Massoud M. Engineering Thermofluids: Thermodynamics, Fluid Mechanics, and Heat Transfer

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322 IIIb. Fluid Mechanics: Incompressible Viscous Flow

Throat

V= V

Venturi

Differential Manometer

V= V

(a) (b)

Figure IIIb.4.4. (a) A venturi equipped with a manometer and (b) an orifice meter

The multiplier in Equation IIIb.4.3 is referred to as the Velocity of Approach Fac-

tor given as

1/1

− . To simplify, we may represent

βα

−=

C to get:

()

212

2V PPAȡm −==

ρα



IIIb.4.4

where

is known as the flow coefficient. The discharge coefficient (C

) corrects

the theoretical equation for the influence of several parameters including velocity

profile, the energy loss between taps, type of meter, and pressure tap locations. In

general, the discharge coefficient is the product of two coefficients, the velocity

coefficient (C

) and the coefficient of contraction (C

). The latter accounts for

losses in the device and the former accounts for the flow area reduction due to the

vena contracta, therefore, C

= V



actual

/ V



theoretical

The velocity coefficient, C

, must be determined experimentally, as there is no

theoretical means of calculating it. The coefficient of contraction is obtained from

the ratio of the flow area at the vena contracta to the flow area at the throat; C

vena

. Finally, we find the discharge coefficient from C

= C

. Note that

Equation IIIb.4.4 is applicable only if changes in the fluid density are negligible.

The applicable equation for cases where the change in density is noticeable is dis-

cussed in Chapter IIIc.

Determination of the Discharge Coefficient

The C

for Bernoulli-type devices are obtained as curve fits to data by ISO and

recommended by ASME, as follows:

Thin Plate Orifice

75.0

5.23

81.2

Re71.910337.0

09.0

184.00312.05959.0

−

+−

−

+−+=

FFC

ββ

In this correlation, F

and F

depend on the location of the pressure taps. For the

upstream tap located at a distance equal to the pipe diameter (D) from the inlet

face and the downstream tap located at a distance of (D/2), these factors are

4. Steady Incompressible Viscous Flow in Piping Systems 323

0.4333 and 0.47, respectively. This is referred to as D: D/2 tap. For the corner

taps where the meter plate meets the pipe wall, F

= F

= 0, hence C

becomes:

75.0

5.281.2

Re71.91184.00312.05959.0

−

+−+=

βββ

IIIb.4.5

Long Radius Nozzle

−=

00653.09965.0

IIIb.4.6

Venturi Nozzle

5.4

196.09858.0

−=

C IIIb.4.7

The above discharge coefficients for orifice (D: D/2 tap), nozzle and venturi are

plotted in Figure IIIb.4.5.

0.59

0.60

0.63

0.64

0.65

0.61

0.62

= 0.8

0.7

0.6

0.5

0.4

0.3

0.2

Orifice

0.93

0.94

0.95

0.96

0.97

0.98

1.00

0.99

(Re)

Nozzle

0.92

0.94

0.96

0.98

0.3 0.4 0.5 0.6 0.7

0.8

Venturi

Figure IIIb.4.5. Discharge coefficient for Bernoulli obstruction meters (ISO-ASME)

4.3. Unrecoverable Head Loss for Bernoulli Obstruction Meters

Let’s consider three pressure taps for the venturi and the orifice meter, as shown in

Figures IIIb.4.4, and IIIb.4.6(a). The first tap is located at least one pipe diameter

upstream of the meter, the second tap located at the throat of the meter and the

third tap located downstream where the flow is fully recovered.

Pressure at these locations may be shown as P

, P

(or P

) and P

, respectively.

Total pressure drop for the meter is given by the differential pressure term P

– P

A portion of the total pressure drop is recoverable when flow reaches location 3

where, for incompressible fluid flow velocity, V

becomes equal to V

. The re-

maining portion of the total pressure drop is the unrecoverable pressure drop given

by P

– P

. The ratio of the unrecoverable head loss and the total pressure differ-

ential, for the square edged orifice, flow nozzle, and venturi with 15-degree cone

angle, is given in Figure IIIb.4.6(b).

324 IIIb. Fluid Mechanics: Incompressible Viscous Flow

vena

- P

: Pressure differential

- P

: Unrecoverable pressure loss

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.1

0 0.1 0.2 0.3 0.4 0.5

0.6

0.7 0.8

Venturi

Nozzle

15 degree

Orifice

- P

)/(P

- P

)

(a) (b)

Figure IIIb.4.6. (a) Schematic of pressure taps for a thin plate orifice and (b) Pressure loss

rate (Miller)

The unrecoverable head loss may also be found from the loss coefficient and

the flow velocity. The loss coefficient for the three Bernoulli obstruction meters is

shown in Figure IIIb.4.7. Expectedly, the flow meter that causes less disturbance

to the flow is associated with smaller loss coefficient. As such, a venturi with a

cone angle of 7 degrees results in the lowest and a thin-plate orifice results in the

highest loss coefficient.

Having the pipe and the throat diameters, we find

and then K for a specific

meter from Figure IIIb.4.7. The unrecoverable head loss is then calculated from K

and the throat velocity

2/Kh

= .

The head obtained from this equation using the loss coefficient from Fig-

ure IIIb.4.7 must agree with the head loss obtained from Figure IIIb.4.6(b).

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.2 0.3

0.4

0.5 0.6 0.7 0.8

Thin Plate Orifice

Nozzle

Venturi

15 degree

7 degree

Figure IIIb.4.7. Frictional loss coefficient for Bernoulli obstruction meters (Bean)

4. Steady Incompressible Viscous Flow in Piping Systems 325

4.4. Types of Problems for Bernoulli Obstruction Meters

Similar to the single-path piping systems, the type of problems for Bernoulli me-

ters can be divided into three groups as shown in Table IIIb.4.2. For a given pipe

diameter (D) and fluid (

), there are 3 parameters in Equation IIIb.4.3. Since we

have only one equation, we can solve for only one unknown. Hence, the other two

parameters must be given. For types I and III, the unknown must be determined

by iteration. The examples that follow examine all of the types listed in Ta-

ble IIIb.4.2. As we shall see in Chapter IIIc, when Equation IIIb.4.34 is applied to

compressible fluids, an additional parameter (Y) will have to be dealt with.

Table IIIb.4.2. Matrix of parameters for Bernoulli-obstruction meters

Type Known Find



∆P

, ∆P



III



,∆P

Type I. Find Pressure Drop, Given Throat Diameter and Flow Rate

Type I problems are generally used in the design process. For example, to size the

pump in a flow loop to provide for the induced pressure drop due to the installa-

tion of the Bernoulli obstruction meter. Problems of this type are straightforward

as no iteration is necessary. This is because, C

can be easily calculated since the

flow rate (hence, the Re number) and

are known. Having C

, we find ∆P from

Equation IIIb.4.4.

Example IIIb.4.8. A long radius nozzle is used to measure the flow rate of water

in a pipeline. The pipe inside diameter is 10 in (25.4 cm) and the nozzle throat di-

ameter is 2.5 in (6.35 cm). The nominal volumetric flow rate measured by this

meter is 500 GPM (31.54 lit/s). Find the recoverable and unrecoverable pressure

drops. Water flows at 100 F (38 C).

Solution: To find ¨P over the nozzle we use Equations IIIb.4.4 and IIIb.4.7:

At 100 F:

= 62 lbm/ft

(993 kg/m

) and v = 7.39E–6 ft

/s (6.86E-7 m

/s)



= 500 GPM = (500/7.481)/60 = 1.11 ft

/s (31.54 lit/s)

D = 10/12 = 0.833 ft, d

= 2.5/12 = 0.208 ft, and

= 2.5/10 = 0.25

(0.833)

/4 = 0.545 ft

, hence, V

= V



= 1.11/0.545 = 2.04 ft/s

(0.208)

/4 = 0.034 ft

, hence, V

= V



= 1.11/0.034 = 32.65 ft/s

= V

D/v = 2.04 × 0.833/7.39E-6 = 230,000

326 IIIb. Fluid Mechanics: Incompressible Viscous Flow

Finding C

from Figure IIIb.4.4 or Equation IIIb.4.6 as:

−=

00653.09965.0

−

000,230

00653.09965.0

= 0.97

−

25.01

97.0

−

= 0.972

Since

ρα

/2V PA

∆=



, solving for ∆P:

∆P =



×××

034.0972.02.322

11.162

1086 lbf/ft

= 7.54 psi

For

= 0.25 from Figure IIIb.4.7 for a nozzle we find K = h

V /2g] ≅ 0.8

2.322

65.32

8.0

×==

= 13.24 ft

This amounts to a pressure drop of:

∆P

= 62 × 32.2 × 13.24/32.2 = 821 lbf/ft

= 5.7 psi

Thus, of 7.54 psi total pressure drop, 5.7 psi is unrecoverable and 1.84 psi is re-

coverable.

Type II. Find Flow Rate, Given Throat Diameter and Pressure Drop

Type II problems are generally used for analysis of already designed and installed

flowmeters. Problems of this type are rather straightforward. Iteration is gener-

ally necessary (except for the venturi nozzle) to find

. Then flow rate is ob-

tained from Equation IIIb.4.4.

Example IIIb.4.9. A venturi is used to measure the flow of oil. Use the follow-

ing data to find the oil flow rate. D = 55 cm, H = 75 cm, h = 175 cm, d

= 22 cm.

Specific gravity (SG) of oil is c

= 0.8 and of the manometer liquid is c

= 1.3.

4. Steady Incompressible Viscous Flow in Piping Systems 327

Throat

Diffuser

SG= c

Solution: Writing the Bernoulli equation between the venturi inlet and venturi

throat, we get:

221

2/2/ gZVPgZVP

ρρρρ

++=++

Expressing velocities in terms of volumetric flow rate and rearranging terms, we

get:

)(V]/1/1[2/)(

121

ZZgAAPP −+−+−

ρρ



Writing in terms of

= d

/D and solving for V



yields:

)]1(/[]H[2V

βρρ

−−∆= gPA



Finally factoring in the discharge coefficient C

and substitutig in terms of the

flow coefficient yields:

ρρα

/]H[2V gPA

−∆=



We find ∆P by writing the Bernoulli equation between the two ends of the ma-

nometer to get:

gδ = P

g(H + δ – h) +

where

and

are densities of fluids in the venturi and the manometer, respec-

tively. Solving for ∆P, we get:

∆P = g[

(H – h) +

Substituting numerical values, we find

∆P = 9.81[(0.8×999)(0.75 – 1.75) + (1.3×999)(1.75)] = 14.455 kPa

Since

= 0.22/0.55 = 0.4, C

= 0.9858 – 0.196(0.4)

4.5

= 0.983. We also find

=−=

4.01/983.0

0.996

(0.22)

/4 = 0.038 m



= 0.996 × 0.038 )8.0999/(]75.0)8.0999(14455[2 ×××− ] = 0.223 m

328 IIIb. Fluid Mechanics: Incompressible Viscous Flow

Type III. Find Throat Diameter, Given Flow Rate and Pressure Drop

Type III problems are used in the design process to calculate the throat diameter

for the intended flow rate and pressure drop. Problems of this type are generally

solved by iteration.

Example IIIb.4.10. We need to simulate a pressure drop of 5.0 psi caused by a

pump with a seized rotor in a test section. To induce this pressure drop to the

flow, we use a thin-plate orifice with corner taps. The test section is a 6 inch pipe

with water flowing at 352 GPM and 60 F. Find the throat diameter of the orifice.

Solution: To find the throat diameter, we should first find C

(

) from

ρβ

/2]1/[V

PCA

∆−=



and set it equal to C

from the ISO curve fit for

thin-plate orifice. Since both of these equations are highly non- linear functions of

, we have to resort to iteration.

D = 6/12 = 0.5 ft hence, A

= 3.14(0.5)

/4 = 0.196 ft

. Since V



= 352/(7.481×60)

= 0.784 ft

/s, therefore V

= 0.784/0.196 = 4 ft/s and Re

= V

D/v = 4× 0.5/1.23E-

5 = 0.163E6. Final results of the iteration are:

= d

(-): 0.485 Throat diameter (in): 2.913

Throat velocity (ft/s): 16.96 Reynolds number (-): 0.16E6

Discharge Coefficient (-): 0.602 Loss coefficient (-): 1.910

Head loss (ft): 8.536 Pressure loss (psi): 3.696

Velocity of

approach factor (-): 1.029 Flow coefficient (-): 0.620

3.2

3.4

3.6

3.8

4.2

4.4

1234567

Pressure Drop (psi)

Throat Diameter (in)

= 5 ft/s

= 6 in

4. Steady Incompressible Viscous Flow in Piping Systems 329

1.5

2.5

3.5

012345678

Flow Velocity (ft/s)

Throat Diameter (in)

∆

P = 5 psi

= 6 in

Having solved the problem, we now want to find how the throat diameter changes

with flow or with pressure drop. Intuitively, we know that for given D

and ∆P,

the larger the flow rate, the larger the throat diameter. This is because pressure

drop increases with flow velocity. That lessens demand on the thin-plate orifice

for pressure drop. Conversely, for given flow velocity and pipe inside diameter,

the higher the pressure drop (∆P), the smaller the throat diameter. These are

shown in the above plots.

Problems involving flow in single-path systems and the Bernoulli obstruction

meters can be easily solved by using the software on the accompanying CD-ROM.

4.4. Flow in Compound Conduits

In common practice, pipes of different lengths and diameters are connected via re-

ducers and enlargers. This constitutes a serial-path system as shown in Fig-

ure IIIb.4.8(a). For example, the riser of a containment spray system consists of a

serial flow path. On the other hand, pipes may be connected at both ends to a

common plena, or header, such that flow entering the inlet plena is divided be-

tween the pipes. This constitutes a system with parallel flow paths. There are two

types of parallel flow path systems. In the first type, flow is mixed only at the

inlet and exit plena, as shown in Figure IIIb.4.8(b). We refer to this type as closed

parallel flow path. This is the type that we study in this chapter. In the second

type, flow is also intermixing between various flow paths. For example, the fuel

assemblies of a BWR core constitute a closed parallel path system where flow is

mixed in the lower plenum prior to entering the core and in the upper plenum after

leaving the core. On the other hand, the core of a PWR consists of many open

parallel flow paths where flows are mixed due to the existing cross flow between

the subchannels, having lateral communication.

We now compare the hydraulic characteristics of the serial and closed parallel

path systems with respect to flow rate and pressure drop. For simplicity, we refer

to the closed parallel flow path as parallel flow path.

330 IIIb. Fluid Mechanics: Incompressible Viscous Flow

h,1

h,2

h,3

h,1

h,2

h,3

(a) (b)

Figure IIIb.4.8. Flow paths connected in series and in parallel and their equivalent flow

path

Comparison of Serial-Path Systems with Parallel-Path Systems

For the steady flow of incompressible viscous fluids, we first consider pipes con-

nected in series as shown in Figure IIIb.4.8(a). Pipes in series have generally dif-

ferent length, diameter, and loss coefficient. Therefore, while flow rate is the

same for all the pipe segments, pressure drops are generally different:

Pipes in Series

321321321

VVV PPPPPPP ∆+∆+∆=∆∆≠∆≠∆==



Similarly, as shown in the right side of Figure IIIb.4.8(b), parallel pipes may also

have different diameter, length, and loss coefficient. In this case however;

Pipes in Parallel

321321321

VVVVVVV PPP ∆=∆=∆++=≠≠



Example IIIb.4.11. A pump delivers water at a steady rate of



/s to a branch

containing two parallel lines. Find flow rate in branch 1 in terms of



and the

known diameters, lengths, flow areas, and loss coefficients.

Branch 2

Branch 1

Solution: We use ∆P

= ∆P

and then substitute from the continuity equation to

obtain:

4. Steady Incompressible Viscous Flow in Piping Systems 331

−

¦¦



If we represent

()

[]

/K/ ADfL

+ = R then the above equation reduces to:

()

0VVV2V

212

121

=−+−



RRRR

From this equation we find branch 1 flow rate as:

()

221

VVV

RRRRR

−

−+±−





()

212121

/V/V RRRRR −±−=



The plus sign is chosen if R

. Otherwise, we choose the minus sign.

System Curve for Serial-Path Systems

We consider a serial-path system, consisting of pipes of various lengths (L), di-

ameters (D), and loss coefficients (K). This piping system has, in general, an ele-

vation change (Z

– Z

). Equation IIIa.3.31 for this system can be written as:

()

ADA

fZZ

System



++−=

−

¦¦

where due to the lack of any pump in the system, h

is dropped and h

is replaced

by Equation IIIb.3.12. Depending on the flow regime, H

system

is either a linear

function of velocity or a parabolic function. If flow is laminar then f = 64/Re =



D. Substituting, for the friction factor, we find:



System

IIIb.4.8-1

where c

= Z

– Z

and c

= Σ64

L’/(2g

DA). For turbulent flow in smooth pipes,

if we use an approximate value for the friction factor as given in Table IIIb.3.2, we

find



System

IIIb.4.8-2

where c

= ΣfL’/(2gDA

). The plot of H

system

versus V



is known as the system

curve.