. Consider CV of



(Fig. 13a)


 CV is bound by six CSs.
 fluid density

Consider CS marked 1:
Rate of massin:
(It is equivalent to to of the Reynold’s transport theorem)
CS marked 2:
Rate of massout:
 Net rate of change in mass in xdirection

 
 
Similar expressions can be obtained for the other directions, for examples, 
 Net rate of accumulation of mass in CV

 
Therefore,  
 
The significance of this expression is the same as that for the total massconservation rule applied to a CV in the earlier lecture 
Rearrange: 
This is the vectorform of continuityequation and readily applied for all coordinate systems. In this course we will focus mostly on Cartesian and cylindrical 1D or 2D geometry. To this end, the above conservation equation may also be written as: 

Recall the difference between 
 if the flow is steady
 if the fluid is incompressible,
In other words, local density my change with time ,as the fluid enters and leaves a ‘CV’ at different rates resulting in the accumulation or depletion of mass of the matter per unit volume of the CV. This is referred as the unsteadystate. However, if the fluid is incompressible, its density will be constant and the material rate of change of density will be zero. 
(incompressible fluid) can be written for 2D geometries as : 
: Cartesian 
: Cylindrical 
Pay special attention to the last term of the aboveequation. For 1D radial flow, 
. 

(Fig. 13b)

const under steadystate.
To sumup, total mass flow rate is constant at all locations in the rdirection. However, mass flux changes because the differential area changes in the rdirection


B. Momentum  conservation equation 

(Fig. 13c)

Reynolds Transport Theorem: 
Property, 

1^{st} Term: local rate of change of momentum in CV
2^{nd} Term: Net rate of momentum in and out of CV through CS.
3^{rd} Term: material or partial –rate of change of momentum, and it may be seen as the external bodyforce or the source term for the generation of momentum by the external forces.

Mathematically,

Rewrite

MomentumTheorem (Example) 
 Consider the steadystate flow of water though a reducing bend. The watervelocity is uniform at the inlet and outlet crosssection . If the water enters the bend at and is discharged at to the atmosphere, determine the force required to hold the bend in place or the force acting on the bend. Neglect the weight of the bend and water in the bend.


(Fig. 14a)

Choose CV so that CS is perpendicular to the velocity field at the inlet and the outlet. Apply momentum theorem: 
Under steadystate, 1 ^{st} term of the RHS is zero.
Note that , which follows from continuity. Apply continuity under SS:
Therefore,

 
Let us make freebody diagram on CV 


There are pressure forces acting on by the water; reaction forces on CV, Rx and Ry (reaction forces from the support to hold bend in place). Atmospheric pressure acts uniformly over the entire CV, except on where the pressure is (absolute). Therefore, add and substract to make the contribution of atmosphere zero on CV. In other words, . 
.
to hold the bend in place 
 Consider the flow of an incompressible fluid under steadystate in an expander:


(Fig. 14c)

Fluidvelocity is uniform at the entrance and at the outlet . Pressures are considered also uniform at sections 1 and 2 of the nozzle. Atmospheric pressure acts uniformly on the nozzle. Calculate the pressure drop .

Choose CV as shown below: 

(Fig. 14d)
(velocity everywhere in the smaller crosssection
Apply continuity over the same CV:
Or,
Substitute,
neglecting viscous forces (friction on wall), neglecting viscous forces.
Equating, Ans. 
 Choice of CV is important. If CV is chosen so that the CS cuts the nozzle, then there will be reaction forces, which will be unknown:


(Fig. 14e)

; However there is no viscous force as in the previous case!
Weight of the nozzle and the water inside. The latter–CV may be chosen when the question posed is: determine the force to hold the nozzle in place.
(Differential form of momentum conservation rate).
Before we apply Reynolds Transport theorem, it is important to understand the forces acting on a fluidelement.
We have earlier noted that the only force acting on a static fluid is the normal force or the pressure acting normal to the surface of the fluid element.




(Fig. 15a)

In flowing fluid, there is a tangential force or shear force acting parallel to the surface of the fluidelement. Thus, there are two types of stress developed in a flowing fluid: one is the normal stress, and the other is the shear stress, : 


(Fig. 15b)

It is shown that, where is the strain rate. 
 In the case of a static fluid, at any location in the fluid. In a moving fluid, however, :


(Fig. 15c)

 If fluid is inviscid,
And, , as in the case of a static fluid
 For a 3D flow, there are two and one on a plane, as shown below



(Fig. 15d)

There are sign conventions:
x: direction of normal to the plane or the direction of the area
y: direction of shear stress
 The Newton’s law of viscosity relates shearstress to velocity gradient via the viscosity of a fluid. For 1D flow of a Newtonian fluid,
is the velocity gradient and can also be represented as the strain rate. In general, when shear stress is applied on a fluidvolume, there are translation, deformation, dilation, rotation and distortion, mathematically expressed by strain rates. Analogous to the representation of shear stresses, strain rates are represented as the combination of the following velocity gradients: 

The Newton’s extended law of viscosity relates . For example, 


Again, it is obvious that if a fluid is stationary or inviscid 


Equation of motion (conservation of momentum in differential form) 
Consider flow of an incompressible fluid 


(Fig. 15f)

Xmomentum balance: 


(Fig. 15g)

See the figure above.
F body force = (due to gravity) 
Newton’s 2^{nd} law of motion can be applied on CV 
 
or, 
or, 
Insert the expressions for and simplify for an incompressible fluid


 
 
 
Similarly, one can write (or develop) ymomentum balance equation 

In vector form 

1^{st} term: Accumulation or transient or unsteady state
2^{nd} term: inertial (convective momentum flux)
3^{rd} term: pressure force
4^{th }term: viscous forces
5^{th} term: External bodyforce
We call this equation NavierStokes (NS) equation
Assumptions: (1) (incompressible fluid)
(2) Newtonian fluid
(3) Laminar flow

Before we apply the NS equation, let us make a note that the NS equation is a 2^{nd} order PDE. Therefore, 2 BCs are required for along each direction to solve for the velocityfield. We should note the common boundary conditions:

 No slip condition at the solid surface: it is an experimental observation that the relative velocity at the surface is zero. So, at a stationary solid surface
 At the interface of liquidgas or liquid–liquid (immiscible fluids), there is no jump in the velocity, shear stress and pressure



(gradients maybe different) 
 Symmetric BC:

At the center line of tube or rectangular channel, velocity gradient is zero. 


(Fig. 16b)
Fully developed flow:
( Note: this statement is equivalent to saying, neglect the end effects)
Apply continuity:

For fully developed flow, first term is zero.
is constant along y direction.
But, at y = ± (no slip condition)
So, everywhere.
Apply NS in xdirection:


1^{st} term is zero for SS for fully developed flow; . Horizontal Therefore, the equation is simplified as

No external pressure was applied and in addition, it is a long plate

 Integrate twice, 
Apply BC (1) (no slip condition) 
BC (2) 
Solve for to obtain
This is the expression for velocity profiles in a Couette flow 
(Note: at ). 


(Fig. 16d)
Example 2: Suppose that an external pressuredrop along xdirection is applied across the channel, in addition to the conditions mentioned in example 1. Now, calculate the velocity profiles in the channel. 
You should check the two BCs.

Special case

 (Same as in example 1)


, where,

 This is called Poiseuilleflow (velocity profile is parabolic)



(Fig. 16e)

 Determine the average velocity in the channel

 
 
Simplify to show that

or, 
Next, plot for Poiseuille and Couette flow
 : Poiseuille flow



Steadystate, laminar flow through a horizontal circular pipe 
Assumptions: (no swirling, circulation)

We are interested in solving 
Continuity: 
 
Fully developed flow: 
Therefore, r 
But
 (Noslip condition) 
 everywhere. 
So, the problem is reduced to solving . 
Note:
Relook at last two equations. 
 and depend on gravity. There is no viscous effects
 depends on viscous forces only. There are no gravitational effects.

Therefore, define nongravitational pressure, so that
or

Check: this definition of satisfies the above momentumbalance equation.
Therefore,
Substitute,

Say, (const ) 
integrate twice to obtain
Apply BC1: (symmetric BC)
(No slip BC)
where,
Note that velocity profile is parabolic in the tube. Calculate, volumetric flow rate

 
You should be able to show that 
Or,
Re–arrange,
This equation to calculate pressuredrop in a horizontal pipe of Length L and insidediameter D, for a viscous incompressible fluid flowing under steadystate fully developed laminar condition, is known as HagenPoiseuille equation
Newtonian and nonNewtonian fluid  macroscopic pressuredrop balance 
In the previous lecture, we applied the NS equation to obtain an expression for the pressure–drop in a tube for the condition of the steadystate laminar flow of a Newtonian fluid. We can also derive the same expression, the Hagen poiseuille equation, by making an integral (macroscopic) force balance.


(Fig. 18a)

If the flow is laminar (Re<2100) the velocity profile is parabolic:

For the Newtonian fluid,

 and, 
Therefore, : shearstress varies linearly in rdirection of the tube. 
Now, make a force balance over

which is the same as that obtained by applying differential momentum balance (NS equation) and then integrating with BCs.
 From the above expression, one can also determine viscosity of a fluid by measuring pressure drop across a tube for different flowrates:



(Fig. 18b)

The aboveplot is known as pseudoshear diagram, which is generally used to determine the viscosity of a slurrymixture. A similar approach can be followed to determine the viscosity or effective viscosity of NonNewtonian fluids.


Such a fluid has the following shearstress vs strain rate characteristics: 

(1)

where is known as the yield stress. Fluid is supposed to be ‘frozen’ if 

or, 
If the flow is laminar, the velocity profile is parabolic near the walls and is uniform in the central core of the tube. Shearstress in such fluid also varies linearly with r:



(Fig. 18c)

 

Equation (1) can be integrated with to obtain: 


Therefore,
The uniform velocity within the core of the tube 


Again, 
Integrate as an exercise to obtain

 


(Fig. 18d)
Energy Conservation Equation 
Similar to the derivation of continuity and momentum conservation equations, one can also apply Reynolds Transport equation to obtain the energyconservation equation. In this course, we will, however, not derive the expression and restrict our discussion to the salient features of the equation and its application to the fluidflow.
For a flowsystem, (specific property) for energy consists of three terms :


where, N = total property 
Recall, for N= mass 
 for N = momentum 
 for N= energy (J) 
  KE PE IE 
The three terms may be recognized as kinetic energy, potential energy, and internal energy, respectively. 

  
 
 



(Fig. 19a)
 If (stagnant fluid) there is no surface work,

 is the shaft work, for example, due to a turbine or a pump or blower, which delivers energy to the fluid or viceversa. if the machine is turbine

 if the machine is pump or blower 

 
Note that is zero if 
 (inviscid fluid)
 (tangential velocity) = 0
 (choice of CS)

The general energybalance equation can be rearranged to obtain the following expression for a CV: 

The terms of the equation are accumulation, conduction, shaftwork, viscous work, and energy efflux across CS, including pressureenergy .
For most engineering applications of steadystate, two–port system (one inlet and one outlet), uniform flow over the ports (1D flow), and being perpendicular to CS (by a suitable choice), one can obtain the following expression for energy conservation:


 
Moreover, if the flow is under isothermal condition without exchange of heat with surrounding (insulated system) and
Note: unit of each term is or 
If (no pump or turbine)
over two ports system, for example, in a tube or pipe or a tank:


(Fig. 19b)


The above conservation equation is also known as mechanical energy balance, consisting of KE, PE, and pressureenergy. Each term is also known as ‘head’. We renote the assumptions made in obtaining the above ME balance equation.
(1) SS (2) isothermal (3) No shaft work (4) negligible viscous work, .

It is important to note that viscous work can be neglected if the fluid is inviscid or a suitable choice of CS so that is perpendicular to CS.
                                                        