- The continuum approach is usually not valid when the gas pressure is very small (few milli-torr 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 Theorem-Pressure 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 2-D(x-z) scenario. Pressure acts on all 3-faces 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 x-direction) 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)
|
2-Now, 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 z-direction

|
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, pressure-variation 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/m3 in comparison to 1000 kg/m3 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 two-immiscible liquids (one is heavier than the other) |
|
(Fig. 5a)
|
|
Consider two liquids of densities . Depth (measured form the top) of the top-fluid is  and that of the bottom-fluid is 
Put the co-ordinate 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 co-ordinate axis at the top of the surface, with z-axis 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 water-barometer |
- 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 flow-field 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 steady-state conditions,  . However, an observer moving with the flow field will experience ‘acceleration' as he moves from the larger diameter-section to the smaller diameter-section 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 1-2) |
The last expression is tantamount to saying that the velocity of a labeled or marked material-particle 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 power-plant (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 balloon-sensor 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 flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. The identified mass moves around in the flow-field. 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 flow-field 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 co-ordinate axis is first fixed. A ‘CV’ is then marked in the flow-field. 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 co-ordinates  . 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 2nd 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 3rd 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 property-flux 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)
|
 |
| (Particle-mass cannot be created or destroyed) |
The 1st term of the right-hand-side is zero: particle-mass 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 fluids-mixture.
The RHS of the above equation is the integral (over the entire cross-section) of the differential volumetric flow rate: |
|
|
 |
 |
|  |
Or Ans.
Apply the conservation law again over larger to show that,

Ans.
Further, (for the 2nd section) (for the 1st section)
or, 
Re-visit, 
or, (volumetric flow rate)
or, , average velocity in the section
Compare to to show that .
Differential form of mass-conservation |
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 mass-in:
Differential form of mass-conservation |
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 mass-in:
(It is equivalent to to of the Reynold’s transport theorem)
CS marked 2:
Rate of mass-out: 
- Net rate of change in mass in x-direction
|
|  |
|  |
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 mass-conservation rule applied to a CV in the earlier lecture |
Re-arrange:  |
This is the vector-form of continuity-equation and readily applied for all co-ordinate 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: |

|
Re-call 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 unsteady-state. 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 above-equation. For 1D radial flow, |
. |
|
(Fig. 13b)
|
const under steady-state.
To sum-up, total mass flow rate is constant at all locations in the r-direction. However, mass flux changes because the differential area changes in the r-direction
|
|
B. Momentum - conservation equation |
|
(Fig. 13c)
|
Reynolds Transport Theorem: |
Property, |
 |
1st Term: local rate of change of momentum in CV
2nd Term: Net rate of momentum in and out of CV through CS.
3rd Term: material or partial –rate of change of momentum, and it may be seen as the external body-force or the source term for the generation of momentum by the external forces.
|
Mathematically, 
|
Re-write
|

Momentum-Theorem (Example) |
- Consider the steady-state flow of water though a
reducing bend. The water-velocity is uniform at the inlet and outlet cross-section . 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 steady-state, 1 st term of the RHS is zero.

Note that  , which follows from continuity. Apply continuity under SS: 
Therefore,
|
|  |
Let us make free-body- 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 steady-state in an expander:
|
|
(Fig. 14c)
|
Fluid-velocity 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 cross-section 
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 fluid-element.
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 fluid-element. 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 shear-stress 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 fluid-volume, 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)
|
X-momentum balance: |
|
|
(Fig. 15g)
|
See the figure above.
F body force = (due to gravity) |
Newton’s 2nd 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) y-momentum balance equation |
|
In vector form |
|
1st term: Accumulation or transient or unsteady state
2nd term: inertial (convective momentum flux)
3rd term: pressure force
4th term: viscous forces
5th term: External body-force
We call this equation Navier-Stokes (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 2nd order PDE. Therefore, 2 BCs are required for along each direction to solve for the velocity-field. 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 liquid-gas 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 x-direction:
|
 |
1st 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 pressure-drop along x-direction 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 Poiseuille-flow (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
|
|
|
Steady-state, laminar flow through a horizontal circular pipe |
Assumptions:  (no swirling, circulation)
|
We are interested in solving |
Continuity: |
|  |
Fully developed flow: |
Therefore, r |
But
| (No-slip condition) |
| everywhere. |
So, the problem is reduced to solving . |
Note: 
Re-look 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 non-gravitational pressure, so that

or
|
Check: this definition of  satisfies the above momentum-balance 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 pressure-drop in a horizontal pipe of Length L and inside-diameter D, for a viscous incompressible fluid flowing under steady-state fully developed laminar condition, is known as Hagen-Poiseuille equation
Newtonian and non-Newtonian fluid - macroscopic pressure-drop 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 steady-state 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, : shear-stress varies linearly in r-direction 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 above-plot is known as pseudo-shear diagram, which is generally used to determine the viscosity of a slurry-mixture. A similar approach can be followed to determine the viscosity or effective viscosity of Non-Newtonian fluids.
|
|
Such a fluid has the following shear-stress 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. Shear-stress 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 energy-conservation 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 fluid-flow.
For a flow-system,  (specific property) for energy consists of three terms :
|
 |
where, N = total property |
Re-call, 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 vice-versa. 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 energy-balance equation can be re-arranged to obtain the following expression for a CV: |
|
The terms of the equation are accumulation, conduction, shaft-work, viscous work, and energy efflux across CS, including pressure-energy .
For most engineering applications of steady-state, two–port system (one inlet and one outlet), uniform flow over the ports (1-D 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 pressure-energy. Each term is also known as ‘head’. We re-note 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.
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