Final Design Project
Optimization of a gas pipeline and pumping facility.
Salim Nasser
Design of Thermal/Fluid Systems
Dr.Y. Cao
EML 4706
Dept. of Mechanical Engineering
Florida International University
Fall Semester 2002
Problem Statement:
The purpose of this design Project is to optimize a gas pipeline and pumping
facility were work can be recovered at the destination point. In order to recover work at
the destination point, a compressor at the entrance point compresses the methane gas
used to much higher pressures so as to increase the density of the gas (reduced the
volume it occupies). A turbine is used at the exit point to expanding gas back to its
original pressure, and therefore, recovering energy in the process. The turbine will cause
the generator to which it is attached to generate this energy as it rotates. By
pressurizing the gas, the diameter of the pipe, which connects the compressor and
turbine, is minimized. This is of great importance since the pipeline has a link of 150km.
On the front end, the system consists of a compressor powered by an electric motor and
the pipeline used to transmit the compressed methane. On the production side of the
system, consisting of a turbine and a generator, we use the energy stored in the
compressed gas to produce electricity. Given a specified mass flow rate we notice that
by using methane at a high pressure permits a high-pressure drop and increases the
density of the gas. This compression does come at an additional cost. In order to
mechanically compress and uncompress the fluid we must have add additional energy
required for compression and not all this energy is recovered at the turbine due to
inefficiencies of the electric motor, compressor, and turbine, as well frictional losses in
the pipeline. See diagram.
Data used in System Optimization:






Methane enters the compressor at 100kPa
Methane leaves the turbine at 100 kPa
All the power generated by the turbine and generator can be used
Gas enters the compressor at 20 C, is cooled after the compression, and
remains at 20 C throughout the pipeline. The distance between the
compressor outlet and the location where the gas is cooled down to 20 C
is negligible compared to the total pipeline length of 150 km.
The methane flow rate is 60 kg/s.
The efficiencies of the electric motor and generator are 95%

The efficiency (with respect to the isentropic process) of the compressor
is 80% and of the turbine is 85%.
The cost related to the system and operation is as follows:


Electric motor and generator first cost, $45 per kilowatt output (
Compressor first cost, $110 per kilowatt input (input designated Wc )

Value of electricity at compressor end of pipeline, 5 cents per kilowatt
hour
Value of electricity at turbine end of pipeline, 6 cents per kilowatt hour
Turbine first cost, $135 per kilowatt output (output designated Wt )




Pipe per cost in dollars per meter length 280D1.6, where D is the pipe
diameter in meters
Life of system is 15 years, the annual interest rate is 8%, and the
operation is continuous.
Diagram:
Related Equations:
:















 
Equations used:
V  AV
eq.2
PV  mRT
Eq.4 P   RT
eq.5
A
Eq.1
Eq.7
dp 
f  mRT  m 

  dl
2 D  PA  A 
Eq.9
 Pdp 
P3
F.2
F.3
4
eq.6
D2
eq.8
dp 
V
V mRT

A
PA
f
V V   dl
2D
eq.10
P2
F.1

m  AV 
eq.3
L
fRTm 2
dl
2 DA2 0
WCin  WMO
WGO
8760  ( 1 
n
i)  1
i  ( 1  i)
n
k 1


mC p   P2  k

T1    T1 

0.75   P1 


1.309 1


kJ  
 P3  1.309 
 kg  
 0.95(0.80)  75   2.2
 300  300 


s 
KgK  

 100, 000 


=((2.931*109
Methane Properties:
 k  1.299
kJ
Kg * K
kJ
R  0.519625
kg * K
 C p  2.21

 F  0.018


Procedure:
The method by which the analysis was done consisted of first constructing an objective
function which produces the total costs of construction. The constraint function is based
primarily on equation f.3, which deals with the pressure drop across the pipeline. This is
the constraint used its dependence on the pipe diameter. In this particular optimization
our ultimate goal is to determine and minimize the price of the compressor, turbine, and
the cost associated with the diameter of the pipe. To do this we must implement tools
learned such as Newton-Raphson method and Lagrange multipliers method.
The following is the order in which the optimization was performed:
1. The objective function and constraint functions are formulated:
Objective function:


Constraint function:
f1 ( p2  p3  d) 
2.931  109
5
2
2
 p2  p3
d
2. The three partial derivatives for the objective and constraint function are
calculated using MathCAD. (See Appendix)
3. Using these derivatives we obtain the Lambda equations:
i1
2953206.5242887092567
p2
.76405867970660146699
   ( 2  p2)
482712020.00603422686
 
 p3 
i2
i3
1
.76405867970660146699
.6
67200000.0  d
constraint i4
   
2.931  109
5
0
   ( 2  p3)
 p3
14655000000.000 


6

d
2
 p2  p3
2
0
2
0
0
d
4. Then we proceed to use the Lambda equations in the Newton-Raphson by taking
their derivatives along with the derivative of the constraints with respect to
P2 , P3 , D, and . See Appendix for calculations.
5. The Newton-Raphson method is applied by creating an iterative program using
MatLAB and applying the values obtained from MathCAD. See appendix for
MatLAB program.
Results and Conclusion:
The final results obtained are as follows:
OBJECTIVE FUNCTION
Minimal cost gas pipeline system
$101,989,236.59
CONSTRAINTS
Work of the generator output
Work of the motor output
Pressure loss
Work of the turbine out
Work of the compressor input
10327.99065kW
31635.70013kW
-1.03137E+12
10871.56911kW
30053.91513kW
Initial cost of compressor
Initial cost of turbine
$3,005,391.51
$1,358,946.14
Optimized size if pipe diameter
0.689m
SYSTEM OPERATION DATA
T1
T3
P1
P2
P3
P4
D
293K
293K
100000Pa
793178.2541Pa
648853.9175Pa
100000Pa
0.689924217m
Based upon the results above we can make some basic conclusions. We see that our
pressure within the system rises to around 800,000 Pa. Given that our output and input
pressure are given at 100,00Pa that leaves us with a compression ratio of around 8:1.
The work and calculations performed for the Lagrange multiplier equations done using
MathCAD are found the appendix section of this document. The Newton-Raphson
method was used to simulate the system and generate the values for P2, P3, and the
pipe diameter D the ultimate solution was found using Excel. The total minimum cost of
the gas pipeline system was found to be $101,989,236.59. This value is based on a
minimum diameter of the pipeline of 0.689 meters, a minimum cost of the compressor
of $3,005,391.51, and a minimum cost of the turbine of $1,358,946.14.
Conclusion:
Analysis of this system concluded that the value found for the pipeline diameter would
minimize both the initial costs of the compressor as well as the turbine, therefore
optimizing the total costs of production of this system.
Appendix:
MATLAB PROGRAM:
% The following program Stimulates a gas pipeline and pumping facility
by means of the
%Newton-Raphson method .
i=0;
p2=600000;
p3=45000;
L=.5;
d=.8;
dp2=.02;
dp3=.02;
dd=.02;
dL=.02;
% after the defined values are set, the program uses a while Loop to
solve for the values
%of d,p3,p2& L. The values are solved for level of accuracy of 0.1%.
while ((abs(dp2) > 0.001) | (abs(dp3) > 0.001) | (abs(dd) > 0.001) |
(abs(dL) > 0.001))
f1=2953206.52/((p2^.76405867970660146699))-(L*(-2*p2));
f2=(-482712020.00603422686/((1/p3)^.76405867970660146699*(p3^2)))(L*(2*p3));
f3= (67200000.0*(d^.6))-(L*(-14655000000.000/(d^6)));
f4 = (2.931E+9/(d^5))-(p2^2) +(p3^2);
B=[f1; f2; f3; f4];
%function 2%
dp2f2= (-2256423.0778489526717/(p2^1.7640586797066014670))+ 2*L;
dp3f2=0;
ddf2=0;
dLf2=2*p2;
%function 3%
dp2f3=0;
dp3f3=(368820308.68/(((1/p3)^1.764))*(p3)^4)+(965424040.01/
(((1/p3)^.764)*(p3)^3))- 2*L;
ddf3=0;
dLf3=-2*p3;
%function 4%
dp2f4=0;
dp3f4=0;
ddf4=(40320000.00/(d^.4))-(87930000000*(L/(d^7)));
dLf4=(14655000000/(d^6));
%function 5%
dp2f5=-2*p2;
dp3f5=2*p3;
ddf5=(-14655000000.000/(d^6));
dLf5=0;
A=[dp2f2 dp3f2 ddf2 dLf2; dp2f3 dp3f3 ddf3 dLf3 ; dp2f4 dp3f4 ddf4
dLf4; dp2f5 dp3f5 ddf5 dLf5];
x=A\B;
dp2=x(1);
dp3=x(2);
dd=x(3);
dL=x(4);
p2=p2-dp2;
p3=p3-dp3;
d=d-dd;
L=L-dL;
i=i+1;
end