Pipeline Design

5

Pipeline Design

Pages

Fluid Flow in Steel Pipes

5.02 - 5.03

Pipeline Sizing - Pressure Loss

5.04 - 5.06

Fittings - Pressure Loss

5.07 - 5.09

Water Flow in Straight Pipes - Pressure Loss

5.10 - 5.19

Useful Pipe Properties

5.20

Valves - Pressure Loss

5.21 - 5.24

Compressible Fluids

5.25 - 5.27

Steam

5.28 - 5.29

Water Hammer

5.30 - 5.32

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GEORGEFISCHER…

5.01

Fluid Flow in Steel Pipes

and is measured in metres squared per second (m 2/s). µ ν= ρ

The flow of fluids is a complex process, the study of which is known as fluid dynamics. Fluid transport is affected by the physical properties of the fluid, the type of flow, the pipe dimensions and the properties of the pipe material. There are very few transport problems which can be completely solved by the purely mathematical equations of fluid dynamics. For everyday situations the solutions are dependent on experimentally determined factors, such as the friction factor. Most real problems can be solved using the Darcy formula, which relies on this experimental friction factor.

Density and Viscosity for Water

Physical Properties of Fluids

Velocity of Fluid

The properties relevant to fluid flow are summarized below.

The mean velocity of a fluid is given by:

Density: This is the mass per unit volume of the fluid and is generally measured in kg/m 3. Another commonly used term is specific gravity. This is in fact a relative density, comparing the density of a fluid at a given temperature to that of water at the same temperature. S=

ν= kinematic viscosity (m 2/s) µ = absolute viscosity (Pa s or Ns/m 2) ρ = density (kg/m 3)

Temperature °C

10 75 150

Density Absolute viscosity Pa s kg/m 3

1000 975 917

1.3 x 10 -3 0.4 x 10 -3 0.2 x 10 -3

Fig. 5.01 Extract from CIBSE Guide C4.3

v= Q A v = velocity of fluid (m/s) Q = volume flow rate (m 3/second) A = pipe cross sectional area (m 2)

ρ ρ water

S = specific gravity (dimensionless) ρ = density of fluid (kg/m 3) ρ water = density of water (kg/m 3) = 1000 at 10° C Viscosity: This describes the ease with which a fluid flows. A substance like treacle has a high viscosity, while water has a much lower value. Gases, such as air, have a still lower viscosity. The viscosity of a fluid can be described in two ways. a) Absolute (or dynamic) viscosity: This is a measure of a fluid's resistance to internal deformation. It is expressed in pascal seconds (Pa s) or newton seconds per square metre (Ns/m 2). [1Pas = 1 Ns/m 2] b) Kinematic viscosity: This is the ratio of the absolute viscosity to the density 5.02

GEORGEFISCHER…

Types of Fluid Flow

Reynolds Number

When a fluid moves through a pipe two distinct types of flow are possible, laminar and turbulent. Laminar flow occurs in fluids moving with small average velocities and turbulent flow becomes apparent as the velocity is increased above a critical velocity. In laminar flow the fluid particles move along the length of the pipe in a very orderly fashion, with little or no sideways motion across the width of the pipe. Turbulent flow is characterised by random, disorganised motion of the particles, from side to side across the pipe as well as along its length. There will, however, always be a layer of laminar flow at the pipe wall - the socalled 'boundary layer'.

A useful factor in determining which type of flow is involved is the Reynolds number. This is the ratio of the dynamic forces of mass flow to the shear resistance due to fluid viscosity and is given by the following formula. vd i ν Re = Reynolds number (dimensionless)

Re =

1

2

d i = pipe inside diameter (m) v = velocity of fluid (m/s) ν = kinematic viscosity (m 2/s)

3

In general for a fluid like water when the Reynolds number is less than 2000 the flow is laminar. The flow is turbulent for Reynolds numbers above 4000. In between these two values (2000
The two types of fluid flow are described by different sets of equations. In general, for most practical situations, the flow will be turbulent.

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5

d

v

6 Disturbed Turbulent Flow Re >2000 and Re <4000

Laminar Flow - Re < 2000

v 7 v 8

Disturbed Turbulent Flow - Re> 4000

Fig. 5.02 Velocity profiles for different types of flow.

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GEORGEFISCHER…

5.03

Pipeline Sizing Pressure Loss

b) changes in size and shape or direction of flow c) obstructions

Pipeline Sizing What size should the pipe be? The following formula can be used as a first approximation for a given flow rate. d = 1.13

Q v

where v = flow velocity (m/s) d = inside pipe diameter (m) Q = flow rate (m 2/s) Normally the flow velocity is unknown and must be approximated. The following are generally accepted design velocities. Liquids v = 1.0-3.0 m/s Gases v= 10-30 m/s

For normal, cylindrical straight pipes the major cause of pressure loss will be friction. Pressure loss in a fitting or valve is greater than in a straight pipe. When fluid flows in a straight pipe the flow pattern will be the same through out the pipe. In a valve or fitting changes in the flow pattern due to factors (b) and (c) will cause extra pressure drops. Pressure drops can be measured in a number of ways. The SI unit of pressure is the pascal. However pressure is often measured in bar. To convert from bar to pascals we use: 1 bar = 10 5Pa Another way of measuring pressure drop is in terms of head loss. To convert head loss to pascals we use : ∆p = Hρg

Example What size steel pipe should be used when the flow rate Q = 8 x 10 -3 m 3/s and the velocity v = 2m/s?

where ∆p = pressure drop in pascals ∆H = head loss in metres ρ = density of fluid (kg/m3)

d =1.13

8 x 10 2

-3

g = acceleration due to gravity = 9.81m/s 2

= 0.071m = 71mm In this case a 3'' (DN80) steel pipe can be used, since it ha a nominal bore of 80.9mm. The next size down is 2 1/2" (DN65) which has a bore of 68.7mm and would cause a reduction in the required flow rate to 7.45 x 10 -3m 3/s. If the flow rate may be slightly less than 8 x 10 -3 m 3/s, it would be more economical to use the smaller pipe. Pressure Loss Whenever fluid flows in a pipe there will be some loss of pressure due to several factors: a) friction- this is affected by the roughness of the inside surface of the pipe, the pipe diameter, and the physical properties of the fluid.

5.04

GEORGEFISCHER…

Fluid Flow in Straight Pipes The pressure drop due to friction is given by Darcy's formula: ∆H pi = 2fv 2 gd i ∆H pi = head loss (m/m) f = friction factor (dimensionless) v = fluid velocity (m/s) g = acceleration due to gravity= 9.81m/s 2 d i = inside pipe diameter (m) The friction factor (f) is a dimensionles constant which is dependent on the Reynolds number (Re) and the roughness of the pipe material. For turbulent flow (Re > 4000 for water) the friction factor has a very complex formula. However, a practical means of calculating the friction factor is provided by Moody's diagram shown in Fig. 5.03 . A value for the friction factor can be read directly from this Example Find the pressure loss per unit length for water flow at 3 l/s through 2 1/ 2" (DN65) medium grade BS 1387 steel pipe at 10°C. Pipe size 65mm Inside diameter d i = 68.7 x 10 -3 m Relative roughness 6.7x10 -4 See page 5.20 Useful Pipe Properties

Cross-sectional area A = π x (68.7x10 3m/2) 2m 2 Volume flow rate Q = 3x10 -3 m 3/s Absolute viscosity µ = 1.3Pa s x 10 -3 Density ρ = 1000kg/m 3 (both from Fig. 5.01)

Step 1 Calculate Reynolds number Re = vd i ν find flow velocity v = Q = volume flow rate A cross-sectional area = 3x10 -3 (π x (68.7x10 -3 ) 2 4 = 0.81m/s find kinematic viscosity GEORGEFISCHER…

diagram provided the Reynolds number and the relative roughness of the pipe material are known. Values for the relative roughness are given in the table on page 5.20. When considering steel pipes it is not only necessary to allow for pipe roughness, but also possible build up of scale and rusting. Pipe roughnesses and correction factors, used to account for these, are given in tables on page 5.20. Equivalent length: this is the length of pipe which will produce a frictional pressure loss of one velocity pressure. It is given by:

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2

3

le = di 4f l e = equivalent length (m) d i = inside pipe diameter (m) f = friction factor (dimensionless)

4

ν = µ = 1.3x10 -3 =1.3 x10 -6 m 2/s ρ 1000

5

Re = vd i = (0.81) x (0.0687) ν 1.3 x10 -6 = 4.3x10 4 Step 2 Use Moody diagram to determine the friction factor. For a Reynolds number of 4.3 x10 4 and a roughness of 6.7 x 10 -4 we find f = 0.006

6

Step 3 Find head loss in m/m

7

∆H pi = 2fv 2 =2 x (0.006) x (0.81) 2 g di (9.81) x (0.0687)

8

= 0.012 m/m = 12mm/m Step 4 If necessary change to pascals. ∆p =∆ H ρ g =(0.012) x (1000) x (9.81) = 118 Pa

9

So the pressure loss in this system is 12mm/m or 118 Pa/m 5.05

Moody’s Diagram

Fig. 5.03

5.06

GEORGEFISCHER…

Fittings - Pressure Loss There are two general types of fittings. Reducing (or expanding) fittings change the cross sectional area of flow, while deflecting fittings alter the direction of flow (e.g. bends and elbows). Couplings or unions offer no appreciable resistance to flow so may be considered as straight pipe. The pressure drop for a valve or fitting consists of two parts: 1. The pressure drop we would obtain for straight pipe of the same length as the fitting. 2. The pressure drop of the fitting itself. The pressure loss factor (ζ - value) of the fitting is determined experimentally and allows us to calculate the total pressure drop due to both of these factors.

Example 1 What is the ζ - value for a system of 1 inch (DN25) diameter with a malleable cast iron elbow and a diverging tee where the flow being considered is along the run of the tee?

1

Step 1 Find the ζ - value for the malleable cast iron elbow. From the table we see this is 0.8. ( Fig. 5.05)

2

Step 2 Find the ζ - value for the tee junction. The run is the important part so we have flow from A 1 to A 3 in Fig. 5.05 . So we read the basic ζ 3 value of 0.2. We do not need to add on a reduction or enlargement factor as the pipe bore does not change. Step 3 Find the total ζ - value

4

Finding the ζ - value

The ζ - value for all the fittings will be:

Fig. 5.05 gives the ζ value for several

ζ TOTAL = 0.8 + 0.2 = 1.0

fittings. For the elbows and bends, the ζ value can be easily read directly from the table. For example a 90° elbow 1 1/ 4 (DN32) has ζ= 0.7 However the situation may be more complex. In some pipe systems we may have a reduction or enlargement in pipe size. This change in size has a ζ - value which depends on the ratio of the two pipe areas (see example below). Tees and junctions have ζ - values which vary depending on which branch is being considered. Once the basic ζ value for the relevant branch has been determined extra factors need to be added on. The first is for the elbow or bend (if the flow is through the straight section of a tee then this is zero). Finally any reduction or enlargement in pipe cross sectional area must be accounted for.

Example 2 What is the ζ - value for the following system? Diverging junction (DN 32)

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1" (DN 25) pipe

Reducer

6 1 1/4 " (DN 32) pipe

Fig. 5.04

The diverging junction has the effect of a 90° elbow.

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Step 1 Find the basic ζ - value for a diverging junction. The table gives ζ 2 = 0.5 (Fig. 5.05) Step 2 Find the contribution due to the 90° elbow. From the elbows and bends table we see for a 32mm 90° elbow we have ζ = 0.7. Step 3 Find the contribution to ζ due to the reduction from 32mm pipe to 25mm pipe.

GEORGEFISCHER…

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5.07

First we find the ratio of the two cross sectional areas. cross sectional area = π x (radius)2 A 1 = π x (32) 2 4

A 2 = π x (25) 2 4

L ef = ζ d i 4f L ef = equivalent length of fitting (m) ζ = pressure loss factor (dimensionless)

A 2 = 0.6 A1

d i = internal pipe diameter (m)

The value corresponding to A 2 = 0.6 A1

For an entire pipe system, the equivalent length of all the fittings is added to the actual length of pipe to give a total effective length. The total pressure loss is then equal to this effective length multiplied by the pressure loss per unit length for the appropriate material and pipe diameter (see Figs. 5.08-5.10) .

is ζ = 0.25. Step 4 Find the total ζ by adding together contributions from the previous three steps.

f = friction factor (dimensionless)

ζ TOTAL = 0.5 + 0.7 + 0.25 = 1.45

Calculating pressure loss Once we have calculated the ζ value we can find the pressure loss in two ways. Method 1 Here we use the following formula. ∆p fi = ζ ρ v 2 2 ∆p fi = pressure change in fittings (Pa) ζ = pressure loss factor (dimensionless) ρ = density (kg / m 3) v = flow velocity (m/s) The flow velocity for a given volume flow rate and pipe dimension can be found in the water flow pressure loss tables. The pressure loss for the valves and fittings is simply added to the pressure loss for the straight pipe in a system to find the total pressure loss. Method 2 Equivalent length The equivalent length of a fitting is the length of pipe, of the same material and diameter as the fitting, which would give the same pressure loss as the fitting itself. It can be calculated as follows.

5.08

GEORGEFISCHER…

Pressure Loss Factors for Pipe Fittings (based on velocity pressure of combined flow. Factors refer to the branch indicated by the subscript, e.g. ζ 2 is for flow from branch 2).

Tees and Junctions ζ2

Diverging

ζ3

ζ2

Diverging

0.5

0.5

+

Converging

(

+

)

factor for bend or elbow as appropriate +

(

1

)

factor for enlargement or reduction where bores differ

0.2

(

+

(

)

factor for enlargement or reduction where bores differ

Converging

factor for bend or elbow as appropriate

2

)

+

(

factor for enlargement or reduction where bores differ

Reductions

Enlargements

Based on velocity pressure in smaller pipe

Based on velocity pressure in smaller pipe

)

3

A2 A1

ζ

A2 A1

ζ

A1 A2

ζ

A1 A2

ζ

0.1 0.2 0.3

0.55 0.50 0.45

0.4 0.6 0.8

0.40 0.25 0.05

0.1 0.2 0.3

0.80 0.65 0.50

0.4 0.6 0.8

0.35 0.15 0.05

Elbows and Bends

5

DN

Type

10-25mm 32-50mm

65-90mm

>100mm

0.6

0.6

Malleable Cast Iron 90° Elbow

0.8

Malleable Cast Iron 45° Elbow

0.6

0.6

0.5

0.5

Malleable Cast Iron Bend

0.7

0.5

0.4

0.4

Malleable Cast Iron Return Bend

0.9

0.8

0.8

–

0.7

4

6

7

8

Fig. 5.05 Extract from CIBSE Guide, Table 4.36.

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GEORGEFISCHER…

5.09

Water Flow in Straight Pipes - Pressure Loss The following tables (Figs. 5.08-5.10) relate a pressure loss per unit length (in pascals) to the volume flow rate in the pipe. The correction tables which precede them allow compensation for rust and a higher temperature. The value read from the flow tables is multiplied by the appropriate factor from the correction tables.

Example The pressure drop for water at 10°C, flowing at 1 x 10 -3 m 3/s through heavy black 1 1/ 2 " (DN40) pipe, can be read from Fig. 5.08 as 220 Pa/m. However if this pipe were rusted we would need to apply a correction factor. As the nominal pipe size required is not actually listed in Fig. 5.06 we use the next size up, 2" (DN 50). Our pressure loss is about 200 Pa so we read off a correction factor of 3.3. We multiply our original pressure loss by this factor to find the loss for rusted pipes. 220 x 3.3 gives us 726 Pa/m. For new pipes the pressure loss is 220 Pa/m, but for rusted pipes it would be 726 Pa/m.

Correction factors for rusted steel pipes Nominal pipe size mm inches 1 15 /2 25 1 50 2 100 4

2 1.0 2.4 2.3 2.3

5 3.2 2.7 2.5 2.4

Pressure loss as read from tables 10 20 50 100 200 3.2 3.5 3.8 4.1 4.3 2.9 3.1 3.4 3.6 3.7 2.7 2.9 3.1 3.2 3.3 2.5 2.7 2.8 2.9 2.9

500 1000 4.5 4.7 3.9 4.0 3.4 3.5 3.1 3.1

Fig. 5.06

Correction factors for water at 150°C Nominal pipe size mm inches 1 15 /2 25 1 50 2 100 4

2 1.00 0.90 0.90 0.95

5 1.00 0.90 0.92 0.96

Pressure loss as read from tables 10 20 50 100 200 0.91 0.95 0.96 0.97 0.99 0.92 0.93 0.95 0.97 0.99 0.94 0.96 0.98 1.0 1.0 0.97 0.99 1.0 1.0 1.0

500 1000 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Fig. 5.07

5.10

GEORGEFISCHER…

Pressure loss in steel pipes The following pages (5.12-5.17) tabulate the pressure loss data for water flowing in different grades of steel pipe at either 10°C or 75°C, for a range of flow rates.

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GEORGEFISCHER…

5.11

Pressure loss for water flow at 10°C in steel pipes with velocity of flow (v), for heavy grade (H) and galvanised (G) steel

Volume flow rate l/s 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.50 2.00 2.50 3.00 3.50

Pressure loss per unit length (Pa/m) and velocity of flow (v) 20 mm 25 mm 32 mm 3 / 4 in 1 in 1 1 / 4 in

15mm 1 / 2 in H 140 420 880 1500

G v 175 600 1500 2250 3500 5000 7000 9000 11000

0.28 0.57 0.85 1.13 1.42 1.70 1.98 2.26 2.55

H 25.0 92.5 200 320 480 660 880 1200 1400 1700

G v 50 120 250 450 700 1000 1250 1750 2000 2750

0.16 0.32 0.48 0.64 0.80 0.95 1.11 1.27 1.43 1.59

3000 3500 4500 5000 5500 6500 7000 8000 9000 10000

1.75 1.91 2.07 2.23 2.39 2.56 2.71 2.87 3.03 3.18

H 6.5 32.5 62.5 120 160 220 280 360 460 540 640 760 880 1000 1200 1300 1500 1600 1800

G v

H

G

40 mm 1 1 / 2 in

v

H

G v 0.04 0.08 0.12 0.16 0.20 0.24 0.28 50 0.32 60 0.36 80 0.40

50 80 130 200 275 400 500 600 800

0.1 0.2 0.3 0.41 0.51 0.61 0.71 0.82 0.92 1.02

2.0 8.0 17.5 27.5 40.0 50 52.5 70 70.0 90 87.5 110 120 140 140 175

0.8 2.5 0.19 0.25 0.31 0.37 0.44 0.50 0.56 0.62

1.5 3.5 7.5 12.5 20.0 25.0 32.5 42.5 50.0 62.5

900 1100 1250 1500 1750 2000 2150 2250 2500 2750

1.12 1.22 1.32 1.43 1.53 1.63 1.73 1.83 1.94 2.04

160 200 220 240 280 320 360 380 420 460

200 225 275 320 350 400 450 500 550 600

0.68 0.75 0.81 0.87 0.93 1.00 1.06 1.12 1.18 1.24

72.5 85.0 97.5 120 140 150 160 180 200 220

7000 11000

3.06 4.08

980 1500 1700 2250 3500 5000 7000

1.87 2.49 3.11 3.73 4.35

460 760 1200 1700

90 100 120 140 160 175 200 225 250 275

0.44 0.48 0.52 0.56 0.60 0.64 0.68 0.72 0.76 0.80

600 1000 1750 2250 3000

1.19 1.60 1.99 2.39 2.79

Fig. 5.08 Data extracted and re-arranged (uses volume flow rate rather than mass flow rate) From CIBSE Guide, tables C4-17, C21.

5.12

GEORGEFISCHER…

Pressure Loss for water flow at 10°C in steel pipes With velocity of flow (v), for heavy grade (H) and galvanised (G) steel.

Volume flow rate l/s

1

Pressure loss per unit length (Pa/m) and velocity of flow (v) 50 mm 2 in H G

0.1 0.2 0.3 0.4 0.5

0.8 4.0 8.0 15.0 20.0

0.6 0.7 0.8 0.9 1.0

27.5 35.0 45.0 55.0 65.0

50.0 55.0 70.0 80.0

v

65 mm 2 1/ 2 in H G

v

80 mm 3 in H G

0.05 0.10 0.15 0.20 0.25

v

100 mm 4 in H G

v

0.3 1.5 2.5 4.0 5.5

0.03 0.06 0.09 0.12 0.15

0.2 0.5 1.0 2.0 2.5

0.02 0.04 0.06 0.08 0.10

0.1 0.3 0.5 0.7

0.03 0.04 0.05 0.06

0.31 0.36 0.41 0.46 0.51

7.5 9.5 12.5 15.0 20.0

0.18 0.21 0.24 0.27 0.3

3.5 4.5 5.5 7.0 8.0

0.12 0.14 0.16 0.18 0.20

1.0 1.3 1.5 2.0 2.5

0.08 0.09 0.10 0.12 0.13

4.5 7.5 12.5 17.5 22.5 27.5 32.5 40.0

50.0

0.19 0.25 0.32 0.38 0.45 0.51 0.57 0.64

2

1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

140 240 340 480 640 820 1100 1300

175 300 450 700 900 1250 1500 1750

0.76 1.02 1.27 1.53 1.78 2.04 2.29 2.55

37.5 62.5 92.5 140.0 180.0 220.0 280.0 340.0

50.0 80.0 110.0 175.0 225.0 275.0 350.0 450.0

0.45 0.60 0.75 0.90 1.06 1.21 1.36 1.51

17.5 27.5 42.5 57.5 77.5 97.5 120.0 160.0

50.0 70.0 90.0 120.0 150.0 175.0

0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10

1500 1800

2250 2500 3000 3500 4000 4500 5000 5500 6000 7000

2.80 3.06 3.31 3.57 3.82 4.08 4.33 4.59 4.84 5.10

400.0 480.0 540.0 620.0 720.0 800.0 900.0 1000.0 1200.0 1300.0

500.0 600.0 700.0 800.0 1000.0 1100.0 1250.0 1350.0 1500.0 1750.0

1.66 1.81 1.96 2.11 2.26 2.41 2.56 2.71 2.87 3.02

180.0 220.0 240.0 280.0 320.0 360.0 400.0 440.0 500.0 540.0

225.0 250.0 300.0 350.0 400.0 450.0 500.0 550.0 650.0 700.0

1.10 1.20 1.29 1.39 1.49 1.59 1.69 1.79 1.89 1.99

47.0 55.0 65.0 72.5 82.5 92.5 110.0 120.0 130.0 140.0

55.0 70.0 80.0 90.0 100.0 110.0 130.0 140.0 160.0 175.0

0.70 0.76 0.83 0.89 0.96 1.02 1.08 1.15 1.21 1.27

8000

5.61

1500.0 1800.0

2000.0 2250.0 2750.0 3200.0 3500.0 4000.0 4500.0 5500.0 6000.0

3.32 3.62 3.92 4.22 4.52 4.82 5.13 5.43 5.73

640.0 760.0 880.0 1000.0 1200.0 1300.0 1500.0 1700.0 1800.0

800.0 1000.0 1250.0 1350.0 1500.0 1750.0 2000.0 2250.0 2500.0

2.19 2.39 2.59 2.79 2.99 3.18 3.38 3.58 3.78

180.0 200.0 240.0 280.0 300.0 340.0 380.0 420.0 480.0

200.0 250.0 300.0 350.0 400.0 450.0 500.0 550.0 600.0

1.40 1.53 1.66 1.78 1.91 2.04 2.17 2.29 2.42

11 12 13 14 15 16 17 18 19

Fig. 5.08 (contd).

GEORGEFISCHER…

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5.13

Pressure loss for water flow at 75°C in steel pipes with velocity of flow (v), for heavy grade (H) and medium grade (M) steel

Volume flow rate litres/sec 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.50 2.00 2.50 3.00 3.50

Pressure loss per unit length (Pa/m) and velocity of flow (v) 15mm 1 / 2 in H 90 360 760 1300 2000

20 mm 3 / 4 in M

25 mm 1 in

M

v

H

v

H

65 240 500 880 1400 1900

0.28 0.57 0.85 1.13 1.42 1.70

20 15 70 55 160 120 260 200 400 300 560 420 760 560 960 720 1200 900 1500 1100

0.16 0.32 0.48 0.64 0.80 0.95 1.11 1.27 1.43 1.59

6.5 22.5 47.5 82.5 140 180 240 300 380 460

1900 1400 1600 1900

1.75 1.91 2.07 2.23 2.39 2.56 2.71 2.87 3.03 3.18

540 660 780 900 1000 1200 1300 1500 1600 1900

M

32 mm 1 1/ 4 in

40 mm 1 1/ 2 in

v

H

M

v

5 20 35 60 90 140 180 220 300 340

0.1 0.2 0.3 0.41 0.51 0.61 0.71 0.82 0.92 1.02

2.0 5.5 12.5 20 30 42.5 55 70 85 120

1.5 5 9 17 25 35 45 55 70 85

0.06 0.12 0.19 0.25 0.31 0.37 0.44 0.50 0.56 0.62

400 480 560 660 740 840 980 1100 1200 1300

1.12 1.22 1.32 1.43 1.53 1.63 1.73 1.83 1.94 2.04

140 160 180 200 240 260 300 320 360 400

100 120 140 160 200 220 240 260 280 320

0.68 0.75 0.81 0.87 0.93 1.00 1.06 1.12 1.18 1.24

840 680 1500 1200 1800

1.87 2.49 3.11

H 0.8 2.5 5.5 9.0 15 20 25 35 40 50 60 65 80 90 100 120 140 160 180 200

M

v

0.7 2.5 4.5 7.5 12.5 17.5 22.5 27.5 35 40

0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.32 0.36 0.40

50 55 67.5 75 85 97.5 120 130 140 160

0.44 0.48 0.52 0.56 0.60 0.64 0.68 0.72 0.76 0.80

320 540 840 1200 1600

1.19 1.60 1.99 2.39 2.79

Fig. 5.09 Data extracted and re-arranged (uses volume flow rate rather than mass flow rate) From CIBSE Guide, tables C4-11, C4-12.

5.14

GEORGEFISCHER…

Pressure loss for water flow at 75°C in steel pipe with velocity flow (v ) for heavy grade (H) and medium grade (M) steel

Volume flow rate l/s

H

1

Pressure loss per unit length (Pa/m) and velocity of flow (v) 50 mm 65 mm 80 mm 2 in 2 1/ 2 in 3 in M v H M v H M v H

100 mm 4 in M v

2

0.1 0.2 0.3 0.4 0.5

0.8 3.0 6.0 9.0 15.0

0.7 2.5 5.0 8.5 12.5

0.05 0.10 0.15 0.20 0.25

0.3 0.8 2.0 3.0 1.0

0.2 0.7 1.5 2.5 3.5

0.03 0.06 0.09 0.12 0.15

0.2 0.4 0.7 1.5 2.0

0.1 0.4 0.7 2.5 3.5

0.02 0.04 0.06 0.08 0.10

0.1 0.2 0.4 0.5

0.1 0.2 0.3 0.5

0.03 0.04 0.05 0.06

0.6 0.7 0.8 0.9 1.0

20.0 27.5 35.0 42.5 55.0

17.5 25.0 30.0 37.5 45.0

0.31 0.36 0.41 0.46 0.51

5.5 7.5 9.5 12.5 15.0

5.0 6.5 8.0 10.0 12.5

0.18 0.21 0.24 0.27 0.3

2.5 3.5 4.5 5.0 6.5

2.0 3.0 4.0 5.0 5.5

0.12 0.14 0.16 0.18 0.20

0.7 0.9 1.2 1.5 2.0

0.7 0.8 1.4 1.5 2.0

0.08 0.09 0.10 0.12 0.13

1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

120 200 300 420 580 740 920 1200

95 180 260 360 480 640 780 960

0.76 1.02 1.27 1.53 1.78 2.04 2.29 2.55

30.0 50.0 77.5 120.0 160.0 200.0 240.0 300.0

27.5 45.0 67.5 95.0 140.0 180.0 220.0 260.0

0.45 0.60 0.75 0.90 1.06 1.21 1.36 1.51

15.0 22.5 35.0 47.5 62.5 82.5 120.0 140.0

12.5 20.0 30.0 42.5 57.5 75.0 92.5 120.0

0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

3.5 6.0 9.0 12.5 17.5 22.5 27.5 32.5

3.5 5.5 8.5 12.5 17.5 20.0 25.0 30.0

0.19 0.25 0.32 0.38 0.45 0.51 0.57 0.64

5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10

1400 1700

1200 1400

2.80 3.06 3.31 3.57 3.82 4.08 4.33 4.59 4.84 5.10

360.0 420.0 480.0 560.0 640.0 720.0 820.0 920.0 1200.0

320.0 360.0 440.0 500.0 560.0 640.0 720.0 820.0 880.0 980.0

1.66 1.81 1.96 2.11 2.26 2.41 2.56 2.71 2.87 3.02

160.0 180.0 220.0 240.0 280.0 320.0 360.0 400.0 440.0 480.0

140.0 160.0 20.0 220.0 260.0 300.0 320.0 360.0 400.0 440.0

1.10 1.20 1.29 1.39 1.49 1.59 1.69 1.79 1.89 1.99

40.0 47.5 55.0 62.5 70.0 80.0 90.0 100.0 120.0 130.0

37.5 42.5 50.0 57.5 65.0 72.5 82.5 90.0 100.0 120.0

0.70 0.76 0.83 0.89 0.96 1.02 1.08 1.15 1.21 1.27

1200.0 1400.0 1700.0 1900.0

3.32 3.62 3.92 4.22

580.0 680.0 820.0 920.0 1100.0 1200.0 1400.0 1500.0 1700.0

520.0 600.0 720.0 820.0 940.0 1100.0 1200.0 1400.0 1500.0

2.19 2.39 2.59 2.79 2.99 3.18 3.38 3.58 3.78

150.0 160.0 200.0 240.0 280.0 300.0 340.0 380.0 420.0

140.0 160.0 200.0 220.0 260.0 280.0 320.0 360.0 400.0

1.40 1.53 1.66 1.78 1.91 2.04 2.17 2.29 2.42

11 12 13 14 15 16 17 18 19

5.61 6.12

3

4

5

6

7

8

Fig. 5.09 (contd.)

9

GEORGEFISCHER…

5.15

Pressure loss for water flow at 75°C in steel pipe with velocity flow (v), for galvanised (G) steel

Volume flow rate litres/second 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0

Pressure loss per unit length (Pa/m) and velocity of flow (v) 15mm 1 / 2in G 140 550 1250 2250 3500 5000 7000 9000 11000

v 0.28 0.57 0.85 1.13 1.42 1.70 1.98 2.26 2.55

20mm 3 / 4 in G v

25mm 1 in G

v

32mm 1 1/ 4 in G v

40mm 1 1/ 2 in G v

50 100 225 400 600 900 1250 1500 2000 2250

0.16 0.32 0.48 0.64 0.80 0.95 1.11 1.27 1.43 1.49

50 70 110 175 250 350 450 550 700

0.2 0.3 0.41 0.51 0.61 0.71 0.82 0.92 1.02

50 55 70 100 120 140

0.31 0.37 0.44 0.50 0.56 0.62

50 70

0.36 0.40

2750 3500 4000 4500 5000 6000 7000 7500 8000 9000

1.75 1.91 2.07 2.23 2.39 2.56 2.71 2.87 3.03 3.18

800 1000 1250 1350 1500 1750 2000 2250 2500 2750

1.12 1.22 1.32 1.43 1.53 1.63 1.73 1.83 1.94 2.04

175 200 250 275 325 350 400 450 500 550

0.68 0.75 0.81 0.87 0.93 1.00 1.06 1.12 1.18 1.24

80 90 110 120 140 165 175 200 225 250

0.44 0.48 0.52 0.56 0.60 0.64 0.68 0.72 0.76 0.80

6000 11000

3.06 4.08

1250 2250 3500 5000 7000 9000 11000

1.87 2.49 3.11 3.73 4.35 4.98 5.60

550 1000 1500 2250 3000 4000 5000 6000 7000 9000 10000 11000

1.19 1.60 1.99 2.39 2.79 3.18 3.58 3.98 4.38 4.78 5.18 5.57

Fig. 5.10 Data extracted and re-arranged (uses volume flow rate rather than mass flow rate) From CIBSE Guide, tables C4-16.

5.16

GEORGEFISCHER…

Pressure loss for water flow at 75°C in steel pipes with velocity of flow (v), for galvanised (G) steel

Volume flow rate litre/ second

1

Pressure loss per unit length (Pa/m) and velocity of flow (v)

G

50mm 2 in v

65mm 2 1/ 2 in G v

80mm 3 in

0.8 0.9 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

50 55 70 150 275 400 600 800 1000 1500 1750

0.41 0.46 0.51 0.76 1.02 1.27 1.53 1.78 2.04 2.29 2.55

50 70 100 140 200 250 350 400

0.45 0.60 0.75 0.90 1.06 1.21 1.36 1.51

50 60 80 110 130 175

0.50 0.60 0.70 0.80 0.90 1.00

5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

2000 2250 2750 3250 3500 4000 4500 5000 6000 6500

2.80 3.06 3.31 3.57 3.82 4.08 4.33 4.59 4.84 5.10

500 550 700 800 900 1000 1100 1250 1350 1500

1.66 1.81 1.96 2.11 2.26 2.41 2.56 2.71 2.87 3.02

200 250 275 350 400 450 500 550 600 700

1.10 1.20 1.29 1.39 1.49 1.59 1.69 1.79 1.89 1.99

50 60 70 80 90 100 120 130 140 160

0.70 0.76 0.83 0.89 0.96 1.02 1.08 1.15 1.21 1.27

11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0

8000

5.61

1850 2250 3750 3000 3500 4000 4500 5000 5500 6000

3.32 3.62 3.92 4.22 4.52 4.82 5.13 5.43 5.73 6.03

800 900 1150 1250 1500 1750 2000 2100 2250 2500

2.19 2.39 2.59 2.79 3.00 3.18 3.38 3.58 3.78 4.00

200 225 275 300 350 400 450 500 550 700

1.40 1.53 1.66 1.78 1.91 2.04 2.17 2.29 2.42 2.55

G

v

100mm 4 in G v

2

3

4

5

6

7

Fig. 5.10 (contd.)

8

9

GEORGEFISCHER…

5.17

Example 1 What is the pressure loss for the system below, made of heavy grade steel pipe, when water flows at 10°C? The flow rate is 1litre/second. Reducer 1 1/4 " (DN 32)

2" (DN 50) pipe

1m

90° Diverging Junction

45° Elbow

/"

1 4

1

Fig 41

Fig 240

(D

Fig 130

N

v= Q = A

flow rate (m 3/s) cross-sectional area (m 2)

for 2" pipe

for 1 1/4" pipe

1m

the diverging junction. We read its basic ζ - value from table 2 as 0.5. Then we add a factor for a 90° elbow which is 0.7 So we have ζ TOTAL = 0.6 + 0.7 = 1.3 The pressure loss is calculated from : ∆p = ζ ρ v 2 2

1x10 -3 π x (57.3 x10 -3 ) 2 4 = 0.39m/s

v=

1x10 -3 π x (34.3 x10 -3 ) 2 4 = 1.08m/s

v=

Pipe losses The total amount of straight pipe in the system is as follows: 3m of 2" pipe 2m of 1 1/4" pipe From Fig. 5.08 we read off the following values for pressure loss per metre: 2" pipe 65Pa/m 460 Pa/m 1 1/4" pipe So the total pressue loss due to straight pipe is ∆p pi = 3x65 + 2x460 = 1115 Pa Fitting losses We need to split the fitting losses into two sections, one for the 2" pipe fittings and the other for the1 1/4" pipe fittings. Reductions or enlargements, connecting pipes of different diameters, are included in the section for the smaller pipe. So here the reduction is included in the 1 1/4" fittings.

)

The internal pipe diameters are (pipe diameter Fig. 5.14 ) For 2" pipe d i = 51.3 x 10 -3 m 1 1 /4 " pipe d i = 34.4 x 10 -3 m First calculate the flow velocities.

32

3m

Fig. 5.11

= 1.3 x (1000 )x (0.39) 2 2 = 99 Pa 1 1/4" Fittings Here we have a reduction and a 45° elbow to consider. For the 45° elbow we have ζ = 0.6. For the reduction we need to find A 2/A 1 A1 = A2

π x (34.4x10 -3) 2 4 = 0.45 π x (51.3x10 -3) 2 4

This is between the values for ζ=0.40 and ζ=0.25, so we take the larger value. ζ TOTAL = 0.6 + 0.4 =1.0 The pressure loss here is ∆p = ζ ρ v 2 = 1.0 x (1000) x (1.08) 2 2 2 = 583 Pa So the total pressure loss due to fittings is ∆p fi = 583 +99 = 682 Pa Total Loss The total pressure loss will be ∆p pi + ∆p fi = 1115 + 682 = 1797 Pa The total pressure loss is 1.8 kPa

2" Pipe Fittings The only fitting to be considered here is 5.18

GEORGEFISCHER…

Example 2

So

Consider part of a system of 2 1/2" (DN65) heavy grade steel pipe as shown below. Water is flowing at a rate of 1 litre per second at 10°C.

Re = 0.28 x (0.067) 1.3x10 -6

1

= 14430 The roughness is given in Fig. 5.13 as 6.9x10 -4 and we combine this with our Reynolds number in the Moody diagram to find the friction factor (f).

5m

2

f = 0.0075 5m

l ef = 2.4 x 0.067 4 x 0.0075

Fig 90

5m

5m

So the equivalent length is:

5m

Fig. 5.12

Let's approach this problem from the perspective of equivalent lengths. Pipe: There is 25m of straight pipe. Fittings: There are four 90° elbow fittings 2 1/2" (DN65), which each have a ζ of 0.6. So, ζ TOTAL = 2.4 To find the equivalent length of these fittings we use

3

= 5.36 Now we add this equivalent length to the real length of straight pipe to get an effective length. Effective length = 25m + 5.36m = 30.36m The pressure loss per unit length for 65mm pipe with a flow rate of 1l/s is 20Pa/m. So we multiply this value by our effective length to get a total pressure loss.

4

5

∆p TOTAL = 20•30.36 = 607.2 Pa So we lose 607.2 Pa of pressure in this part of the system.

l ef = ζd i 4f

6

We need to find the Reynolds number to use this formula. Re = vd i ν

7

The flow velocity for 2 1/2" (DN65) pipe with a flow rate of 1 litre per second is found below. v = Q A

=

flow rate (m 3/s) cross-sectional area (m 2)

=

1 x 10 -3 π x (67 x 10 -3) 2 4

8

= 0.28 m/s The internal diameter is 67mm from Fig. 5.14 and the kinematic viscosity (ν) is 1.3x10 -6 m 2/s.

GEORGEFISCHER…

9

5.19

Useful Pipe Properties Relative Roughness Fig. 5.13 Extract from CIBSE Guide Table C4.5.

Nominal pipe size mm 10 15 20 25 32 40 50 65 80 100

inches 3

/8 1 /2 3 /4 1 1 1/ 4 1 1/ 2 2 2 1/ 2 3 4

Relative roughness Mild steel BS 1387 Medium black

Heavy black

Heavy Galvanised

3.7x10 -3 2.9 x 10 -3 2.1x10 -3 1.7x10 -3 1.3x10 -3 1.1x10 -3 8.7x10 -4 6.7x10 -4 5.7x10 -4 4.4x10 -4

4.1x10 -3 3.1 x 10 -3 2.2x10 -3 1.8x10 -3 1.3x10 -3 1.1x10 -3 9.0x10 -4 6.9x10 -4 5.8x10 -4 4.5x10 -4

1.4x10 -2 1.4 x 10 -2 7.5x10 -3 5.9x10 -3 4.4x10 -3 3.8x10 -3 2.9x10 -3 2.3x10 -3 1.9x10 -3 1.5x10 -3

Internal Diameters of Pipes Fig. 5.14 Extract from CIBSE Guide Table C4.4.

Nominal pipe size mm 10 15 20 25 32 40 50 65 80 100

inches 3

/8 /2 3 /4 1 1 1 /4 1 1/ 2 2 1 2 /2 3 4 1

Mean internal diameter/mm Mild Steel BS 1387 Medium black

Heavy black

12.4 16.1 21.6 27.3 36.0 41.9 53.0 68.7 80.7 105.1

11.3 14.9 20.4 25.7 34.4 40.3 51.3 67.0 79.1 103.5

Heavy Galvanised 10.8 14.4 19.9 25.2 33.9 39.8 50.8 66.5 78.6 102.8

Pipe Wall Thicknesses Fig. 5.15

5.20

Nominal pipe size mm

inches

10 15 20 25 32 40 50 65 80 100

3

/8 /2 3 /4 1 1 1 /4 1 1/ 2 2 1 2 /2 3 4 1

Wall Thickness/mm Mild Steel BS 1387 Medium black Heavy black 2.3 2.6 2.6 3.2 3.2 3.2 3.6 3.6 4.0 4.5

2.9 3.2 3.2 4.0 4.0 4.0 4.5 4.5 5.0 5.4

GEORGEFISCHER…

Valves - Pressure Loss Flow rate/ Flow factor The flow value or k v factor is a convenient means of calculating flow rates in hydraulics. It allows for all internal resistances and for practical purposes is regarded as reliable.

where ν = kinematic viscosity (centistokes) k v = flow factor for water (dimensionless) Q = flow rate (litres/minute)

1

Gases ∆p x p2

Qg = 30.8 kv

2

ρxT The k v factor is defined as the flow rate of water in litres per minute with a pressure drop of 1 kg/cm 2 across the valve. The relationships between k v factor, flow rate (Q) and pressure drop (∆p) are given in the following formulae. Liquids with kinematic viscosity less than 22 centistokes (22 x 10 -6 m 2/s) e.g. water, hydraulic oil kv = Q

ρ ∆p

or

Q = kv

∆p ρ

or ∆p = ρ x Q 2 k v2 where Q = flow rate (litres per minute) ρ = density of the liquid (kg/dm 3) ∆p = pressure drop (kg/cm 2 )

valid for ∆p <

The correction factor is given by: kv 200 x Q

only

4

5

Where there are several flow factors in series, the resultant k v factor is 1

=

1 k v1 2

+

1 k v2 2

+

1 k v3 2

+ …+

1

6

k vn 2

Where the flow factors are in parallel, the resultant k v factor is

7

k vx = k v1 + k v2 + k v3 + …… + k vn All references are in metric (k v) units. For imperial (f) or american (cv) units the following conversions may be used:

valid for c ≤ 3 only

Flow Value Conversion Table kv Cv f kv Cv f

1 14.28 17.09 0.07 1 1.1966 0.0585 0.8357 1

Fig. 5.16

GEORGEFISCHER…

3

8

k vn = k v x c

c=1 + ν

2

where Q g = flow rate (m 3 / hour) k v = flow factor for water (dimensionless) ∆p = pressure drop (kg / cm 2) = p 1-p 2 p 1 = absolute inlet pressure (bar) p 2 = absolute outlet pressure (bar) ρ = density of gas at 0°C (kg/m 3) T = absolute temperature (Kelvin) = 273 + t t = temperature in Celsius (°C)

k vx2 Liquids with kinematic viscosity greater than 22 centistokes The effect of viscosity, caused by friction between the particles of the fluid, is no longer negligable, and the flow rate is reduced. The flow factor must be multiplied by a correction factor, c, to give a new flow factor, k vn.

p1

5.21

9

Example 2 Part1 What is the k v factor for a 1 1/4 " water pipeline with a flow of 300 litres/min, an inlet pressure of 0.5 bar and an outlet pressure of 0 bar? Part 2 If a valve has to be fitted and the minimum acceptable flow rate in the pipline is 250 litres/min, which type of valve should be used?

Fig. 5.18

Given Q =250 litres/min ρ = 1kg/dm 3 ∆p = p 1 – p 2 = 0.5 –0 = 0.5 kg/cm 2 kvt = ?

Part 3 Having established which type of valve should be used, what will be the true flow rate of the system?

1 0 .5

kvt = 250

kvt = 354 Solution to part 1 Calculate the k v for the pipeline (k vp)

The k v factor for the valve (k vv) can now be established by subtracting the k v factor for the pipline (k vp ) from the k v factor for the total system (k vt). For this purpose, the formula for calculating flow factors in series should be used, which is 1 k vx thus

Fig. 5.17

Given

Q ρ ∆p k vp

= 300 litres/min = 1 kg/dm3 = p1 – p2 = 0.5 – 0 = 0.5 kg/cm2 = ?

kvp = Q = 300

ρ ∆p 1 0.5

= 424

Solution to part 2 First it is necessary to calculate the k v factor for the total system (k vt).

5.22

2

1

=

k v1

2

+

1 k v2 2

1 1 1 = 2 2 k vv k vt k vp 2 1 1 1 = 2 2 k vv 354 424 2 =7.98 ×10 -6 – 5.56 ×10 -6 = 2.42 ×10 -6 k vv =

1 2.42 ×10 -6

= 643 In this example, the k v factor for the valve has been determined by calculation. It may also be found by the simpler method of reading off the nomograph, Fig. 5.20 .

GEORGEFISCHER…

by using the nomograph: 1 = 6 x10 -6 + 1.5 x 10 -6 k vt2 = 7.5 x10

1

-6

k vt = 365 The true flow for the system can now be calculated as follows: Q

= kv

2

∆p ρ

Fig. 5.19

= 365 The section of the nomograph reproduced above allows us to read off the reciprocals of the squares of the flow factors. k vt=354 gives 8.5 x10 -6 and k vp=424 gives 6 x10 -6

0.5 1

3

= 258 litres/min

4

We find the reciprocal of k vv by subtracting these two values. 8.5 x10 -6 - 6 x10 -6 = 2.5 x10 -6

5

The value for the flow factor corresponding to 2.5×10 -6 can also be read from the nomograph. Thus, k vv = 640

6

The valve used must therefore be one with a minimum k v100 factor of 640. A typical value for a ball valve, which could be used in this example is k vv = 800 (k v valves are supplied by valve manufacturers).

7

Solution to part 3 First, calculate the k v factor for the pipline with the 1 1/ 4 " ball valve installed. 1 = kvt2 =

GEORGEFISCHER…

1 + kvp2

8

1 kv 2(ball valve)

9

1 1 + 2 424 800 2

5.23

Nomogram for valve losses

ρ (kg/dm 3 ) (kg/cm 2 )

Fig. 5.20

5.24

GEORGEFISCHER…

Compressible Fluids Fluids such as air and steam are compressible. A force acting on them may decrease the fluid volume (increasing its density) rather than causing movement of the fluid. This density change will lead to a pressure drop. If this drop in pressure is less than 10% of the inlet pressure we can treat the fluid as incompressible. If this is not the case then new equations must be used to describe the flow depending on the type of fluid and the surrounding conditions.

Compresssed Air

3. Connect the point on reference line F to the working pressure (line E). Extend this line to line G. 4. Read the value of the pressure drop at this intesection point.

2

Example The problem below is solved by the dotted lines on the nomogram. Pipe length = 300m Free air flow = 175 litres/second Pipe inner diameter = 90mm Working pressure = 9 bar

3

Pipe Losses For compressed air the pressure drop can be found from the following equation

Following the steps above we draw the 3 lines shown on the nomogram. The final line intersects line G at approximately 0.04. So the pressure drop is 0.04 bar.

4

∆p = 1.6 x 10 8 x V 1.85 x L d5 x P

A general rule is that the velocity of compressed air must be less than 6 m/s . Using this restriction we find maximum flow values through medium grade steel as given in Fig. 5.21 .

5

∆p= pressure drop (bar) V = free air flow (m 3/s) L = pipe length (m) d = inside pipe diameter (mm) P = initial pressure (bar) Values can be inserted directly into this formula, or the nomogram (Fig. 5.22) can be used. The nomogram gives the free air flow in litres/second and the pressure drop in bar. Conversions to cubic metres per second and pascal are given below. m 3/s = l/s x 10 -3 bar = 10 5 Pa To use the nomogram, follow these steps: 1. Connect pipe length (line A) to free air flow (line B). Extend this line to the first reference line (line C).

Maximum Recommended Flow of Compressed Air at 7 bar Nominal bore (mm) 10 15 20 25 32 40 50 65 80 100

Rate of air flow (litres/second) 5 10 17 25 50 65 100 180 240 410

6

7

8

Fig. 5.21

2. Connect the point found on this reference line to the inner pipe diameter (line D) and extend to the second reference line (line F).

GEORGEFISCHER…

1

9

5.25

Air Flow through Black Iron Pipes

Fig. 5.22

5.26

GEORGEFISCHER…

Fitting Losses For compressed air the pressure loss for a system is most easily calculated using equivalent lengths. The following table gives the equivalent lengths for fittings of various pipe sizes.

1

Pressure Loss - Equivalent Pipe Lengths Fitting

2

Equivalent pipe length (m) pipe diameter Nom. Size DN

1

/4 20

1 25

1 1/ 2 40

2 50

3 80

4 100

3

/2 15

90° Elbow

0.2

0.3

0.4

0.6

0.8

1.3

1.6

Bend

0.1

0.2

0.3

0.5

0.6

1.0

1.2

Run of Tee Junction

0.2

0.3

0.5

0.8

1.0

1.6

2.0

Branch of Tee Junction

0.6

1.0

1.5

2.4

3.0

4.8

6.0

Reducer

-

0.3

0.5

0.7

1.0

2.0

2.5

3

4

Fig. 5.23

Example A compressed air system where air flows at 0.2m 3/s has a working pressure of 10 bar (10 x10 5Pa) and consists of 20m of 1 1/2" (DN 40) straight pipe with the following fittings: 2 x 90° elbow 1 x tee: air flows through the branch of the tee

l ef = 2 x 0.6 + 2.4 = 3.6m Adding the length of straight pipe to this value gives the effectve length of the system. 3.6 + 20 = 23.6 Now we can use the nomogram (Fig. 5.22) to find the pressure drop.

The 90° elbow has an equivalent length of 0.6m. So the total equivalent length of the fittings

6

Effective pipe length = 23.6 Free air flow = 0.2m 3/s = 200 l/s Nominal Pipe diameter = 40mm Working pressure = 10 bar

First we find the equivalent length of the fittings.

The tee has an equivalent length of 2.4m.

5

7

The pressure drop is found to be approximately 0.4 bar.

8

5m Fig 130 1 1 / 2

DN 40

5m

5m

Fig 90 11/2

9

DN 40 Fig 90 11/2

Fig. 5.24

GEORGEFISCHER…

5m

5.27

Steam Steam is another compressible fluid which behaves in a similar way to compressed air. Two important factors to remember when laying steam piping are that temperature changes can lead to expansion or contraction, which must not put any excessive stress on the system, and that no water should be allowed to collect anywhere in the system. Each time the system is started from cold there will be water of condensation to dispose of. Another consideration is that if the velocity exceeds about 60 m/s the system may become intolerably noisy. The volume flow rate corresponding to a velocity of 60m/s is given below for various pipe sizes. Maximum Flow rates corresponding to a velocity of 60m/s at 100°C Nominal size mm inches 10 20 25 32 40 50 65 80 100

Max. volume flow rate (l/s)

3

/8 3 /4 1 1 1/4 1 1/2 2 1 2 /2 3 4

6.1 19.6 31.0 55.6 76.3 121.2 211.3 294.1 504.2

d i = pipe inside diameter (m) V = volume flow rate (m 3/s) f = friction factor Applying the method used in the analysis of water flow we calculate the friction factor (f) from the Reynolds number. Re = vd i ν Values for the friction factor are given in the table Fig.5.26 below: Friction Factors for different Re Pipe size mm in 10 25 50 80 100

3

/8 1 2 3 4

10

4

Reynolds no. 10 5 10 6

0.035 0.033 0.032 0.032 0.031

0.029 0.025 0.022 0.021 0.020

0.027 0.023 0.020 0.018 0.017

10 7 0.027 0.023 0.019 0.017 0.016

Fig. 5.26

The pressure loss can now be calculated. Fitting losses The pressure loss due to the fitting is found using: ∆p = ζρ v 2 2

Fig. 5.25

The density and viscosity of steam at various temperatures is given in Fig. 5.27 (Properties of Saturated Steam).

Pressure Loss

where ∆p = pressure loss (Pa) ζ = pressure loss factor (dimensionless) ρ = density (kg/m 3) v = velocity (m/s)

Pipe losses The pressure loss for steam flowing in a straight pipe is given by: ∆p = 6.48 fρ V2 d i 100000 ∆p = pressure loss per unit length (Pa/m) 3 ρ = density (kg/m )

5.28

GEORGEFISCHER…

Properties of Saturated Steam bar

Pressure kPa

Temperature °C

Density kg/m 3 ρ

Kinematic Viscosity (m 2/s) ν

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0 12.0 14.0 18.0 20.0 22.0 24.0 28.0

50 100 150 200 250 300 350 400 450 500 600 700 800 900 1000 1200 1400 1800 2000 2200 2400 2800

81.35 99.63 111.37 120.23 127.43 133.54 138.88 143.63 147.92 151.85 158.84 164.96 170.41 175.36 179.88 187.96 195.04 207.10 212.37 217.24 221.78 230.04

0.309 0.590 0.863 1.129 1.391 1.650 1.908 2.165 2.415 2.667 3.165 3.663 4.167 4.651 5.155 6.135 7.092 9.091 10.000 10.989 12.048 14.085

5.2 x 10 -5 2.7 x 10 -5 1.9 x 10 -5 1.4 x 10 -5 1.2 x 10 -5 1.0 x 10 -5

1

2

8.4 x 10 -6 6.6 x 10 -6 6.0 x 10 -6 5.1 x 10 -6 4.4 x 10 -6 3.8 x 10 -6 3.4 x 10 -6 3.1 x 10 -6 2.6 x 10 -6 2.3 x 10 -6 1.8 x 10 -6 1.6 x 10 -6 1.5 x 10 -6 1.3 x 10 -6 1.1 x 10 -6

3

4

5

Fig. 5.27

6

7

8

9

GEORGEFISCHER…

5.29

Water Hammer Water hammer, or surge pressure, is a term used to describe dynamic surges caused by pressure changes in a piping system. They occur whenever there is a deviation from the steady state, i.e. when the velocity of the fluid is increased or decreased, and may be transient or oscillating. Waves of positive or negative pressure may be generated by any of the following: 1. opening or closing of a valve 2. pump start up or shut down 3. change in pump or turbine speed 4. wave action in a feed tank 5. entrapped air

The pressure waves travel along at speeds limited by the speed of sound in the medium, causing the pipe to expand and contract. The energy carried by the wave is dissipated and the waves are progressively damped. (see Figure ?) The pressure excess due to water hammer must be considered in addition to the hydrostatic load, and this total pressure must be sustainable by the pipng system. In the case of oscillatory surge pressures extreme caution is needed as surging at the harmonic frequency of the system could lead to catastrophic damage.

Wavelength Fig. 5.28

Pressure Change

Damped pressure wave The maximum positive or negative addition of pressure due to surging is a function of fluid velocity, bulk modulus of elasticity of the fluid, pipe dimensions and the modulus of elasticity of the pipe material. It can be calculated using the following steps.

E = modulus of elasticity of pipe wall (Pa) d i = pipe inner diameter (mm) t = pipe wall thickness (mm)

Step 1 Determine the velocity of the pressure wave Vw =

K ρ x (1 +

K x di ) txE

where V w = velocity of pressure wave (m/s) K = bulk modulus of elasticity of fluid (Pa) ρ = fluid density (kg/m 3) 5.30

GEORGEFISCHER…

Step 2 Calculate maximum pressure change due to surging. ∆p = Vw x ∆v x ρ x 10 -5 where ∆p = maximum pressure change (bar) V w = velocity of pressure wave (m/s) (see step 1) ∆v = change in fluid velocity (m/s) = (v 1-v 2) v1 = velocity of fluid before change (m/s) v2 = velocity of fluid after change ( m/s) ρ = density of fluid (kg/m 3) NB. All pressure rises induced by a flow reduction will have a corresponding reflected pressure drop (vacuum). If this exceeds the expected static minimum operation pressure it must not exceed the collapsing pressure for safe operation of the system.

Step 3 Calculate the maximum and minimum total pressures p max = P + ∆p p min = P – ∆p where p max = maximum total pressure (bar)

withstand this water hamer pressure instantaneously. As long as the calculated maximum total pressure due to water hammer is within the maximum allowable pressure for a system the effects of water hammer will not be serious. The exception to this rule is when the pressure surges are oscillatory (e.g. from a positive displacement pump). In this case the system must be treated as if a load equal to the maximum total pressure, P max , exists throughout the lifetime of the pipe. If the total pressure due to water hammer does not fall within these limits an increase in pipe diameter should be considered, or measures should be taken to reduce surge occurance (i.e. actuated valves, surge tanks, slow start-up pumps). When using actuated valves it is common to design valves with closure times greater than the critical period T c to reduce water hammer. The critical period is the time taken by the pressure wave to complete one circuit of the pipeline.

2

3

4

5

Tc = 2L Vw

6

where Tc = critical period L = pipe length (m) V w = pressure wave velocity (m/s)

p min = minimum total pressure (bar) P = expected operating pressure (bar) ∆p = change due to surge pressure (calculated in step 2)

7

8

Step 4 Compare the maximum total pressure due to water hammer, calculated in step 3, with the maximum allowable pressure in the system. The maximum allowable pressure is determined by the pressure rating of the weakest component in a system. As most surges last for a matter of seconds the system need only GEORGEFISCHER…

1

9

5.31

Example Water piping from a storage tank is connected to a primary shut-off valve, which is hydraulically actuated with an electrical remote control. The water flow rate is Q= 10m 3/h. The working pressure is 6 bar. The pipe details are: material medium grade steel Nom. size 100 inner diameter 105 mm wall thickness 4.5 mm pipeline length 500m operating temp. 40°C Modulus of elasticity E = 200 x 10 9 other information: water density bulk modulus of water

ρ = 1000 kg/m 3 K = 2.05 GPa

Step 3 Find maximum pressure P max = 6 + 4.12 = 10.12 bar ∆p < p so the minimum pressure is positive. If the minimum pressure was negative (i.e. ∆p > p), then we would need to ensure that all the components in the system could withstand the negative pressure. Step 4 The maximum total pressure due to water hammer is 10.12 bar. If this value is less than the maximum allowable instantaneous pressure in our system then the effects of water hammer are acceptable. If it is not, adjustments need to be made to pipe dimensions, or valve closing time, to reduce water hammer.

Step 1 Find velocity of pressure wave Vw

What is the critical period?

2.05 x10 9

=

(

1000 x 1 +105 x 2.05 x 10 9 4.5 x 200 x 10 9

)

= 1286 m/s Step 2 Find velocity of fluid before change

v1 =

=

T c = 2L = 2 x 500 V w 1286 = 0.78s If we use an actuated valve with a closing time greater than this critical period ( e.g. a closing time of 1.5s) this will help to reduce water hammer.

volume fluid flow cross-sectional area 10/60 x 60 m 3/s π (0.105 / 2 ) 2 m 2

= 0.32 m/s assume water velocity goes to zero after the valve is closed i.e. ∆v = 0.32m/s Find pressure change ∆p = 1286 x (0.32) x 1000 x 10 -5 = 4.12 bar

5.32

GEORGEFISCHER…