Bird R.B., Stewart W.E., Lightfoot E.N. Transport Phenomena

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57.5 Estimation of the Viscous Loss

207

Table

7.5-1

Brief Summary of Friction Loss Factors for Use with Eq. 7.5-10

(Approximate Values for Turbulent Flow)"

Disturbances

Sudden changes in cross-sectional areab

Rounded entrance to pipe 0.05

Sudden contraction

Sudden expansionc

Orifice (sharp-edged)

Fittings and valves

90" elbows (rounded) 0.4-0.9

90" elbows (square) 1.3-1.9

45" elbows 0.3-0.4

Globe valve (open) 610

Gate valve (open) 0.2

Taken from

Kramers,

Physische

Transportverschijnselen,

Technische Hogeschool Delft, Holland

(19581,

pp.

53-54.

Here

(smaller cross-sectional area)/(larger cross-sectional area).

See derivation from the macroscopic balances in Example

7.6-1.

then

:(v)',

where

(v)

is the

velocity

upstream

from the enlargement.

Most flow systems contain various "obstacles," such as fittings, sudden changes

diameter, valves, or flow measuring devices. These also contribute to the friction loss

ED.

Such additional resistances may be written in the form of

Eq.

7.5-4,

with

determined by

one of two methods:

(a)

simultaneous solution of the macroscopic balances, or

(b)

experi-

mental measurement. Some rough values of

are tabulated in Table 7.5-1 for the conven-

tion that

(v)

is the average velocity

downstream

from the disturbance. These

values are

for

turbulent flow

for which the Reynolds number dependence

not too important.

Now we are in a position to rewrite

Eq.

7.4-7 in the

approximate

form frequently used

for

turbulent flow

calculations in a system composed of various kinds of piping and addi-

tional resistances:

12L

:(z$

v:)

~(1~

zli

jiY

Rif)i

(v2e)

(7.5-10)

sum over

all

sum over

all

sections of fittings, valves,

straight conduits meters, etc.

Here

is the mean hydraulic radius defined in

Eq.

6.2-16,

is the friction factor defined

Eq.

6.1-4, and

is the friction loss factor given in Table 7.5-1. Note that the

and

the first term refer to the velocities at planes 1 and

the

in the first sum is the average

velocity in the ith pipe segment; and the v in the second sum is the average velocity

downstream

from the ith fitting, valve, or other obstacle.

What is the required power output from the pump at steady state in the system shown

Fig.

7.5-l? Water at 68OF

62.4 lb,/ft3;

1.0 cp) is to be delivered to the upper tank at a rate

Requirement

of 12 ft3/min. All of the piping is 4-in. internal diameter smooth circular pipe.

for

Pipeline

Flow

208

Chapter 7 Macroscopic Balances for Isothermal Flow Systems

SOLUTION

Fig.

7.5-1.

Pipeline flow

with friction losses be-

cause of fittings. Planes

and

are just under

Plane

the surface of the liquid.

The average velocity in the pipe is

and the Reynolds number is

Hence the flow is

turbulent.

The contribution to

from the various lengths of pipe

will

The contribution to

from the sudden contraction, the three

90"

elbows, and the sudden ex-

pansion (see Table 7.5-1) will be

($v2eJi

$(2.30)~(0.45

3(;)

8 ft2/s2

(7.5-14)

Then from Eq. 7.5-10 we get

(32.2)(105

20)

wrn

(7.5-15)

Solving for

wrn

we get

This is the work (per unit mass of fluid) done

the fluid

the pump. Hence the pump does

2830 ft2/s2 or 2830/32.2

88 ft lbf/lbrn of work on the fluid passing through the system.

The

mass rate of flow is

Consequently

Wrn

(12.5)(88)

1100 ft lbf/s

2 hp

1.5 kW

(7.5-18)

which is the power delivered by the pump.

s7.6

Use of the Macroscopic Balances for Steady-State Problems

209

Table

7.6-1

Steady-State Macroscopic Balances for Turbulent Flow in Isothermal Systems

Mass:

CW,

Zw2

(A)

Momentum:

X(vlwl

plS1)ul

C(v2~2

p2S2h2

mtot!4

Ff-s

(B)

Mechanical energy:

(D)

Notes:

(a) All formulas here assume flat velocity profiles.

(b)

Zwl

w,,

wlb

wlc

where

w,,

p,,v,,S,,,

etc.

(c)

and

are elevations above an arbitrary datum plane.

(d) All equations are written for compressible flow;

for

incompressible flow,

57.6

USE OF THE MACROSCOPIC BALANCES

FOR STEADY-STATE PROBLEMS

53.6

we saw how to set up the differential equations to calculate the velocity and pres-

sure profiles for isothermal flow systems by simplifying the equations of change. In this

section we show how to use the set of steady-state macroscopic balances to obtain the al-

gebraic equations for describing large systems.

For each problem we start with the four macroscopic balances. By keeping track of

the discarded or approximated terms, we automatically have a complete listing of the as-

sumptions inherent in the final result. All of the examples given here are for isothermal,

incompressible flow. The incompressibility assumption means that the velocity of the

fluid must be less than the velocity of sound in the fluid and the pressure changes must

be small enough that the resulting density changes can be neglected.

The steady-state macroscopic balances may be easily generalized for systems with

multiple inlet streams (called la,

Ib,

lc,

. .

and multiple outlet streams (called 2a, 2b,

2c,

.).

These balances are summarized in Table 7.6-1 for turbulent flow (where the ve-

locity profiles are regarded as flat).

An incompressible fluid flows from a small circular tube into a large tube in turbulent flow,

as shown in

Fig.

7.6-1. The cross-sectional areas of the tubes are

and

S2.

Obtain an expres-

Pressure Rise and

sion for the pressure change between planes

and

and for the friction loss associated with

Friction Loss in

the sudden enlargement in cross section. Let

S,/S2,

which is less than unity.

Sudden Enlargement

Plane

area

surface of area Cylindrical hibe

cross-sectiona~

Fig.

7.6-1.

HOW through a sudden

area

enlargement.

210

Chapter 7 Macroscopic Balances for Isothermal Flow Systems

SOLUTION

(a)

Mass balance.

For steady flow the mass balance gives

For a fluid of constant density, this gives

(b)

Momentum balance.

The downstream component of the momentum balance is

The force

Ff,,

is composed of two parts: the viscous force on the cylindrical surfaces parallel

to the direction of flow, and the pressure force on the washer-shaped surface just to the right

of plane

and perpendicular to the flow axis. The former contribution we neglect (by intu-

ition) and the latter we take to be

p,(S2

S,) by assuming that the pressure on the washer-

shaped surface is the same as that at plane

We then get, by using Eq. 7.6-1,

Solving for the pressure difference gives

or, in terms of the downstream velocity,

Note that the momentum balance predicts (correctly) a

rise

in pressure.

(c)

Angular momentum balance.

This balance is not needed. If we take the origin of coor-

dinates on the axis of the system at the center of gravity of the fluid located between

planes

and 2, then [r,

u,l

and

[r2

u21

are both zero, and there are no torques on the

fluid system.

(dl

Mechanical energy balance.

There is no compressive loss, no work done via moving

parts, and no elevation change, so that

,(v:

v:)

(p,

p2)

Insertion of Eq. 7.6-6 for the pressure rise then gives, after some rearrangement,

which is an entry in Table 7.5-1.

This example has shown how to use the macroscopic balances to estimate the friction loss

factor for a simple resistance in a flow system. Because of the assumptions mentioned after

Eq. 7.6-3, the results in Eqs. 7.6-6 and

are approximate. If great accuracy is needed, a correc-

tion factor based on experimental data should be introduced.

diagram of a liquid-liquid ejector is shown in Fig. 7.6-2. It is desired to analyze the mixing

of the two streams, both of the same fluid, by means of the macroscopic balances. At plane

Petfowance of

the two fluid streams merge. Stream la has a velocity

and a cross-sectional area is,, and

LiPid-LiPid Ejector

stream

has a velocity

iv,

and a cross-sectional area $5,. Plane

is chosen far enough down-

stream that the two streams have mixed and the velocity is almost uniform at

v,.

The flow is

57.6 Use of the Macroscopic Balances for Steady-State Problems

211

SOLUTION

Stream lb

Fig.

7.6-2.

Flow in a liquid-liq-

uid ejector pump.

turbulent and the velocity profiles at planes

and

are assumed to be flat. In the following

analysis

F+,

is neglected, since it is felt to be less important than the other terms in the mo-

mentum balance.

(a)

Mass balance.

At steady state, Eq.

(A)

of Table 7.6-1 gives

Hence, since

S2, this equation gives

for the velocity of the exit stream. We also note, for later use, that

w,,

wlb

$w2.

(b)

Momentum balance.

From Eq. (B) of Table 7.6-1 the component of the momentum bal-

ance in the flow direction is

or using the relation at the end of (a)

from which

$Po

This is the expression for the pressure rise resulting from the mixing of the two streams.

(c)

Angular momentum balance.

This balance is not needed.

(d)

Mechanical energy balance.

Equation

(D)

of Table 7.6-1 gives

or, using the relation at the end of (a), we get

Hence

is the energy dissipation per unit mass. The preceding analysis gives fairly good results for

liquid-liquid ejector pumps. In gas-gas ejectors, however, the density varies significantly and

it is necessary to include the macroscopic total energy balance as well as an equation of state

in the analysis. This is discussed in Example 15.3-2.

212

Chapter 7 Macroscopic Balances for Isothermal Flow Systems

EXAMPLE

7.6-3

Thrust

Pipe

Bend

SOLUTION

Water at 95°C is flowing at a rate of 2.0 ft3/s through a 60" bend, in which there is a contrac-

tion from 4 to 3 in. internal diameter (see Fig. 7.6-3). Compute the force exerted on the bend if

the pressure at the downstream end is 1.1 atm. The density and viscosity of water at the con-

ditions of the system are 0.962 g/cm3 and 0.299 cp, respectively.

The Reynolds number for the flow in the 3-in. pipe is

At this Reynolds number the flow is highly turbulent, and the assumption of flat velocity pro-

files is reasonable.

(a)

Mass balance.

For steady-state flow,

w,.

If the density is constant throughout,

in which

is the ratio of the smaller to the larger cross section.

(b)

Mechanical energy balance.

For steady, incompressible flow,

Eq.

(d) of Table 7.6-1

be-

comes, for this problem,

According to Table 7.5-1 and

Eq.

7.5-4, we can take the friction loss as approximately g(4v:)

iv:. Inserting this into

Eq.

7.6-20 and using the mass balance we get

This is the pressure drop through the bend in terms of the known velocity v2 and the known

geometrical factor

(c)

Momentum balance.

We now have to consider both the

and y-components of the mo-

mentum balance. The inlet and outlet unit vectors will have

and y-components given

ulw

uly

cos

8, and

u,,

sin 8.

Fluid

out

internal

520

Fig.

7.6-3.

Reaction force at a reducing

diameter

bend in a pipe.

37.6 Use of the Macroscopic Balances for Steady-State Problems

213

The x-component of the momentum balance then gives

where

is the x-component of

F,+.

Introducing the specific expressions for

and u12, we

get

v,(pvlS,)

vJPv~SJ

cos

plS1

p2S2

cos

pv;S2(p

cos 0)

(pl

p2)S1

p2(S1

S2 cos 6)

(7.6-23)

Substituting into this the expression for p,

p2 from Eq. 7.6-21 gives

pv;S2(@

cos

pv;s2pp1(&

$p2)

pg(h2

h,)S2p-'

p2S2(p-'

cos 0)

w2(ps2)-'(p

COS

pg(h2

h1)S2p-'

p2S2(p-'

cos

The y-component of the momentum balance is

-(v2w2

p2S2) sin

m,,,g

-w2(pS2)-I sin 0

p2S2 sin 0

vR2Lpg

in which

and

are the radius and length of a roughly equivalent cylinder.

We now have the components of the reaction force in terms of known quantities. The nu-

merical values needed are

60 lb,/ft3

(2.0)(60)

1201b,/s

cos

;

sin 6

+fi

16.2 lbf/in.'

With these values we then get

(120)2

(d)

(16.2)(0.049)(144)

2(0.049)(32.2)

Hence the magnitude of the force is

d304'

2342

384 Ibf

1708

(7.6-29)

The angle that this force makes with the vertical is

arctan(F,/Fy)

arctan 1.30

52"

(7.6-30)

In looking back over the calculation, we see that all the effects we have included are impor-

tant, with the possible exception of the gravity terms of 2.6 Ibf in

and 2.5 Ibi in

F,.

214

Chapter 7 Macroscopic Balances for Isothermal Flow Systems

rectangular incompressible fluid jet of thickness

emerges from a slot of width c, hits a flat

plate and splits into two streams of thicknesses

bZa

and

bZb

as shown in Fig. 7.6-4. The emerg-

The

Impinging Jet

ing turbulent jet stream has a velocity

and a mass flow rate w,. Find the velocities and mass

rates of flow in the two streams on the plate.'

SOLUTION

We neglect viscous dissipation and gravity, and assume that the velocity profiles of all three

streams are flat and that their pressures are essentially equal. The macroscopic balances then

give

Mass balance

W2a

W2b

Momentum balance

(in the direction parallel to the plate)

VlWl

COS

'U2aW2a

v2bW2b

(7.6-32)

Mechanical energy balance

iv?wl

&$b~2b

(7.6-33)

Angular momentum balance

(put the origin of coordinates on the centerline of the jet and

an altitude of

gb,;

this is done so that there will be no angular momentum of the incoming jet)

This last equation can be rewritten to eliminate the b's in favor of the w's. Since w1

pv,b,c

and w,,

pvzab2,c, we can replace

b,,

by (w,/pv,c)

(w2,/pv2,c) and replace

bZb

cor-

respondingly. Then the angular momentum balance becomes

Velocity

ass rate of flow

city

Mass rate of flow

b2b

Plate

b2a

Mass rate of flow

W2b

W2a

Fig.

7.6-4.

Jet impinging on a wall and splitting into two streams. The point

which is the origin of coordinates for the angular momentum balance, is

taken to be the intersection of the centerline of the incoming jet and a plane

that is at an elevation

ib,.

For alternative solutions to this problem, see

Batchelor,

Introduction

Fluid

Dynamics,

Cambridge University Press (1967), pp. 392-394, and

Whitaker,

Introduction to Fluid Dynamics,

Prentice-Hall, Englewood Cliffs,

N.J.

(1968), p. 260. An application of the

compressible

impinging jet

problem has been given by

Foa,

U.S.

Patent 3,361,336 Uan. 2,1968). There, use is made of the fact

that if the slot-shaped nozzle moves to the left in Fig. 7.6-4 (i.e., left with respect to the plate), then, for

compressible fluid, the right stream will

cooler than the jet and the left stream will

warmer.

s7.6 Use of the Macroscopic Balances for Steady-State Problems

215

Now Eqs. 7.6-31

solved we find that

,32,33, and 36 are four equations with four unknowns. When these are

Hence the velocities of all three streams are equal. The same result is obtained by applying

the classical Bernoulli equation for the flow of an inviscid fluid (see Example 3.5-1).

common method for determining the mass rate of flow through a pipe is to measure the pres-

sure drop across some "obstacle"

the pipe. An example of this

the orifice, which is a thin

~so~h~~al Flow

plate with a hole

the middle. There are pressure taps at planes

and

upstream and dom-

Liquid

Through

stream of the orifice plate. Fig. 7.6-5(a) shows the orifice meter, the pressure taps, and the gen-

Orifice

era1 behavior of the velocity profiles as observed experimentally. The velocity profile at plane

cross section of pipe

Plane

PlaAe

hhmneter ~laAe

I I

Plane

Fig.

7.6-5.

(a)A sharp-edged orifice, showing the approximate velocity

profiles at several planes near the orifice plate. The fluid jet emerging

from the hole is somewhat smaller than the hole itself. In highly turbu-

lent flow this jet necks down to a minimum cross section at the vena con-

tracts.

The extent of this necking down can be given by the contraction

coefficient,

(S,,,,

c,,,ac,,/S,). According to inviscid flow theory,

T/(T

0.611 if So/Sl

[H.

Lamb, Hydrodynamics, Dover,

New York (1945), p. 991. Note that there is some back flow near the wall.

(b)

Approximate velocity profile at plane

used to estimate (vi)/(v2).

216

Chapter 7 Macroscopic Balances for Isothermal Flow Systems

SOLUTION

will be assumed to be flat. In Fig. 7.6-5(b) we show an approximate velocity profile at plane

which we use in the application of the macroscopic balances. The standard orifice meter equa-

tion is obtained by applying the macroscopic mass and mechanical energy balances.

(a)

Mass balance.

For a fluid of constant density with a system for which

the

mass balance in Eq. 7.1-1 gives

With the assumed velocity profiles this becomes

and the volume rate of flow is

pv,S.

(b)

Mechanical energy balance.

For a constant-density fluid in a flow system with no eleva-

tion change and no moving parts,

Eq.

7.4-5 gives

The viscous loss

is neglected, even though it is certainly not equal to zero. With the as-

sumed velocity profiles, Eq. 7.6-43 then becomes

;cv;

v:,

(7.6-44)

When Eqs. 7.6-42 and 44 are combined to eliminate

u,,

we can solve for

to get

We can now multiply by pS to get the volume rate of flow. Then to account for the errors

in-

troduced by neglecting

and by the assumptions regarding the velocity profiles we include

a discharge coefficient,

Cd,

and obtain

Experimental discharge coefficients have been correlated as a function of

So/S

and

the

Reynolds n~mber.~ For Reynolds numbers greater than lo4,

approaches about 0.61 for

all

practical values of

So/S.

This example has illustrated the use of the macroscopic balances to get the general form

the result, which is then modified by introducing a multiplicative function of dimensionless

groups to correct for errors introduced by unwarranted assumptions. This combination

macroscopic balances and dimensional considerations is often used and can be quite useful.

57.7

USE OF THE MACROSCOPIC BALANCES

FOR UNSTEADY-STATE PROBLEMS

In the preceding section we have illustrated the use of the macroscopic balances for solv-

ing steady-state problems. In this section we turn our attention to unsteady-state prob-

lems. We give two examples to illustrate the use of the time-dependent macroscopic

balance equations.

Tuve and

Sprenkle,

Instruments,

6,202-205,225,232-234 (1935);