150A Review Session 2/13/2014 Fluid Statics

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 Pressure acts in all directions, normal to the surrounding surfaces

 or

 Whenever a pressure difference is the driving force, use gauge pressure

o Bernoulli equation

o Momentum balance with Patm everywhere outside

 Whenever you are looking for the total pressure, use absolute pressure

o Force balances

 Hydrostatic head = pressure from a static column of fluid

 From a z-direction force balance on a differential cube:

o Differential form: use when a variable is changing continuously

. Compressible (changing density) fluid, i.e. altitude

. Force in x or y direction (changing height), i.e. column support

o Integrated form: use when all variables are constant

. Total pressure in z direction, i.e. manometer

 Buoyancy

o Recommended to NOT use a “buoyant force”

. “Buoyant force” is simply a pressure difference

o Do a force balance including all the different pressures & weights

. Downward: gravity (W=mg) and atmospheric pressure (Patm*A)

. Upward: pressure from fluid below submerged container (ρgΔz*A) 150A Review Session 2/13/2014  Manometers

o Pressures are equal at the two points that are at the same height and

connected by the same fluid

o Work up or down from that equality point, including all fluids (& Patm)

 Compressible fluids

o At constant elevation, from ideal gas law,

o 3 ways to account for changing elevation using the ideal gas law

. Isothermal ( )

. Linear temperature gradient ( )

. Isentropic ( )

Macroscopic Momentum Balances

 [Accumulation = In – Out + Generation] for momentum of fluid

o Flow of momentum in (acceleration):

o Flow of momentum out (acceleration):

o Generation of momentum due to forces acting on fluid

o No momentum accumulation at steady state, so

 This is the same as Newton’s second law

 Sum of the forces on the fluid accelerate fluid

o Force due to pressure gradients (careful w/ sign needed for the geometry!)

o Friction forces by the wall on the fluid, opposite direction of flow 150A Review Session 2/13/2014

o Body forces due to gravity

rd  By Newton’s 3 law, there is an equal and opposite force of the fluid on the wall

 Therefore, the force of the fluid on the wall is

o Signs: fluid pushes forward as it enters, pushes behind it as it leaves

 Divide into x-direction and y-direction

o Calculate magnitude & direction

 If in laminar flow, =4/3 (turbulent flow ~1)

 Derivation of Hagan-Poiseuille flow below starts from here

Macroscopic Mechanical Energy Balances

 Bernoulli equation:

 The units of all terms are:

 Include α when have laminar flow (α=2 for laminar, α~1 for turbulent)

 Approach for solving Bernoulli problems

1. Make a drawing

2. Label points A (“inlet”) and B (“outlet”)

 Pick points where you know the properties

3. Write entire Bernoulli equation

4. Eliminate terms that are zero or very small compared to other terms

5. Determine absolute sign of each term

 Energy you start with needs to balance the energy you convert to

6. Plug in values, check units, solve for unknown 150A Review Session 2/13/2014 Friction Losses Through Conduits

 Viscous dissipation due to flow through a horizontal, straight, constant D pipe:

o By energy balance, equal to the pressure drop along the length:

o Plus momentum balance gives in terms of shear stress (τw):

o Definition of the Fanning friction factor:

o Combining gives in terms of Fanning friction factor:

o Expressions for valid for laminar & turbulent flow, Newtonian & non fluids

o For non-circular conduits, use same correlations replacing D with

 Reynold’s number defines ratio of inertial forces to viscous forces:

 Laminar Newtonian flow

o Velocity profile can be calculated exactly (H-P flow):

o Average velocity is:

o Volumetric flow rate:

o Friction factor is known exactly:

 Turbulent Newtonian flow

o Transition to turbulent flow for Re > 2100

o Friction factor correlated empirically where k is roughness

. Note that k has dimensions, make sure they match the D

o Use the Moody Chart or Churchill Correlations to find

o Often have to iterate to solve these problems 150A Review Session 2/13/2014 Non-Newtonian Fluid

 Same momentum and energy balances (Bernoulli), just new

 Do not obey Newton’s law of viscosity where viscosity μ is constant

 Power law fluids

o Non-linear relation between shear stress and shear rate

o Pseudoplastic (shear-thinning, most common) i.e. paint, polymer solutions

o Dilatant (shear-thickening) i.e. cornstarch + water

o Effective viscosity must be determined experimentally

o New Reynold’s number

o New critical Reynold’s number

o Laminar:

o Turbulent: use power law chart

 Bingham plastic

o Acts as a solid for , flows as Newtonian fluid for

o i.e. ketchup, mayo

o Parameters must be determined experimentally

o Two important dimensionless parameters: &

o Laminar (non-linear equation):

o Turbulent: use Bingham plastic chart

150A Review Session 2/13/2014 Conversions

Mass: Distance: Volume:

Pressure:

Temperature:

Viscosity:

Force:

Power:

Other useful information

From F=ma,

150A Review Session 2/13/2014 Hagen-Poiseuille Flow in a Pipe – Derivation Using the Navier-Stokes Equations

As we saw in class and in the macroscopic momentum balance abvoe, Newton’s 2nd law and a momentum balance give:

Note: Think about the signs every time based on how you define what’s going in and out of your system!

This is analogous to looking at a microscopic scale, where the left-hand side is a density multiplied by acceleration. This is called the Cauchy momentum equation.

In class we saw Newton’s law of viscosity in 2D,

Expanding this to a differential form in all directions and using it in the previous equation, we get the Navier-Stokes equations for constant and :

Now to solve the Hagen-Poiseuille problem for flow in a pipe, we can make a few simplifications. In an horizontal, constant diameter pipe, there will be no acceleration terms and no gravity terms, so that:

Expanding this in Cartesian coordinates for the z-component of the velocity:

For our pipe problem though, Cylindrical coordinates will be more convenient:

Don’t get overwhelmed with these expressions, the problems we’ll do simplify and drop many of these terms. Also, the complete Navier-Stokes equations in Cartesian, Cylindrical, and Spherical coordinates are given in Appendix E of your book. 150A Review Session 2/13/2014

Figure: Flow in a straight, horizontal, constant diameter (D) pipe. The velocity profile is as a function of the radius (r) and the overall volumetric flow rate is Q. Pressures are denoted po and pL at the start z=0 and end z=L of the pipe, respectively.

For flow in a pipe, as shown in the diagram, we have only a z-component of velocity along the length of the pipe. This is only a function of where the particle is in regards to the radius ( ) of the pipe. It is not a function of the angle ( ) due to symmetry or the location along the length of the pipe (z) as we have said there is no acceleration and the flow is fully developed. Therefore, we have only and we can simplify the above equation:

With a constant pressure gradient along the length (L) of the pipe, we can now integrate this equation twice to get the velocity profile, .

150A Review Session 2/13/2014

So now we have an expression for , which as expected from our sketch is parabolic, and we just need to use the boundary conditions to solve for the constants.

Boundary conditions:

at  (1) no-slip condition at the surface of the pipe

finite at or  ( 2) here we know that the velocity is at its maximum in the middle of the pipe

at r= 0

Using boundary condition (2) in either form, we can immediately see that , because and .

Now using boundary condition (1),

Putting this all together, we get the velocity profile we saw in class:

From this we can easily derive the shear stress of the fluid on the wall ( ) at the pipe surface, as well as the volumetric flow rate (Q) and the average velocity ( ). Let’s start with the volumetric flow rate, by integrating over the cross-sectional area of the pipe:

150A Review Session 2/13/2014

Where we’ve defined pressure drop as in the expression above. Dividing by the cross-sectional area over which we determined the volumetric flow rate, we get the average velocity.

Now we can also determine the shear stress ( ) of the fluid on the wall at the pipe surface (r=R). This is in the direction of the fluid’s motion.

Or equivalently, the shear stress exerted by the wall on the fluid ( ), again at the surface. This is opposing the fluid’s motion.