ﺔﺜﻟﺎﺜﻟﺍ ﺔﻠﺣﺮﻤﻟﺍ · poundal = ibm ×32.17 ft/sec2 f= lbw m=slug lbw =...

الجامعة التكنولوجية

قسم الهندسة الكيمياوية الثالثةالمرحلة

خزن ونقل النفط ومشتقاته

م.د. علي عبد الرحمن

Save from: http://www.uotechnology.edu.iq/dep-chem-eng/index.htm

Storage & Transport of Crude Oil and Petroleum Products

1- Units

2- Classification of process

3- Fluids – Definitions , properties

4- Classification of fluids

5- Forms of Energy

6- Energy balance on closed system

7- Energy balance on flow system

8- Bernoulli's Equation

9- Examples on applied energy equation

10- Pressure drop in a pipe

11- Pressure drop at points in system

12- Pressure drop in non-isothermal liquid flow

13- Fittings and valves

14- Calculation of pressure drop in short gas lines

15- Calculation of pipe diameter for given pressure drop and flow

rate

16- Complex liquid gathering systems

17- Design gathering system for liquid hydrocarbon

2

Dr. Ali Abdulrahman Al-Ezzi

Storage & Transport of Crude Oil and Petroleum Products

The transfer of fluids through piping and equipment is accompanied by

friction and may result in changes in pressure, velocity, and elevation. These effects

require input of energy to maintain flow at desired rates.

Any think that is derived from the primary

Means

TANKS

Piping &Pumps,

Compressor

Liquid Gas

Fluids

Process

PROCESS

Physical Chemical

M1 M2

E1

U1, EK1, EP1, P1V1

W

Surrounding

Surrounding

Q E2

U2, EK2, EP2, P2V2

Liquid + Gas + Solid

Physical System

Dimension or Quantities or Fundamental

QqQuantities or Secondary

Primary

Force (F)

Time (T)

Temp (∅)

Mass (M)

Length (L)

3


Units: Can be defined as the means of expressing dimensions according to the systems of

measurements.

MEASUREMENTS SYSTEMS

S.I (international or metric) ENGLISH

CgS MKS Absolute Poundal

FPS

Gravitational

F (primary) mass (secondary)

American British

gc : Correction Factor (for Newton's law)

Fα m × a F=c × m × a C=F/m × a , In (SI unit ) C= N × sec2/ Kg × m

In (English unit) C= lbf × sec2/32.17. lbm.ft [ gC = 1/c] = Kg. m/N. sec or 32.17 lbm .ft /

lbf.sec2, F= (m/gc).a

Note: g/gc=lbm

lbf

g

g

ftlbm

lbfft

lbfftlbm

ft

C

1sec

secsec/

sec/ 2

22

2

Note: kg

Jm

2

2

sec Prove:

2

2

2 secsec

m

Kg

mmkg

Kg

mN

Kg

J

cm

Pound

Kg

F= dyne

M= gm

Dyne=gm× cm/sec2

F= N

M=Kg

N= Kg × m/sec2

F= poundal

M= Ibm

poundal = Ibm × ft/sec2

F= lbf

M=lbm

poundal = Ibm ×32.17 ft/sec2

F= lbw

M=slug

lbw = slug ×ft/sec2

4


For example: EK= mNJm

KEg

VmgK

C

2

22

sec2

Note: Btu→Cal→ft. lbf→J

1 Btu = 778 ft.lbf 1 Cal = 4.184 J KW = 737.56 ft.lbf/sec

1 Btu = 252 Cal 1hp = 550 ft.lbf/sec

PROCESS

Unsteady State Steady Stat

d parameter/ d time ≠ zero d parameter/ d time = zero

Reversible Irreversible

It’s the process that doesn't that takes in consideration

Take into consideration the the loss by energy due to

Loss of energy fraction and fraction

Its imaginary process and the Efficiency may be equal 100% Throttling Adiabatic

Reversible Reversible

Open system, flow, continuous closed, non-flow

Turbine Boiler Condenser Pump

T=constant polytrophic Q= 0 V=constant p=constant

Isothermal adiabatic isometric isobaric

Insulated system

isochoric

H,T,P,U,S,V

5


1-Definition of Fluids

A fluid is defined as a substance that deforms continuously whilst acted upon

by any force tangential to the area on which it acts. Such a force is termed a shear

force, and the ratio of the shear force to the area on which it acts is known as the shear

stress (see fig. 1). The characteristic that distinguishes a fluid from a solid is its

inability to resist deformation under an applied shear stress (a tangential force per unit

area). When a fluid is at rest neither shear forces nor shear stresses exist in it. A solid,

on the other hand, can resist a shear force while at rest. In a solid, the shear force may

cause some initial displacement of one layer over another, but the material does not

continue to move indefinitely and a position of stable equilibrium is reached.

Definition 1 Fluid is any substance that deforms continuously when

subjected to a shear stress, no matter how small.

Shear forces are possible only while relative movement between layers is taking

place.

Fluids may be sub-divided into liquids and gases. A fixed amount of a liquid

has a definite volume which varies only slightly with temperature and pressure. If the

capacity of the containing vessel is greater than this definite volume, the liquid

occupies only part of the container, and it forms an interface separating it from its

own vapor, the atmosphere or any other gas present.

Gas- a fixed amount of a gas, by itself in a closed container, will always

expand until its volume equals that of the container. Only then can it be in

equilibrium.

In the analysis of the behavior of fluids an important difference between

liquids and gases is that, whereas under ordinary conditions liquids are so difficult to

compress that they may for most purposes be regarded as incompressible, gases may

be compressed much more readily. Where conditions are such that an amount of gas

undergoes a negligible change of volume, its behavior is similar to that of a liquid and

it may then be regarded as incompressible. If, however, the change in volume is not

negligible, the compressibility of the gas must be taken into account in examining its

behavior.

6


Figure 1: (A) The displacement and the corresponding shear strain increase linearly

with time. For a fluid, the relationship between shear stress and shear strain is

proportional. (B) The fluid velocity in the x direction, u, is a function of the y

coordinate. The velocity u(y) varies linearly from 0 at the bottom plate to U0 at the

top plate.

Liquids have much greater densities than gases. As a consequence, when considering

forces and pressures that occur in fluid mechanics, the weight of a liquid has an

important role to play. Conversely, effects due to weight can usually be ignored when

gases are considered. The different characteristics of solids, liquids and gases result

from differences in their molecular structure.

Figure 2: The liquid occupies only part of the container, and it forms an interface

separating it from its own vapor.

7


Figure 3: A fixed amount of a gas, by itself in a closed container, will always expand

until its volume equals that of the container.

All substances consist of vast numbers of molecules separated by empty space.

The molecules have an attraction for one another, but when the distance between them

becomes very small (of the order of the diameter of a molecule) there is a force of

repulsion between them which prevents them all gathering together as a solid lump.

2-Fluid properties Definition 2 The density of a fluid (or any other form of matter) is the amount of

mass per unit volume.

V

M

or the density at a point in fluid as

V

MLim

0

The unit of density is kg/m3.

The density r (rho) or more strictly, mass density, of a fluid is its mass per unit

volume, while the specific weight 𝛾 (gamma) is its weight per unit volume. In the

British Gravitational (BG) system density 𝜌 will be in slugs per cubic foot (kg/m3 in

SI units), which can also be expressed as units of lb_sec2/ft4 (N_s2/m4 in SI units)

Specific weight 𝛾 represents the force exerted by gravity on a unit volume of fluid,

and therefore must have the units of force per unit volume, such as pounds per cubic

8


foot (N/m3 in SI units). Density and specific weight of a fluid are related as:

Since the physical equations are dimensionally homogeneous, the dimensions of

density are

Note that density 𝜌 is absolute, since it depends on mass, which is independent of

location. Specific weight 𝛾, on the other hand, is not absolute, since it depends on the

value of the gravitational acceleration g, which varies with location, primarily

latitude and elevation above mean sea level.

Specific volume v is the volume occupied by a unit mass of fluid. We

commonly apply it to gases, and usually express it in cubic feet per slug (m3/kg in

SI units).

Specific volume is the reciprocal of density. Thus

Specific gravity s of a liquid is the dimensionless ratio

Physicists use 4°C (39.2°F) as the standard, but engineers often use 60°F

(15.56°C). In the metric system the density of water at 4°C is 1.00 g/cm3 (or 1.00

g/mL), equivalent to (1000 kg/m3, 62.4 lb/ft3, 8.34 lb/gal) and hence the specific

gravity (which is dimensionless) of a liquid has the same numerical value as its

density expressed in g/mL or Mg/m3.

The specific gravity of a gas is the ratio of its density to that of either

hydrogen or air at some specified temperature and pressure, but there is no general

9


agreement on these standards, and so we must explicitly state them in any given case.

Since the density of a fluid varies with temperature, we must determine and specify

specific gravities at particular temperatures.

Compressible and Incompressible Fluids

Fluid mechanics deals with both incompressible and compressible fluids, that

is, with liquids and gases of either constant or variable density. Although there is no

such thing in reality as an incompressible fluid, we use this term where the change in

density with pressure is so small as to be negligible. This is usually the case with

liquids. We may also consider gases to be incompressible when the pressure variation

is small compared with the absolute pressure.

Ordinarily we consider liquids to be incompressible fluids, yet sound waves,

which are really pressure waves, travel through them. This is evidence of the elasticity

of liquids. In problems involving water hammer we must consider the compressibility

of the liquid.

The flow of air in a ventilating system is a case where we may treat a gas as

incompressible, for the pressure variation is so small that the change in density is of

no importance. But for a gas or steam flowing at high velocity through a long

pipeline, the drop in pressure may be so great that we cannot ignore the change in

density. For an airplane flying at speeds below (100 m/s), we may consider the air to

be of constant density. But as an object moving through the air approaches the

velocity of sound, which is of the order of (1200 km/h) depending on temperature, the

pressure and density of the air adjacent to the body become materially different from

those of the air at some distance away, and we must then treat the air as a

compressible fluid.

Compressibility of Liquids

The compressibility (change in volume due to change in pressure) of a liquid

is inversely proportional to its volume modulus of elasticity, also known as the bulk

modulus. This modulus is defined as

K) =)

10


where v = specific volume and p= pressure. As v/dv is a dimensionless ratio, the units

of Ev and p are identical. The bulk modulus is analogous to the modulus of elasticity

for solids; however, for fluids it is defined on a volume basis rather than in terms of

the familiar one-dimensional stress–strain relation for solid bodies. The bulk modulus

is a property of the fluid and for liquids is a function of temperature and pressure. At

any temperature we see that the value of Ev increases continuously with pressure, but

at any one pressure the value of Ev is a maximum at about 120°F (50°C). Thus water

has a minimum compressibility at about 120°F (50°C). By rearranging the definition

of Ev, as an approximation we may use for the case of a fixed mass of liquid at

constant temperature

where Ev is the mean value of the modulus for the pressure range and the

subscripts 1 and 2 refer to the before and after conditions.

VISCOSITY

Is a measure of the resistance of a fluid which is being deformed by either

shear stress or tensile stress. In everyday terms (and for fluids only), viscosity is

"thickness" or "internal friction". Thus, water is "thin", having a lower viscosity,

while honey is "thick", having a higher viscosity. Put simply, the less viscous the fluid

is, the greater its ease of movement (fluidity). Viscosity escribes a fluid's internal

resistance to flow and may be thought of as a measure of fluid friction. For example,

high-viscosity felsic magma will create a tall, steep stratovolcano, because it cannot

11


flow far before it cools, while low-viscosity mafic lava will create a wide, shallow-

sloped shield volcano. All real fluids (except super fluids) have some resistance to

stress and therefore are viscous, but a fluid which has no resistance to shear stress is

known as an ideal fluid or inviscid fluid.

NOTE: ideal fluids don't have (compressibility, viscosity, surface tension)

In general, in any flow, layers move at different velocities and the fluid's

viscosity arises from the shear stress between the layers that ultimately oppose any

applied force. The relationship between the shear stress and the velocity gradient can

be obtained by considering two plates closely spaced at a distance y, and separated by

a homogeneous substance. Assuming that the plates are very large, with a large area

A, such that edge effects may be ignored, and that the lower plate is fixed, let a force

F be applied to the upper plate. If this force causes the substance between the plates to

undergo shear flow with a velocity gradient u (as opposed to just shearing elastically

until the shear stress in the substance balances the applied force), the substance is

called a fluid. The applied force is proportional to the area and velocity gradient in the

12


fluid and inversely proportional to the distance between the plates. Combining

these three relations results in the equation:

where μ is the proportionality factor called viscosity.

This equation can be expressed in terms of shear stress.

Thus as expressed in differential form by Isaac Newton for straight, parallel

and uniform flow, the shear stress between layers is proportional to the velocity

gradient in the direction perpendicular to the layers:

Laminar shear, the non-constant gradient, is a result of the geometry the fluid is

flowing through (e.g. a pipe).

13


Hence, through this method, the relation between the shear stress and the velocity

gradient can be obtained. Note that the rate of shear deformation is which

can be also written as a shear velocity,

Viscosity coefficients can be defined in two ways:

• Dynamic viscosity, also absolute viscosity, the more usual one (typical units

Pa·s, Poise, P);

• Kinematic viscosity is the dynamic viscosity divided by the density (typical

units m2/s, Stokes, St).

Intensive and Extensive Properties

Thermodynamic properties can be divided into two general classes, intensive

and extensive properties. An intensive property is independent of the amount of mass.

The value of an extensive property varies directly with the mass. Thus, if a quantity of

matter in a given state is divided into two equal parts, each part will have the same

value of intensive property as the original and half the value of the extensive property.

Soap, sludge

14


State Functions and State Properties

The state of a system can be changed, for example by increasing its

temperature or changing its composition. Properties of the system, whose change

depends only on the initial (before) and final states of the system, but not on the

manner used to realize the change from the initial to the final state, are referred to as

state properties or state functions. In other words, the change in a state function or

state property X, between some final (state 2) and initial (state 1) situations, can be

expressed as

In equation 1, Xfinal only depends on the final state of the system, and Xinitial only

on the initial state of the system. Equation 1 does not require any information

whatsoever as to how the system got from the initial to the final state, since X does

not depend on the details of the path followed.

ENERGY FORMS

A fluid may possess several forms of energy. All fluids possess energy due to

their temperature and this is called INTERNAL ENERGY. All possess

GRAVITATIONAL or POTENTIAL ENERGY due to their elevation relative to

some datum level. If the fluid is moving it will possess KINETIC ENERGY due to its

velocity. If it has pressure then it will possess FLOW ENERGY. Often pressure and

temperature are the main two governing factors and we add internal energy to flow

energy in order to produce a single entity called ENTHALPY. Let us look at each in

more detail.

GRAVITATIONAL or POTENTIAL ENERGY

In order to raise a mass m kg a height z meters, a lifting force is required

which must be at least equal to the weight mg.

The work done raising the mass is as always, force x distance moved so

Work = mgz

Since energy has been used to do this work and energy cannot be destroyed, it follows

that the energy must be stored in the mass and we call this gravitational energy or

15


potential energy P.E. There are many examples showing how this energy may be got

back, e.g. a hydro-electric power station. P.E. = mgz

KINETIC ENERGY

When a mass m kg is accelerated from rest to a velocity of v m/s, a force is needed to

accelerate it. This is given by Newton's 2nd Law of Motion F= ma.

After time t seconds the mass travels x meters and reaches a velocity

V= m/s.

The laws relating these quantities are

a = v/t and x = vt/2

The work done is W = Fx = max = mv2/2

Energy has been used to do this work and this must be stored in the mass

and carried along with it. This is KINETIC ENERGY. K.E. = mv2/2W

16


FLOW ENERGY

When fluid is pumped along a pipe, energy is used to do the pumping. This

energy is carried along in the fluid and may be recovered (as for example with an air

tool or a hydraulic motor). Consider a piston pushing fluid into a cylinder.

The fluid pressure is p N/m2. The force needed on the piston is

F= PA

The piston moves a distance x meters. The work done is

W = Fx = PAx

Since Ax =V and is the volume pumped into the cylinder the work done is

W = PV

Since energy has been used doing this work, it must now be stored in the fluid and

carried along with it as FLOW ENERGY.

F.E. = PV

INTERNAL ENERGY

This is covered in more detail later. The molecules of a fluid possess kinetic

energy and potential energy relative to some internal datum. Usually this is regarded

simply as the energy due to the temperature and very often the change in internal

energy in a fluid which undergoes a change in temperature is given by

ΔU = mcΔT

17


The symbol for internal energy is U kJ or u kJ/kg. Note that a change in

temperature is the same in degrees Celsius or Kelvin. The law which states internal

energy is a function of temperature only is known as JOULE'S LAW.

ENTHALPY

When a fluid has pressure and temperature, it must possess both flow and

internal energy. It is often convenient to add them together and the result is

ENTHALPY. The symbol is H kJ or h kJ/kg.

H = F.E. + U

Next you need to study the properties of fluids and the laws relating them.

GAS LAWS

A gas is made of molecules which move around with random motion. In a

perfect gas, the molecules may collide but they have no tendency at all to stick

together or repel each other. In other words, a perfect gas in completely inviscid. In

reality there is a slight force of attraction between gas molecules but this is so small

that gas laws formulated for an ideal gas work quite well for real gas.

Each molecule in the gas has an instantaneous velocity and hence has kinetic

energy. The sum of this energy is the internal energy U. The velocity of the molecules

depends upon the temperature. When the temperature changes, so does the internal

energy. The internal energy is for all intents and purposes zero at - 273o C. This is the

absolute zero of temperature. Remember that to convert from Celsius to absolute, add

on 273. For example 40oC is 40 + 273= 313 Kelvins.

PRESSURE

If a gas is compressed it obtains pressure. This is best explained by

considering a gas inside a vessel as shown. The molecules bombard the inside of the

container. Each produces a momentum force when it bounces. The force per unit area

is the pressure of the gas. Remember that Pressure = Force/area

p = F/A N/m2 or Pascal's: Note that 103 Pa = 1 kPa 106 Pa = 1 MPa 105 Pa = 1 bar

18


Absolute pressure: The actual pressure at a given position. It is measured relative

to absolute vacuum (i.e., absolute zero pressure).

Gage pressure: The difference between the absolute pressure and the local

atmospheric pressure. Most pressure-measuring devices are

calibrated to read zero in the atmosphere, and so they indicate gage

pressure.

Vacuum pressures: Pressures below atmospheric pressure.

Pressure instruments

A- Piezometer

B- Manometers

1. Simple Manometer

2. Micro manometer

3. Differential manometer

4. Inverted differential manometer

C- Mechanical Gauge

1. Bourdon

2. Diaphragm

3. Dead weight

19


Specific gravity for ideal gas = S = 𝜌 𝑎𝑛𝑦 𝑔𝑎𝑠

𝜌 𝑎𝑖𝑟 for ideal gas PV = NRT

PV = Wt/Mwt×RT but 𝝆 =𝑾𝒕

𝑽=

𝒑 𝑴𝑾𝒕

𝑹𝑻

S = [𝒑 𝑴𝑾𝒕

𝑹𝑻]𝐺𝐴𝑆/[

𝒑 𝑴𝑾𝒕

𝑹𝑻]𝐴𝐼𝑅 if same temp &pressure

S = [𝑴𝑾𝒕]𝒂𝒏𝒚 𝒈𝒂𝒔

[𝑴𝑾𝒕]𝒂𝒊𝒓 =

[𝑴𝑾𝒕]𝒂𝒏𝒚 𝒈𝒂𝒔

𝟐𝟖.𝟗𝟔

If non ideal gases 𝝆 = 𝒑 𝑴𝑾𝒕

𝒛𝑹𝑻 where z = compressibility factor

Z is function of Pr, Tr, Vr where Pr=𝑷

𝑷𝑪 , Tr=

𝑻

𝑻𝑪 , Vr=

𝑽

𝑽𝑪

PC: critical pressure, TC: critical temperature, VC: critical volume

According “API” Gravity

Specific gravity and API (American Petroleum Institute) gravity are

expressions of the density or weight of a unit volume of material.

The specific gravity is the ratio of the weight of a unit volume of oil to the weight of

the same volume of water at a standard; both specific gravity and API gravity refer to

these constants at 60 ⁰F (16 ⁰ C).

• Light Crude Oil >31

Mixed Based 22-31

Heavy crude<22 API is a major factor for Crude pricing

For liquids lighter than water

𝒔𝒑𝒈 = ( 𝟏𝟒𝟎)/(𝟏𝟑𝟎 − 𝑩𝒆𝟎) 𝑩𝒆𝟎 Baume dgree

For liquids heavier than water

𝒔𝒑𝒈 = 𝟏𝟒𝟓

𝟏𝟒𝟓 − 𝑩𝒆𝟎

NOTE: IF TC < Troom GAS

IF TC > 𝑇𝑟𝑜𝑜𝑚 Vapor

20


Many P-V-T equations of state are available for example:

1- Van der Waals equation was the first equation capable of representing vapor-

liquid coexistence. The pressure (p) is related to the temperature (T), ideal gas

constant (R) and molar volume (V) via:

where

2-

EXAMPLE

Density of a Nonideal Gas from Its Equation of State the Redlich-Kwong equation

of carbon dioxide is

(P + 63.72(106)/ √𝑇V2))(V - 29.664) = 82.05T

with P in atm, V in mL/gmol and T in K. The density will be found at P = 20 and

T = 400. Rearrange the equation to

V = 29.664 + (82.05) (400)/ (20 + 63.72(106)/√𝟒𝟎𝟎V

2).

Substitute the ideal gas volume on the right, V = 1641; then find V on the left;

substitute that value on the right, and continue. The successive values of V are

V = 1641, 1579, 1572.1, 1571.3, 1571.2, . . . mL/gmol and converge at 1571.2.

Therefore, the density is p = l/V = 111571.2, or 0.6365gmol/L or 28.00g/L.

21


Temperature: - It’s the measure of the average (thermal) Kinetic Energy of the

molecules of the substance as a result of the random motion of the

molecules at thermal equilibrium.

Absolute scale: - It is selected as the lowest possible temperature that could be

reached depending on the ideal gas law.

Relative scale: - Is based on a reference points.

22


23


Systems and boundaries

Any engineering device can be denoted as a

system, separated from the rest of the universe by a

system boundary.

Interactions between the system and the rest of the

universe can be exchange of mass and/or energy.

The part of the universe that interacts with the

system is called the surroundings, or environment.

Isolated systems: no exchange of mass or energy with the surroundings

for example: thermos bottle (Dewar flask).

Closed systems: exchange of energy with the surroundings, but no exchange of

mass for example: light bulb, space station.

Open systems: exchange of energy and mass with the surroundings for example:

combustion engines, pumps, distillation columns, living organisms.

← An isolated system

is a special kind of

closed system

Q = heat W = work

24


Work

Definitions

From Newtonian mechanics, we know going from state 1 to state 2, which the work

1W2 is done by a force moving through a distance. The word “work” was first used in

this sense by the French mechanician Gaspard-Gustave Coriolis

Note that we have anticipated that the work differential is inexact. This is an

important point, as work integrals will be path-dependent, and work will not be a state

variable for a system. Here, F is a three-dimensional force vector, x is a three-

dimensional distance vector, and · is the dot product operator. Recall that the dot

product of two vectors yields a scalar.

The terms F and x are scalar equivalents valid for one-dimensional systems. The units

of force are N, those of distance are m, so the units of work are N m, which have been

defined as Joules (J).

Work is done by a system if the sole effect on the surroundings (i.e. everything

external to the system) could be the raising of a weight. We take the following sign

convention:

• + work done by the system,

• − work done on the system.

This sign convention is not universal. Many physicists use precisely the opposite

convention. Probably the reason for this convention is that thermodynamics is a

science that was invented by engineers in the nineteenth century. And those engineers

wanted to produce work from steam engines. Systems doing work were viewed

favorably and endowed with a positive sign. We associate energy with the ability to

do work. We define

• Power: the time rate of doing work = δW/dt.

25


• Specific work: the work per unit mass w = W/m. Because work is path-dependent,

the intensive quantity w is not a thermodynamic state variable.

Work for a simple compressible substance

Consider the scenario sketched in Fig. below In state 1, we have a material at pressure

P confined in a cylinder of cross-sectional area A. The height of the piston in the

cylinder is x. The pressure force of the material on the piston is just balanced by

weights on top of the piston.

The figure above represent sketch of piston-cylinder arrangement as work is done as

the material expands when weights are removed.

Now, remove one of the weights. We notice a motion of the piston to a new

height x+dx.We let a long time elapse so the system comes to rest at its new

equilibrium. We notice there is a new pressure in the chamber, P + dP sufficient to

balance the new weight force. Obviously work was done as a force acted through a

distance. Let us calculate how much work was done. The differential work is given

from this Eq

δW = Fdx.

Now, F varies during the process. At state 1, we have F = PA. At state 2, we

have F = (P + dP)A. Let us approximate F by its average value:

26


Let us only retain terms which are differential and neglect the square of differential

terms, so δW = PAdx.

Now, since Adx = dV, the differential volume, we get the important formula:

δW = PdV.

We can integrate this and get the equally important

Note we employ the unusual notation 1W2 to emphasize that the work

depends on the path from state 1 to state 2. We are tempted to write the incorrect form

R 2 1 δW = W2 −W1, but this would imply the work is a state function, which it is

not, as shown directly.

EXAMPLE

An ideal gas undergoes a two-step process. Beginning at state 1, it is

isothermally compressed to state 2. Then it is isobaric ally compressed to state 3. Find

the work. The process is sketched in Fig. below

The figure represents Sketch of two-step, isothermal-isobaric, compression of an ideal

gas.

27


Note that 1W3 < 0, since V2 < V1 and V3 < V2. So work is done on the system in

compressing it. Note also that even though we do not know T, we can solve the

problem. This is because work only requires information about the P-V state of the

system and the nature of the process.

Heat

Let us make the following definition:

• Heat: a form of energy transferred across the boundary of a system at a given

temperature to another system (or the surroundings) at a different temperature by

virtue of the temperature difference between the two.

We adopt the notion that bodies do not contain heat, but that heat only has relevance

as a type of energy that crosses system boundaries. Note that work is in a similar

class; it is not contained within a system, but can be identified when it crosses system

boundaries. We will make a distinction between heat and work energy transfers.

We also note that when two bodies are at the same temperature, there can be no heat

transferred between the two bodies. The subject of heat transfer considers the details

of the heat transfer process. There are three fundamental classes of heat transfer:

• heat diffusion, also called conduction. Physically this is due to local effects. Bacon

is fried via conduction effects as a culinary example.

This is characterized by Fourier’s law

q = −k∇T,

28


where q is the heat flux vector with units J/s/m2 = W/m2, k is the thermal

conductivity with units J/s/m/K = W/m/K, and ∇T is the vector representing the

gradient of temperature. Recall that ∇T is a vector pointing in the direction in which T

rises most rapidly. Because of the minus sign, we see then that the thermal energy

flows in the direction of most rapid temperature decrease. This law was developed by

Joseph Fourier, who built an elegant and correct theory of a special case of non-

equilibrium thermodynamics before the laws of equilibrium thermodynamics were

formulated, let alone fully understood.

In one dimension, we get

If we multiply by the local cross-sectional area, we find ˙Q = qA, and

Here, ˙Q has units J/s or W (Watts).

• convection. This is actually a version of conduction, albeit enhanced by fluid flow.

For some systems, convective effects are well modeled by Newton’s law of cooling :

Here, σ is the Stefan-Boltzmann constant, σ = 5.67 × 10−8 W/m2/K4.

We adopt the traditional engineering sign convention for heat:

• + heat enters the system,

• - heat leaves the system.

The sign convention again is motivated by nineteenth century steam engines. Heat

had to be added into a system to get work out of the system. Since engineers were and

are concerned with this problem, this convention is taken.

29


Energy Balance on Closed System

For a closed system, the energy balance relates two states of the

system, an initial state and a final state. The changes in energy between

initial and final states of the system are brought about by additions of

energy through heat (Qin) and additions of energy through work done on

the system (Won).

in

30


Energy Balance on open System

A mole balance can be written as the following

for a nonreacting system at steady-state,

Flow work is the work the inlet fluid

must do on the system to displace fluid within

31


32


Summary of Calculating First Law Quantities at Steady-State when Shaft-

Work, Kinetic Energy, and Potential Energy are ignored for an Ideal Gas

33


34


Mechanical Energy Balance and the Bernoulli Equation

The general energy balance derived previously can be recast into a

“mechanical energy balance.” The mechanical energy balance is most

useful for processes in which changes in the potential and kinetic energies

are of primary interest, rather than changes in internal energy or heat

associated with the process. Thus the mechanical energy balance is

mainly used for purely-mechanical flow problems -i.e. problems in which

heat transfer, chemical reactions, or phase changes are not present. We

start from equation 1, in which we have reverted from specific enthalpy

notation back to internal energy and PV work using H j ˆ = U j ˆ + Pj V j ˆ

..(1)

Now we make several additional assumptions. First, we assume steady

state, so that all terms on the left hand side become zero. Second, we

assume that the system has only a single inlet and a single outlet.

Moreover, steady state implies that the inlet mass flowrate must equal the

outlet mass flowrate, in order to avoid accumulation of material in the

system. These adjustments lead to

..(2)

In the above equation, subscript "in" refers to the inlet port, and

subscript "out" to the outlet port. Next, we divide the entire equation by

m&, and write specific volume (volume/mass) as Vˆ = 1/, where is the

density (mass/volume) of the flowing material. Moreover, we assume that

the flow is incompressible, so that density is constant. A constant density

35


means that Vˆ in = Vˆ out = 1/. Also, we define U ˆ = U ˆout - U ˆ in, P =

Pout - Pin, etc. With these changes, equation 2 becomes

…….(3)

The term (DUˆ −Q/m ) – in the absence of chemical reactions, phase

changes, or other sources of large amounts of heat transfer; i.e. under

typical conditions of usage for the mechanical balance equation – will

mostly represent heat generated due to the viscous friction in the fluid. In

such situations, this term is called the friction loss and we will write it as

F ˆ. With this last change, equation 3 assumes the usual form of the

mechanical energy balance

……(4)

Note that WS/m is the shaft work performed by the system on the

surroundings, per unit mass of material passing through the system.

In many instances, the amount of energy lost to viscous dissipation

in the fluid is small compared to magnitudes of the other terms in

equation 4. In such a case, F ˆ = 0. Moreover, many common flows (e.g.

fluid flow through a pipe; what is the system in this case?) do not have

any appreciable shaft work associated with them, so that _WS/m=0 for

such inviscid (i.e. frictionless) flows with no shaft work, the mechanical

energy balance simplifies to Bernoulli’s equation:

The formal derivation holds that in steady flow the sum of the kinetic,

potential, and flow energies of a fluid particle is constant along a

streamline, when compressibility and frictional effects can be neglected.

Also: total pressure along a streamline is constant.

36


Bernoulli’s equation has a wide range of application, despite its

simplifying assumptions.

Fluid Flow in Pipes

We will be looking here at the flow of real fluid in pipes – real

meaning a fluid that possesses viscosity hence loses energy due to friction

as fluid particles interact with one another and the pipe wall.

The shear stress induced in a fluid flowing near a boundary is

given by Newton's law of viscosity:

𝜏∝ du /dy

This tells us that the shear stress, τ, in a fluid is proportional to the

velocity gradient - the rate of change of velocity across the fluid path. For

a “Newtonian” fluid we can write:

𝜏 = μ du/dy

Where the constant of proportionality, μ, is known as the coefficient of

viscosity (or simply viscosity).

Recall also that flow can be classified into one of two types, laminar or

turbulent flow (with a small transitional region between these two). The

non-dimensional number, the Reynolds number, Re, is used to determine

which type of flow occurs:

Re = ρ u d /μ

For a pipe

Laminar flow: Re < 2000

Transitional flow: 2000 < Re < 4000

Turbulent flow: Re > 4000

37


It is important to determine the flow type as this governs how the amount

of energy lost to friction relates to the velocity of the flow. And hence

how much energy must be used to move the fluid.

Q: - Derive Darcy or Weisbach Equation

Q: - Derive General Equation for head loss in pipe due to friction

Q: - Derive an Equation for pressure drop along circular pipe

Consider a cylindrical element of incompressible fluid flowing in the

Figure: Element of fluid in a pipe

The pressure at the upstream end, 1, is p, and at the downstream

end, 2, the pressure has fallen by Δp to (p-Δp).

The driving force due to pressure (F = Pressure x Area) can then be

written driving:

Force = Pressure force at 1 - pressure force at 2

𝑝𝐴 − (𝑝 − 𝛥𝑝)𝐴 = 𝛥𝑝𝐴 = 𝛥𝑝 ×𝜋𝑑2

4

The retarding force is that due to the shear stress by the walls

= shear stress × area over which it acts

= τw × area of pipe wall

= τw π dL

As the flow is in equilibrium,

𝝉𝒘 𝝉𝒘

dL

P1 FLUID

U average velocity

P2 Direction of

flow

38


driving force = retarding force

𝛥𝑝 ×𝝅𝒅𝟐

𝟒 = τw π dL

𝛥𝑝 =𝟒𝜏𝑤 𝑳

𝒅………………… (1)

This equation (1) giving an expression for pressure loss in a pipe in terms

of the pipe diameter and the shear stress at the wall on the pipe.

𝛥𝑝 =𝟒𝜏𝑤 𝑳

𝒅 Multiply and divided

𝝆𝑼𝟐/𝟐

𝝆𝑼𝟐/𝟐

𝛥𝑝 =𝟖𝝉𝒘

𝝆𝑼𝟐 ×𝑳

𝑫×

𝝆𝑼𝟐

𝟐

𝛥𝑝 = 𝟖𝒋𝑭 ×𝑳

𝑫×

𝝆𝑼𝟐

𝟐

𝒋𝑭 𝑴𝒐𝒅𝒂𝒚 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓 =𝟏

𝟐𝒇(𝒇𝒂𝒏𝒏𝒊𝒏𝒈 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏,𝐃𝐚𝐫𝐜𝐲−𝐖𝐞𝐢𝐬𝐛𝐚𝐜𝐡 =

𝝉𝒘

𝝆𝑼𝟐

𝛥𝑝 = 𝟒𝒇 ×𝑳

𝑫×

𝝆𝑼𝟐

𝟐

but 𝛥𝑝 = 𝝆 × 𝑔 × Δℎ

𝛥ℎ = 𝟒𝒇 ×𝑳

𝑫×

𝑼𝟐

𝟐𝒈

𝛥𝑝 = 𝟒𝒇 ×𝑳

𝑫×

𝝆𝑼𝟐

𝟐𝒈𝒄

SI readers may ignore gc in all equations,

if they wish.

39


When a fluid flows in a pipe, some of its mechanical energy is

dissipated by friction. The ratio of this frictional loss to the

kinetic energy of the flowing fluid is defined as the Fanning

friction factor, ft. Thus

For flow through noncircular

pipes, the Reynolds number is based on the hydraulic

diameter

40


Table: Flow Quantities, Reynolds Number, and Friction

Factor

41


Fig.1 This figure is useful for finding the pumping requirement or

frictional loss when you are given the flow rate of fluid in a pipe

(Adapted from Moody (1944)) or given d and u, find Ws or Δp

42


Fig.2 This figure is useful for finding the flow rate when you are

given the driving force for flow (gravitational head, pumping energy

input, etc.) (Adapted from H. Rouse; see discussion after Moody

(1944)) or given d and Δp or Ws, find u

The pipe roughness, needed in these charts, is given in Table below

for various common pipe materials.

Table : Roughness of clean pipe

43


Fig. Development of the mechanical energy balance for flow in pipes

For laminar flow (Re<2,100), the friction factor and the frictional

loss can be found either from Fig. 1 or 2 or from the following simple

theoretical expressions derived by Poiseuille:

Two different friction factors are in common use today:

(i) fF, the Fanning friction factor

(ii) fD, the Darcy friction factor

44


For the range of flows from Re=4,000 to 10

8, these expressions were

cleverly combined by Colebrook (1939) to give

The above expressions reduce to a number of special cases. Thus for fully

developed turbulence in rough pipes, where fF is independent of Re,

equations above becomes

Piping systems have contractions, expansions, valves, elbows, and all

sorts of fittings. Each has its own particular frictional loss. A convenient

way to account for this is to put this loss in terms of an equivalent length

45


of straight pipe. Thus, the equivalent length of a piping system as a whole

is given by

In turbulent flow the equivalent lengths of pipe fittings are independent of

the Reynolds number, and Table below shows these values for various

fittings. Unfortunately, in laminar flow the equivalent length varies

strongly with the Reynolds number is distinctive for each fitting. Thus,

simple generalizations such as Table below cannot be prepared for the

laminar flow regime.

FITTINGS AND VALVES

Friction due to fittings, valves and other disturbances of flow in pipe

line is accounted for by the concepts of either their equivalent lengths of

pipe or multiples of the velocity head. Accordingly, the pressure drop

equation assumes either of the forms

Values of equivalent lengths Li and coefficients K, are given in Table

below. Another well-documented table of Ki is in the Chemical

Engineering Handbook (McGraw-Hill, New York, 1984 p. 5.38).

Comparing the two kinds of parameters,

O R I F I C E S

In pipe lines, orifices are used primarily for measuring flow rates but

sometimes as mixing devices. The volumetric flow rate through a thin

plate orifice is

46


TABLE: Equivalent pipe length for various pipe fittings (turbulent flow only)

Pressure Drop in Non isothermal Liquid Flow

EXAMPLE: Oil is pumped at the rate of 6000 lb/hr through a reactor

made of commercial steel pipe 1.278in. ID and 2000ft long. The inlet

condition is 400°F and 750psia. The temperature of the outlet is 930°F

and the pressure is to be found. The temperature varies with the distance,

L ft, along the reactor according to the equation

T = 1500 - 1100 exp (-0.0003287L) (oF) ……… (1)

The viscosity and density vary with temperature according to the equations

…… (2)

…… (3)

47


48


49


50


51


52


53


54


55


56


57


58


59


60


TABLE 2

61


62


63


64


65


EXAMPLE: Find the mass rate of water at 60 0F can be delivered

through 1320 ft length of smooth 6-in pipe under pressure

deference of 0.25 psi. Assume that the pipe is smooth (Hydraulically smooth).

Given that 𝜌 = 62.4 lb/ft3

,

µ = 6.9× 10-4

lb/ft.sec,

0.25 psi = 025 lbf/in2= 36 lbf/ft

2×32.17 lbm.ft/lbf.sec

2 = 1.158×10

3

lbm/ft.sec2

66


A more flexible and convenient solution is afforded by determining a

single-diameter line that is equivalent to the system shown. From the

equation of

𝛥𝑝 = 𝟐𝒇 ×𝑳

𝒅×

𝝆𝑼𝟐

𝒈𝒄

But 𝑣 = 𝑄/𝝅𝒅𝟐

𝟒 = 4𝑄/𝜋𝑑2 substitute the value of velocity in above

𝜟𝒑 = 𝟐𝒇 ×𝑳

𝒅×

𝝆𝟏𝟔𝑸𝟐

𝒅𝟒𝝅𝟐𝒈𝒄 = 𝒇 ×

𝑳

𝒅𝟓×

𝟑𝟐𝑸𝟐𝝆

𝝅𝟐𝒈𝒄

Another form

𝜟𝒑𝒇 = 𝑪× 𝑳

𝒅𝟓 Where C = 𝟑𝟐𝑸𝟐𝝆

𝝅𝟐𝒈𝒄

𝜟𝒑𝑭𝑨− = C ×

𝑳−

𝒅𝑨𝟓 = 𝜟𝒑𝒇𝑩= C ×

𝑳

𝒅𝑩𝟓

67


Where 𝑳−= length of line of diameter dA that will give the same friction

loss as LB feet of line dB.

The term C could be taken constant if diameter difference is not large

(Since same fluid and f function of d).

Diameter exponent of 4.828 should be used instead of 5 highest precision

Now if constants are canceled from equation and rearranged it

𝑳− = 𝐿𝑒 = 𝐿𝐵 {𝑑𝐴

𝑑𝐵}

5

If 𝑑𝑒 is defined as the diameter of line of length 𝐿𝐴 that will give the

same friction loss as 𝐿𝐵 feet of line𝑑𝐵. The above equation may be

rearranged to yield

𝒅𝒆 = 𝒅𝑩 {𝑳𝑨

𝑳𝑩}

𝟏/𝟓

When 𝐿𝑒 is calculated the system become: equivalent total length

𝐿𝐵+ 𝐿𝐴 with diameter 𝑑𝐴 then the 𝜟𝒑𝒇 of the system could be calculated.

68


EXAMPLE:

Therefore, the system shown may be considered as equivalent to one

consisting of 1423 m of 10 cm pipe.

69


ﺔﺜﻟﺎﺜﻟﺍ ﺔﻠﺣﺮﻤﻟﺍ · poundal = ibm ×32.17 ft/sec2 f= lbw m=slug lbw =...

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