INTER AMERICAN UNIVERSITY OF PUERTO RICO

BAYAMON CAMPUS

Hardy-Cross Method

MECHANICAL ENGINEERING DEPARTMENT

Prof. Rafael Salgado Mangual

4/17/2012

By: Josean Otero, Erik T. Rosado, Ramón Y. Marrero, Joel Ruiz

2

Table of Contents Introduction: ........................................................................................................................ 3

Procedure Discussion and Results: ...................................................................................... 4

Conclusion: ........................................................................................................................ 11

Appendix: .......................................................................................................................... 12

Data and Results, MatLab Command Windows: ........................................................... 12

MatLab Program: ........................................................................................................... 15

Cleaning Pipeline: .......................................................................................................... 21

High Pressure Cleaning .............................................................................................. 21

Robotic Pipe Cleaning ................................................................................................ 21

Hazen-Williams Coefficient: ......................................................................................... 22

3

Introduction:

This project consists of using the Hardy-Cross method to analyze different aspects

of a pipeline network. This pipeline network presents a Category II problem, in which

analyzing is the purpose of the project. The objectives of this project are, firstly

determining the volumetric flows in each pipeline of the network, and as a secondary

objective the pressure at each node is to be calculated. Afterwards, two valves and four

heat exchangers are installed on the network which demand another set of calculations

for the network. The complete iteration and the additional calculations were done using

the Mat-Lab programing software, taking into account that to begin the process, the

volumetric flows have to be initially estimated.

4

Procedure Discussion and Results:

To begin the analysis of this program it was initially assumed that the directions of

the flows in each mesh of piping are clockwise. The following table illustrates the pipe

specifications provided in the book, which facilitate the analysis of the project.

Table 1: Pipe Specification

Pipe Information

Pipe L (ft) D(in) D(ft)

1 50 6 0.5

2 20 3 0.25

3 20 3 0.25

4 20 3 0.25

5 50 6 0.5

6 40 3 0.25

7 40 3 0.25

8 40 3 0.25

9 40 3 0.25

10 20 3 0.25

11 20 3 0.25

12 20 3 0.25

13 50 6 0.5

14 50 6 0.5

As with every Hardy-Cross situation the volumetric flow through every pipeline

must be initially estimated so that the actual flows can be iterated with MatLab. These

estimated flows were found using a clockwise loop like figure 1 below.

Figure 1: Loop orientations

5

Table 2: Hardy-Cross Estimated Values

Initial Volumetric Flow Values (CFS)

Loop I Loop II Loop III Loop IV Loop V

Q1 -0.6 Q4 -0.3 Q3 -0.15 Q2 -

0.075 Q10 0.3

Q2 0.075 Q6 0.3 Q7 0.15 Q8 0.075 Q11 0.15

Q3 0.15 Q7 -

0.15 Q8

-

0.075 Q9

-

0.675 Q12 0.075

Q4 0.3 Q10 -0.3 Q11 -0.15 Q12 -

0.075 Q13 0.6

Q5 0.6 Q14 -0.6

These were the initial values entered into the MatLab program available in the

Appendix. After several iterations in MatLab the flows in each line yielded the following

values.

Table 3: Iterated flows (with MatLab)

Final Flow Values (CFS)

Q1 0.6000

Q2 0.2445

Q3 0.0000

Q4 0.2445

Q5 0.6000

Q6 0.3555

Q7 0.2445

Q8 0.2445

Q9 0.3555

Q10 0.2445

Q11 0.0000

Q12 0.2445

Q13 0.6000

Q14 0.6000

6

Figure 2: Direction of the volumetric flows in each line

As can be seen from table 3 and figure 2, pipelines 3 and 11 do not have any water

flowing through them. This is mainly due because of the higher resistance the pipelines 3

and 11 exert on the volumetric flow, also due to the overall symmetry of the system.

After the flows were iterated the problem requested that the pressure at each node be

determined. This was done using the head loss in each pipeline and the energy equation.

The figure below displays the node nomenclature that was used in the program code and

a consequent table depicting the exact pressure at each node.

Figure 3: Node Nomenclature

A B

7

Table 4: Pressure at each node of the network

Node

Pressure

(psig)

A 100

A1 99.77

A2 99.26

A3 99.26

A4 99.77

B1 97.73

B2 97.22

B3 97.22

B4 97.73

B 97.5

As can be observed not much pressure was lost in this system, mainly because of the

relative short length of the network from point A to point B, and no fittings or restrictions

causing major loss in head. Now that these calculations were made we proceeded to add

two valves into the outer pipelines that restrict the flow to said lines, so that four heat

exchangers in each longitudinal line have the same volumetric flow. The “C” value for

the valve in the outer pipelines was obtained by recalling that the head loss for two pipes

in parallel must be the same therefore, the major and minor losses are known and all that

remains is to solve for “C”. This loss coefficient was simply calculated with MatLab and

yielded a value of 5.93 as seen on the command window on the appendix. Afterwards, a

pump was installed after point B and its purpose was to raise the pressure of the fluid at

this point to 100 psig. This again was done using MatLab, and to make it possible the

pressure at each node had to be recalculated because of the volumetric flow difference.

The aforementioned pressures are visible on the following table.

8

Table 5: Pressure at each node (With devices)

Node

Pressure

(psig)

A 100

A1 99.77

A2 99.03

A3 99.03

A4 99.77

B1 94.34

B2 93.6

B3 93.6

B4 94.34

B 94.11

Table 5 shows how more pressure was lost once the devices were installed in the

network. The energy equation was then used to determine the power of the pump which

yielded values of 0.7845 HP when no devices (two valves and four heat exchangers) are

in the network, and 1.8501 HP when the devices are installed, this shows a substantial

increase in power due to the valves and the heat exchanger causing a higher head loss.

Additionally, the pressure increase that the pump must give for the network with no

devices is just 2.5 psig, while with devices the pump must increase 5.89 psig. Figure 4

illustrates this difference in power by just applying the mentioned devices to the network.

Figure 4: Comparison of pump power between the two systems

Without Heat Exchanger andValves

With Heat Exchanger andValves

Series1 0.7845 1.8501

0

0.5

1

1.5

2

HP

Pump Power

9

Lastly, we took the liberty to change the Hazen-Williams factor to illustrate what

benefits does a better quality material have over a lower quality one have in terms of

efficiency. By changing the Hazen-Williams factor in our program we noticed that the

pressure drops due to friction were smaller and the pump at the end of point B used less

power. For instance, when a material such as aluminum which has an average Hazen-

Williams factor of 140 is used as a pipe material, the pump power was as low as 0.4207

HP (without devices) which is a substantial decrease below the 0.7845 HP that yielded

earlier with the Hazen factor of 100. To better visualize this behavior two graphs were

constructed that demonstrate how the pump power increases as the Hazen factor

increases.

Figure 5: Hazen Factor versus pump power (Without Devices)

Figure 6: Hazen Factor versus pump power (With Devices)

0

0.5

1

1.5

2

2.5

708090100110120130140150

Pum

p P

ow

er [

HP

]

C

C versus Pump Power (Without Devices)

0

0.5

1

1.5

2

2.5

708090100110120130140150

Pum

p P

ow

er [

HP

]

C

C versus Pump Power (With Devices)

10

By observing figure 5 and figure 6 it is clearly visible that when higher the Hazen-

Williams Factor is the less power will be needed and the system will be more efficient.

Furthermore, as is typical with Cast Iron pipes that deteriorate over time the Hazen factor

decreases and makes the pipes less efficient, which at the time of design the most

efficient material should be taken into consideration.

11

Conclusion: To begin our analysis of piping system was necessary to assume some initial

volumetric flow values. After making the necessary assumptions and enter this data into a

MatLab programming, we obtained the actual data from our system. So being able to

calculate with the head loss in each pipeline and the energy equation, the pressures in

each of the nodes of the system. In which shows that our system has only a loss of 2.5 psi

between node A and B. As low pressure loss is due because our pipeline network consists

of small distances and are without any restrictions (fittings, elbow, valve, etc.).

Now, to perform the second part was indicates that to the system will be added

two valves into the outer pipelines that restrict the flow to said lines, so that four heat

exchangers in each longitudinal line that have the same volumetric flow through

them.The problem question what is the value of "C" for the valve in the outer pipelines,

and the same was obtained by recalling that the head loss for two pipes in parallel must

be the same. Therefore, the major and minor losses are known and all that remains is to

solve for “C”. This loss coefficient as a value of 5.93. Also worth mentioning, that the

difference in pressure for this system is 5.89 psig.

Afterwards, a pump was installed after point B and its purpose was to raise the

pressure of the fluid at this point to 100 psig (pressure of node A). Analyzing for the

pump we obtain that the power for the first system is 0.7845 HP, and for the second

system is 1.8501 HP. These increase in power due to the valves and the heat exchanger

causing a higher head loss.

Finally it should be mentioned that, for the realization of this project the program

MatLab has been an invaluable tool. This allowed us to make the relevant calculations,

including the iterative process, and display graphics for the comparison of data without

any difficulty.

To reduce the internal friction of the pipe is recommended that it be cleaned

occasionally, using various methods, see appendix (Cleaning Pipeline). Also the pipe can

be replaced by other pipe with a Hazen-Williams constant much larger, see Appendix

(Hazen-Williams Coefficient). Moreover clean a pipe internally improves efficiency,

decreasing its pressure and change in power consumption.

12

Appendix:

Data and Results, MatLab Command Windows:

The following are the data and results used in the discussion, displayed by the MatLab Command

Window.

The volumetric flows in each line are (CFS):

Q1=

0.6000

Q2=

0.2445

Q3=

0

Q4=

0.2445

Q5=

0.6000

Q6=

0.3555

Q7=

0.2445

Q8=

0.2445

Q9=

0.3555

Q10=

0.2445

13

Q11=

0

Q12=

0.2445

Q13=

0.6000

Q14=

0.6000

The pressure at each node is (psig):

PA =

100

PA1 =

99.8769

PA2 =

99.6037

PA3 =

99.6037

PA4 =

99.8769

PB1 =

98.7840

PB2 =

98.5108

PB3 =

98.5108

14

PB4 =

98.7840

PB =

98.6609

The "C" value of the valves is:

C=

3.1805

The following values are when the piping network does not have any devices

The pressure increase that the pump must overcome is [psi]:

DP=

1.3391

The power the pump delivers to the fluid is:

Whp=

0.4207

The pressure at each node with valves and heat exchangers are (psig):

PAD =

100

PA1D =

99.8769

PA2D =

99.4778

PA3D =

99.4778

PA4D =

99.8769

15

PB1D =

95.8277

PB2D =

95.4286

PB3D =

95.4286

PB4D =

95.8277

PBD =

95.7046

The following values are when the piping network has valves and heat exchangers:

The pressure increase that the pump must overcome is [psi]:

DP=

4.2954

The power the pump delivers to the fluid is:

Whp=

1.3496

>>

MatLab Program: %Program Created on 3/26/13 %This Program uses the Hardy-Cross Method to determine the volumetric %flow of a pipe network. clc clear all close all %Known Data Pa=100*144; % (lbf/ft^2) Q=1.2; % (CFS) %Loop I

Q1=-0.6; Q2=0.075; Q3=0.15;

16

Q4=0.3; Q5=0.6; %Loop II Q4II=-0.3; Q6=0.3; Q7=-0.15; Q10=-0.3; %Loop III Q3III=-0.15; Q7III=0.15; Q8=-0.075; Q11=-0.15; %Loop IV Q2IV=-0.075;

Q8IV=0.075; Q9=-0.675;

Q12=-0.075; %Loop V Q10V=0.3; Q11V=0.15; Q12V=0.075; Q13=0.6; Q14=-0.6; %Other Known Data N=1.852; N1=0.852; m=4.8704; k=4.727; K1=(k*50)/((100^N)*((6/12)^m)); K2=(k*20)/((100^N)*((3/12)^m)); K3=(k*20)/((100^N)*((3/12)^m)); K4=(k*20)/((100^N)*((3/12)^m)); K5=(k*50)/((100^N)*((6/12)^m)); K6=(k*40)/((100^N)*((3/12)^m)); K7=(k*40)/((100^N)*((3/12)^m)); K8=(k*40)/((100^N)*((3/12)^m)); K9=(k*40)/((100^N)*((3/12)^m)); K10=(k*20)/((100^N)*((3/12)^m)); K11=(k*20)/((100^N)*((3/12)^m)); K12=(k*20)/((100^N)*((3/12)^m)); K13=(k*50)/((100^N)*((6/12)^m));

K14=(k*50)/((100^N)*((6/12)^m)); DQI=1; DQII=1; DQIII=1; DQIV=1; DQV=1; %Iteration

while abs(DQI)>1*10^-100 && abs(DQII)>1*10^-100 && abs(DQIII)>1*10^-100 && abs(DQIV)>1*10^-100 && abs(DQV)>1*10^-100 %Loop I X1=abs(Q1^N1); X2=abs(Q2^N1); X3=abs(Q3^N1); X4=abs(Q4^N1); X5=abs(Q5^N1); Y1=K1*Q1*X1; Y2=K2*Q2*X2; Y3=K3*Q3*X3; Y4=K4*Q4*X4; Y5=K5*Q5*X5;

17

Z1=K1*X1; Z2=K2*X2; Z3=K3*X3; Z4=K4*X4; Z5=K5*X5; AI=Y1+Y2+Y3+Y4+Y5; BI=Z1+Z2+Z3+Z4+Z5; DQI=(-AI)/(N*BI); %Loop II X4II=abs(Q4II^N1); X6=abs(Q6^N1); X7=abs(Q7^N1); X10=abs(Q10^N1); Y4II=K4*Q4II*X4II;

Y6=K6*Q6*X6; Y7=K7*Q7*X7;

Y10=K10*Q10*X10; Z4II=K4*X4II; Z6=K6*X6; Z7=K7*X7; Z10=K10*X10; AII=Y4II+Y6+Y7+Y10; BII=Z4II+Z6+Z7+Z10; DQII=(-AII)/(N*BII); %Loop III X3III=abs(Q3III^N1); X7III=abs(Q7III^N1); X8=abs(Q8^N1); X11=abs(Q11^N1); Y3III=K3*Q3III*X3III; Y7III=K7*Q7III*X7III; Y8=K8*Q8*X8; Y11=K11*Q11*X11; Z3III=K3*X3III; Z7III=K7*X7III; Z8=K8*X8; Z11=K11*X11; AIII=Y3III+Y7III+Y8+Y11; BIII=Z3III+Z7III+Z8+Z11; DQIII=(-AIII)/(N*BIII); %Loop IV

X2IV=abs(Q2IV^N1); X8IV=abs(Q8IV^N1); X9=abs(Q9^N1); X12=abs(Q12^N1); Y2IV=K2*Q2IV*X2IV; Y8IV=K8*Q8IV*X8IV; Y9=K9*Q9*X9;

Y12=K12*Q12*X12; Z2IV=K2*X2IV; Z8IV=K8*X8IV; Z9=K9*X9; Z12=K12*X12; AIV=Y2IV+Y8IV+Y9+Y12; BIV=Z2IV+Z8IV+Z9+Z12; DQIV=(-AIV)/(N*BIV); %Loop V X10V=abs(Q10V^N1); X11V=abs(Q11V^N1); X12V=abs(Q12V^N1); X13=abs(Q13^N1);

18

X14=abs(Q14^N1); Y10V=K10*Q10V*X10V; Y11V=K11*Q11V*X11V; Y12V=K12*Q12V*X12V; Y13=K13*Q13*X13; Y14=K14*Q14*X14; Z10V=K10*X10V; Z11V=K11*X11V; Z12V=K12*X12V; Z13=K13*X13; Z14=K14*X14; AV=Y10V+Y11V+Y12V+Y13+Y14; BV=Z10V+Z11V+Z12V+Z13+Z14; DQV=(-AV)/(N*BV);

%Q Calculations %Loop I

Q1=Q1+DQI; Q2=Q2+(DQI-DQIV); Q3=Q3+(DQI-DQIII); Q4=Q4+(DQI-DQII); Q5=Q5+DQI; %Loop II Q4II=Q4II+(DQII-DQI); Q6=Q6+DQII; Q7=Q7+(DQII-DQIII); Q10=Q10+(DQII-DQV); %Loop III Q3III=Q3III+(DQIII-DQI); Q7III=Q7III+(DQIII-DQII); Q8=Q8+(DQIII-DQIV); Q11=Q11+(DQIII-DQV); %Loop IV Q2IV=Q2IV+(DQIV-DQI); Q8IV=Q8IV+(DQIV-DQIII); Q9=Q9+DQIV; Q12=Q12+(DQIV-DQV); %Loop V Q10V=Q10V+(DQV-DQII); Q11V=Q11V+(DQV-DQIII); Q12V=Q12V+(DQV-DQIV); Q13=Q13+DQV;

Q14=Q14+DQV; end if Q3 < 1*10^-10 && Q11 < 1*10^-10 Q3=0; Q11=0; end %Part A

Q1=abs(Q1); Q2=abs(Q2); Q3=abs(Q3); Q4=abs(Q4); Q7=abs(Q7); Q8=abs(Q8); Q9=abs(Q9); Q10=abs(Q10); Q11=abs(Q11); Q12=abs(Q12); Q14=abs(Q14); %The head losses in each line with no devices H1=K1*Q1^N;

19

H2=K2*Q2^N; H3=K3*Q3^N; H4=K4*Q4^N; H5=K5*Q5^N; H6=K6*Q6^N; H7=K7*Q7^N; H8=K8*Q8^N; H9=K9*Q9^N; H10=K10*Q10^N; H11=K11*Q11^N; H12=K12*Q12^N; H13=K13*Q13^N; H14=K14*Q14^N; %The pressure at each node with no devices (Lbf/ft^2)

Pa1=(Pa-H5*(62.3)); Pa2=(Pa1-H4*(62.3));

Pa3=(Pa2-H3*(62.3)); Pa4=(Pa-H1*(62.3)); Pb1=(Pa1-H6*(62.3)); Pb2=(Pb1-H10*(62.3)); Pb3=(Pb2-H11*(62.3)); Pb4=(Pa4-H9*(62.3)); Pb=(Pb1-H13*(62.3)); disp('The volumetric flows in each line are (CFS): '); fprintf('\n'); disp('Q1= '),disp(Q1); disp('Q2= '),disp(Q2); disp('Q3= '),disp(Q3); disp('Q4= '),disp(Q4); disp('Q5= '),disp(Q5); disp('Q6= '),disp(Q6); disp('Q7= '),disp(Q7); disp('Q8= '),disp(Q8); disp('Q9= '),disp(Q9); disp('Q10= '),disp(Q10); disp('Q11= '),disp(Q11); disp('Q12= '),disp(Q12); disp('Q13= '),disp(Q13); disp('Q14= '),disp(Q14); %The pressure at each node with no devices (PSI) disp('The pressure at each node are (psig): ');

PA=Pa/144 PA1=Pa1/144 PA2=Pa2/144 PA3=Pa3/144 PA4=Pa4/144 PB1=Pb1/144 PB2=Pb2/144

PB3=Pb3/144 PB4=Pb4/144 PB=Pb/144 %Part B %"C" Values HfY=K4*(0.3^N); HfX=K7*(0.3^N); HfT=(2*(HfY))+HfX+63*(0.3^2); %Hft must be the same as the one with the valve V=(4*0.3)/(pi*(3/12)^2); C=(HfT-(K6*0.3^N)-63*(0.3^2))/((V^2)/(2*(32.2))); disp('The "C" value of the valves is: '); fprintf('\n');

20

disp('C= '),disp(C); %Part C %Necesary pressure increase for the pump (without any devices) DP=PA-PB; W=DP*144*Q; % [lbf-ft/s] Whp=W/550; %[HP] disp('The following values are when the piping network does not have any devices'); disp('The pressure increase that the pump must overcome is [psi]: '); fprintf('\n'); disp('DP= '),disp(DP); disp('The power the pump delivers to the fluid is: '); fprintf('\n'); disp('Whp= '),disp(Whp); %The Pressure at each node with devices

H33=0; H44=K4*((0.3)^N);

H22=K2*((0.3)^N); H66=HfT; H77=HfX+63*(0.3^2); H88=H77; H99=HfT; H1010=K10*(0.3^N); H1111=0; H1212=K12*(0.3^N); Pa22=(Pa1-H44*(62.3)); Pa33=(Pa22-H33*(62.3)); Pb11=(Pa1-H66*(62.3)); Pb22=(Pb11-H1010*(62.3)); Pb33=(Pb22-H1111*(62.3)); Pb44=(Pa4-H99*(62.3)); Pbd=(Pb11-H13*(62.3)); %In [psi] disp('The pressure at each node with valves and heat exchangers are (psig): '); PAD=Pa/144 PA1D=Pa1/144 PA2D=Pa22/144 PA3D=Pa33/144 PA4D=Pa4/144 PB1D=Pb11/144 PB2D=Pb22/144 PB3D=Pb33/144

PB4D=Pb44/144 PBD=Pbd/144 %Necesary pressure increase for the pump (without any devices) DPD=PAD-PBD; WD=DPD*144*Q; % [lbf-ft/s] WhpD=WD/550; %[HP] disp('The following values are when the piping network has valves and heat exchangers: ');

disp('The pressure increase that the pump must overcome is [psi]: '); fprintf('\n'); disp('DP= '),disp(DPD); disp('The power the pump delivers to the fluid is: '); fprintf('\n'); disp('Whp= '),disp(WhpD);

21

Cleaning Pipeline:

High Pressure Cleaning

Standard high pressure cleaning services for industrial pipes and sewer and drain pipe

cleaning utilize combination flusher and vacuum units with high flow rates for flushing and

removing debris. During the high-pressure cleaning process, high-pressure water pumps and

hoses with flushing head attachments clean debris from inside of the pipe, while pipe walls are

pressure-cleaned as the hose is pulled through the pipe. High thrust capabilities are sometimes

needed for pulling the hose long distances.

After debris is loosened from pipe walls, the water then pulls the suspended debris back

to the manhole where it can be vacuumed or otherwise removed. For sewer and drain pipe

cleaning, special filtering systems can remove debris from the water, with filtered water then

pumped back into the sewer.

Specialized units with larger hoses, higher water output and higher pressure are also

available for jobs that require it. Industrial pipe cleaning and tube cleaning typically require

higher pressures and custom spray patterns (available with customizable flushing head

attachments) for successful cleaning.

Specialized nozzles are also available for navigating elbows and tees, cleaning laterals, and for

high thrust angles.

Robotic Pipe Cleaning

For some industrial pipe cleaning jobs, a robotic cutter is used to detach a blockage, such

as a root mass or hardened calcite deposit, from the wall of a pipe. The range of movement and

versatility of a robotic cutter provides an important function for difficult pipe cleaning job such

as these. Robotic pipe cleaning utilizes CCTV technology and specialized reaming attachments

propelled by hydraulic motors. Reaming and cutting is guided and monitored by CCTV cameras

while trained field technicians assess progress. After blockages are detached from pipe walls, the

debris is read to be flushed and removed.

22

Hazen-Williams Coefficient:

Material Hazen-Williams Coefficient

- c -

ABS - Acrylonite Butadiene Styrene 130

Aluminum 130 - 150

Asbestos Cement 140

Asphalt Lining 130 - 140

Brass 130 - 140

Brick sewer 90 - 100

Cast-Iron - new unlined (CIP) 130

Cast-Iron 10 years old 107 - 113

Cast-Iron 20 years old 89 - 100

Cast-Iron 30 years old 75 - 90

Cast-Iron 40 years old 64-83

Cast-Iron, asphalt coated 100

Cast-Iron, cement lined 140

Cast-Iron, bituminous lined 140

Cast-Iron, sea-coated 120

Cast-Iron, wrought plain 100

Cement lining 130 - 140

Concrete 100 - 140

Concrete lined, steel forms 140

Concrete lined, wooden forms 120

Concrete, old 100 - 110

Copper 130 - 140

Corrugated Metal 60

Ductile Iron Pipe (DIP) 140

Ductile Iron, cement lined 120

Fiber 140

Fiber Glass Pipe - FRP 150

Galvanized iron 120

Glass 130

Lead 130 - 140

Metal Pipes - Very to extremely smooth 130 - 140

Plastic 130 - 150

Polyethylene, PE, PEH 140

Polyvinyl chloride, PVC, CPVC 150

Smooth Pipes 140

Steel new unlined 140 - 150

Steel, corrugated 60

Steel, welded and seamless 100

Steel, interior riveted, no projecting rivets 110

Steel, projecting girth and horizontal rivets 100

Steel, vitrified, spiral-riveted 90 - 110

Steel, welded and seamless 100

Tin 130

Vitrified Clay 110

Wrought iron, plain 100

Wooden or Masonry Pipe - Smooth 120

Wood Stave 110 - 120