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

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302 IIIb. Fluid Mechanics: Incompressible Viscous Flow

[25 / (

/4)] = 0.59 D

2/3

(250/5 × 5280)

0.5

/0.012

Solving for D, we find D

= 2 ft or 24 inches. The reader should now try the

Darcy-Weisbach formula.

Finding the pipe diameter, D, from the Darcy equation requires iteration as we

should solve Equations IIIb.3.4 and IIIb.3.5 simultaneously. Another way to

avoid iteration is to use the Swamee and Jain’s correlation:

04.0

2.5

9.4

75.4

25.1

66.0



νε

IIIb.3.10

where

is in ft, V



is in ft

/s, L is in ft, g = 32.2 ft/s

, h

in ft,

in ft

/s, and D is

in ft. Note that Equation IIIb.3.10 is applicable for 10

– 6

≤

≤ 0.02 and 3 × 10

≤ Re ≤ 3 × 10

Example IIIb.3.5. Solve Example IIIb.3.4 using the Swamee and Jain’s correla-

tion. Use water at T = 65 F.

Solution: From Table IIIb.3.1 we find

= 0.00015 ft. Also for water, v = 0.041

/hr = 1.13E-5 ft

/s. We now substitute in Equation IIIb.3.10:

1.25 4.75 9.4 5.2 0.04

26,399.7(25) 26,399.7

0.66[0.00015 ( ) 1.13E 5 (25) ( ) ] 1.854 ft

32.2 250 32.2 250

D =+−×=

××

= 22.3 in

Pressure Drop Associated with Fittings and Valves

Earlier we studied pressure drop in straight pipes due to the viscosity of the fluid

and the surface condition of the pipe wall. This is known as skin friction. We

now consider other conditions that also result in pressure drop by disturbing the

flow such as twist, turn, or partial obstruction of the flow. These may result in

flow separation and consequently irreversible energy loss to the fluid flow. Simi-

lar to the Darcy pressure drop, this also results in unrecoverable pressure drop.

Losses due to twist, turn, and flow obstruction are generally due to the presence

of valves and fittings such as reducers, enlargers, bends, T for flow division, and

invasive flowmeters. The losses due to the fittings and valves are referred to as

the minor losses. The term minor should not be taken literally as depending on the

flow condition the pressure drop due to the presence of pipe fittings, flowmeters,

and valve may by far surpass the pressure drop caused by the skin friction.

Using an electrical engineering analogy, fluid flow is similar to electric current,

pressure loss is equivalent to voltage difference and disturbance to the flow is

equivalent to electrical resistance. The term “hydraulic resistance” for such dis-

turbances to the flow is therefore well suited. To be consistent with Darcy pres-

3. Pressure Drop in Steady Internal Incompressible Viscous Flow 303

sure drop (Equation IIIb.3.7), ∆P due to fittings and valves is also expressed in

terms of specific kinetic energy:

ρρ



===∆ IIIb.3.11

where K is referred to as loss coefficient. Since combined molecular and turbulent

viscosity is the mechanism that converts mechanical work into heat, it is difficult

to find analytical solutions to the hydraulic resistance-induced pressure drops. For

this reason, each specific hydraulic resistance must be evaluated separately. Ex-

perimental data are obtained for various hydraulic resistance configurations.

These data are generally correlated in terms of dimensionless numbers to account

for the type of fluid (density and viscosity), flow properties (flow velocity), and

the hydraulic resistance geometry (diameter, thickness, etc.) These correlations

can be found in such handbooks as Idelchik, Lyons, and Crane.

Vena

Contracta

= 0.5[1 - (d/D)

]

min

= [1 - (d/D)

]

Figure IIIb.3.2. Loss coefficient associated with sudden change in flow area

As shown in Figure IIIb.3.2, a sudden expansion occurs in a flow when a pipe

is connected to a larger diameter pipe without gradual increase in diameter. This

results in flow separation downstream of the connection edge. In this case, the

loss coefficient is derived as (Example IIIb.3.6):

])/(1[K Dd

−=

This figure also shows a sudden contraction at a junction between two pipe sizes.

In a sudden contraction, flow separation in the downstream pipe causes the main

stream to contract through a minimum diameter, referred to as vena contracta.

The loss coefficient for sudden contraction is:

])/(1[5.0K

−=

For both cases of sudden expansion and contraction, the corresponding head loss

is found in terms of the velocity in the smaller pipe (i.e.,

2/Kh

= ).

304 IIIb. Fluid Mechanics: Incompressible Viscous Flow

Example IIIb.3.6. Derive the loss coefficient for a sudden expansion.

Solution: To find K for a sudden expansion, we write the steady-state form of the

momentum equation. To do this, we consider a control volume the extent of

which is from cross section 1 to cross section 2. The surface forces acting on this

control volume are pressure and shear forces. Hence, we can write:

1 j

)()(

221221

VAVAVVmFFFF

−=−=+−−



where F

= P

, F

= P

, F

is a force due to the interaction between the fluid

and the channel wall at the expansion plane j, acting on area A

– A

and F

is the

shear force. To reduce the number of unknowns, let’s assume that the friction

force in the short distance between planes 1 and 2 is negligible (i.e., F

§ 0).

Hence, the momentum equation simplifies to:

)()(

22122211

VAVAAAPAPAP

−=−+−

Still, we have more unknowns than equations. To get rid of P

, we note that pres-

sure at 1 is practically the same as pressure at cross section j due to the short dis-

tance between cross sections 1 and j, the constant flow area, and the zero shear

force assumption. Substituting for P

= P

we get:

)()(

22212

VAVAPPA −=−

Relating velocities by the continuity equation, V

= V

and substituting, we

get:

[

]

2221

/)/(/)( AAAAVAVAVAPP −=−=−

ρρ

(1)

We also note that Equation IIIa.3.30 gives the relation between P

– P

and head

loss with Z

= Z

and h

= 0:

gVVPP

h2/)(

221

ρρ

+−=−

(2)

We substitute for P

– P

from Equations (2) into Equation (1) and solve for h

()()

[

]

()

gVAAAA

2///21h

2121

+−=

(3)

By comparing with

2/Kh

we find K

≡

(1 – A

)

. Equation (3) is

known as the Borda-Carnot equation.

3. Pressure Drop in Steady Internal Incompressible Viscous Flow 305

In a sudden contraction, if the upstream diameter approaches infinity, the loss

coefficient is associated with “pipe entrance”. As shown in Figure IIIb.3.3, the

smoother the entrance to a pipe, the lower the loss coefficient. For example, in the

reentrant type, the pipe penetrates the flow field, causing perturbation of stream-

lines, some flow separation, and large head loss.

R = 0.2 d

Reentrant

K = 0.78

Sharp - Edged

K = 0.5

Slightly Rounded

K = 0.25

Well Rounded

K = 0.05

Figure IIIb.3.3. Loss coefficient associated with various pipe entrances

Table IIIb.3.2 (Crane) gives the friction factor for clean commercial steel pipe

for fully turbulent flow in terms of nominal pipe size. To find loss coefficient in

various valves and fittings (Table IIIb.3.3), we can either calculate the friction fac-

tor or use approximate value from Table IIIb.3.2 based on the pipe diameter.

Table IIIb.3.2. Turbulent flow friction factor for clean commercial steel pipe

Nominal Size (inch) 1/2” 3/4” 1” 1 ¼” 1 ½” 2” 2 ½” , 3”

Nominal Size (mm) 15 20 25 32 40 50 65 - 80

f 0.027 0.025 0.023 0.022 0.021 0.019 0.018

Nominal Size (inch) 4” 5” 6” 8” - 10” 12” - 16” 18” - 24”

Nominal Size (mm) 100 125 150 200 - 250 300 - 400 450 - 600

f 0.017 0.016 0.015 0.014 0.013 0.012

The loss coefficient (K) for a variety of valves and fittings is presented in Ta-

ble IIIb.3.3. In this table, the loss coefficient is given in terms of an equivalent

length, K = fL

/D. Dependence of K on D, for flow through valves and fittings, is

similar to the dependence of f on D, for flow in straight pipes. Hence, the L

term tends towards a constant value for various pipe sizes of a given type of fit-

ting. For example, consider flow through a fully opened gate valve installed on a

2 inch diameter pipe (K = 8 from Table IIIb.3.3). The L

associated with this

valve is found from L

/D = K/f = 8/0.019 = 421. Thus L

= 421D = 70 ft where f is

estimated from Table IIIb.3.2 as f = 0.019.

Pipe and tube data are provided in Tables A.III.1 through A.III.3. In summary,

total pressure drop is the summation of that given by Equations IIIb.3.7 and

IIIb.3.11:

306 IIIb. Fluid Mechanics: Incompressible Viscous Flow



+=∆

¦¦

IIIb.3.12

Equation IIIb.3.12 can alternatively be written as:

ρρρ





+=∆

IIIb.3.13

The frictional losses in fittings and valves are often referred to as form losses.

Note that we may write Equation IIIb.3.13 as:

∆



IIIb.3.14

where K = fL/D +

ΣK

where index i refers to various fittings and valves on the

piping system. As discussed in Chapter IIIc, Equation IIIb.3.14 may also be ap-

plied to compressible flow with some modifications.

Application and Various Types of Valves

Schematics of various control and check valves are shown in Figure IIIb.3.4. De-

sign details of a chemical, food processing, or power plant including the inter-

relationship of various systems and components are generally documented in so

called Piping & Instrumentation Diagrams (P&IDs). To simplify drafting and the

application of the P&IDs, symbols representing various components are devised.

For example, Figure IIIb.3.5 shows symbols used to represent various types of

valves. We can divide valves into two categories, flow control valves and pres-

sure control valves. We first discuss flow control valves.

Gate Valve

Butterfly Valve

Globe Valve

Ball Valve

Swing Check Valve Push Check Valve

Figure IIIb.3.4. Schematics of various valves

3. Pressure Drop in Steady Internal Incompressible Viscous Flow 307

A gate valve is generally used to isolate components. As such, a gate valve is

either fully open or fully closed. On the other hand, to throttle the flow, globe

valves are utilized. Expectedly, globe valves are associated with much higher-

pressure drop for the flow than open gate vales. Other flow control valves include

butterfly valves, ball valves, and needle valves. Directional valves include various

types of check valves. For example, in a tilting disc and a swing check valve, the

disc is readily lifted in the flow direction due to the flow momentum acting on the

disc. However, if flow reverses, the pressure exerted on the disc would keep the

valve tightly shut.

Safety valves (SVs), to control pressure, are generally spring loaded. If the

pressure exceeds the high-pressure set point, the valve will lift and is then reset

when pressure drops below the low-pressure set point. Pilot Operated Relief

Valves (PORVs) can be manually or remotely operated (lifted) to relieve pressure.

PORVs operate on a minimum and maximum pressure differential. In remotely

operated valves, the action of obstructing or allowing the flow of fluids can be ac-

complished by a driver to change the position of the gate, disc, plunger, etc.

Valves equipped with such drivers are generally known as motor operated valves

(MOVs). Depending on the type and size of a valve, the driver is an electric mo-

tor, an air operated (pneumatic) system, or electromagnetic (solenoids). For ex-

ample, PORVs are generally solenoid valves.

Table IIIb.3.3. Loss coefficient (K) in L

/D for valves and fittings

Category Item Loss Coefficient (K/f)

Elbow 90 Threaded 30

45 Threaded 16

90 Welded 14

45 Welded 10

Tee Line Flow 20

Branch Flow 60

Valve Gate (fully open) 8

Gate (75% open) 35

Gate (50% open) 160

Gate (25% open) 900

Swing check (fully open) 50

Lift check (fully open) 600

Globe (fully open) 340

Angle (fully open) 150

Ball (fully open) 3

Butterfly (fully open) 50

Valve flow rate is also defined in terms of the flow coefficient C

(also known

as valve sizing coefficient) being the ratio of the theoretical and the actual flow

rates. In applications however, C

for the flow of incompressible fluid is defined

as the flow rate of water in GPM at 60 F and at a pressure drop of 1 psi across a

valve (Crane). The metric equivalent of flow coefficient is flow factor K

, defined

as the flow rate of water in m

/h at 20 C, which results in a pressure drop of 1 bar.

308 IIIb. Fluid Mechanics: Incompressible Viscous Flow

Having the C

of a valve, flow rate through the valve at temperatures other than

60 F (or liquid other than water) and pressure drops other than 1 psi may be cal-

cualted from:

SPC /V ∆=



IIIb.3.15

where S

is the liquid specific gravity, ∆P is in psi, and the volumetric flow rate is

in GPM.

Example IIIb.3.7. A valve with a flow coefficient of 1000 results in a pressure

drop of 2 psi (13.8 kPa) across the valve for water flowing at 110 F (43 C). Find

the corresponding flow rate.

Solution: At 15 psia, we find

(60 F) = 62.4 lbm/ft

and

(110 F) = 61.88

lbm/ft

. The flow rate is found from Equation IIIb.3.15 as V



= 1000 ×

[2/(61.88/62.4)]

1/2

= 1420 GPM (89.6 lit/s).

The valve loss, or resistance coefficient, can be expressed as a function of the

valve C

and the pipe inside diameter by substituting for the definition of C

into

Equation IIIb.3.11, which results in

/891K

Cd= where pipe inside diameter d

is in inches. We may also find C

for a given K and d from C

= 29.84d

0.5

Example IIIb.3.8. A valve of C

= 500 is installed on a 5-inch pipe. Find the

valve loss coefficient.

Solution: The corresponding loss coefficient for this valve is found as

23.2500/)5(891/891K

2424

===

Example IIIb.3.9. Find the pressure drop over a half-open gate valve located on

a horizontal 6-inch pipe carrying SAE 10W oil at 92 F and 1000 GPM. (

= 0.025

in,

= 0.00018 ft

/s, and S

= 0.88).

Solution: In the absence of a pump and for steady flow in a horizontal pipe,

Equation IIIa.3.31 simplifies to Equation IIIb.3.12. For a half-open gate valve, K

= 160f. We find f from:

[][ ]

029.0)025.0/6log(214.1)/(log214.1

=+=+=

−−

Therefore, K = 0.029 × 160 = 4.64. Flow area is: A =

(6/12)

/4 = 0.196 ft

Flow rate is: V



= 1000/(60 × 7.481) = 2.228 ft

/s. Flow density is:

= 62.4 ×

0.88 = 54.9 lbm/ft

3. Pressure Drop in Steady Internal Incompressible Viscous Flow 309

Pressure drop becomes ∆P = 4.64× 54.9 (2.228)

/[2 × 32.2 × 0.196

] = 511 lbf/ft

= 3.5 psi.

Checking on Re to ensure flow is turbulent; Re = [2.228 × (6/12)]/[0.00018 ×

0.196) = 3.2E4 > 4000

Note that from Figure IIIb.3.1, for

/D = 0.004 and Re = 3.2E4, we find f § 0.03.

Diaphragm valve

Needle valve

Plug valve

Angle valve

Open Gate valve

Closed Gate valve

Open Globe valve

Closed Globe valve

Check valve

Safety/Relief valve

Control valve

Motor Operrated valve

Closed Ball valve

Open Ball valve

Closed Butterfly valve

Open Butterfly valve

Figure IIIb.3.5. Symbols for various valves

Hydraulic diameter (D

) is defined for channels having a flow area, different

than pipes and tubes. Since flow area and perimeter for a circle are given as A

flow

/4 and P =

D, respectively we conclude that A

flow

= PD/4. We use the same

concept of dividing four times flow area by the wetted perimeter for other conduits

to obtain an equivalent diameter D

= 4 A

flow

wetted

Example IIIb.3.10. Consider an annular conduit between two concentric cylin-

ders of diameters D

and D

. Find an equivalent diameter for this conduit.

Solution: As shown in the figure, the flow area and the wetted perimeter are

found as:

310 IIIb. Fluid Mechanics: Incompressible Viscous Flow

flow

)(

DD − /4 and P

wetted

+ D

). Therefore,

(

)

()

−=

−

Example IIIb.3.11. In a BWR, fuel rods are arranged in a square array within a

fuel assembly. Find the hydraulic diameter for an interior, side, and corner sub-

channel as shown in the figure.

Solid

boundary

Interior subchannel

1: Interior subchannel

Corner subchannel

2: Side subchannel

Solution: With respect to the interior subchannel, we note that the control surface

is bounded by the surface of the four surrounding fuel rods and the gap between

the rods. The wetted perimeter consists only of the solid surfaces (i.e., the four

quadrants of the four surrounding rods, as the gap is only an imaginary boundary).

These are summarized in the table where D is the fuel rod outside diameter and s

is the rod pitch.

No. Type Perimeter Flow Area Hydraulic diameter

1 Interior

–

/4 4[s

–

/4]/

2 Side (

D/2) + s [s

–

/4]/2 2[s

–

/4]/

[(

D/2) + s]

3 Corner (

D/4) + s [s

–

/4]/4 [s

–

/4]/

[(

D/4) + s]

where we have assumed the rod to wall distance to be half of the rod to rod pitch.

4. Steady Incompressible Viscous Flow in Piping Systems

In this section we study the internal flow of incompressible viscous fluids in pipe-

lines under steady-state conditions. We first begin with the flow analysis in a sin-

gle-path system, the most familiar example of which is pumping liquids from a

suction reservoir to a discharge reservoir, as shown in Figure IIIb.4.1. Gas and oil

pipelines are examples of single-path systems. The flow path may consist of any

number of valves, fittings, and instruments such as pressure gages, flowmeters,

temperature probes, etc.

4. Steady Incompressible Viscous Flow in Piping Systems 311

Suction Head

and P

are maintained

throughout the pumping process

Suction Lift

and P

are maintained

throughout the pumping process

Figure IIIb.4.1. Pumping liquid in a single path system

To analyze such systems, we may apply Equation IIIb.3.31 between points A and

B, located on the liquid surface of the suction and discharge reservoirs, respec-

tively. Alternatively, we may use Equation IIIa.3.44 in steady-state resulting in:

accvelfricgravstatpump

PPPPP

−

∆+∆+∆+∆=∆ IIIb.4.1

The definition of various differential pressure terms in Equation IIIb.4.1 is as fol-

lows:

ABstat

PPP −=∆

()

ABgrav

ZZgP −=∆

)K(

fric



+=∆



−=∆

]

[

]

[

acc

vel



ρρ

Writing the momentum equation in terms of differential pressures makes compre-

hension of the momentum equation intuitively simple. The left side of Equa-

tion IIIb.4.1 represents pressure increase in the flow while moving from pump

suction to pump discharge due to momentum transfer from the pump impeller to

the flow. In steady-state operation, this increase in pressure must overcome fric-

tional losses and provide for pumping the liquid to a higher elevation. The first

term on the right side,

∆P

stat

= P

– P

, is the difference in static pressure of point

A and point B. If pressure at the suction reservoir is higher than pressure in the

discharge reservoir, this term would assist the pump head and if there is no pump,

this term would provide the driving force.

The second term in the right side is pressure difference due to gravity (

∆P

grav

A significant portion of ∆P

pump

is used to lift liquid from point A and deliver it to