FLIP THERMOFLUID

FUNDAMENTAL OF FLUID MECHANICS

Note that the contribution from C to D, ρLgh2, cancelled out with the contribution from E
to D. In fact, the pressure at C is the same as that at E, since they are both the same
vertical distance above D and are connected by a continuous column of the same fluid.

Example 6.2 :

Question :

A manometer containing liquid mercury is used to measure the pressure of a mass of
nitrogen gas. The density of the gas is known to be 1.6 kg/m3. If the heights h1 and h3 in
Figure 6-1 are 1.5 cm and 4.32 cm, respectively, what is the pressure of the gas?
Assume the manometer is in an environment where the temperature is 20°C and the
atmospheric pressure is 101 kPa.

Solution:
Rearranging Eq. 6-2, the pressure of the nitrogen is

PA = Patm + ρLgh3 - ρggh1
Using the density of mercury at 20°C from Table A-6, the pressure is

PA = 101kPa (1000 Pa / 1 kPa ) + (13,579 kg/m3) (9.81 m/s2) (4.3cm) ( 1m / 100 cm ) -
( 1.6 kg/m3 ) ( 9.81 m/s2 ) (1.5 cm) ( 1 m/ 100cm)

PA = 101, 000 Pa + 5,755 Pa - 0.24Pa
PA = 1.07 X 105 Pa = 107 kPa

In Example 6-2 the ρgh term due to the gas was much smaller than the ρgh term due to
the liquid, because the density of the gas is so much smaller than that of the liquid.In
fact, this will nearly always be the case, since gases, in general, are so much less dense
than liquids. As a result, Eq. 6-4 is usually approximated as

. PA - Patm ≈ ρLgh3 (Equation 6.3)

When gas pressure is measured with a U-tube manometer, this approximation will be
used unless otherwise noted. If a manometer is used to measure the pressure of a
liquid, then Eq. 6-4 applies.

THINK : Do you know what is the difference between gage pressure and
absolute pressure ?

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The left-hand side of Eq. 6-3 is the difference between measured pressure and
atmospheric pressure. This pressure is called gage pressure.

Pgage = PA - Patm = Pabsolute - Patm

In British units, the gage pressure is indicated by "psig" to distinguish it from the
absolute pressure, which is called "psia."

Many types of pressure gauges, including the manometer described above, actually
measure only the "gage pressure." In the manometer, the height of fluid is the
measured quantity. The atmospheric pressure must be known from some other
measuring device before the absolute pressure can be found. If "psi" is used, it
should be clear from the context which is meant. In the SI system, pressure is
typically measured in kPa. Both absolute pressures and gage pressures are
expressed in kPa, and the engineer must recognize whether absolute or gage is
meant.

EXERCISE 6.1

A manometer is attached to a rigid tank containing gas at pressure P. The
manometer fluid is mercury at 20°C. Using data on the figure, find pressure in
the tank.

6.1.2 Specific Gravity

In fluid statics, the density of a liquid is calculated using the specific gravity. By
definition, specific gravity is the ratio of the density of a liquid to the density of water at
4°C:

SG = ρliquid /ρwater

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where SG is the specific gravity. The density of water varies slightly with temperature,
reaching a maximum value of 1000 kg/m3 at 4°C, so this is chosen as the reference
value. Specific gravity is a dimensionless quantity. For example, the specific gravity of
kerosene is 0.817; therefore, in SI units, the density of kerosene is 817 kg/m3.

Specific gravity applies to solids as well as to liquids. The specific gravity of a solid is
the ratio of its density to that of water at 4°C. If the specific gravity of a solid is less than
1, then the solid will float in water; if specific gravity is greater than 1, the solid will sink.
If the specific gravity of a liquid is less than 1 and the liquid is immiscible with water,
then the liquid will form a layer on top of the water. If the specific gravity of an
immiscible liquid is greater than 1, then the liquid will form droplets and sink to the
bottom of the water, ending up in a layer under the water.

6.1.3 Mechanical application of fluid statics

The principles of fluid statics can be used to gain mechanical advantage. A hydraulic
jack, as shown in Figure 6- 2, is used to lift heavy objects, such as cars, parts of
buildings, packing crates, and so on. In each leg of the lift shown, the hydraulic fluid
comes to the same level, h ; therefore, the fluid pressure is the same at points I and 2.
Since pressure is force per unit area,

F1 / A1 = F2 / A2

Solving for F1 ,

F1 = ( A1 / A2 ) F2

If A1 is much smaller than A2, then a small force F1 can counterbalance a large force F2.
Typically, the large force is due to the weight of the object being lifted.

KKTM Figure 6.2 A hydraulic jack

96

FUNDAMENTAL OF FLUID MECHANICS

Mechanical advantage never comes without a price. If force F1 in the small-diameter leg
pushes the fluid downward a small amount, ∆h1, the level in the large tube will rise,
lifting the load. However, the volume of the hydraulic fluid stays the same, so the rise in
the large tube will be

V1 = A1 ∆h1= V2 = A2∆h2
∆h2 = ( A1/ A2 ) ∆h1

Since A1 is much smaller than A2, the load is lifted a small distance compared to the
decrease in fluid level in the small tube.

In some applications, the transmission of force is all that is required. For example, in
cars and trucks, the force applied by the driver to the brake pedal is magnified by the
hydraulic brake system, so that a person of normal strength can stop a moving vehicle.

6.1.4 Forces on Submerged Plane Surfaces

If a surface is immersed in a fluid, a force is exerted on the surface due to the
pressure of the surrounding fluid. It is more difficult to remove the plug from a sink
filled with water than it is to remove a plug from an empty sink. Pressure forces from
the water wedge the plug into place and aid in forming a seal between the plug and
the drain.

Figure 6.3 An arbitrarily shaped flat plate immersed in a static fluid

In Figure 6.3, an arbitrarily shaped flat plate is submerged in a liquid. An infinite plane
that extends in the x- and y-directions and contains the plate makes an angle θ with
the surface of the liquid, as shown. The x-direction is perpendicular to the y- and z-
directions; it is not shown in the figure. The y-coordinate is the distance along the
imaginary plane from the liquid surface.

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The pressure in the liquid is a function of the depth, z, which is the coordinate
perpendicular to the liquid surface. The pressure is
P(z) = Patm + ρgz
The pressure may be expressed as a function of y as
P(y) = Patm + ρgy sinθ

Remember that pressure is always normal to the submerged plate. The force on a
differential area of the plate is the pressure at that location multiplied by the differential
area. To find the total force on the plate, integrate over the plate area to obtain
FR = ∫ PdA = ∫ [ Patm + ρgy sinθ ] dA

where F R represents the resultant force on the plate due to liquid pressure. Integrating
the first term and removing constants from the integral in the second term gives
FR = PatmA + ρgy sinθ ∫y dA
Therefore, the magnitude of the resultant force on one side of a submerged plane
surface is

FR = PatmA + ρgsinθ yc A

where yc is the obliqued depth of the centroid of the surface.
In addition to the magnitude, it is often necessary to know the location at which the
resultant force is applied.

Figure 6.4 : Point of application of the resultant force, FR

In Figure 6.4, point P is the location at which FR acts. Point P always lies below point C,
the centroid of the surface. The reason is simple. If a horizontal line is drawn through
point C, as shown in Figure 6.4, then this line bisects the surface into two equal areas.
The area below the line is at a greater depth than that above the line, so there is more
total force on the area below the line; thus the resultant force must lie in the lower half of
the area

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The resultant force is applied at

Where Ixx,C is the area moment of inertia of the surface. The location of the centroid and
the area moments of inertia for some common shapes are given in Table 6.1

Table 6.1 Centroids and area moment of inertia for common surfaces

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EXERCISE 6.2
Consider a heavy car submerged in water in a lake with a flat bottom (refer to Figure
4). The driver's side door of the car is 1.1 m high and 0.9 m wide, and the top edge
of the door is 8 m below the water surface. Determine the net force acting on the
door (normal to its surface) and the location of the pressure center if :
a) the car is well-sealed and it contains air at atmospheric pressure, and
b) the car is filled with water.

8m

Door, 1.1 m  0.9 m

6.1.5 Buoyancy

The principle of buoyancy can be simply stated as:

The buoyant force on an immersed object is equal to the weight of the fluid displaced by
the object. This force acts upward through the center of gravity of the object.

This principle was known to the ancient Greeks and is attributed to Archimedes (287-212
BC). In equation form, it is

FB = ρf Vg

where FB is the buoyant force, ρf is the density of the fluid in which the body is
immersed, V is the immersed volume and g is the gravitational force.

An object in water will experience a buoyant force that may be greater than, less than,
or equal to the weight of the object. If the object is less dense than water, its weight is
not sufficient to overcome the buoyant force, and it will rise. Conversely, if the object is
more dense than water, it will sink. Wood rises in water and comes to rest floating on
the surface, while iron sinks and comes to rest on the bottom. If the object has exactly

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FUNDAMENTAL OF FLUID MECHANICS

the density of water, it will float, suspended, totally immersed, at least in theory. In
practice, this is a position of unstable equilibrium, and the object would tend to drift
either upward, downward, or laterally due to small currents or to slight differences
between its density and that of water.

In many applications, we are interested in a body floating at the interface between two
fluids. An example is a ship floating on the ocean, where the upper fluid is air and the
lower fluid is water (refer Figure 6.4).

Figure 6.4: A body floating in water

For this situation, the buoyant force is

FB = ( ρa Va + ρw Vw)g
where ρa is the density of air and ρw is the density of water. Since the density of air is
so much smaller than that of water, we may write

FB = ρw Vw g

The buoyant force is equal to the weight of the object, so

ρo Vtot g = ρw Vw g
where ρo is the density of the object, Vw is the volume submerged in the water, and Vtot
is the total volume of the object. An alternate form is

ρo / ρw = Vw / Vtot
This equation can be used, for example, to calculate the submerged volume of an
iceberg or a raft.

Example 6.3:

Question:

A cylindrical float has a 254 mm diameter and is 305 mm long. What should be the
density of the float material if it is to have 9/10 of its volume below the surface of fluid
water.

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Solution:

Cylinder : diameter (D) = 0.245m, lenghth (L) = 0.305 m

Thus, the cylinder volume
V cylinder = (πD2/4 ) x L

= (π (0.254)2 /4 )x (0.305)
= 0.0155 m3

9/10 of the cylinder volume submerge

so, V cylinder submerge = 0.9 ( 0.0155) = 0.014

The buoyant force is equal to the weight of the object
ρo Vtot g = ρw Vw g
ρo (0.0155)(9.81) = (1000) (9.81)(0.014)
ρo = 903.2 kg/m3

EXERCISE 6.3

If the “tip of the iceberg” ( the volume of the iceberg above the water surface)
is 79 m3, what is the volume of the submerged iceberg? For seawater
density use ρseawater = 1.027 g/cm3.

6.2 FLUIDS DYNAMICS

Fluids Dynamics involves moving fluids. The flow of air over a truck, the flow of oil in a
pipeline, and the flow of water issuing from a fire hose are just a few of the many
examples of fluid flow that can be analyzed using fluid dynamics. In this section,
Bernoulli’s equation and equations for conservation of mass of a moving fluid will be
introduced which can be applied to a very wide range of processes and phenomena

6.2.1 BERNOULLI EQUATION

Bernoulli equation, first presented by Daniel Bernoulli (1700-1782). It is one of the most
famous and useful equations in fluid mechanics. It applies for a steady, incompressible,
inviscid, and isothermal flow with no work.

P1/ρ + gz1 + V1 2 / 2 = P2/ρ + gz2 + V2 2 / 2

Bernoulli Equation

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By definition, an incompressible flow is a flow with constant density. Most common
liquid flows are virtually incompressible. Liquids strongly resist changes in volume under
mild pressure.
The next assumption is the inviscid (zero viscosity) assumption. There are, in general,
three forces that act on flowing fluids-gravity, pressure, and friction. In an inviscid flow,
frictional effects are zero. This is often valid if the viscosity of the fluid is very low and
the flow channel is short. Other sources of frictional losses in pipe flow include sudden
contractions, serpentine passages through valves, and flow through porous media. If
the flow has a smooth route with rounded comers and no major flow restrictions, it can
often be approximated as inviscid.
The final assumption is the isothermal assumption. This simply means that the fluid
does not change temperature.
The Bernoulli equation has been used for finite control volumes with one inlet and one
outlet. It can also be applied to infinitesimal control volumes and to control volumes with
more than one inlet and/or outlet. In Figure 6.5 a flow through a nozzle is shown

Figure 6.5 : Streamlines in a nozzle

The figure includes so-called streamlines. A streamline indicates the path that a fluid
particle takes as it accelerates through the nozzle. Because a streamline is always
tangent to the velocity vector, no mass flow crosses a streamline.
Figure 6.5 shows a control volume aligned with the streamline and extending from
point 1 to point 2. This control volume is finite in length but infinitesimally thin in the
dimension perpendicular to the streamline. Such a control volume is sometimes called
a streamtube.

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Fluid enters the left side of the streamtube at point 1 and leaves through the
right side at point 2. No fluid flows out the lateral sides. The flow within the streamtube
meets the criteria for the Bernoulli equation. That is, it is incompressible, inviscid, and
isothermal. Therefore, Bernoulli's equation applies along a streamline.

Figure 6.6: Water issuing from a tank through two different exits

The Bernoulli equation can sometimes be used if there is more than one inlet or outlet.
For example, in Figure 6.6 water drains from a tank through two different outlets. The
dotted line is a streamline that divides the flow into two regions. The upper region
contains all the fluid that leaves through the upper outlet, and the lower region contains
all the fluid that leaves through the lower outlet. It is possible to calculate the shape
and placement of the dividing streamline, but that is beyond the scope of this text. The
dividing streamline is parallel to the velocity vector. No mass flows across the dividing
streamline. As a result, we may write the BernouIli equation for the upper region as

P1/ρ + gz1 + V1 2 / 2 = P2/ρ + gz2 + V2 2 / 2
For the lower region, the Bernoulli equation

P1/ρ + gz1 + V1 2 / 2 = P3/ρ + gz3 + V32 / 2

Both of these equations apply simultaneously.

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Example 6.4 :
Question:
Water drains at a steady rate from a very large tank through a pipe of diameter 4 cm.
Assume frictionless, incompressible flow. Because the area of the top surface of the
water is large compared to the outlet pipe diameter, we also assume that the velocity of
the receding top surface is negligible. Find the rate at which mass drains from the tank
Solution:

Define station 1 at the top surface of the water in the tank and station 2 at the exit.
Assume the flow is frictionless and incompressible. The Bernoulli equation is

P1/ρ + gz1 + V1 2 / 2 = P2/ρ + gz2 + V2 2 / 2
The velocity at station 1 is assumed to be zero. With this simplification, and setting P1
and P2 equal to atmospheric pressure,

Patm /ρ + gz1 = Patm /ρ + gz2 + V2 2 / 2

which reduces to
g (z1 - z2) = V2 2 / 2

or

The mass flow rate is found from

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Using the density of water at 25°C from Table A-6,

EXERCISE 6.4
Water flows at 25 kg/s through a gradual contraction in a pipe. The upstream
diameter is 8 cm and the downstream diameter is 5.6 cm. If the exit pressure
is 60 kPa, find the entrance pressure. Assume frictionless flow.

6.2.2 CONSERVATION OF MASS FOR AN OPEN SYSTEM
In an open system, we often need to keep track of how much mass is in the control
volume at a given time. To do this, consider the control volume in Figure 6.7.

Figure 6.7: Differential masses entering and exiting a control volume

Figure 6.7a shows the control volume at time t. Here ∆mi is a small mass that will enter
the control volume during the time period ∆t. Likewise, ∆me is a small mass that will exit
the control volume in the time period ∆t (∆mi is not necessarily equal to ∆me). The mass
in the control volume, mcv, is all the mass within the dotted line. Figure 6.7b shows the
control volume at time t + ∆t. The mass at the inlet, ∆mi, has entered the control volume
and the mass at the exit, ∆me has left the control volume. If we equate all the mass
shown in Figure 6.7a with all the mass shown in Figure 6.7b, then

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∆mi + mcv(t)= ∆me + mcv(t+∆ t)
where mcv(t) is the mass in the control volume at time t, and mcv (t + ∆t) is the mass in
the control volume at time t + ∆t. The quantity mcv (t) contains within it the differential
mass ∆me Similarly the quantity mcv(t + ∆t) includes the differential mass ∆mi. The
equation may be rearranged to the form

mcv(t+∆ t ) - mcv(t) = ∆mi - ∆me

Divide both sides by the time period ∆t to get
[ mcv (t + ∆ t ) - mcv(t)] /∆ t = [ ∆mi - ∆me ] /∆ t

Take the limit as ∆t approaches zero so that the equation becomes
dmcv/dt = dmi /dt - dme /dt

The term dmi/dt is called the inlet mass flow rate and is often abbreviated as

Similarly,

the resulting open system mass balance equation in rate form is

Open systems are often operated in steady state; that is, dmcv/dt = O. In this case, total
mass neither increases nor decreases in the control volume, and the mass in the control
volume is constant with time. If there is only one stream flowing in and one stream
flowing out, then, in steady state, their flow rates are equal.

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There are useful alternative expressions for the flow rates on the right-hand side of open
system mass balance equation. Refer to Figure 6.8, which shows a small quantity of
mass ∆m, which will enter the control volume in time period ∆t.

Figure 6.8 : A differential mass entering a control volume

This mass is contained in a differential volume that has a cross-sectional area A and a
height ∆x. The velocity with which the mass enters th control volume is also shown. This
differential mass ∆m is related to a differential volume ∆V by

∆m =ρ∆V ( eq. 6.4.2 a)
where ρ is the density of the fluid entering the control volume. The differential volume is
given by

∆V = ∆xA

The differential mass flows into the control volume with velocity of
v = ∆x / ∆t

where the definition of velocity has been used, and the velocity is assumed uniform over
the area A. Combining the last two equations gives

∆V = v∆tA ( eq. 6.4.2 b)
Substituting Eq. 6.4.2b in Eq. 6.4.2.a produces

∆m = p v ∆t A
or

∆m/∆t = p v A ( eq. 6. 4.2 c)

It is also common to describe flow in terms of a volumetric flow rate. To derive the
volumetric flow rate, start with Eq. 6.4.2.b, which is

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∆V = v∆tA
Dividing by ∆t gives
∆V/ ∆t = vA
In the limit as ∆t approaches zero

The quantity V is called the volumetric flow rate. If velocity varies over the area, A, then
the volumetric flow rate may be written as

Combining this with Eq. 6.4.2c produce

Example 6.4:
Question:
Air enters a nozzle at 90°C, 180 kPa with an average velocity of 60 m/s. The duct has
an inlet diameter of 10 cm. The exit has a diameter of 6.3 cm and is open to the
atmosphere. If the velocity at the exit is 249 m/s, what is the temperature there?

Solution: 109
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FUNDAMENTAL OF FLUID MECHANICS

Define the control volume to follow the inside surface of the nozzle and cut across the
entrance and exit. This is a steady flow with one inlet and one outlet. By conservation of
mass,
Because the flow is steady, the mass in the control volume does not change with time

and
Therefore

This may also be written as

Assuming air is an ideal gas
or, since

Solving for p and substituting gives

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FUNDAMENTAL OF FLUID MECHANICS
which simplifies to

Solving for the exit temperature,

At the exit, the nozzle is open to the atmosphere; therefore, pressure is atmospheric.
Substituting values,

Note that absolute temperatures were necessary in this example because equations
were derived using the ideal gas law.

EXERCISE 6.5

Water flowing in a horizontal pipe branches into two pipes as shown
and issues into the atmosphere. Neglecting all viscous effects, find
the volumetric flow rate in each pipe and the diameter, D .

3

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SUMMARY

1. The pressure force on a surface immersed in a fluid is always normal to the
surface. In a static fluid of constant density (incompressible), the pressure is a
function of depth .

2. If the submerged surface is exposed to the atmosphere on one side and to a fluid
whose upper surface is at atmospheric pressure, the equation for FR and yP
simplify to

FR = PatmA + ρgsinθ yc A

yP = yc + Ixx,C / yc A

3. The principle of buoyancy can be simply stated as:

“The buoyant force on an immersed object is equal to the weight of the fluid
displaced by the object. This force acts upward through the center of gravity of
the object”.

4. In an open system, conservation of mass is given by

5. The Bernoulli equation, which applies to an isothermal, incompressible,
frictionless flow with no work or heat transfer is

P1/ρ + gz1 + V1 2 / 2 = P2/ρ + gz2 + V2 2 / 2

REFERENCES

a. Kaminsky D. A, Jensen M. K., (2005), “Introduction to Thermal and Fluid
Engineering”, John Wiley & Sons, Inc.

b. Eastop, T.D, McConkey, A., (2004), 5th Edition, “Applied Thermodynamics for
Engineering Technologist”, Longman.

c. Yunus, A.C, Michael, A.B., (2002), 4th Edition, “Thermodynamics, an
Engineering Approach”, Mc Graw Hill, New York.

d. D.F. Young, B.R. Munson, T.H. Okiishi, (2004), “Fundamental of Fluid
Mechanics”, 4th Edition, John Wiley & Sons, Inc.

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CHAPTER 7

INTERNAL FLOW

INTERNAL FLOW

CHAPTER 7 INTERNAL FLOW

INTRODUCTION

When we turn on a water faucet or open the valve on a propane-fired grill or fill car with
gasoline, we use a system in which a fluid (liquid or gas) flows inside a closed conduit.
The proper design of that system requires a means to move the fluid from one place to
another (pump or compressor) and a determination of pressure differences, flow rate and
velocities.

The power required by the pump or compressor also must be determine the needed
information is calculated from an analysis of fluid flow through the pipes comprise the
system.

In previous chapters, we dealt with one-dimensional flows. We did not take into account
the effects of fluid friction in pipe flows, and the average velocity in the pipe was used to
describe the flow.

In real flows, we still use the average velocity to describe the flow but recognize that the
velocity is not the same at every distance from the wall. Fluid near the wall moves more
slowly than fluid near the center of the pipe, and at the wall the fluid velocity is zero. The
variation in velocity is a direct result of fluid friction. Fluid friction also causes pressure
losses as fluid flows through a conduit.

In this chapter, we use principles of conservation of mass and momentum to develop
equations for the velocity a function of position and apply that information to the prediction
of pressure changes' internal flows.

Fluid friction can be an important contributor to pressure losses. Import frictional losses
also occur in many common pipeline elements, such as valves, elbow junctions, and so
on. In addition, pressure and gravity forces must be taken into account.

Figure 7.1 shown Baseballs and softballs are examples of sports projectiles. The airflow
around the ball is science that must be mastered by every baseball or softball pitcher. The
airflow around a smooth ball is much different than the airflow around a ball with stitches.
In the flight of a smooth ball the air molecules travel around the ball to the back where they
meet and mingle and combine to push the ball forward.

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Figure 7.1 Baseballs and softballs are examples

The pressure behind the ball is less than the pressure in front. When the ball has stitches
as in a baseball or softball, turbulence occurs where the stitches are. The turbulence
causes the air to stick to the ball just a little longer and reduces the wake (as in a boat's
wake), which reduces drag. These stitches can also change the direction of the ball. A
good pitcher uses the spin and the stitch alignment to throw curve balls.

To deal with frictional effects, the fundamental fluid property called viscosity is introduced.
Most people have some experience with viscous fluids and would describe honey or oil as
more viscous than water. In this chapter, viscosity is defined on mathematical basis and
used in the development of equations for internal flow.

Think: Less obvious is the magnitude of the viscosity of gases. Is air more or less
viscous than water?

LEARNING OBJECTIVES

At the end of this chapter, you should be able to:

1. Describe the concepts involved in a internal flows system
2. Explain the flow regime in a typical pipe, Laminar, Transitional, and Turbulent.
3. Apply the Laminar and the turbulent into a simple pipe.

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7.1 VISCOSITY

When a solid is placed under a shear stress, it deforms. For example, if you hold a
cube of gelatin between your hands and move one hand slightly, the cube will
change shape (see Figure 7-2); you must maintain the force to keep the cube
deformed. If too much force is applied, the cube will eventually tear. Solids typically
behave this way, but most solids are so strong that large forces are needed to
produce noticeable deformation.

Figure 7-2 Deformation of a gelatin cube under the action of a shear stress

Fluids behave differently than solids and, under a shear stress, deform
continuously. One way to visualize this behavior is by analogy to a stack of papers
on a horizontal sum subjected to a force. If one presses downward on the stack,
the papers do not move.

However if one slides the top paper horizontally, then it will move forward and may
drag along papers beneath it. As long as a shearing force is applied, the sheets of
paper will continue to slide over each other. Fluid in low-speed flow can be thought
of as sliding in "laminar,” or sheets that rub against each other.

The formal definition of a fluid is a substance that deforms continuously under the
application of a shear stress. Both liquids and gas behave this way, and both are
classified as fluids.

The tangential force, Ft (also called a shear force), is due to internal friction as
layers of fluid slide over each other. The magnitude of the shear force per unit area
is called the shear stress, T. The shear stress is

F  t
A

The quantity,  is called the viscosity, which is property of the fluid. It is an
indication of how much internal friction is present. The unit viscosities are (N.s)/m2
or (lbf.s)/ft2.
The Kinematic viscosity,  , is defined the dynamic viscosity divided by the density,




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INTERNAL FLOW

In general, liquid viscosity decrease exponentially with increasing temperature, as
shown in Figure 7-3, but gas viscosity increases with temperature.

Figure 7-3 Dynamic viscosity for a variety of liquids and gases as a function of
temperature. (Source: Munson, B. R., D. F. Young, and T. H. Okiishi,

Fundamentals of Fluid Mechanics, 4th ed., Wiley, New York, 2002, p. 829. Used
with permission.)

Example 7.1:

Question:

What is the viscosity?

Solution:

A fluid is a substance that deforms continuously under the application of a shear
stress, or an internal property of a fluid that offers resistance to flow.

Example 7.2:
Question:
What is the shear stress?
Solution:
Shear stress is the magnitude of shear force per unit area.

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Example 7.3:
Question:
Why the fluids behave differently than solid, under a shear stress will deform
continously?
Solution:
Because, this behavior is on a horizontal sum subjected to a force. If one presses
downward on the objects, the few objects do not move. A shearing force is applied
on the object over each other.

EXERCISE 7.1

1. Calculate the shear stress of gelatin cube with shear force of 20 N, and area of
0.05 m2.

2. Explain why an increase in temperature generally results in a decrease in
viscosity.

3. Calculate the kinetic viscosity of engine oil with 900 kg/m3 density and viscosity
dynamic of 38,500x10-4 N.s/m2.

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7.2 FULLY DEVELOPED LAMINAR FLOWN IN PIPES

Flow in various types of round tubes is a topic of great practical importance. Water
is distributed to homes via pipes, blood flows in the body through blood vessels,
gas is piped from remote sites for thousands of miles, and oil is pumped from wells
through circular pipes. These are only a few of the numerous applications of pipe
flow in industrial, commercial, residential, military, and natural systems. In electric
circuits, current, I, is driven by an applied voltage difference,  , and is limited by
the electrical resistance, Relectric:

I 

Relectric

In heat transfer, the heat transfer rate ( Q ) is proportional to the driving temperature
difference, T, and is, inversely proportional to the thermal resistance:

Q  T

Rthermal

For internal fluid flows, a driving pressure difference, P, causes flow (V ) and is
resisted by the hydraulic resistance:

V  T

Rhydraulic

In all three cases, the flow (electric~ current, heat transfer, or fluid flow) is
proportional a driving potential (voltage drop, temperature drop, or pressure drop,
respectively) and divided by a resistance to the flow.

In the entrance region, fluid is forced away from the wall and into the center region
of the pipe. Figure 7-4 shows the velocity profile in the axial direction; however,
there is also a component of velocity in the radial direction in the entrance region.
Eventually, the boundary layer extends to the center of the pipe, and the flow
profile takes on a rounded shape.

At a point farther downstream, the flow profile stops changing, and no further flow
occurs in the radial direction. All flow is in the axial direction. This is called the fully
developed flow region, where the velocity profile is independent of the distance
from the pipe entrance.

KKTM Figure 7-4 Development of the internal flow field in a pipe.
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To analyze flow in a pipe, we begin with a fully developed, steady, incompressible
laminar flow. The term laminar implies that the fluid moves in sheets, or "lamina”,
that slip relative to each other. Under other conditions, flow becomes unstable and
turbulent, and the velocity at a given location fluctuates.

Formula for pressure drop gives,

8L
RVP  2 m  gL sin  ; Laminar, fully developed flow, inclined pipe

8L
RVP  m
2 ; Laminar, fully developed flow, horizontal pipe

Example 7.4:

Question:

What is the fully developed flow region?

Solution:

Fully developed flow region is when at a point farther downstream, the flow profile
stops changing, and no further flow occurs in the radial direction. In addition, all
flow is in the axial direction.

Example 7.5:

Question:

List of pipe flow application in industrial, commercial, residential, military, and
natural systems.

Solution:

a. Water is distributed to homes via pipes
b. Blood flows in the body through blood vessels
c. Gas is piped from remote sites for thousands of miles
d. Oil is pumped from wells through circular pipes

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EXERCISE 7.2

1. Water flows at an average of 1 m/s in a horizontal pipe of diameter 1.5 cm and
length 2 m, assuming a fully developed laminar flow; calculate the pressure drop
from inlet to outlet. Water temperature is 20o C.

2. Water flow at an average of 2.5 m/s in a inclined pipe of diameter 2 cm,  =25o,

and length 4 m, assuming a fully developed laminar flow; calculate the pressure
drop from inlet to outlet. Water temperature is 25o C.

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7.3 LAMINAR AND TURBULENT FLOW

In the previous section, we assumed a steady, laminar flow. Under certain
conditions, flow is no longer laminar but becomes unstable and chaotic. Figure 7-5
shows the trajectories of fluid "particles" in laminar and turbulent flow in laminar
flow, fluid particles follow a smooth trajectory, with no oscillations about the
average velocity.

By contrast, in turbulent flow, the particles follow a rough trajectory with many
random oscillations about the average velocity. You may have felt the effect of
turbulence during an airplane flight. Normally, the flight is very smooth; however, if
the plane hits a pocket of turbulence, the plane bounces around.

Laminar flow is analogous to an army marching off to war in neat, straight row and
columns. Turbulent flow is like an army in retreat-disorganized, somewhat random,
and moving quickly. The faster people move, the more difficult it is to maintain a
marching formation. Fluid particles behave just this way.

Figure 7-5 Trajectory of a particle in different flow regimes:
(a) Laminar, (b) Turbulent, (c) Transition.

Now we can define the Reynolds Number as

Re  DV


Where D is a diameter and V is a velocity appropriate for the flow. Osborn
Reynolds identified this parameter in 1883 as being important in fluid mechanics,
and it is named in his honor. Hence, the Reynolds number is a non-dimensional
parameter, as can be easily verified. A non-dimensional pressure drop per unit
length of pipe, we define the Darcy friction factor, f, as

f  P
L

Formula for fully developed laminar flow

64
f  ReD

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Formula for laminar or turbulent:

VP f L  2
D
 m

2

The Reynolds number is important because flows with the same Reynolds number
behave similarly. At lower Reynolds numbers, flows remain laminar. At high
Reynolds numbers, the flow becomes turbulent. Between these two regimes the
flow is transitional; that is, the flow becomes more and more unstable with
increasing Reynolds number until, finally, the flow is fully turbulent.

It is possible to perform a formal, mathematical, stability analysis of the flow to
show that Reynolds number determines stability, but that analysis is beyond the
scope of this text. We must be satisfied with recognizing that since Re is the only
parameter on which the solution depends; Re must determine the character of the
flow.

Experiments show that the flow regime in a typical pipe is a function of Reynolds
number according to

Re < 2100 Laminar
2100 < Re < 4000 Transitional
4000 < Re Turbulent

These ranges are approximate and depend on various factors such as the
roughness the pipe wall and the nature of the inlet flow.

Example 7.6:

Question:

What is the difference between laminar and turbulent flow?

Solution:

In laminar flow, fluid particles follow a smooth trajectory, with no oscillations about
the average velocity. By contrast, in turbulent flow, the particles follow a rough
trajectory with many random oscillations about the average velocity.

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Example 7.7:
Question:
What is Reynolds number represented?
Solution:
Reynolds number is a number that represent the flow regime. At lower Reynolds
numbers, flows remain laminar. At high Reynolds numbers, the flow becomes
turbulent.

EXERCISE 7.3

1. The biomedical device uses compressed air to drive the plunger in a piston-
cylinder assembly with velocity 0.8 m/s that will push the drug (viscosity and
density similar to water at 10°C) through the hypodermic needle (inside diameter
0.25 mm and length 50 mm).
a. Calculate the Reynolds number.
b. Determine the flow regime.

2. Calculate the Darcy friction factor for fully developed laminar flow of the
biomedical device uses compressed air to drive the plunger in a piston-cylinder
assembly that will push the drug (viscosity and density similar to water at 10°C)
through the hypodermic needle (inside diameter 0.25 mm and length 50 mm).

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7.4 HEAD LOSS

In the last two sections, we investigated pipeline flow using the momentum
equation. In this section, we apply the energy equation to viscous flow in a pipe. In
Figure 7-6, an incompressible, viscous fluid flows through an inclined pipe of length
L.

Figure 7-6 Flow in an inclined pipe.

We define a head loss caused by frictional effect:

h P P z z L 1 2  Inclined pipe
  g   1 2

 

The quantity DP/ g is called the pressure head. In the absence of friction (i.e.,

z zwhen hL=0), the pressure head would raise the fluid a distance 1 .

2

When friction is present, some of the pressure head is consumed in overcoming

viscous friction and only the remainder is available to raise the fluid to a higher

z zelevation. If, for example the pressure head is 15 m but the fluid rise ( 1 ) is

2

only 12 m, then head loss is 3 m. Head loss provides a convenient way to visualize

the effects of friction.

z zIn a horizontal pipe, 1 , and head loss is

2

h P PL 1 2   P Horizontal pipe
 g  g
 

Head loss is related to friction factor.

hL  f L V2 Laminar or turbulent, any pipe orientation
D 2g

L V 2
D 2
P f The pressure drop of horizontal conduit

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Example 7.8:
Question:
What is the factor causes the head loss?
Solution:
Head loss is caused by fluid friction effect.

Example 7.9:
Question:
Why the head loss is very important in a pipe flow?
Solution:
Because is applied the energy equation to viscous flow in a pipe. A viscous fluid
flow through an inclined pipe is totally different with viscous fluid flows through a
horizontal pipe.

EXERCISE 7.4

1. The biomedical device uses compressed air to drive the plunger in a piston-
cylinder assembly with 0.8 m/s velocity that will push the drug (viscosity and
density similar to water at 10°C) through the hypodermic needle (inside diameter
0.25 mm and length 50 mm).
a. The pressure drop.
b. The head loss.

2. Calculate the volume flow rate and the pumping power requirements for question
no.1 (Exercise 7.4).

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7.5 FULLY DEVELOPED TURBULENT FLOW IN PIPES

Analytical solutions for turbulent flows are not possible, even for the simplest
geometries. Rather, turbulent flow is studied using a combination of simplifying
assumptions, numerical methods, and experiments. In this section, the results of
experimental study are presented.

The non-dimensional Reynolds number and friction factor are especially helpful in
guiding experimental work in turbulent flow. For a given geometry, the friction
factor depends only on the Reynolds number and the relative wall roughness,
which is the roughness height divided by the pipe diameter.

One need only measure pressure drop, velocity, and so on for the original case,
not for both cases, to determine the friction factor at the given Reynolds number.

The friction factors for circular tubes with fully developed turbulent flow have been
determined by many experiments. Those data were curve-fit using a regression
analysis to produce the following correlation, was published by Haaland in 1983:

1  1.8 log   /D 1.11  6.9  Turbulent flow
f  3.7  
Re 

The Colebrook equation is presented in graphical form in Figure 7-7, the so-called
Moody chart, which includes both laminar and turbulent flow regimes. In the
turbulent regime, the chart contains a family of curves corresponding to different
values of relative roughness,  / D .

The higher the relative roughness, the greater the friction factor becomes, as
expected. In the laminar regime, surface roughness is unimportant. Friction factor
decreases with increasing Reynolds number.

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Figure 7-7 Friction factor as a function of Reynolds number and relative roughness
for round pipes – the Moody chart.
(Source: Munson et al., Fundamentals of Fluid Mechanics, 4th ed., Wiley, NY,
2002, p., 477. Used with permission.)

Example 7.10:
Question:
What are the important factors in guiding experimental work in turbulent flow?
Solution:
The non-dimensional Reynolds number and friction factor are especially helpful in
guiding experimental work in turbulent flow.

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Example 7.11:
Question:
What are the methods used to calculate the friction factor of circular tubes with fully
developed turbulent flow?
Solution:
The methods used in order to calculate friction factors for circular tubes with fully
developed turbulent flow are :
a. Colebrook equation, presented in Moody Chart
b. Haaland equation

EXERCISE 7.5

1. Consider a heat exchanger that has 1000 2.5-cm diameter smooth tubes in
parallel, each 6-m long. The total water flow of 1 m3/s at 10 °C flows through the
tubes. Neglecting entrance and exit losses, determine:
a. the pressure drop (in kPa)
b. the pumping power required (in kW)
c. the pumping power for the same flow rate if solid deposits from the water build
up on the inner surface of the pipe with a thickness of 1-mm and an
equivalent roughness of 0.4 mm.

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7.6 ENTRANCE EFFECTS

By definition, in a fully developed flow, the velocity profile does not change with
downstream position and the only component of velocity is in the axial direction.
Recall that the fiction factor is a non dimensional pressure drop per unit length.

While the friction factor is constant in the fully developed region, it varies in the
entrance region, where the velocity profile is changing. At the pipe entrance, large
frictional effect result from the difference in velocity between the wall and the core
of the flow, and the friction factor is high.

The friction factor diminishes throughout the entrance region as the flow develops
and reaches an asymptotic value in the fully developed region, as illustrated in
figure 7-8. Consequently, the pressure drop per unit length is greater in the
entrance region than in the fully developed region.

Figure 7-8 Friction factor in the entrance region of a pipe.

The entrance length can be determined by advance analyses that are beyond the
scope of this text. Some useful relations for entrance length are

Lent,h  0.065 Re D laminar, Re  2100
L  ent,h  4.4 Re 1 / 6 D turbulent, Re  4000

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The subscript h refers to hydrodynamic entrance length. (Later, we will also
introduce a thermal entrance length.) Entrance effects tend to be important in
laminar flow for short pipes.

For example, if Re = 1000, then Lent, h = 65 D. If the pipe diameter were 1 cm and
the total pipe length 1 m (100 cm), then the first 65 cm would be in the entrance
region and only the last 35 cm would have a fully developed flow. Using
relationship for fully developed flow for the whole pipe would result in an
underestimate of the pressure drop.

Turbulent flow is less sensitive to entrance effects. For example, if Re = 10,000,
then Lent, h = 20.4D. Nevertheless, one should always check to detect the
presence of entrance effect for both laminar and turbulent flow.

Example 7.12:

Question:

How to justify that the flow is actually fully developed?

Solution:

In a fully developed flow, the velocity profile does not change with downstream
position and the only component of velocity is in the axial direction

Example 7.13:

Question:

What happens to the the pressure drop upon entering the pipe?

Solution:

The friction factor diminishes throughout the entrance region as the flow develops
and reaches an asymptotic value in the fully developed region, as consequently,
the pressure drop per unit length is greater in the entrance region than in the fully
developed region.

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EXERCISE 7.6

1. The pipe diameter were 1 cm and the total pipe length 1 m:
a. Calculate Lent, if Re=1,000.
b. Analysis the entrance region.

2. The pipe diameter were 1 cm and the total pipe length 1 m:
a. Calculate Lent, if Re=10,000.
b. Analysis the entrance region.

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7.7 STEADY-FLOW ENERGY EQUATION

Pipeline systems frequently include components such as pump, fans, turbines, and
other device that add or remove energy from a fluid flow. We refer to all this energy
flows as shaft work. In this section, we consider systems that involve both shaft
work and frictional losses in pipes.

The formula of steady, incompressible flow energy equation in the form,

P1 V2 P2 V2
 1  z1  hP   2  z2  hT  hL
g g
2g 2g

Each term in this equation has the units of length and can be visualized as a
“head”. By definition:

P  Pressure head
g

V2

2g  Velocity head
hL = Head loss
hP = Pump head
hT = Turbine head

Head is a sort of equivalent elevation. For example, if the pump head supplied is 4
m, then the pump can lift a frictionless fluid a net elevation of 4 m provided there is
no change in pressure or velocity.

Our sign convention for work is that work is positive if it is done by the fluid and
negative if it is done on the fluid.

By such a definition, pump work is negative and turbine work is positive. In
practice, it is often convenient to define new variables, Wp and Wt, which are
always positive. The control volume work is related to these new variables by

w w w 
cv T P

We assume the flow enters at the same temperature as the surroundings. The fluid
temperature increases (usually slightly) due to frictional heating, and as result,
there may be some heat transfer to the surrounding.

The quantity wP is the power that the pump actually supplies to the fluid. If the
pump operation were ideal, less work would be needed to raise the pressure of the
fluid.

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Example 7.14:
Question:
List the components that frequently used in pipeline system.
Solution:
a. Pump
b. Fans
c. Turbines, and
d. other device that could add or remove energy from a fluid flow

Example 7.15:
Question:
a. What does it means by positive and negative value for work?
b. What are the sign conventions of work for pump and turbine?
Solution:
a. The sign iconvention s positive if the work is done by the fluid, and negative if

the work is applied to the fluid.
b. The work of pump is negative and the work of turbine is positive.

EXERCISE 7.7

1. At an oil tank farm, an engineer opens a valve at the end of a 5-cm diameter, 50-
m long horizontal pipe from the bottom of a large diameter oil tank. The oil tank is
open to the atmosphere, and the oil depth is 6.5 m. The oil has a SG = 0.85 and a
kinematics viscosity of 6.8×10-4 m2/s. Neglecting minor losses, determine the
initial flow rate from the tank (in m3/s).

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7.7 STEADY-FLOW ENERGY EQUATION

Pipeline systems frequently include components such as pump, fans, turbines, and
other device that add or remove energy from a fluid flow. We refer to all this energy
flows as shaft work. In this section, we consider systems that involve both shaft
work and frictional losses in pipes.

The formula of steady, incompressible flow energy equation in the form,

P1 V2 P2 V2
 1  z1  hP   2  z2  hT  hL
g g
2g 2g

Each term in this equation has the units of length and can be visualized as a
“head”. By definition:

P  Pressure head
g

V2

2g  Velocity head
hL = Head loss
hP = Pump head
hT = Turbine head

Head is a sort of equivalent elevation. For example, if the pump head supplied is 4
m, then the pump can lift a frictionless fluid a net elevation of 4 m provided there is
no change in pressure or velocity.

Our sign convention for work is that work is positive if it is done by the fluid and
negative if it is done on the fluid.

By such a definition, pump work is negative and turbine work is positive. In
practice, it is often convenient to define new variables, Wp and Wt, which are
always positive. The control volume work is related to these new variables by

w w w 
cv T P

We assume the flow enters at the same temperature as the surroundings. The fluid
temperature increases (usually slightly) due to frictional heating, and as result,
there may be some heat transfer to the surrounding.

The quantity wP is the power that the pump actually supplies to the fluid. If the
pump operation were ideal, less work would be needed to raise the pressure of the
fluid.

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7.8 MINOR LOSES

In addition to viscous losses due to friction in straight section of pipe, there are
often other sources of losses in pipeline flow. For example, if the flow turns through
an elbow or passes through a valve, there is an added frictional loss.

All pipeline losses from sources other than wall friction are traditionally called
minor losses. Do not be fooled by the name. Sometimes the so-called minor
losses are greater than the head loss in straight sections of pipe, and these “minor”
losses than dominate the flow situation.

The head loss due to minor losses can be determined experimentally and

correlated as

hL  K L V2

2g

Where KL is called the loss coefficient. The loss coefficient is dimensionless. The
head loss due to minor losses is added to the head loss in straight pipe sections so
that the steady-flow energy equation becomes

P1 V2 P2 V2
 1  z1  hP   2  z2  hT  hL
g g
2g 2g

Where EhL represents the sum of all frictional losses, whatever the source.

In fact, loss coefficients depend to some extent on Reynolds number and on pipe
diameter. The number listed should be used only as guidelines. Figure 7-9 give
loss coefficients for pipeline entrance and exits. Figure 7-10 plots the loss
coefficients in sudden expansions and contractions.

Figure 7- 9 Entrance flow conditions and loss coefficient: (a) reentrant, KL = 0.8,
(b) sharp-edged, KL = 0.5, (c) slightly rounded, KL = 0.2,
(d) well-rounded, KL = 0.04.

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Figure 7-10 (a) Loss coefficient in a sudden contraction.
(b) Loss coefficient in a sudden expansion

Example 7.16:
Question:
What is the minor losses?
Solution:
Minor losses is a losses from sources other than wall friction

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Example 7.17:
Question:
What is the difference between sudden contraction and sudden expansion?
Solution:
Sudden contraction: the diameter at the entrance is bigger than the exit.
Sudden expansion: the diameter at the entrance is smaller than the exit.

EXERCISE 7.8

1. Air at 105 kPa and 25 °C flows from a 7.5-cm circular duct into a 22.5-cm circular
duct. The downstream pressure is 6.5-mm of water higher than the upstream
pressure. Determine:
a. the average air velocity approaching the expansion (in m/s)
b. the volumetric flow rate (in m3/s)
c. the mass flow rate (in kg/s).

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SUMMARY

In this chapter we have studied :

1. The principles of conservation of mass and momentum, also developing equations
for the velocity as a function of position and apply that information to the prediction
of pressure changes in internal flows.

2. The fundamental fluid property (viscosity) in order to deal with frictional effects.
3. The regime of Laminar and Turbulent flows
4. The application of internal flows system into a simple pipe.

REFERENCES

a. Kaminsky D. A, Jensen M. K., (2005), “Introduction to Thermal and Fluid
Engineering”, John Wiley & Sons, Inc.

b. Eastop, T.D, McConkey, A., (2004), 5th Edition, “Applied Thermodynamics for
Engineering Technologist”, Longman.

c. Yunus, A.C, Michael, A.B., (2002), 4th Edition, “Thermodynamics, an Engineering
Approach”, Mc Graw Hill, New York.

d. D.F. Young, B.R. Munson, T.H. Okiishi, (2004), “Fundamental of Fluid Mechanics”,
4th Edition, John Wiley & Sons, Inc.

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