fluid and particle mechanics - hmaddah.com

CHEN-231 – Spring 2021

Fluid and Particle Mechanics

Week #5: Lecture Notes

Chemical and Materials Engineering Department

King Abdulaziz University (KAU) at Rabigh

Instructor: Dr. Hisham Maddah

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Example 1 in the text.

Chapter 5:- 5.2: Applications of the Integral Expression for Linear Momentum

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Forces imposed on the fluid include: 1) Pressure forces at sections (1) and (2).2) Body force due to the weight of fluid in the

control volume.3) Forces due to pressure and shear stress, Pw and

𝜏w, exerted on the fluid by the pipe wall.

Assume that the resultant force on the fluid (due to Pw and 𝝉w) by the pipe is symbolized as B, and its x and y components as Bx and By, respectively.

The overall momentum balance, the external forces acting on the fluid in the control volume are:


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Each component of the unknown force B is assumed to have a positive sense.

The actual signs for these components, when a solution is obtained, will indicate whether or not this assumption is correct.

Evaluating the surface integral in both the x and y directions, we have …


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Another C.V. approach …This control volume is bounded simply by the straight planes cutting through the pipe at sections (1) and (2).

The fact that a control volume such as this can be used indicates the versatility of this approach, that is,that “the results of complicated processes occurring internally may be analyzed quite simply by considering only those quantities of transfer across the control surface.”


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The force having components Bx and By is exerted on the control volume by the section of pipe cut through at sections (1) and (2).

The pressures at (1) and (2) are gage pressures, as the atmospheric pressures acting on all surfaces cancel.

Note that the resulting equations for the two control volumes are identical. Thus, “a correct solution may be obtained from various chosen C.V.s” as long as they are analyzed carefully and completely.


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We were to evaluate the force exerted on the pipe rather than that on the fluid.

The force sought is the reaction to B and has components equal in magnitude and opposite in sense to Bx and By.

The components of the reaction force, R, exerted on the pipe by the fluid …


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Example 2 in the text.The control-volume expression for linear momentum (the momentum theorem), consider the steam locomotive tender, which obtains water from a trough by means of a scoop. The force on the train dueto the water is to be obtained.


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Solution:

The logical choice for a control volume in this case is the water-tank/scoop combination.

Our control-volume boundary will be selected as the interior of the tank and scoop.

As the train is moving with a uniform velocity, there are two possible choices of coordinate systems:

1) Fixed in space.2) Moving with the velocity of the train, v0.

Let us first …analyze the system by using a moving coordinate system.


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Let us first, analyze the system by using a moving coordinate system.

with the xy coordinate system moving at velocity v0

(inertial coordinate system). All velocities are determined with respect to the x and y axes.

Velocity of the fluid is going to be in the opposite direction of the velocity of the moving train, scooping.

Fx is the total force exerted on the fluid by the train and scoop, neglecting forces due to pressure and shear … “The momentum flux term” (per unit length)

The rate of change of momentum within the control volume is zero, as the fluid in the control volume has zero velocity in the x direction.

“fluid on train”

“train on fluid”


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Now let us consider the same problem with a stationary coordinate system, Employing once again the control-volume relation for linear momentum …

where the momentum flux is zero, as the entering fluid has zero velocity. There is, no fluid leaving the C.V.

“The momentum flux term” = Zero

v is the velocity relative to fixed coordinates, v∙n is the velocity relative to the control-volume boundary.

“fluid on train”

“train on fluid”

ሶ𝒎 = mass of fluid entering the C.V.


Chapter 5:- Tutorial: Practice Problems

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(a) Find the area at section 2,

𝐴2 = 𝐴𝑗 + 𝐴𝑠𝐴𝑠 = 𝐴2 − 𝐴𝑗 = 0.54 𝑓𝑡2

Assume:

Steady state (st. st.) →𝑑

𝑑𝑡=0

Incompressible flow → 𝜌=constant

−𝑣𝑗𝐴𝑗 + −𝑣𝑆𝐴𝑠 + 𝑣2𝐴2 = 0

−90 × 0.06 + −10 × 0.54 + 0.6𝑣2 = 0

𝒗𝟐= 𝟏𝟖 𝒇𝒕/𝒔

Given, 𝐴𝑗 = 0.06 𝑓𝑡2 𝑣𝑗 = 90 𝑓𝑡/𝑠

𝐴𝑠 = ? ? 𝑓𝑡2 𝑣𝑠 = 10 𝑓𝑡/𝑠𝐴2 = 0.6 𝑓𝑡2

inflow outflowinflow


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(b) Assume steady state, apply “integral momentum”

𝑃1𝐴𝑗 + 𝑃1𝐴𝑠 − 𝑃2𝐴2 = …

𝑣𝑗𝜌 −𝑣𝑗 𝐴𝑗 + 𝑣𝑠𝜌 −𝑣𝑠 𝐴𝑠 + 𝑣2𝜌 +𝑣2 𝐴2

𝑃1 𝐴𝑗 + 𝐴𝑠 − 𝑃2𝐴2 = −𝜌𝑣𝑗2𝐴𝑗 − 𝜌𝑣𝑠

2𝐴𝑠 + 𝜌𝑣22𝐴2

𝑃2 − 𝑃1 =𝜌

𝑔𝑐 𝐴2𝑣𝑗2𝐴𝑗 + 𝑣𝑠

2𝐴𝑠 − 𝑣22𝐴2

𝑃2 − 𝑃1 = 1116.23 Ibf/ft2

𝑷𝟐 − 𝑷𝟏 = 𝟕. 𝟕𝟓 𝐩𝐬𝐢Given, 𝐴𝑗 = 0.06 𝑓𝑡2 𝑣𝑗 = 90 𝑓𝑡/𝑠

𝐴𝑠 = ? ? 𝑓𝑡2 𝑣𝑠 = 10 𝑓𝑡/𝑠𝐴2 = 0.6 𝑓𝑡2

𝐴2

[ gc=32.2 (Ibm) (ft)/(Ibf) (s2) ]


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Assume steady state, apply “integral momentum”

Momentum in x-direction:

𝐹𝑥 = 𝑣𝑗 cos 𝜃 𝜌Aj −𝑣𝑗 + 𝑣𝑗𝜌Aj 𝑣𝑐

𝐹𝑥 = −𝜌Aj𝑣𝑗 𝑣𝑗 cos 𝜃 − 𝑣𝑐

𝐹𝑥 = − 998.2𝜋

40.12 (20)(20 cos 45 − 4.5)

𝐹𝑥 = −1511 N “w.r.t the water jet”

∴ Force exerted on the car is +1511 N in x-directionGiven, 𝑣𝑐 = 4.5 𝑚/𝑠 𝑣𝑗 = 20 𝑚/𝑠

𝐹𝑥 = ? ? 𝑁 𝐹𝑦 = ? ? 𝑁

outwardinward

20 cos 45

20 sin 45


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Assume steady state, apply “integral momentum”

Momentum in y-direction:

𝐹𝑦 = −𝑣𝑗 sin 𝜃 𝜌Aj −𝑣𝑗 + 0

𝐹𝑦 = 𝜌Aj𝑣𝑗2 sin 𝜃

𝐹𝑦 = 998.2𝜋

40.12 20 2(sin 45)

𝐹𝑦 = +2217 N “w.r.t the water jet”

∴ Force exerted on the car is −2217 N in y-directionTotal for on the car: 𝑭 = +𝟏𝟓𝟏𝟏 𝐞𝐱 − 𝟐𝟐𝟏𝟕 𝐞𝐲

Given, 𝑣𝑐 = 4.5 𝑚/𝑠 𝑣𝑗 = 20 𝑚/𝑠

𝐹𝑥 = ? ? 𝑁 𝐹𝑦 = ? ? 𝑁

outwardinward

20 cos 45

20 sin 45


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For frictionless flows, no drag (shear stress) on plate

Assume steady state flow, and then apply “integral momentum”

Momentum in x-direction or y- direction:

𝐹𝑛 =ඵ𝑐.𝑠.

𝑣𝑛𝜌 𝑣 ∙ 𝑛 𝑑𝐴 = 0

𝐹𝑛 = (𝑣 cos 𝜃) 𝜌Aj −𝑣𝑗

𝐹𝑛 = −(𝑣 cos 𝜃) ሶ𝑚 = − 1003

52 = −120 Ibf


𝐹𝑥 = ? ? 𝑁 𝐹𝑦 = ? ? 𝑁

inflow

StationaryIbf s2/ft = slugs

C.V.

Fn


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For frictionless flows, no drag (shear stress) on plate

Assume steady state flow, and then apply “integral momentum”

Momentum in x-direction or y- direction:

𝐹𝑥 = −1203

5= −72 Ibf

𝐹𝑦 = −1204

5= −96 Ibf


𝐹𝑥 = ? ? 𝑁 𝐹𝑦 = ? ? 𝑁

C.V.

Fn

Ibf s2/ft = slugs

Revision Slides: Midterm #1 Materials

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Chapter 1:- 1.1: Fluids and the Continuum

• Momentum Transfer: Study of the motion of a fluid and the forces that produces these motions.

• Newton’s 2nd Law: [ Force (F) = mass (m) x acceleration (a) ]➔ F = ma1. Force is directly related to “the rate of change of momentum”.

2. Rate of change of momentum: a = dv/dt = [m/s2].

3. Forces: e.g. [gravity, pressure, stress].

Fluid Mechanics = Momentum Transfer

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Chapter 1:-1.3: Point-to-point Variation of Properties in a Fluid

• Point-to-point variation in pressure:• From dP/ds zeroth and maximum paths, the

maximum directional derivative for pressure along the path s is a suggested vector as …

• e is a unit vector in x, y, or z directions (Cartesian coordinates).

• Gradient = grad = 𝜵 = pronounced “del”, that is “a vector with maximum rate of change of dependent variable with respect to distance”where ..

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Chapter 1:- 1.5: Compressibility

• Compressibility• Incompressible (liquids): 𝜌 = constant• Compressible (gases): 𝜌 ≠ constant

• Bulk Modulus (𝜷)• Fluid property that characterizes compressibility, N/m2

• A measure of how resistant to compression a substance is.• Ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume.

• Mach number (M)• Ratio of fluid velocity (v) to the speed of sound or acoustic velocity (C) in a fluid.• Dimensionless number to identify fluid compressibility (M<0.2 incompressible).

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−

Chapter 1:- 1.6: Surface Tension

• Capillary effects• Forces Balance on a Wetting Liquid (Water) in a Tube

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Note that volume of water is equal to its mass since its

density is 1 g/cm3 = 1000 kg/m3

Chapter 2:- 2.1: Pressure Variation in a Static Fluid

• Specific gravity (S.G.)• Every material has a specific gravity value (e.g. S.G.Hg = 13.6).• 𝝆: Density (kg/m3) or (Ibm/ft3), 𝝆𝑯𝟐𝑶 = 1000 kg/m3 = 62.4 Ibm/ft3

• One can get … 𝝆𝑯𝒈= S.G.Hg x 𝜌𝐻2𝑂 = 13600 kg/m3

• Conversions:• Inch → ft , ÷12• In2 → ft2 , ÷144• psi (Ibf/in2) → Ibf/ft2 , x144• 1 atm = 1.013x105 Pa (N/m2) = 14.7x144 Ibf/ft2

• 1 atm = 14.7 psi• 32.2 Ibm ft/s2 = Ibf

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𝐒. 𝐆. =𝛒

𝛒𝐇𝟐𝐎

x1000

x106

x105

x1.013x105

kPa

MPa

Bar

atm

Pa

Chapter 2:- 2.1: Pressure Variation in a Static Fluid

• Pressure = Force/Area = F/A , In Pascal (Pa = N/m2)• The pressure increases when going downwards (+).

• The pressure decreases when going upwards (–).

• The pressure is constant horizontally.

• in P=𝜌gh, always keep gc in the denominator. ➔

• In SI system, gc=1 (kg) (m)/(N) (s)2 , g=9.8 m/s2 ➔ g/gc=9.8 N/kg

• In AE system, gc=32.2 (Ibm) (ft)/(Ibf) (s2), g=32.2 ft/s2

➔ g/gc=1 Ibf/Ibm

• Example, Find PA – PB in the following …

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𝐏 =𝜌𝒈𝒉

𝒈𝒄

Chapter 2:- 2.4: Buoyancy

• Buoyancy is the force that causes objects to float.

• An upward force exerted by a fluid that opposes the weight of a partially or fully immersed object.

• Buoyancy is caused by the differences in pressure acting on opposite sides of an object immersed in a static fluid.

In Equilibrium:

• Buoyancy force (N) = Weight force (N) ➔ Fb=W

1. If W > Fb: Body will be completely immersed.

2. If W < Fb: Body floats on top or might be submerged.

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gc gc

=

Chapter 2:- 2.4: Buoyancy

• For a body submerged in a fluid with density, the resultant force (F) holds the body in equilibrium.

• Use given fluid S.G. to get density according to fluid temperature.

• The resultant force F is composed of two parts:

• Archimedes principle: The body experiences an upward force equal to the weight of the displaced fluid.

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(W)

(Fb)

V: volume of the body.

Chapter 4:- 4.1: Integral Relation {Conservation of Mass: Control-Volume Approach}

• From the net outward flow of mass across the control surface, or the net mass efflux from the control volume.

• Positive (+): Net efflux

• Negative (−): Net influx

• Zero (0): Constant mass within the C.V.

• Rate of accumulation:

• Thus, integral expression (mass balance over C.V.) …

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Chapter 4:- 4.1: Integral Relation {Conservation of Mass: Control-Volume Approach}

• Flowrates:

Volumetric flowrate = ሶ𝒗 [m3/s = ft3/s]

A=area [m2]

v=velocity [m/s]

Mass flowrate = ሶ𝒎 [kg/s = Ibm/s]

𝜌=density [kg/m3]

A=area [m2]

v=velocity [m/s]

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ሶ𝒗 = 𝒗𝑨

ሶ𝒎 = 𝝆 ሶ𝒗 = 𝝆𝒗𝑨

Chapter 4:- 4.2: Specific Forms of the Integral Expression

• With such assumption … the integral expression can be simplified,

• Steady flow → C.V.=0 , (since 𝑑

𝑑𝑡= 0 and t →∞)

• Incompressible → 𝜌=constant

• v∙n sign from n (outward) and v directions:

Input: 𝜃=180° → v∙n = −v1 (opposite)

Output: 𝜃=0° → v∙n = +v2 (Similar)

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Density can be taken out of integration, cancelled if C.V.=0

n: normal vector (perpendicular)

n nv1 v2

Chapter 5:- 5.1: Integral Relation for Linear Momentum

• Newton’s 2nd Law of Motion:“The time rate of change of momentum of a system is equal to the net force acting on the

system and takes place in the direction of the net force.”

• What is momentum? Momentum= mv = mass*velocity

• The control volume approach:

Divide system into 3 parts I, II, III as the following …• Region I is occupied by the system only at time t.• Region II is occupied by the system at t + ∆t. • Region III is common to the system both at t and at t + ∆t.

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[overall linear-momentum balance for a control volume]“Integral form of Momentum Theorem”

Note: v=velocity , V=volume

• In Rectangular coordinates …

Single-Vector Equations ..

Assumptions: as before,

Steady state (St. St.) ➔𝑑

𝑑𝑡=0

Incompressible ➔ 𝜌=constant


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• Signs of velocity components and scalar product in the C.S.

[in (−) , out (+)] , based on the n-vector analysis and similar/opposite directions➔ normal vector (n) is always coming outward of a plane.

Sign is based on axis direction and can analyzed if velocity inclined to an angle …

Similarly, for F ➔ its sign is based on axis and the force can be analyzed analytically with (sin) or (cos) of the given angle.

+

+

−

−

x

y

𝜃x

v

v cos𝜽

v s𝐢𝐧𝜽


Dr. HA Maddah 33Note: v=velocity , V=volume

• Analysis of inclined velocity or force components from the given angle ..

𝜃

v

v cos𝜽

v s𝐢𝐧𝜽

𝜃

F

F sin𝜽

F cos𝜽


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• Important notes for problem solving:1. Define the C.V. carefully before start solving.

2. Solve accordingly and choose the easiest path.

3. Remember that F and v (not with normal vector) can be analyzed.

4. If pressure given, F=PA .. and remember that pressure is always an inward force.

5. Use ሶ𝑣 = 𝑣1𝐴1 = 𝑣2𝐴2 [m3/s] to check your solution correctness.

6. Use ሶ𝑚 = 𝜌1 ሶ𝑣1 = 𝜌2 ሶ𝑣2 [kg/s] to check your solution correctness.

7. Weights are always downward forces (e.g. water or pipe weight)

F=w=mg/gc

8. Gauge (relative) vs. absolute pressure: Pabs = Pgauge + Patm

(P can also be angled)n: normal vector (perpendicular)

n nP1A1 P2A2

Chapter 5:- Magnitude of a Vector

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