﻿ Hydraulic calculations of pipelines. Calculation of pipeline diameter. Selection of pipelines
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# Hydraulic calculations of pipelines. Calculation of pipeline diameter. Selection of pipelines

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## General description of pipelines

Pipes interconnect various apparatuses of chemical plants. They are used for transfer of substances between separate apparatuses. As a rule, several separate pipes with the help of connections make up single pipeline system.

Pipeline is a system of pipes connected by connecting elements and used for transport of chemical substances and other materials. As a rule, buried pipelines are used at chemical plants for transport of substances. In respect to self-contained and isolated parts of a plant, they also pertain to pipeline system or network.

Configuration of self-contained pipeline system may include:

1. Pipes.
2. Connecting elements of pipes.
3. Seal packing connecting two detachable sections of pipeline.

All abovementioned elements are manufactured individually, after which they are connected as a single pipeline system. Moreover pipelines can be equipped with heating and necessary insulation made of different materials.

Size of pipes and materials for their manufacture is selected on the basis of process and resign requirements set in each individual case. But for standartization of pipelines sizes their classification and unification were performed. The key criterion was permissible pressure under which pipe operation is possible.

### Nominal inside diameter DN

Nominal inside diameter DN (nominal diameter) is a parameter used in pipeline systems as characterization factor with the help of which alignment of pipeline parts, such as pipes, valves, fittings, etc, is performed.

Nominal diameter is a nondimensional quantity, but numerically it approximately equals to pipe inside diameter. Example of nominal inside diameter designation: DN 125.

Figure 1.1 Nominal inside diameter

Also nominal inside diameter is not denoted in drawings and does not substitute real diameters of pipes. It approximately corresponds to the clear diameter for certain pipeline sections (Figure 1.1.). If numerical values of nominal inside diameters are implied, they are selected so as to increase pipeline transmission capacity in the range from 60 to 100% when passing from one nominal inside diameter to the next one.

Generally accepted nominal diameters:

3, 4, 5, 6, 8, 10, 15, 20, 25, 32, 40, 50, 65, 80,

100, 125, 150, 200, 250, 300, 350, 400, 450, 500, 600, 700, 800, 900, 1,000,

1,200, 1,400, 1,600, 1,800, 2,000, 2,200, 2,600, 2,800, 3,000, 3,200, 3,400, 3,600, 3,800, 4,000.

Sizes of these nominal inside diameters are set with a view to avoid any problems with mutual alignment of parts. When determining nominal diameter on the basis of pipeline inside diameter value, the value of nominal inside diameter, which is the nearest to the pipe clear diameter, is selected.

### Nominal pressure PN

Nominal pressure PN is a quantity corresponding to maximal pressure of pumped medium at 20° C, making possible long-term operation of pipeline having specified dimensions.

Nominal pressure is a nondimensional quantity.

Like a nominal diameter, a nominal pressure was calibrated on the basis of accumulated operation experience (Table 1.1).

Table 1.1. Standard nominal pressures (DIN 2401)
1 10 100 1,000
1.6 16 160 16,000
2.5 25 250 2,500
4 40 400 4,000
6 63 630 6300

Nominal pressure for a particular pipeline is selected on the basis of pressure practically created in it, by selecting nearest larger value. Moreover, fittings and valves in this pipeline should also correspond to the same level of pressure. Thickness of pipe walls is calculated on the basis of nominal pressure, and provide pipe operation ability with pressure value equal to nominal (Table 1.1).

### Permissible operating overpressure pe,zul

Nominal pressure is used only for operation temperature 20°C. With rise of temperature pipe load capability drops. At the same time permissible overpressure reduces correspondingly. Value pe,zul shows maximal overpressure that can be in the pipeline system upon rise of operation temperature value (Figure 1.2).

Figure 1.2 Plot of permissible overpressures

### Piping materials

During selection of materials to be used for pipelines manufacture characteristics, such as parameters of medium to be transported through the pipeline, and tentative operation pressure in this system are taken into account. Possibility of corrosive action of the pumped medium on material of pipe walls should also be taken into account.

Practically all pipeline systems and chemical plants are made of steel. Gray cast iron or unalloyed constructional steels are used to manufacture pipelines in case of absence of high mechanical loads and corrosive action.

In the case of higher operation pressure and absence of loads with corrosion active action, a pipeline made of improved steel or steel castings is used.

If medium corrosion action is great or strict requirements are set for product pureness, the pipeline is made of stainless steel.

If pipeline should resist sea water influence, copper-nickel alloys are used for its manufacture. Also aluminum alloys and metals, such as tantalum or zirconium, can be used.

Different types of plastic are often used as piping materials, which is stipulated by its high resistance to corrosion, light weight and easiness of processing. Such material is suitable for sewage water pipelines.

### Pipeline shaped elements

Pipelines made of plastic are suitable for welding and are assembled at the erection site. Such materials include steel, aluminum, thermoplastic, copper, etc. For connection of  straight sections of pipes specially manufactured shaped elements, for example, elbows, branches, gates and diameter reducers (Figure 1.3) are used. Such fittings can be part of any pipeline.

### Pipe connections

Special connections are used to mount individual parts and fittings. They are also used for connection of necessary valves and apparatuses to the pipeline.

Connections are selected (Figure 1.4) depending on:

1. materials used for manufacture of pipes and fittings. The main selection criterion is welding ability.
2. operation conditions: low or high pressure and low or high temperature.
3. production requirements set for the pipeline system.
4. availability of fixed or detachable connections in the pipeline system.
Figure 1.4 Types of pipe connections

### Linear expansion of pipes and its compensation

Geometrical shape of objects can be changed both by force action and with variation of their temperature. These physical phenomena result in pipeline, which is mounted in the unstressed conditions and free of temperature action, is subjected to some linear expansion or contraction negatively affecting its functional performance, while in service due to pressure and temperature exposure.

In case when expansion is not necessary to compensate, pipeline system deformation takes place. In this case flanged packing and pipes interconnections can be damaged.

### Thermal linear expansion

During pipelines mounting potential change of length due to temperature rise or the so called thermal linear expansion, denoted as ΔL, should be taken into account. This quantity depends on the pipe length denoted as Lo and temperature differential Δϑ =ϑ2-ϑ1 (figure 1.5).

Figure 1.5 Linear thermal expansion of pipe

In the abovementioned formula a is a coefficient of the thermal linear expansion of this material. This quantity equals to the value of linear expansion of 1 m long pipe upon temperature rise by 1°C.

### Elements of pipes expansion compensation

Pipe branches

Thanks to special branches, which are welded in the pipeline, it is possible to compensate natural linear expansion of pipes. For this compensating U-shaped, Z-shaped, and angular branches, as well as harp compensators are used (figure 1.6).

Figure 1.6 Compensating pipe branches

They take up pipes linear expansion by means of their own deformation. But such method is possible only with some limitations. Expansion is compensated by different angled elbows in high-pressure pipelines. Due to pressure in such branches corrosion can be enhanced.

### Wavy pipe compensators

This device consists of thin-walled metal corrugated pipe which is called bellows and stretches towards pipeline (Figure 1.7).

Such devices are built in the pipeline. Preload is used as a special expansion compensator.

What concerns axial compensators, they only can compensate the linear expansions that occur along pipe axis. In order to avoid lateral displacement and internal staining the inside guide ring is used. In order to protect pipeline against external damage, as a rule, special lining is used. Compensators, which are not equipped with inside guide ring, absorb lateral shifts and vibration which can be originated from the pumps.

### Pipe insulation

In case when high-temperature medium is transported through the pipeline, it should be insulated in order to avoid losses of heat. If low-temperature medium is transported through the pipeline, the insulation is used to prevent its heating by outside environment. In such cases insulation is arranged using special insulation materials wrapped around pipes.

As a rule, the following materials are used:

1. Under low temperatures up to 100°C rigid foam plastic, like polystyrene or polyurethane.
2. Under medium temperatures around 600°C shaped shells or mineral fiber, like rock wool or glass felt.
3. Under high temperatures approx 1200°C – ceramic fiber, for example, aluminum silicate.

Pipes with nominal inside diameter below DN 80, and thickness of insulation layer less than 50 mm are insulated, as a rule, by means of insulating shaped elements. To this end two shells are wrapped around the pipe and fastened with metal band, and then are closed with tin-plate housing (figure 1.8).

Figure 1.8 Thermal insulation by means of shaped elements

Pipelines with nominal inside diameter over DN 80 should be fitted with thermal insulation having bottom shell (Figure 1.9). Such shell comprises clamp rings, braces and metal lining made of galvanized soft steel or stainless sheet steel. Space between pipeline and metal housing is filled with insulation material.

Figure 1.9 Thermal insulation with bottom shell

Thickness of insulation is calculated be determination of production costs and losses, which are incurred due to waste of heat, and is from 50 to 250 mm.

Thermal insulation should be applied throughout the length of pipeline system, including zones of branches and elbows. It is very important to control so as unprotected parts, which can cause heat losses, do not appear. Flange connections and valves should be provided with shaped insulation elements (Figure 1.10). It provides unobstructed access to the connections points without the need to remove insulation material from the whole pipeline system in case if break of air-tightness is occurred.

Figure 1.10 Thermal insulation of flange connection

Correct selection of pipeline system insulation solves numerous problems, such as:

1. Avoidance of dramatic drop of leaking medium temperature and, as a consequence, energy saving.
2. Prevention of temperature drop in gas transmission systems below dew point, thus excluding condensate formation which may result in major corrosion destructions.
3. Avoidance of condensate emission in steam pipelines.

## Pressure drop in pipeline systems and calculations of pipeline hydraulic resistance

Calculations of pipeline are carried out in order to determine the head necessary for overcoming hydraulic resistance which, in its turn, is necessary for correct selection of machines for liquid or gaseous media pumping.

In the general case drop of pipe pressure can be calculated by the following formula:

Δp=λ·(l/d1)·(ρ/2)·v²

Δp – pressure drop on a pipe section, Pa
l – length of pipe section, m
λ - friction coefficient
d1 – pipe diameter, m
ρ – density of pumped medium, kg/m3
v – flow rate, m/s

Hydraulic resistance may occur due to different factors, and two major groups are distinguished: friction resistances and local resistances.

Friction resistance is caused by all sorts of unevenness and roughness on pipeline surface being in contact with the pumped medium. Friction, having braking effect and requiring additional energy consumption for its overcoming, occurs during fluid flow between it and pipeline walls. Created resistance to a great extent depends on the mode of pumped medium flow.

With laminar flow and small values of Reynolds number (Re) corresponding to it, characterized by uniformity and absence of mixing between fluid or gas adjacent layers, influence of roughness is minor. It is explained by the fact that extreme viscous underlayer is often thicker than the layer formed by unevenness and bulges on the pipeline surface. Under such conditions the pipeline is regarded as hydraulically smooth.

With the increase of Reynolds number thickness of viscous underlayer reduces, and interrupts overlap of irregularities by underlayer, and influence of roughness on hydraulic resistance increases and becomes dependent on both Reynolds number and average height of bulges on the pipeline surface.

Further increase of Reynolds number converts the pumped medium into turbulent flow mode, in which viscous underlayer is completely destroyed and created friction depends only on the degree of roughness.

Calculation of friction loss is made by formula:

HТ=[(λ·l)/dэ]·[w2/(2g)]

HТ – head losses due to friction resistance, m
λ – friction coefficient
l – pipeline length, m
dЭ – pipeline equivalent diameter, m
w – flow rate, m/s
g – gravity acceleration, m/s2

Flow range Range of Reynolds number Formula of friction coefficient λ
Smooth flow 2,320<Re<10/e λ=(0,316/Re0,25)
Mixed flow 10/e<Re<560/e λ=0,11·[e+(68/Re)]0,25
Turbulent flow Re>560/e λ=0,11·e0,25

In table:

e – pipe relative roughness
Δ – pipe absolute roughness (mm)
dЭ – pipe equivalent diameter (mm)
Re – Reynolds number
w – flow rate (m/s)
dЭ – pipe equivalent diameter (mm)
ρ – medium density (kg/m3)
μ – dynamic viscosity (Pa·s)

### Equivalent diameter in calculations of pipelines

Equivalent diameter is used in calculations of non-cylindrically shaped pipelines (oval, rectangular), and corresponds to the diameter of the round pipeline which creates friction losses similar to that of non-cylindrically shaped pipeline with the similar length. There are various formulae for calculation of pipelines having different geometrical shapes, but in the general case the following formula is used:

dэ = 4F/P

dЭ – pipeline equivalent diameter, m
F – pipeline cross-sectional area, m2
Р – inside perimeter of pipeline cross-sectional area, m

For cylindrically shaped pipelines equivalent and inside diameters will evidently coincide. In the case of open conduits the formula for equivalent diameter calculation is another:

dэ = 4F/Pс

dЭ – conduit equivalent diameter, m
F – cross-sectional area of fluid flow, m
Рc – wetted perimeter, m

Wetted perimeter is a length of line of flow contact with conduit or pipe walls that restrain this flow.

Local resistances are created by different pipeline elements in which flow of the pumped medium is subjected to sudden deformations with change of direction, velocity or vortex effects. These can be gates, valves, pipeline turns, junctions, etc.

Head losses in local resistance are calculated in the following way:

Hмсмс·[w2/(2g)]

HМС – losses of head in local resistance, m
ζМС – local resistance coefficient
w – flow rate, m/s
g – gravity acceleration, m/s2

As can be seen from the formula, head losses in local resistance depend not only on the rate and local resistance coefficient, values of which are summarized in the table for different types of local resistances to simplify calculations.

In the majority of cases local resistance coefficients do not depend on the flow rate of the pumped fluid and are determined depending on characteristics of the local resistance. Values of resistance coefficients for most wide-spread cases are given below:

Initial and end pipe sections
Pipe inlet Pipe outlet
Sharp edges Rounded edges
0.5 0.2 1
Resistance coefficient is obtained by multiplying two quantities K1 and К2
К1 depends on the angle of flow change
Angle 20 30 45 60 90 110 130 150 180
К1 0.31 0.45 0.60 0.78 1.00 1.13 1.20 1.28 1.40
К2 depends on turn radius R and pipe inside diameter d
R/d 1 2 4 6 15 30 50
K2 0.21 0.15 0.11 0.09 0.06 0.04 0.03
Elbow (angle 90°)
Pipe diameter, mm 12,5 25 37 50 Over 50
Local resist. coefficient 2.2 2 1.6 1.1 1.1
Normal valve (full opening)
Pipe diameter, mm 13 20 40 80 100 150 200 250 350
Local resist. coefficient 10.8 8.0 4.9 4.0 4.1 4.4 4.7 5.1 5.5
Straight-through valve (full opening)
At Re > 3·105
Pipe diameter, mm 25 38 50 65 76 100 150 200 250
Local resist. coefficient 1.04 0.85 0.79 0.65 0.6 0.5 0.42 0.36 0.3
At Re < 3·105
(relevant local resistance coefficient is multiplied by coefficient k the value of which depends on the Reynolds number)
Re 5,000 10,000 20,000 50,000 100,000 200,000
k 1.4 1.07 0.94 0.88 0.91 0.93
Pipeline sudden expansion
Re F 1/F2
0.1 0.2 0.3 0.4 0.5 0.6
10 3.1 3.1 3.1 3.1 3.1 3.1
100 1.7 1.4 1.2 1.1 0.9 0.8
1,000 2.0 1.6 1.3 1.05 0.9 0.6
3,000 1.0 0.7 0.6 0.4 0.3 0.2
3,500 and over 0.81 0.64 0.5 0.36 0.25 0.16
In the table:
F1 – the smallest of pipeline cross-sections
F2 – the largest of pipeline cross-sections
Re – Reynolds number
Pipeline abrupt contraction
Re F1/F2
0.1 0.2 0.3 0.4 0.5 0.6
10 5.0 5.0 5.0 5.0 5.0 5.0
100 1.3 1.2 1.1 1.0 0.9 0.8
1,000 0.64 0.5 0.44 0.35 0.3 0.24
10,000 0.5 0.4 0.35 0.3 0.25 0.2
100,000 and over 0.45 0.4 0.35 0.3 0.25 0.2
In the table:
F1 – the smallest of pipeline cross-sections
F2 – the largest of pipeline cross-sections
Re – Reynolds number

By summing up all the above given equations we obtain general equation for calculation of the pump head:

Hоб = HТ+HМС = (λ·l)/dэ·[w2/(2g)]+∑ζМС·[w2/(2g)] = ((λ·l)/dэ+∑ζМС)·[w2/(2g)]

∑ζМС – sum of all coefficients of local resistances

### Calculation of pipeline optimal diameter

Calculation of pipeline optimal diameter is a complicated problem, which requires technical and economic assessments and consideration of numerous particular factors. This is attributed to close interrelation between parameters of pipeline under design and characteristics of flow of medium pumped through it. Increase of the pumped medium velocity makes it possible to reduce pipeline diameter required to maintain set flow rate, thus reducing consumption of materials and making system erection cheaper and easier. At the same time, velocity increase inevitably involves losses of head, which requires additional energy consumption for medium pumping. Excessive reduction of velocity also can result in unwanted aftermath.

Formula for calculation of the pipeline optimal diameter is based on the flow rate formula (for circular pipe):

Q = (Πd²/4)·w

Q – flow rate of pumped fluid, m3/s
d – pipeline diameter, m
w – flow velocity, m/s

The flow rate is most often a set quantity in problems on pipeline design. In such case the unknown quantities are only a pipeline diameter and flow velocity. Comprehensive technical and economic calculation may be very labor-intensive and complicated, so optimal values of pumped medium velocity, taken from reference materials, drawn up on the basis of experimental findings, are used in practice:

Pumped medium Pipeline optimal velocity, m/s
FLUIDS Gravity flow:
Viscous fluids 0.1 – 0.5
Low-viscosity fluids 0.5 – 1
Pumping:
Suction pipeline 0.8 – 2
Delivery pipeline 1.5 – 3

GASES Natural draught 2 – 4
Low pressure (fans) 4 – 15
High pressure (compressor) 15 – 25

VAPORS Overheated 30 – 50
Saturated vapors under pressure of:
Over 105 Pa 15 – 25
(1-0.5)·105 Pa 20 – 40
(0.5-0.2)·105 Pa 40 – 60
(0.2-0.05)·105 Pa 60 – 75

Final calculation formula for pipeline optimal diameter is the following:

d = √(4Q/Πw)

Q – flow rate of pumped fluid, m3/s
d – pipeline diameter, m
w – flow velocity, m/s

## Examples of problems and solutions for calculation and selection of pipelines:

### Example No. 1

What are the head losses for local resistances in horizontal pipeline having diameter of 20 x 4 mm, through which water is pumped from open reservoir to reactor with pressure of 1.8 bar? Distance between reservoir and reactor is 30 m. Water flow rate is 90 m3/h. Total head equals to 25 m. Friction coefficient is taken equal to 0.028.

Solution:

Water flow velocity in pipeline equals to:

w=(4·Q) / (π·d2) = ((4·90) / (3,14·[0,012]2))·(1/3600) = 1,6 m/s

We find head friction losses in the pipeline:

HТ = (λ·l) / (dэ·[w2/(2·g)]) = (0,028·30) / (0,012·[1,6]2) / ((2·9,81)) = 9,13 m

Total losses are:

hп = H - [(p2-p1)/(ρ·g)] - Hг = 25 - [(1,8-1)·105)/(1000·9,81)] - 0 = 16,85 m

Losses on local resistance fall within:

16,85-9,13=7,72 m

### Example No. 2

Water is pumped by centrifugal pump across horizontal pipeline at velocity of 1.5 m/s. Total created head equals to 7 m. What is the pipeline maximal length, if water is taken from open reservoir, pumped across horizontal pipeline with one gate valve and two 90° elbows and flows out from pipe to another reservoir? Pipeline diameter equals to 100 mm. Relative roughness is taken equal to 4·10-5.

Solution:

For pipe with diameter of 100 mm coefficients of local resistances will equal to:

For 90° elbow – 1.1; gate valve – 4.1; pipe outlet – 1.

Then we determine value of velocity head:

w2 / (2·g) = 1,52 / (2·9,81) = 0,125 m

Head losses for local resistances will equal to:

∑ζМС · [w2/(2·g)] = (2·1,1+4,1+1) · 0,125 = 0,9125 m

We find total head losses for friction resistance and local resistances from the formula of the pump total head (geometrical lift head under these conditions equals to 0):

hп = H - (p2-p1)/(ρ·g) - Hг = 7 - ((1-1)·105)/(1000·9,81) - 0 = 7 m

Then friction head losses will amount to:

7-0,9125 = 6,0875 m

We calculate value of the Reynolds number for the flow in the pipeline (water dynamic viscosity is taken as 1·10-3 Pa·s, and density – 1,000 kg/m3):

Re = (w·dЭ·ρ)/μ = (1,5·0,1·1000)/(1·10-3) = 150000

In accordance with this number using the table we calculate a friction coefficient (the arithmetic formula is selected on the principle that value Re falls within the range of 2,320<Re<10/e, corresponding to laminar flow):

λ = 0,316/Re0,25 = 0,316/1500000,25 = 0,016

We express and find a maximal pipeline length from the formula of head friction losses:

l = (Hоб·dэ) / (λ·[w2/(2g)]) = (6,0875·0,1) / (0,016·0,125) = 304,375 m

### Example No. 3

The pipeline with the inside diameter of 42 mm is given. It is connected to the water pump with flow rate of 10 m3/h and creating head of 12 m. Temperature of the pumped medium is 20° C. Pipeline configuration is given in the figure below. It is necessary to calculate the head losses and check whether this pump is capable of pumping water at pipeline set parameters. Absolute roughness of pipes is taken as equal to 0.15 mm.

Solution:

We calculate the velocity of the fluid flow in the pipeline:

w = (4·Q) / (π·d2) = (4·10) / (3,14·0,0422)·1/3600 = 2 m/s

Velocity head corresponding to the found velocity will equal to:

w2/(2·g) = 22/(2·9,81) = 0,204 m

Friction coefficient should be found before the calculation of friction losses in pipes. In the first place we determine pipe relative roughness:

e = Δ/dЭ = 0,15/42 = 3,57·10-3 mm

Reynolds criterion for water flow in the pipeline (water dynamic viscosity at 20° C is 1·10-3 Pa·s, and density is 998 kg/m3):

Re = (w·dЭ·ρ) / μ = (2·0,042·998) / (1·10-3) = 83832

We find out the water flow mode:

10/e = 10/0,00357 = 2667

560/e = 560/0,00357 = 156863

The found value of Reynolds criterion falls within the range of 2667<83,832<156,863 (10/e < Re < 560/e), hence, friction coefficient should be calculated by the following formula:

λ=0,11·(e+68/Re)0,25 = 0,11·(0,00375+68/83832)0,25 = 0,0283

Head friction losses in the pipeline will equal to:

HТ = (λ·l)/dэ · [w2/(2·g)] = (0,0283·(15+6+2+1+6+5))/0,042 · 0,204 = 4,8 m

Then it is necessary to calculate head losses for local resistances. It is followed from the pipeline diagram that local resistances are represented by two gate valves, four rectangular elbows and one pipe outlet.

Tables do not contain values of coefficient of local resistances for normal gate valves and rectangular elbows with pipe diameter of 42 mm, so we will use one of the methods of approximate calculation of values we are interested in.

We take table values of coefficients of local resistances of normal gate valve for diameters of 40 and 80 mm. We assume that plot of values of coefficients represents straight line in this range. We set up and solve system of equations in order to find a plot of dependence of the local resistance coefficient on the pipe diameter:

{
4,9 = a·40+b
4 = a·80+b
=
{
a = -0,0225
b = 5,8

Sought equation has the view:

ζ = -0,0225·d+5,8

With diameter of 42 mm local resistance coefficient will equal to:

ζ = -0,0225·42+5,8 = 4,855

Similarly we find the value of local resistance coefficient for rectangular elbow. We take table values for diameters of 37 and 50 mm and solve system of equations, making similar assumption on the nature of plot at this section:

{
1,6=a·37+b
1,1=a·50+b
=
{
a = -0,039
b = 3,03

Sought equation has the view:

ζ = -0,039·d+3,03

With diameter of 42 mm local resistance coefficient will equal to:

ζ = -0,039·42+3,03 = 1,392

For pipe outlet local resistance coefficient is taken as equal to one.

Head losses for local resistances will equal to:

∑ζМС · [w2/(2g)] = (2·4,855+4·1,394+1) · 0,204 = 3,3 m

Total head losses in the system will equal to:

4,8+3,3 = 8,1 m

According to the data obtained we conclude that this pump is suitable for water pumping through this pipeline, as the head, it creates, is larger than total head losses in the system, and fluid flow velocity stay within the optimum margin.

### Example No. 4

Section of straight horizontal pipeline with inside diameter of 300 mm was subjected to repair by way of replacing 10 m long pipe section with the inside diameter of 215 mm. The total length of pipeline section under repair is 50 m. Section to be replaced is 18 m away from the beginning. Water flows at 20°C at velocity 1.5 m/s though the pipeline. I is necessary to find out how hydraulic resistance of the pipeline section under repair will change. Friction coefficients for pipes with diameter of 300 and 215 mm are taken equal to 0.01 and 0.012 correspondingly.

Solution:

Initial pipeline created head loss only for fluid friction with walls during pumping. Replacement of the pipe section resulted in occurrence of two local resistances (abrupt contraction and abrupt expansion of passage conduit), and section with changed pipe diameter, where friction losses will be different. The remaining pipeline section was not changed and, consequently, can not be considered as part of this problem.

We calculate the pipeline water flow rate:

Q = (π·d²) / 4·w = (3,14·0,3²) / 4·1,5 = 0,106 m³/s

As the flow rate does not change throughout pipeline length, we can determine the flow velocity on the pipe section subjected to repair:

w = (4·Q) / (π·d²) = (4·0,106) / (3,14·0,215²) = 2,92 m/s

The obtained value of the flow velocity in the pipe replaced section stay within the optimal range.

In order to determine the local resistance coefficient firstly we calculate Reynolds criterion for different diameters of pipes and ratio of cross-sectional areas of these pipes. Reynolds criterion for pipe with diameter of 300 mm (water dynamic viscosity at 20° C is 1·10-3 Pa·s, and density – 998 kg/m3):

e = (w·dЭ·ρ) / μ = (1,5·0,3·1000) / (1·10-3) = 450000

Reynolds criterion for pipe with diameter of 215 mm (water dynamic viscosity at 20° C is 1·10-3 Pa·s, and density – 998 kg/m3):

Re = (w·dЭ·ρ) / μ = (1,5·0,215·1000) / (1·10-3) = 322500

Ratio of pipe cross-sectional areas equals to:

((π·d1²)/4) / ((π·d2²)/4) = 0,215² / 0,3² =5,1

Using the tables we will find values of coefficients of local resistances, rounded the ratio of areas to 5. For sudden expansion it will equal to 0.25, and for sudden contraction will also equal to 0.25.

Head losses for local resistances will equal to:

∑ζМС·[w²/(2g)] = 0,25·[1,5²/(2·9,81)] + 0,25·[2,92²/(2·9,81)] = 0,137 m

Now we calculate friction losses in replaced pipeline section for initial and new pipe sections. For pipe with diameter of 300 mm they will equal to:

HТ = (λ·l)/dэ · [w²/(2g)] = (0,01·10)/0,3 · [1,5²/(2·9,81)] = 0,038 m

For pipe with diameter of 215 mm:

HТ = (λ·l)/dэ · [w²/(2g)] = (0,012·10)/0,215 · 2,92²/(2·9,81) = 0,243 m

Therefrom we conclude that friction losses in the pipeline will increase by:

0,243-0,038 = 0,205 m

Total increase of friction losses in the pipeline will equal to:

0,205+0,137 = 0,342 m