How to access published papers for free

February 08, 2020 Add Comment
how to access published papers for free
Source: The Verge

Accessing a published paper in the right way is too costly. So Alexandra Elbakyan an Under Graduate student from Kazakhstan developed a website that can access all the published papers from the top publishers. The pirate queen was sued for infringing copyrights.

Where can you access?

 https://sci-hub.tw/

If the link is not working. Please let me know in the comments. I will update the working link.

How can you access?

You need to enter DOI of the published papers to download.

What is DOI?


Is This Legal?

No

Then why am I promoting it?

Right to education to everyone is part of the Indian Constitution. Even though it is wrong to infringe copyrights, I encourage this project as its main aim is to have an Open Access to all scientific published work. This blog post is the last thing I can do to support a good cause. 

Got More Time?

Read this article on the verge to find out more about pirate queen Alexandra Elbakyan! 


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PIPELINE STIMULATION LAB ANALYSIS

February 05, 2019 Add Comment


PIPELINE STIMULATION LAB ANALYSIS notes for m.tech pipeline engineering Jawaharlal Nehru technological university Kakinada.


Download Lab observation here

https://mega.nz/#!i0cUTQzD!jOS89yB4wp9X5l9wqPgGvSsFD370KbSsl-OhIuPbY04


download record here

https://drive.google.com/file/d/1UbkWvrOlpknR_D6bAKSu_STEIxi782ZX/view?usp=sharing



Aim: To solve the Q-Equations and obtain required flow rates
q(1)+q(2)-q(3)-0.06=0
[1469*(q(1)^1.974)]-[2432*(q(2)^1.927)]-15=0
[1469*(q(1)^1.974)]+[5646*(q(3)^1.971)]-40=0
Software required: Matlab
Matlab Code:
Clc
clear
syms q1 q2 q3
q=[q1 q2 q3];
f(1)=q(1)+q(2)-q(3)-0.06;
f(2)=[1469*(q(1)^1.974)]-[2432*(q(2)^1.927)]-15;
f(3)=[1469*(q(1)^1.974)]+[5646*(q(3)^1.971)]-40;
for i=1:3
for j=1:3
d(i,j)=diff(f(i),q(j));
end
end
x=0.1;
y=0.05;
z=0.09;
for i=1:10
D=subs(d,q,[x(i) y(i) z(i)]);
F=subs(f,q,[x(i) y(i) z(i)]);
Q=subs(q,q,[x(i) y(i) z(i)]);
X=inv(D)*F';
M=Q'-X;
x(i+1)=M(1);
y(i+1)=M(2);
z(i+1)=M(3);
end
fprintf('required flow rates are')
M1=eval(M)

Output:
Command Window
Required flow rates are
M1 =

0.1022
0.0200
0.0622
Aim: To solve the Q-Equations and obtain required flow rates
q(1)-q(2)-q(4)-0.06=0
q(2)-q(3)-0.08=0
q(3)+q(4)-q(5)-0.05=0
[196*(q(1)^1.819)]+[3520*(q(2)^1.955)]+[2380*(q(3)^1.895)]+[192*(q(5)^1.834)]-60=0
[748.56*(q(4)^1.892)]+[192*(q(5)^1.834)]-5=0
Software required: Matlab
Matlab Code:
clc
clear all
syms q1 q2 q3 q4 q5
q=[q1 q2 q3 q4 q5];
f(1)=q(1)-q(2)-q(4)-0.06;
f(2)=q(2)-q(3)-0.08;
f(3)=q(3)+q(4)-q(5)-0.05;
f(4)=[196*(q(1)^1.819)]+[3520*(q(2)^1.955)]+[2380*(q(3)^1.895)]+[192*(q(5)^1.834)]-60;
f(5)=[748.56*(q(4)^1.892)]+[192*(q(5)^1.834)]-5;
for i=1:5
for j=1:5
d(i,j)=diff(f(i),q(j));
end
end
x=0.2;
y=0.1;
z=0.02;
r=0.04;
s=0.01;
for i=1:10
D=subs(d,q,[x(i) y(i) z(i) r(i) s(i)]);
F=subs(f,q,[x(i) y(i) z(i) r(i) s(i)]);
Q=subs(q,q,[x(i) y(i) z(i) r(i) s(i)]);
X=inv(D)*F';
M=Q'-X;
x(i+1)=M(1);
y(i+1)=M(2);
z(i+1)=M(3);
r(i+1)=M(4);
s(i+1)=M(5);
end
fprintf('required flow rates are')
M1=eval(M)

Output:
Command Window
Required flow rates are
M1 =

0.2320
0.1056
0.0256
0.0664
0.0420

Aim: To solve the H-Equations and obtain required heads
[[(500-h1)/k(1)]^(1/n(1))]-[[(h1-h2)/k(2)]^(1/n(2))]-[[(h1-h3)/k(4)]^(1/n(4))]-0.5=0
[[(h1-h2)/k(2)]^(1/n(2))]-[[(h2-h4)/k(3)]^(1/n(3))]-0.35=0
[[(h1-h3)/k(4)]^(1/n(4))]-[[(h3-h4)/k(5)]^(1/n(5))]-0.5=0
[[(h2-h4)/k(3)]^(1/n(3))]+[[(h3-h4)/k(5)]^(1/n(5))]-[[(h4-h5)/k(6)]^(1/n(6))]-0.5=0
[[(h4-h5)/k(6)]^(1/n(6))]-0.25=0
Software required: Matlab
Matlab Code
clc
clear all
k=[5.6845 16.4967 24.3685 24.7450 19.0411 126.3843];
n=[1.9381 1.9185 1.8858 1.9185 1.8611 1.8970];
syms h1 h2 h3 h4 h5
h=[h1 h2 h3 h4 h5];
f(1)=[[(500-h1)/k(1)]^(1/n(1))]-[[(h1-h2)/k(2)]^(1/n(2))]-[[(h1-h3)/k(4)]^(1/n(4))]-0.5;
f(2)=[[(h1-h2)/k(2)]^(1/n(2))]-[[(h2-h4)/k(3)]^(1/n(3))]-0.35;
f(3)=[[(h1-h3)/k(4)]^(1/n(4))]-[[(h3-h4)/k(5)]^(1/n(5))]-0.5;
f(4)=[[(h2-h4)/k(3)]^(1/n(3))]+[[(h3-h4)/k(5)]^(1/n(5))]-[[(h4-h5)/k(6)]^(1/n(6))]-0.5;
f(5)=[[(h4-h5)/k(6)]^(1/n(6))]-0.25;
for i=1:5
for j=1:5
d(i,j)=diff(f(i),h(j));
end
end
p=476;
q=465;
r=460;
s=458;
t=450;
for i=1:10
D=subs(d,h,[p(i) q(i) r(i) s(i) t(i)]);
F=subs(f,h,[p(i) q(i) r(i) s(i) t(i)]);
H=subs(h,h,[p(i) q(i) r(i) s(i) t(i)]);
X=inv(D)*F';
M=H'-X;
p(i+1)=M(1);
q(i+1)=M(2);
r(i+1)=M(3);
s(i+1)=M(4);
t(i+1)=M(5);
end
fprintf('Required Heads are:')
M1=eval(M)
Output:
Command Window
Required flow rates are
M1 =
476.0566
464.7825
460.6951
458.9139
449.8025


Exp. No: Date
Aim: To solve the H-Equations and obtain required heads
[[(100-h1)/k(1)]^(1/n(1))]-[[(h1-h2)/k(2)]^(1/n(2))]-[[(h1-h3)/k(4)]^(1/n(4))]-1.0=0
[[(h1-h2)/k(2)]^(1/n(2))]+[[(h3-h2)/k(3)]^(1/n(3))]-1.5=0
[[(h1-h3)/k(4)]^(1/n(4))]-[[(h3-h2)/k(3)]^(1/n(3))]+[[(90-h3)/k(5)]^(1/n(5))]-0.8=0

Software required: Matlab
Matlab Code
clc
clear all
k=[7.59 9.63 48.6 39.7 16.5];
n=[1.936 1.901 1.882 1.768 1.935];
syms h1 h2 h3
h=[h1 h2 h3];
f(1)=[[(100-h1)/k(1)]^(1/n(1))]-[[(h1-h2)/k(2)]^(1/n(2))]-[[(h1-h3)/k(4)]^(1/n(4))]-1.0;
f(2)=[[(h1-h2)/k(2)]^(1/n(2))]+[[(h3-h2)/k(3)]^(1/n(3))]-1.5;
f(3)=[[(h1-h3)/k(4)]^(1/n(4))]-[[(h3-h2)/k(3)]^(1/n(3))]+[[(90-h3)/k(5)]^(1/n(5))]-0.8;
for i=1:3
for j=1:3
d(i,j)=diff(f(i),h(j));
end
end
p=68;
q=56;
r=67;
for i=1:10
D=subs(d,h,[p(i) q(i) r(i)]);
F=subs(f,h,[p(i) q(i) r(i)]);
H=subs(h,h,[p(i) q(i) r(i)]);
X=inv(D)*F';
M=H'-X;
p(i+1)=M(1);
q(i+1)=M(2);
r(i+1)=M(3);
end
fprintf('Required Heads are:')
M1=eval(M)

Output:
Command Window
Required flow rates are
M1 =

67.5171
56.7927
67.2361


Aim: To solve the Delta Q-Equations and obtain required Delta Qs are
K(1)*((q0(1)+dq(1)+dq(2))^n(1))-K(2)*((q0(2)-dq(1))^n(2))-ws1+ws2=0
K(1)*((q0(1)+dq(1)+dq(2))^n(1))+K(3)*((q0(3)+dq(2))^n(3))-ws1+ws3=0

Software required: Matlab
Matlab Code
clc
clear all
syms dq1 dq2
dq=[dq1 dq2];
K=[1469 2432 5646];
n=[1.974 1.927 1.971];
q0=[0.10 0.05 0.09];
ws1=100;
ws2=85;
ws3=60;
f(1)=K(1)*((q0(1)+dq(1)+dq(2))^n(1))-K(2)*((q0(2)-dq(1))^n(2))-ws1+ws2;
f(2)=K(1)*((q0(1)+dq(1)+dq(2))^n(1))+K(3)*((q0(3)+dq(2))^n(3))-ws1+ws3;
for i=1:2
for j=1:2
d(i,j)=diff(f(i),dq(j));
end
end
x=0;
y=0;
for i=1:10
D=subs(d,dq,[x(i) y(i)]);
F=subs(f,dq,[x(i) y(i)]);
Q=subs(dq,dq,[x(i) y(i)]);
X=inv(D)*F';
M=Q'-X;
x(i+1)=M(1);
y(i+1)=M(2);
end
fprintf('Required Delta Qs are:')
M1=eval(M)

Output:
Command Window
Required Delta Qs are:
M1 = 0.0300
-0.0278




Exp. No: Date
Aim: To solve the Delta Q-Equations and obtain required Delta Qs are
K(1)*((q0(1)+dq(1))^n(1))+K(2)*((q0(2)+dq(1)-dq(2))^n(2))-K(3)*((q0(3)-dq(1))^n(3))-K(4)*((q0(4)-dq(1))^n(4))=0;
-k(5)*((q0(5)dq(2))^n(5))+K(6)*((q0(6)+dq(2))*n(6))+K(7)*((q0(7)+dq(2))^
n(7))-K(2)*((q0(2)+dq(1)-dq(2))^n(2))=0

Software required: Matlab
Matlab Code
clc
clear all
syms dq1 dq2
dq=[dq1 dq2];
K=[1.793 0.497 4.108 2.717 0.755 2.722 1.628];
n=[1.929 1.938 1.921 1.945 1.917 1.942 1.878];
q0=[1.75 3.55 1.05 1.75 1.8 1.5 0.4];
ws1=100;
ws2=85;
ws3=60;
f(1)=K(1)*((q0(1)+dq(1))^n(1))+K(2)*((q0(2)+dq(1)-dq(2))^n(2))-K(3)*((q0(3)-dq(1))^n(3))-K(4)*((q0(4)-dq(1))^n(4));
f(2)=-K(5)*((q0(5)dq(2))^n(5))+K(6)*((q0(6)+dq(2))*n(6))+K(7)*((q0(7)+
dq(2))^n(7))-K(2)*((q0(2)+dq(1)-dq(2))^n(2));
for i=1:2
for j=1:2
d(i,j)=diff(f(i),dq(j));
end
end
x=0;
y=0;
for i=1:10
D=subs(d,dq,[x(i) y(i)]);
F=subs(f,dq,[x(i) y(i)]);
Q=subs(dq,dq,[x(i) y(i)]);
X=inv(D)*F';
M=Q'-X;
x(i+1)=M(1);
y(i+1)=M(2);
end
fprintf('Required Delta Qs are:')
M1=eval(M)

Output:
Command Window
Required Delta Qs are:
M1 =

0.0592
0.0071

Exp No: Date:
Aim: To solve the following Darcy Weisbach equation and Colebrook white equation by using ‘f solve’ command and obtain the required diameter and friction factor.
Software Required: Matlab
Matlab Code:
function output1=sowji(ig)
D=ig(1);
f=ig(2);
e=0.002/12;
hf=40;
Q=2;
g=32.2;
v=0.0000141;
L=3000;
A=pi*(D^2)/4;
eq1=(1/sqrt(f))-1.14+2*log10((e/D)+((7.3434728*v*D)/(Q*sqrt(f))));
eq2=hf-(f*L*(Q^2)/(D*2*g*A));
output1=[eq1 eq2];
end
Output:
Command Window
>> fsolve (@sowji,[0.2 0.6])
Equation solved:
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
ans =
0.4654 0.0170

Exp No:
Date:
Aim: To solve the following equations using ‘fsolve’ command
flow*(1-c)-rate=0
flow*(1-T)+(b*rate)=0
Software Required: Matlab
Matlab Code:
function output=prob(ig)
c=ig(1);
T=ig(2);
b=0.25;
g=30;
flow=25;
rate=c*exp(g*(1-(1/T)));
eq1=flow*(1-c)-rate;
eq2=flow*(1-T)+(b*rate);
output=[eq1;eq2];
end
Output:
Command Window
>> fsolve(@prob, [0.2 0.6])

Equation solved:
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
and =

0.9422 1.0144

PIPELINE ECONOMIC LAWS AND POLICIES

February 05, 2019 Add Comment




Download unit 3



Download 2nd mid-email attachments




PIPELINE ECONOMIC LAWS AND POLICIESnotes for m.tech pipeline engineering Jawaharlal Nehru technological university Kakinada.

PIPELINE SYSTEM AND AUTOMATION CONTROL

February 05, 2019 Add Comment

PIPELINE SYSTEM AND AUTOMATION CONTROL notes for m.tech pipeline engineering Jawaharlal Nehru technological university Kakinada.






Pipeline System automation And Control text book chapter wise download




Notes Chapter wise Downlaod


PIPELINE RISK MANAGEMENT

PIPELINE RISK MANAGEMENT

February 05, 2019 Add Comment

Pipeline Risk Management complete Notes

Download complete notes here 

PIPELINE PROJECT EVALUATION AND MANAGEMENT

September 18, 2018 Add Comment



PIPELINE PROJECT EVALUATION AND MANAGEMENT TEXTBOOK DOWNLOAD HERE

PIPELINE ROUTE SELECTION AND SURVEYING (PRSS)

September 01, 2018 Add Comment

download here





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