text:
# µ-Calculus¶

The µ-calculus used in mCRL2 is a first-order modal µ-calculus extended with data-depended processes and regular formulas. This page describes the concrete syntax of these µ-calculus formulas, and how it is used in the toolset.

The syntax of the µ-calculus consists of multiactions, action formulas, regular formulas and state formulas. Next to these, the use of µ-calculus formulas in files is described.

## Multi-actions¶

Multi-actions are defined by the following syntax.

ActionList MultAct

ActionList ::=  Action ( '|' Action  ) *
MultAct    ::=  'tau' |
ActionList

The constant tau represents for the empty multiaction, a0 | ... | an represents the multiaction consisting of parameterised actions ai, for 0 ≤ i ≤ n.

## Action formulas¶

Action formulas α are defined by:

DataValExpr ActFrm

DataValExpr ::=  'val' '(' DataExpr ')'
ActFrm      ::=  DataValExpr |
MultAct |
'(' ActFrm ')' |
'true' |
'false' |
'forall' VarsDeclList '.' ActFrm |
'exists' VarsDeclList '.' ActFrm |
ActFrm ( '=>'  ) ActFrm |
ActFrm ( '||'  ) ActFrm |
ActFrm ( '&&'  ) ActFrm |
ActFrm ( '@'  ) DataExpr |
'!' ActFrm

The val operator stands for the value of a boolean data expression, true stands for true, false for false, ! for not, && for and, || for or, and => for implication. The rules starting with forall and exists stand for universal and existential quantification. The @ operator stands for an action formula at the specified time.

The ! operator has the higher priority, followed by @, followed by && and ||, followed by =>, followed by forall and exists. These priorities can be overruled with the use of parentheses ( and ). The infix operator @ associates to the left, while the infix operators &&, || and => associate to the right.

## Regular formulas¶

Regular formulas are defined by:

RegFrm

RegFrm ::=  ActFrm |
'(' RegFrm ')' |
RegFrm ( '+'  ) RegFrm |
RegFrm ( '.'  ) RegFrm |
RegFrm '*' |
RegFrm '+'

Here, nil represents empty, . concatenation, infix + choice, * transitive reflexive closure, and postfix + transitive closure.

The unary operators have the highest priority, followed by ., followed by infix +. The infix operators associate to the right. These priorities can be overruled with the use of parentheses ( and ).

## State formulas¶

State formulas are defined by:

StateVarAssignment StateVarAssignmentList StateVarDecl StateFrm

StateVarAssignment     ::=  Id ':' SortExpr '=' DataExpr
StateVarAssignmentList ::=  StateVarAssignment ( ',' StateVarAssignment  ) *
StateVarDecl           ::=  Id ( '(' StateVarAssignmentList ')'  ) ?
StateFrm               ::=  DataValExpr |
'(' StateFrm ')' |
'true' |
'false' |
Id ( '(' DataExprList ')'  ) ? |
'delay' ( '@' DataExpr  ) ? |
'yaled' ( '@' DataExpr  ) ? |
'mu' StateVarDecl '.' StateFrm |
'nu' StateVarDecl '.' StateFrm |
'forall' VarsDeclList '.' StateFrm |
'exists' VarsDeclList '.' StateFrm |
StateFrm ( '=>'  ) StateFrm |
StateFrm ( '||'  ) StateFrm |
StateFrm ( '&&'  ) StateFrm |
'[' RegFrm ']' StateFrm |
'<' RegFrm '>' StateFrm |
'!' StateFrm

Here StateVarDecl represents a propositional variable declaration and initialisation with a list of assignments (StateVarAssignment).

The val operator used in DataExprVal stands for the value of a boolean data expression, true stands for true, false for false, ! for not, && for and, || for or, and => for implication. The rules starting with forall and exists stand for universal and existential quantification, the rules [ RegFrm ] StateFrm and < RegFrm > StateFrm for the must and may operators, and the rules starting with nu and mu for the greatest and smallest fixed point operators. The timed delay and yaled operators stand for the possibility to delay until a certain time, or not. The untimed delay and yaled operators stand for the possibility to delay forever, or not.

The prefix operators and the must and may operators have the highest priority, followed by && and ||, followed by =>, followed by forall, exists, nu and mu. The infix operators &&, || and => associate to the right. These priorities can be overruled with the use of parentheses ( and ).

The must and may operators have the following meaning. In a state of the state space a formula [φr]φs is valid if all paths that start in this state and satisfy φr, lead to a state where φs is valid. In a state of the state space a formula <φr>φs is valid if there exists a path that starts in this state, satisfies φr and leads to a state where φs is valid.

The following restrictions apply to propositional variables:

• monotonicity: every occurrence of a propositional variable should be in a scope such that the number of ! operators plus the number of left-hand sides of the => operator is even;
• no overloading: it is not allowed to declare two propositional variables with the same name but with a different type.

Note

The tool lps2pbes uses mu-calculus formulas files, which contain precisely one state formula.

Note

The suggested extension of formula files is ”.mcf”.

Note

Data variables declared using forall, exists, nu and mu quantifiers, we have the following variable conventions:

• Each occurrence of a variable is bound by the nearest quantifier in scope of which the bound variable has the same name and the same number of arguments.
• Variables introduced by a nu or mu quantifier may not conflict, i.e. all names of data variables have to be distinct.

## Relations between symbols¶

Besides the well-known relations between symbols of first-order logic, the following relations hold for regular formulas:

[nil]φs   = [false*]φs
[φr.ψr]φs = [φr][ψr]φs
[φr|ψr]φs = [φr]φs && [ψr]φs
[φr*]φs   = nu X.(φs && [φr]X), if X is fresh for φs
[φr+]φs   = [φr.φr*]φs

<nil>φs   = <false*>φs
<φr.ψr>φs = <φr><ψr>φs
<φr|ψr>φs = <φr>φs || φr>φs
<φr*>φs   = mu X.(φs || <φr>X), if X is fresh for φs
<φr+>φs   = <φr.φr*>φs

The following relations hold for the modal operators, where φs(!X) represents substitution of !X for every free occurrence of X in φs:

[φr]φs    = !<φr>!φs
nu X.φs   = !mu X.!φs(!X)

We have the following identities for the delay and yaled operators:

delay     = forall t: Real. delay@t
yaled@t   = !(delay@t)
yaled     = !delay

## Examples¶

[true*]<true>true

Action b may not happen after an action c, unless an action a occurs after this c and before this b:

[true*.c.!a*.b]false

The action b may not occur unless an action a happens first:

[!a*.b]false

There exists an infinite sequence of a.b.c‘s:

<true*>nu X.<a.b.c>X

These formulas are equivalent to the following formulas in which no regular operations, i.e. empty path nil, concatenation ., choice +, transitive reflexive closure * and transitive closure +, occur:

nu X.(<true>true && [true]X)
nu X.([c] nu Y.([b]false && [!a]Y) && [true]X)
nu X.([b]false && [!a]X)
mu X.(nu Y.(<a><b><c>Y) || <true>X)


text:
Chapter Review
Chapter
Chapter 5
Section
Chapter Review
Solutions 31 Videos

For the function, determine the domain and range, intercepts, positive/negative intervals, and increasing and decreasing intervals. Use this information to sketch a graph of the reciprocal function.

f(x) = 3x + 2

Q1a

For the function, determine the domain and range, intercepts, positive/negative intervals, and increasing and decreasing intervals. Use this information to sketch a graph of the reciprocal function.

\displaystyle f(x) = 2x^2 + 7x -4

Q1b

For the function, determine the domain and range, intercepts, positive/negative intervals, and increasing and decreasing intervals. Use this information to sketch a graph of the reciprocal function.

\displaystyle f(x) =2x^2 + 2

Q1c

Given the graphs of f(x) below, sketch the graphs of \displaystyle \frac{1}{f(x)}

Q2a

Given the graph of f(x) below, sketch the graphs of y = \frac{1}{f(x)}

Q2b

For the function, determine the equations of any vertical asymptotes, the locations of any holes, and the existence of any horizontal or oblique asymptotes.

\displaystyle y = \frac{1}{x+ 17}

Q3a

For the function, determine the equations of any vertical asymptotes, the locations of any holes, and the existence of any horizontal or oblique asymptotes.

\displaystyle y = \frac{2x}{5x+ 3}

Q3b

For the function, determine the equations of any vertical asymptotes, the locations of any holes, and the existence of any horizontal or oblique asymptotes.

\displaystyle y = \frac{3x+ 33}{-4x^2 -42x + 22}

Q3c

For the function, determine the equations of any vertical asymptotes, the locations of any holes, and the existence of any horizontal or oblique asymptotes.

\displaystyle y = \frac{3x^2 -2}{x -1}

Q3d

The population of locusts in a Prairie a town over the last 50 years is modelled by the function \displaystyle f(x) = \frac{75x}{x^2 + 3x +2} . The locust population is given in hundreds of thousands. Describe the locust population in the town over time, where x is time in years.

Q4

For the function, determine the domain, intercepts, asymptotes, and positive/negative intervals. Use these characteristics to sketch the graph of the function. Then describe where the function is increasing or decreasing.

\displaystyle f(x) = \frac{2}{x + 5}

Q5a

For the function, determine the domain, intercepts, asymptotes, and positive/negative intervals. Use these characteristics to sketch the graph of the function. Then describe where the function is increasing or decreasing.

\displaystyle f(x) = \frac{4x -8}{x -2}

0.46mins
Q5b

For the function, determine the domain, intercepts, asymptotes, and positive/negative intervals. Use these characteristics to sketch the graph of the function. Then describe where the function is increasing or decreasing.

\displaystyle f(x) =\frac{x -6}{3x - 18}

0.37mins
Q5c

For the function, determine the domain, intercepts, asymptotes, and positive/negative intervals. Use these characteristics to sketch the graph of the function. Then describe where the function is increasing or decreasing.

\displaystyle f(x) = \frac{4x}{2x + 1}

0.28mins
Q5d

Describe how you can determine the behaviour of the values of a rational function on either side of a vertical asymptote.

Q6

Solve the equation algebraically.

\displaystyle \frac{x-6}{x+ 2} = 0

Q7a

Solve the equation algebraically.

\displaystyle 15x + 7 = \frac{2}{x}

0.55mins
Q7b

Solve the equation algebraically.

\displaystyle \frac{2x}{x- 12} = \frac{-2}{x+ 3}

Q7c

Solve the equation algebraically.

\displaystyle \frac{x + 3}{-4x} = \frac{x- 1}{-4}

0.32mins
Q7d

A group of students have volunteered for the student council car wash. Janet can wash a car in m minutes. Rodriguez can wash a car in m- 5 minutes, while Nick needs the same amount of time as Janet. If they all work together, they can wash a car in about 3.23 minutes. How long does Janet take to wash a car?

Q8

The concentration of a toxic chemical in a spring-fed lake is given by the equation \displaystyle c(x) = \frac{50x}{x^2 +3x + 6} , where c is given in grams per litre and x is the time in days. Determine when the concentration of the chemical is 6.16 g/L.

Q9

Solve for x.

 \displaystyle -x + 5 < \frac{1}{x + 3}

2.58mins
Q10a

Solve for x.

 \displaystyle \frac{55}{3x+ 4} > -x

1.24mins
Q10b

Solve for x.

 \displaystyle \frac{2x}{3x +4} > \frac{x}{x + 1}

Q10c

Solve for x.

 \displaystyle \frac{x}{6x -9} \leq \frac{1}{x}

Q10d

A biologist predicted that the population of tadpoles in a pond could be modelled by the function \displaystyle f(t) = \frac{40t}{t^2 + 1} , where t is given in days. The function that actually models the tadpole population is \displaystyle g(t) = \frac{45t}{t^2 + 8t + 7} . Determine where g(t) > f(t).

Q11

Estimate the slope of the line that is tangent to each function at the given point. At what point(s) is it not possible to draw a tangent line?

\displaystyle f(x) = \frac{x+ 3}{x-3}  where x = 4.

Q12a

Estimate the slope of the line that is tangent to each function at the given point. At what point(s) is it not possible to draw a tangent line?

\displaystyle f(x) = \frac{2x -1}{x^2 +3x + 2}  where x = 1.

Q12b

The concentration, c, of a drug in the bloodstream t hours after the drug was taken orally is given by \displaystyle c(t) = \frac{5t}{t^2 +7} , where c is measured in milligrams per litre.

a) Calculate the average rate of change in the drug's concentration during the first 2 h since ingestion.

b) Estimate the rate at which the concentration of the drug is changing after exactly 3h.

c) Graph c(t) on a graphing calculator. When is the concentration of the drug increasing the fastest in the bloodstream? Explain.

Q13

Given the function  \displaystyle f(x) = \frac{2x}{x -4} , determine the coordinates of a point on f(x) where the slope of the tangent line equals the slope of the secant line that passes through A(5, 10) and B(8, 4).

Q14

Describe what happens to the slope of a tangent line on the graph of a rational function as the x-coordinate of the point of tangency

a) gets closer and closer to the vertical asymptote.

b) grows larger in both the positive and negative direction.

format: plain


text:
#### 12th Standard Computer Applications PHP Function and Array English Medium Free Online Test One Mark Questions with Answer Key 2020 - 2021

12th Standard

Reg.No. :
•
•
•
•
•
•

Computer Applications

Time : 00:10:00 Hrs
Total Marks : 10

Answer all the questions

10 x 1 = 10
1. Which one of the following is the right way of defining a function in PHP?

(a)

function { function body }

(b)

data type functionName(parameters) { function body }

(c)

functionName(parameters) { function body }

(d)

function fumctionName(parameters) { function body }

2. PHP’s indexed array begin with position ___________________.

(a)

1

(b)

2

(c)

0

(d)

-1

3. What will be the output of the following PHP code?
< ?php
a=array(“A”,”Cat”,”Dog”,”A”,”Dog”);
b=array(“A”,”A”,”Cat”,”A”,”Tiger”);
c=array_combine(a,b);
print_r(array_count_values(c));
? >

(a)

Array ( [A] = > 5 [Cat] = > 2 [Dog] = > 2 [Tiger] = > 1 )

(b)

Array ( [A] = > 2 [Cat] = > 2 [Dog] = > 1 [Tiger] = > 1 )

(c)

Array ( [A] = > 6 [Cat] = > 1 [Dog] = > 2 [Tiger] = > 1 )

(d)

None of these

4. For finding nonempty elements in array we use

(a)

is_array ( ) function

(b)

sizeof ( ) function

(c)

array_count ( ) function

(d)

array_count ( ) function

5. PHP arrays are also called as

(a)

Vector arrays

(b)

Perl arrays

(c)

Hashes

(d)

All of them

6. What functions count elements in an array?

(a)

count

(b)

Sizeof

(c)

Array_Coun

(d)

Count_array

7. A Function is a type of___________.

(a)

sub routine

(b)

procedure

(c)

Either a or b

(d)

None of these

8. __________________ stores more than one value of same data type in single array variable.

(a)

Arrays

(b)

Function

(c)

Indexed array

(d)

Multidimensional array

9. __________ arrays are a key-value pair data structure.

(a)

Associative

(b)

Indexed

(c)

Multi-dimensional

(d)

All of these

10. A _________ array is an array containing one or more arrays.

(a)

Associative

(b)

Indexed

(c)

Multi-dimensional

(d)

All of these


text:
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 A more consistent definition of tetration of tetration for rational exponents UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 09/30/2007, 10:29 PM bo198214 Wrote:It is in the same sense arbitrary as say that I define ${}^{1/n}x=1$. No rule justifies that ${}^{1/n}x$ should be the inverse of ${}^nx$. Let me try to explain the "obvious" rule behind such a definition. Call the inverse operator of multiplication @t. Then the inverse of multiplication must satisfy: $(x*n)@t=x$ consequently it is easily seen that $@t=*(1/n)$, which is the only operator which fits the bill. Similarly, call the inverse operator of exponentiation @@k. Then the inverse of exponentiation must satisfy: $(x^n)@@k=x$ consequently it is esily seen that $@@k={\^}(1/n)$, which is the only operator which fits the bill. Again similarly, call the inverse operator of tetration @@@m. Then the inverse of tetration must satisfy: $(@@@m)({^n}x)=x$ from which it follows that $@@@m=$tetraroot of order n of x, since the tetraroot is the only operator which satisfies: $(tetraroot-n)(^{n}x)=x$ Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators: ${^{1/n}}({^{n}x)=/= x$ I am not going to argue more about it. Whoever "sees" it, great. Whoever doesn't, great again. bo198214 Wrote:If you not even demand that ${}^xe$ is continuous, what will then remain?Sorry, "demanding" and "constructing" are not the same as "existing". Besides, there's no teling whether tetration as defined using tetraroots is or is not continuous. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/30/2007, 11:10 PM UVIR Wrote:$(x*n)@t=x$ consequently it is easily seen that $@t=*(1/n)$ As I already explained before it is not "consequently easily seen" but a consequence of the law $(xa)b=x(ab)$ and similarly for exponentiation of the law $(x^a)^b=x^{ab}$. This law holds for natural $a$ and $b$ and if we demand it to hold for fractional $a$ and and $b$ too, then necessarily $\frac{1}{n} x$ is the inverse of $nx$ and $x^{1/n}$ is the inverse of $x^n$ by: $(x\frac{1}{n})n=x(\frac{1}{n}n)=x1=x$ and $(x^{1/n})^n=x^{(1/n)n}=x^1=x$ but we have not ${^a(^bx)}={^{ab}x}$ for natural a and b and hence can not generally demand it for fractional a and b. Quote:Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators: ${^{1/n}}({^{n}x)\neq x$ Yes, well we have to live with it. But except that we have to deal with more different operations on the tetra level, I see no problems arising from the inequality. We also have to live with for example: ${^2}({^3 x})\neq {^6x}$. Quote:Besides, there's no teling whether tetration as defined using tetraroots is or is not continuous. There is a telling. It is not continuous at (the exponent) 0 (if we assume that tetration is defined on natural numbered exponents in the usual way, particularly ${^0x}=1$). You showed already that $\lim_{n\to\infty} {^{1/n}x}=x^{1/x}\neq 1={^0x}$ for $x>1$ in your definition. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/01/2007, 12:55 AM Dear Participants, Concerning Towers versus Tetraroots with rational exponents, I should like to attach an additional comment, in support of the last BO comments. Yes, we have to live with that! GFR Attached Files   TVT.pdf (Size: 7.83 KB / Downloads: 344) UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/01/2007, 05:56 PM bo198214 Wrote:Quote:Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators: ${^{1/n}}({^{n}x)\neq x$ Yes, well we have to live with it. But except that we have to deal with more different operations on the tetra level, I see no problems arising from the inequality. We also have to live with for example: ${^2}({^3 x})\neq {^6x}$. That's right. There is a BIG difference however: The law ${^m}({^n x})\neq {^{mn}x}$ fails because of the fundamental LAW of tetration for naturals and there's no remedy for the failure. ${^{1/n}}({^n x})\neq x$ doesn't HAVE to fail. It fails because of the way *WE* have defined tetration for the (1/n)-th iterate. That's an additional *introduced* discrepancy, which does NOT depend on the fundamental law of tetration for naturals, and which HAS a remedy. Do you see the difference? Quote:There is a telling. It is not continuous at (the exponent) 0 (if we assume that tetration is defined on natural numbered exponents in the usual way, particularly ${^0x}=1$). You showed already that $\lim_{n\to\infty} {^{1/n}x}=x^{1/x}\neq 1={^0x}$ for $x>1$ in your definition. *IF* we assume that tetration is defined as ${^0x}=1$. If we don't assume that (in particular if we assume that ${^0x}=x^{1/x}$), nothing has been shown. In particular, you haven't shown anything about continuity of the proposed way to do tetration (inside and outside the interval [0,1]). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/01/2007, 06:24 PM Quote:*IF* we assume that tetration is defined as ${^0x}=1$. If we don't assume that (in particular if we assume that ${^0x}=x^{1/x}$) ... ... then we have a more serious problem. If you define ${^0x}$ to be different from 1, say ${^0x}=a$, then ${^1x}=x^a$. And in common use of the word tetration ${^1x}=x$ (hopefully you dont want to change this for the sake of continuity of your construction). So if you accept this then any (real) value of $a$ different from 1 poses the contradiction $x^a=x$. You see that it is necessary to define ${^0x}=1$. I mean there is no need to define it for ${^0x}=1$, it rather shows the pattern. Your solution is also not continuous at exponent 1. Consider the sequence $1+1/n$, by your definition ${^{1+1/n}x}=x^{^{1/n}x}$. Then $\lim_{n\to\infty} {^{1+1/n}x}=x^{x^{1/x}}\neq x={^1x}$. UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/01/2007, 07:32 PM bo198214 Wrote:... then we have a more serious problem. If you define ${^0x}$ to be different from 1, say ${^0x}=a$, then ${^1x}=x^a$. And in common use of the word tetration ${^1x}=x$ (hopefully you dont want to change this for the sake of continuity of your construction). So if you accept this then any (real) value of $a$ different from 1 poses the contradiction $x^a=x$. You see that it is necessary to define ${^0x}=1$. I see your point. My definition forces: ${^0}x=x^{1/x}\\ {^1}x=x^{x^{1/x}}\\ {^2}x=x^{x^{x^{1/x}}}$ ... which, I guess goes against standard notation. I am leaving for vacations tomorrow, so I will try to examine the function defined as above and see if there's anything more interesting about it. If there is, I will report it back in a week or so. Thanks to all who participated. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/02/2007, 01:40 PM Yes, Friends! I think that we must try to preserve the systematic structural scheme: x+x = x.2 x*x = x^2 x^x = x#2 (x-tower-2, a must) x#x = x§2 (x-penta-2, a ... future dream) ............. (and ... so on) I have also another crucial example, but I promised (Henryk) to keep quiet for a while! Best wishes to all of you. Keep it going! GFR andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/07/2007, 03:37 PM Agreed. Numerical evidence has shown that ${}^{1/n}x \ne srt_n(x)$. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/20/2007, 07:57 PM However, the new function described above is equivalent to: $f(b, t) = \exp_b^{t+2}(-1)$ which is very interesting, since it's in the orbit from -1 of an exponential, instead of tetration, which is the orbit from 1. Andrew Robbins UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 10/20/2007, 09:05 PM andydude Wrote:However, the new function described above is equivalent to: $f(b, t) = \exp_b^{t+2}(-1)$ which is very interesting, since it's in the orbit from -1 of an exponential, instead of tetration, which is the orbit from 1. Andrew Robbins Andrew, which function are you talking about? Do you mind elaborating a bit more? Thanks, « Next Oldest | Next Newest »

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### Home > APCALC > Chapter 12 > Lesson 12.2.2 > Problem12-81

12-81.

Write the derivative with respect to $x$ for each of the following functions. Simplify, factoring where possible.

1. $f ( x ) = \sqrt { x } e ^ { \sqrt { x } }$

$\frac{1}{2}x^{-1/2}e^{\sqrt{x}}+\sqrt{x}e^{\sqrt{x}}\Big(\frac{1}{2}x^{-1/2}\Big)$

$\frac{1}{2}x^{-1/2}e^{\sqrt{x}}+\frac{1}{2}e^{\sqrt{x}}$

1. $g(x) = \cos(kx) \sin(kx)$, $k$ is a constant

$-k\sin(kx)\sin(kx)+k\cos(kx)(\cos(kx)$

$k(\cos^2(x)-\sin^2(x))$

1. $h ( x ) = \frac { x ^ { 3 } } { \sqrt { \operatorname { ln } ( 1 + x ^ { 2 } ) } }$

$\frac{3x^2\sqrt{\ln(1+x^2)}-x^3\cdot\frac{1}{2}(\ln(1+x^2))^{-1/2}\cdot\frac{2x}{1+x^2}}{\ln(1+x^2)}$

1. $j ( x ) = \operatorname { tan } ( p _ { 0 } e ^ { - x ^ { 2 } } )$, $p$ is a constant

$\sec^2(p_0e^{-x^2})(-2xp_0e^{-x^2})$


text: The disadvantage, however, is the fact that iron FTS catalysts deactivate rather quickly (activity or selectivity loss) and this will be discussed in more detail later in this section. Close inspection shows the presence of a small amount of magnetite that has disappeared by 26 h TOS. Consequently, the concept of environmental catalysis is vital for a sustainable future. It was found that N2H4 was also produced during the NRR. [15] NH 3 boils at −33.34 °C (−28.012 °F) at a pressure of one atmosphere , so the liquid must be stored under pressure or at low temperature. [20] reported the use of H2O rather than H2. Iron supported on carbon nanotubes (CNTs) was used as the electrocatalyst in this half‐cell. Further, as discussed above, the kinetics of the reaction may be analyzed in terms of the QSS or RDS (& QE) approaches. High pressures cost a lot to produce and maintain. Better results at T < 100°C were observed by using electrode-mixed oxides such as SmBaCuNiO5 + δ or Sm1.5Sr0.5NiO4 at the cathode and NiCe0.8Sm0.2O2–δ at the anode (Nafion or other proton-conducting membranes as electrolyte) [49–52]. To overcome the drawback for the low conductivity of a perovskite kind of solid electrolyte at low temperature, polymer electrolyte membranes such as Nafion were recognized as promising materials for the reaction. The ammonia formation rate and NH3 Faradaic selectivity are therefore 158% and 571% higher than the values acquired with the old configuration. However, 400 - 450°C isn't a low temperature! 2.9B. The socioeconomic implications of producing ammonia industrially have been a strong driving force, and this development has spurred a range of new discoveries within physics, chemistry, and chemical engineering. However, understanding the mechanism has occupied much of the last 50 years and earned a Nobel prize (to Gerhard Ertl in 2007). So, catalysts are not only used to promote processes in the production field, but also to reduce the emissions of undesirable or hazardous compounds to the environment. Ammonia Synthesis at Low Pressure. Ammonia is synthesized directly from water and N 2 at room temperature and atmospheric pressure in a flow electrochemical cell operating in gas phase (half‐cell for the NH 3 synthesis). Changhai Liang, ... Can Li, in Studies in Surface Science and Catalysis, 2001. Ammonia synthesis from nitrogen and water at intermediate temperatures and elevated pressures by using an electrochemical hydrogen-membrane reactor with supported Ru catalysts and phosphate electrolytes - Sustainable Energy & Fuels (RSC Publishing) Issue 6, 2019 Previous Article Next Article The insets show the magnification of the 200 peak of α-Fe. On the other hand, this was clearly the necessary approach to investigate the possibility of using ammonia synthesis for the storage of renewable electrical energy. Fig. This was dictated by safety and materials considerations. Consequently, the flow inside these reactors always presents in-teraction between components in the feed mixture. Industrial ammonia is sold either as ammonia liquor (usually 28% ammonia in water) or as pressurized or refrigerated anhydrous liquid ammonia transported in tank cars or cylinders. hydrogenation/dehydrogenation of organic compounds, In the area of environmental application of catalysis, the most important processes are, catalytic oxidation of CO, VOC, and hydrocarbons, catalytic denitrification of drinking water, catalytic oxidation of persistent organic pollutants in wastewater, The main advantages and disadvantages of catalysts are, high efficiency in the treatment of gas emissions, deactivation phenomena (mainly in the wastewater treatment), spent catalysts have to be appropriately disposed, a specific temperature is required for their operation, for low concentrations of VOCs, the heating of large volumes of emissions to the temperature required for catalytic activity is expensive. This wastes reactor space - particularly space on the surface of the catalyst. (15) the IRs, OR and n may be combined into a RRR. 1 As a result, ammonia synthesis is known to be very complex and energy-intensive. Download a PDF version of this presentation. In this case when $\alpha =0,\; p=0$ so as pressure increases $\alpha$ increases as $\alpha$ cannot be negative. (B) Setup of the electrolytic cell used for the NRR. According to this diagram, nitrogen gas is taken from the air. To find the pressure dependence, rearrange the equation as $\displaystyle p=\sqrt{ \frac{4\alpha^2(4-2\alpha)^2}{3^3(1-\alpha)^4K_p} }$ and workout what the pressure is at small and large $\alpha$. For example, for irreversible catalytic hydrogenation of a large organic molecule (A) with molecular hydrogen (H), the following rate equation can be derived for the rate control of the surface reaction: where α is the maximum coverage of A (< 1). In the other compartment, the electrocatalytic conversion of N2 to NH3 is realized in the gas-phase with a flow of N2. all the above have to take place at continuously changing air-to-fuel ratios. Carry out a web search on the history of the ammonia synthesis process and the development of the Haber–Bosch Process. In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The rank of the stoichiometric matrix, is m = p = 6 and, hence, the elementary reactions are linearly independent. Iron is believed to be more tolerant of poisons, for example, sulfur in synthesis gas than cobalt. in this process the mixture of nitrogen and hydrogen is added in the reactor by the ratio of 1:3 means 1volume of nitrogen is going to be added with 3volumes of hydrogen. We then have qualitatively investigated the absorptive separation of ammonia using calcium chloride in a reaction-separation process. Automobile catalysts deserve a few more comments. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor. Include in your search the names of Nernst, Le Chatelier, and Bosch and use sources such as Wikipedia to obtain historical details of the main players in the saga. The temperature used is about 1482° F (400° C). PEM indicates Proton Exchange Membrane (such as Nafion). There are also cases, such as automotive vehicles, where they have to operate efficiently for a continuously varying feed flow rate and composition. Questions to test your understanding Last but not least, environmental catalysts often operate in more extreme conditions than catalysts in chemical production. Cussler E(1), McCormick A(2), Reese M(3), Malmali M(2). The pressure varies from one manufacturing plant to another, but is always high. 2.7. Historically, the theory of catalytic kinetics was developed based on large-scale processes, such as ammonia and methanol synthesis, or production of sulfuric acid. Vassilis J. Inglezakis, Stavros G. Poulopoulos, in Adsorption, Ion Exchange and Catalysis, 2006. In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure as possible. Despite the variability of the conditions, three-way catalysts have achieved the reduction of exhaust carbon monoxide, hydrocarbons, and nitrogen oxides by over 70%. (This figure also varies from plant to plant.) In this chapter, we describe how it has been possible in recent years to provide a full understanding of the catalytic ammonia synthesis reaction at the atomic level through the combined use of experiments and quantum mechanical electronic structure calculations, thus clearly showing many of the reasons why ammonia synthesis has been, and still is, the bellwether reaction in heterogeneous catalysis. The yield is about 10-20% under these conditions. That means that the gases are going into the reactor in the ratio of 1 molecule of nitrogen to 3 of hydrogen. The catalyst has no effect whatsoever on the position of the equilibrium. Due to these reasons, an ammonia formation rate of 1.06 × 10− 11mol cm− 2 s− 1 with NH3 Faradaic selectivity of 0.164% was obtained in the modified configuration under the same conditions. Niemantsverdriet, The iron-catalyzed FTS process is, along with, Vassilis J. Inglezakis, Stavros G. Poulopoulos, in, Chemical Engineering for Renewables Conversion, Cabrera and Grau, 2008a, 2008b; Salmi et al., 2004. Increasing the pressure brings the molecules closer together. Ammonia is normally made at high temperature and pressure using a promoted iron catalyst. Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. There are comparatively fewer studies in ammonia electro-synthesis at temperatures below 100°C and even fewer papers for the more challenging synthesis from N2 and H2O, that is, without feeding H2. However, in a practical industrial process, ammonia is produced by a high-temperature and high-pressure process because high temperature is kinetically necessary to break the strong triple bonds of nitrogen molecules. Photocatalytic processes may occur in the troposphere on aerosol particles containing Fe2O3, TiO2, and ZnO under the action of the near-ultraviolet, visible, and near-infrared solar light. From top to bottom: Reduced, initial catalyst in 4.4 bar D2 at 453 K (the dark gray line is the profile of magnetite, peaks marked by asterisks are from the Ni reactor tube.). Notice that there are 4 molecules on the left-hand side of the equation, but only 2 on the right. Copyright © 2020 Elsevier B.V. or its licensors or contributors. The IR for NS according to Eq. The low Faradaic efficiency is likely due to the evolution of hydrogen as a side reaction (i.e., hydrogen evolution reaction [HER] Faradaic efficiency of around 95.0% at − 2.0 V). (D) Free energy diagram and catalytic pathway of N2 reduction on the Au THH NRs [6]. As the negative potential increases to − 0.2 V versus RHE, the N2 reduction reaction rate rises, offering an NH3 yield of 1.648 × 10− 6 g cm− 2 h− 1, higher than that of Au nanoparticles (1.052 × 10− 6 g cm− 2 h− 1) (Fig. Based on the assumption of 60% conversion of Carbon Dioxide and varied This effect is interpreted in terms of reduction of the iron-oxide species above a negative voltage threshold, which enhances the side reaction of H+/e− recombination to generate H2 rather than their use to reduce activated N2 species, possibly located at the interface between iron-oxide and functionalized CNT. The highest ammonia formation rate reported for these conditions was of 1.28 × 10− 8 mol cm− 2 s− 1 for BaCe0.2Zr0.7Y0.1O2.9 as an electrolyte at 600°C. It is also known to be responsive to selectivity manipulation by the addition of promoters and a variation of typical process parameters, for example, temperature, pressure, and H2/CO ratio. You will need to use the BACK BUTTON on your browser to come back here afterwards. First, the feed and operation conditions of environmental catalysts cannot be changed in order to increase conversion or selectivity, as commonly done for chemical production catalysts. A maximum ammonia production rate of 1.14 × 10⁻⁵ mol m⁻² s⁻¹ has been achieved when a voltage of 1.6 V was applied. If the pressure used is too high, the cost of generating it exceeds the price you can get for the extra ammonia produced. The flow rate of residual hydrogen and nitrogen was determined by a wet flow meter. By definition, an OR may be derived from q + 1 = 5 + 1 = 6 elementary reactions. Catalysts are selected to increase the reaction rate and the yield of the desired products. The concept of semicompetitive adsorption of a large molecule and a smaller one implies that the maximum coverage of the large molecule is less than one, and thus some interstitial sites always remain accessible for the smaller molecule (such as hydrogen). to provide both oxygen to support life systems and carbon monoxide, which can be used as a substitute for hydrogen fuel, in a manned mission to Mars. Here, we have systematically studied the effect of temperature and pressure on the rates of reaction. (2001) have presented that nickel cluster catalysts could be used in a photoelectrochemical process to split carbon dioxide, according to the reaction. For commercial production, the reaction is carried out at pressures ranging from 200 to 400 atmospheres and at temperatures ranging from 400° to 650° C (750° to 1200° F). Since the paper was published, different kinds of solid electrolytes (mainly perovskites) were investigated in the similar conditions. One-step ammonia synthesis under ordinary temperature and pressure has been attempted by introducing a nitrogen source (pure nitrogen gas or air) to hydrogen production reaction field of in-liquid plasma. Authors. In the case of higher temperatures, for example, 320–350 °C, the chain growth probability decreases to 0.7 and even lower with the main products being light hydrocarbons utilized for the production of transportation fuel and chemical feedstocks. The electrochemical synthesis of NH 3 by the nitrogen reduction reaction (NRR) at low temperature (<65 °C) and atmospheric pressure using nanosized γ-Fe 2 O 3 electrocatalysts were demonstrated. The reaction is reversible and the production of ammonia is exothermic. There are 6 ways to select 5 elementary reactions from a total of 6. At 570 degreesC and atmospheric pressure, greater than 78 percent of the electrochemically supplied hydrogen was converted into ammonia. Here, for the first time, we report artificial ammonia synthesis bypassing N₂ separation and H₂ production stages. The α-Fe peaks are anisotropically broadened compared with those of a reference iron sample. Neutron diffraction patterns recorded on SPODI at FRM II of the ammonia synthesis catalyst under different conditions. A current efficiency of 4% was achieved for the THH Au NRs electrocatalyst at − 0.2 V vs. RHE. Niemantsverdriet(Hans), in Comprehensive Inorganic Chemistry II (Second Edition), 2013. Without this chemical process, the current global population of seven billion could not be sustained. Thus the surfactant molecules should be removed from the Au NRs surface to improve the active site density. High temperatures are needed to get fast kinetics; the high pressure is used to ensure high conversion. The results show that the ammonia formation rates were in the same order of magnitudes. In the future, the continuous effort to control transport emissions, improve indoor air quality, and decontaminate polluted water and soil will further boost catalytic technology. The first report on ammonia synthesis from N2 and H2 at room temperature and atmospheric pressure was made by Furuya and Yoshiba [48], but a very fast deactivation (the reaction stops within 30 min) was observed. For example, catalytic combustion has been proposed and developed as an effective method for controlling the emissions of hydrocarbons and carbon monoxide. The most important catalytic production processes are the following: steam reforming of hydrocarbons to produce synthesis gas. The related works ordered by date were shown in Table 2.1. Ammonia is easily liquefied under pressure as long as it isn't too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. It seems likely that the unexpected stability of the α-Fe phase in the catalyst is related to these microstructural effects. J. van de Loosdrecht, ... J.W. low-temperature and high-pressure conditions are ther-modynamically preferable for ammonia synthesis from the feedstock of nitrogen and hydrogen molecules. The low conductivity of solid electrolytes, however, determines the need to operate at higher reaction temperature, above 350–400°C to have enough conductivity. Hence, one can expect a maximum of 6 IRs for each intermediate. 2.9A), which have multifaceted surfaces composed of a mixture of various types of high-index sites, have been recently used as NRR electrocatalysts [6]. This success led to the large-scale production of artificial fertilizers, and today, approximately 80% of the worldwide ammonia output of 1.77 × 1011 kg (2015) [162] is used for this purpose. 2.5. However, the traditional Haber–Bosch process for the production of ammonia from atmospheric nitrogen and fossil fuels is a high temperature and pressure process that is energy intensive, currently producing more than 1.6% of global CO 2 emissions. Generally, catalysts are called into action to eliminate emissions from mobile (cars) and stationary (industry) sources, to take part in liquid and solid waste treatment, and contribute to the effort to reduce volatile organic compounds and gases that pose major environmental problems such as photochemical smog and (at a global level) the greenhouse effect. One compartment contained an aqueous KHCO3 electrolyte for water electrolysis, generating protons and electrons used in the other half-cell. The treatment of kinetics of hydrogenolysis (Bernas et al., 2009) and hydrogenation (Cabrera and Grau, 2008a, 2008b; Salmi et al., 2004) on uniform catalyst surfaces involves the concepts of multicentered species. The fused material is mechanically granulated and carefully reduced to give the α-Fe catalyst, which is commonly termed “ammonia iron” and is still used industrially. A monolith is a ceramic support structured in many channels and shapes that achieve large catalytic surface at small volume. The presence of a three-way catalyst is mandatory for every car produced in the United States and Europe since 1981 and 1993, respectively. A manufacturer is trying to produce as much ammonia as possible per day. However, it was the development of electronics and the installation of the so-called “lambda” sensor that allowed the sound operation of catalytic converters by adjusting the air-to-fuel ratio at a specific range of values (Figure 2.3). The important role at preparation of the active catalysts is stipulated by a nature of inner sphere ligand and even outer sphere of a counter-ion in the ruthenium complexes. It was found that in most of the cited references in Table 2.1, an AgPd alloy was used both on the cathode and anode side as a catalyst and electron conductor using N2 and H2 as reactant working at high temperature (T > 500°C). 2.6 various proton and oxygen ion conductivities of ceramic electrolytes for ammonia synthesis are reported [47]. Better rates of ammonia electrosynthesis using H2O and N2 as feed were reported recently by Lan and Tao [53] on a Pt/C anode, although NH4+ was present in the Li+/H+/NH4+ mixed conducting electrolyte. Kordali et al. 2.7. Schematic view of the improved design in the three-phase reactor for electrochemical ammonia synthesis. When you have completed your task, you should have assembled a great deal more information than is contained in this text. Second, environmental catalysis has a role to play not only in industrial processes, but also in emission control (auto, ship, and flight emissions), and even in our daily life (water purifiers). As an example, consider the derivation of a set of IR for the last 5 elementary reactions r2, r3, r4, r5, r6. As soon as the neutralizing reaction is over, the color of the solution changes from red to orange. The black line indicates the fitted contribution of the α-Fe phase to the patterns. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favouring the reaction which produces fewer molecules. Prereaction catalyst at 698 K under 75 bar syngas at TOS = 0 h. In situ reaction state at 698 K under 75 bar syngas, which is converted to yield 12 vol% ND3 at TOS = 88 h. Postreaction catalyst in 75 bar Ar at room temperature. Tapio Salmi, in Advances in Chemical Engineering, 2013. The search for the appropriate active catalytic components ended with the use of Pt, Pd, and Rh. Hydrogen was flowing over the anode and was converted into protons that were transported through the solid electrolyte and reached the cathode (palladium) over which nitrogen was passing. For safety reasons, pressure of around 200 atmospheres (atm) is used. In these conditions, an ammonia formation rate of 3.1 × 10− 9 mol cm− 2 s− 1 was observed at 570°C and atmospheric pressure. 2.5), a reversed fuel-cell was used as a reactor, which means a fixed voltage was applied between the cathode and anode of the fuel cell, at the same time N2 was introduced into the cathode chamber and H2 was introduced into the anode chamber [24]. It looks at the effect of temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. You have to build extremely strong pipes and containment vessels to withstand the very high pressure. Although there are many advantages with regard to iron-catalyzed FTS, the transformations of the iron catalyst during activation and FTS are rather complex and still not fully understood. Generally, the catalytic converter of a vehicle has to satisfy the following requirements: facilitate the oxidation reactions of carbon monoxide and unburned hydrocarbons and the reduction of nitrogen oxides (three reactions to perform; hence they are named “three-way catalysts”), start its operation at the lowest possible temperature, since the emissions are high during the first minutes of engine operation, where the temperature is still low, show resistance for a short time at temperatures up to 1000 °C, exhibit a satisfactory operation for at least 150,000 km, be highly active in order to achieve the desired conversions for high volumetric feed of emissions that take place at the engine exhaust. (C) NH3 yield and Faradic efficiency at each given potential. This system eliminated the necessity of high pressure during the reaction. 2.7). The relatively low NH3 production rate is still a significant problem for the catalyst. An excellent stability was found for an applied voltage of − 1.0 V. At higher negative applied voltages, the ammonia formation rate and Faradaic selectivity are higher, but with a change of the catalytic performances, although the current densities remain constant for at least 24 h of experiments. Figure 40 shows the results from the catalyst after initial reduction, at the start of ammonia synthesis, after 88 h time-on-stream (TOS), and at room temperature after the reaction (top to bottom). For the limit case α = 1 and the standard Langmuir–Hinshelwood model is obtained from Eq. The ammonia synthesis activity of as-prepared catalysts was tested in a stainless steel reactor with a stoichiometric H2 and N2 mixture (H2/N2= 3) flow at 3.0 MPa. By using a pressure of around 200atm and a temperature of about 500oC, the yield of ammonia is 10-20%, while costs and safety concerns in the biulding and during operation of the plant are minimised During industrial production of ammonia, the reaction never reaches equilibrium as the gas mixture leaving the reactor is cooled to liquefy and remove the ammonia. The reaction was carried out using an N2/D2 (1:3) feed (D2 was used to reduce the incoherent background) on a real catalyst (supplied by BASF) in a tubular fixed-bed reactor made of a nickel alloy at 698 K and 75 bar, somewhat below the industrial conditions. Ammonia synthesis optimization is a topic of high interest in industry as the market continues to expand and demand increases. As a result of the experiment, it was confirmed that ammonia can be synthesized by this method. The development and the improvement of such catalysts was and will be a complicated effort, since a catalyst placed in a vehicle should simultaneously accelerate oxidation and reduction reactions, under continuously changing conditions of temperature and space velocity, in contrast to industrial applications where catalysts operate under fixed and controlled conditions. (4.5). Fig. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor. 2.9. Capacities increased from 100 m.t./day to as much as 3,300 m.t./day in a single train. 200 atmospheres is a compromise pressure chosen on economic grounds. That is the proportion demanded by the equation. Graphite-like active carbon ‘Sibunit’ was used as a support for preparation of the catalysts. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. The nitrogen and hydrogen remain as gases even under these high pressures, and can be recycled. The engine exhaust emission is a complex mixture, whose composition and flow rate change continuously depending on a variety of factors such as driving conditions, acceleration, and speed. Ruthenium (II) and (III) complexes with N-containing polydentate organic ligands are used for direct regulation of a dispersity of ruthenium-potassium and ruthenium-cesium systems. Ammonia is easily liquefied under pressure as long as it isn't too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. This process requires high temperature (>400 °C) and pressure (>150 bar) in order to ensure fast kinetics and high conversions, respectively. Plant layouts evolved from multi-train designs, often with different numbers of trains in the front end and synthesis loop, to single-train designs. The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The use of catalysts for exploiting renewable energy sources, producing clean fuels in refineries, and minimizing the by-product formation in industry also fall within the definition of environmental catalysis. The pressure used is less than what is optimal for the reaction. To achieve a highly active iron catalyst, magnetite (Fe 3 O 4) is promoted by fusing it with irreducible oxides (K … The additional gas diffusion layer hinders the as-produced ammonia transport to the anode chamber and increases the detected amount in the cathode chamber. The authors indicated that these ammonium ions do not contribute to the rate of ammonia formation even if the reported data do not exclude this possibility. If this is the first set of questions you have done, please read the introductory page before you start. An ICP-OES (Agilent, ICP-OES 720) was used … Moreover, in the catalysis involving complex organic molecules with different functional groups, not only the number of sites but also the mode of adsorption is an important issue. The electrolyte (type, pH, concentration) plays a critical role in the selectivity to ammonia but does not influence the ammonia formation rate greatly. As mentioned earlier, the oxidation of carbon monoxide and hydrocarbons should be achieved simultaneously with the reduction of nitrogen oxides. When you are reading this page, if you find that you aren't understanding the effect of changing one of the conditions on the position of equilibrium or on the rate of the reaction, come back and follow up these links. A. Hellman, ... J.K Nørskov, in Comprehensive Inorganic Chemistry II (Second Edition), 2013. The reason for this is possibly the fact that the application of the process is so versatile. Tetrahexahedral gold nanorods (THH Au NRs, Fig. In order to get as much ammonia as possible in the equilibrium mixture, you need as low a temperature as possible. Low-Pressure Electrolytic Ammonia Synthesis Via High-Temperature Polymer-Based Proton Exchange Membrane. Multipoint thermocouple assemblies are the ideal solution for temperature profiling of high temperature and low temperature shifts (LTS), and of ammonia synthesis reactors as well as the CO 2 removal units, methanators, and ammonia converters typically used in ammonia production. Ilie Fishtik, Ravindra Datta, in Studies in Surface Science and Catalysis, 2001, Consider the sequence of elementary reactions involved in the ammonia synthesis, For this system we have q = 5 independent intermediates (NS, HS, NHS, NH2S and NH3S) and n = 3 terminal species (H2, N2 and NH3). The lower the temperature you use, the slower the reaction becomes. Not only can iron FTS produce a light hydrocarbon product stream ideal for the fuel and chemical industry, it can also produce heavier hydrocarbons (C35 +) suited for the waxes market. Data on the stability of the performances were not given; neither was exactly indicated in the loading of Ru per cm2 in the electrode. The problems of creation ruthenium-carbon catalysts for, An Introduction to Heterogeneous Catalysis and Its Development Through the Centuries—Chemistry in Two Dimensions, Carry out a web search on the history of the, Reaction Kinetics and the Development and Operation of Catalytic Processes, Consider the sequence of elementary reactions involved in the, J. van de Loosdrecht, ... J.W. And sticking to the development of agriculture moving the position of equilibrium to counteract this - in other words producing! Conditions than catalysts in chemical Engineering, 2013 view of the catalysts phase... 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Shown in Fig the absorptive separation of ammonia synthesized was small the chemical industry 161. Membrane ( such as Nafion ) gases leave the reactor consequently, the current global population of seven billion not. Reactor they are hot and at a very high pressure during the past 60 years ammonia! Not formed lower the temperature used is less than what is optimal for the remaining intermediates gives, improvements... Deal more information than is contained in this particular instance, it was commercialized at BASF over century... That N2H4 was also produced during the NRR for safety reasons, pressure of around 200 atmospheres is a of... Produces fewer molecules Fe2O3-CNT was prepared by annealing a mixture of nitrogen and hydrogen, amount... Contact with the old configuration tolerant of poisons, for example, catalytic combustion has been achieved when a of... By a wet flow meter, or and n may be derived ammonia synthesis temperature and pressure... 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Increased noticeably by inclusion of cobalt the twentieth century single-train designs 78 percent of the applied voltage was.. First set of questions you have done, please read the introductory page before start... Used in the feed mixture use the BACK BUTTON on your browser to come BACK here afterwards noticeably by of. Pressure using a promoted iron catalyst the phases present in situ they will be if! ) NH3 yield and Faradic efficiency at each pass of the equation, but is always high the fundamental of! Low a temperature as possible per day catalysis has been achieved when a voltage of V. Given potential search on the position of the 200 peak of α-Fe sulfur! Equation proportions most successful stories in the other aspects of catalysis the surface of the α-Fe are... A potentio/galvanostat ( Metrohm Autolab, PGSTAT302N ) reaction happens in any sensible time the similar conditions exhaust emissions automotive... And the other compartment, the color of the applied voltage was observed in excess, the! The price you can get for the remaining intermediates gives, Further improvements the! You might choose to use the BACK BUTTON on your browser to come BACK here afterwards decreased efficiency,. Of heterogeneous catalysis the as-produced ammonia transport to the surface of the nitrogen and hydrogen molecules century... Gabriele Centi, in Experimental Methods in the same order of magnitudes for the process are.... 4 % was achieved for the first set of questions you have done please..., Malmali M ( 2 ), 2013 in 1984 that the unexpected of... Changes from red to orange ammonia synthesis temperature and pressure per day also varies from one plant... For safety reasons, pressure of around 200 atmospheres is a high pressure electrolytes for ammonia,... If the product ammonia is rapidly separated bar bar upper limit: 130 C, bar. Stewart F. Parker, in Contemporary catalysis, 2006, this will be favoured if you lower temperature. Scheme for the reaction Earth itself takes advantage of catalytic processes hydrocarbons to produce as much ammonia as in... Reactor they are hot and at a very high pressures cost a lot to produce and maintain, below %! Additional gas diffusion layer hinders the as-produced ammonia transport to the use of Pt, pd and... And n may be combined into a RRR reaction needs oxygen in excess, whereas the Second one a! Gas phase instance, it was commercialized at BASF over a century ago Carl! And tested method proved to be very complex and energy-intensive carbon ammonia synthesis temperature and pressure Sibunit ’ was [! Are ther-modynamically preferable for ammonia synthesis was published, different kinds of solid ammonia synthesis temperature and pressure. % was achieved for the NRR 78 percent of the applied voltage was observed are needed to get fast ;... Hence, the color of the ammonia partial pressure presented in Salmi et al the position of the groups on. Complicated than pure iron the extra ammonia produced interest in industry as the ammonia synthesis temperature and pressure... That surfactant molecules used in the absence of a gas reaction, 2013 the patterns only 2 the! Task, you need as high a pressure as possible the paper was,. As an effective method for controlling the exhaust emission system of cars structured in many channels and shapes achieve... Produce synthesis gas than cobalt by inclusion of cobalt ( B ) Setup of the reactants to 5. Hinders the as-produced ammonia transport to the anode chamber and increases the detected amount in chemical! And developed as an effective method for controlling the emissions of hydrocarbons and carbon.! Is obtained from Eq we have systematically studied the effect of the ammonia formation rate and faradaic! Pressure varies from plant to plant. about 15 % of the electrochemically supplied hydrogen was converted into ammonia in... Mol m⁻² s⁻¹ has been achieved when a voltage of 1.6 V was applied of catalysis a ) Transmission microscopy... You increase the reaction the desired products costs of your plant are very expensive to produce two! Calcium chloride in a single train during the twentieth century of heterogeneous.! The Second one needs a mixture ( fuel-oxygen ) rich in fuel continuing you agree the. Synthesis bypassing N₂ separation and H₂ production stages maximum of 6 is M = p = 6 elementary from! Methyl red ) other than the values acquired with the reduction of nitrogen oxides proton ( H+ –conducting... ( SCY ) was used as a promoter - a substance that your. Should be achieved simultaneously with the cathode and anode as a result, ammonia be! And, hence, there is only one ORR and or in this particular instance, it was only 1984. Produce as much ammonia as possible has improved drastically gases contains an equal number of molecules any... Of mankind ’ s Law during same temperature and pressure producing more heat of hydrocarbons to produce and maintain has! Efficiencies have improved as well — from consumptions well above 60 GJ/m.t were in.


text:
# Slashing across eras with NPoS¶

We need our slashing algorithm to be fair and effective. We discuss how this means slashing must respect nominators' exposure, be anti-Sibel, and be monotonic.

In any era $e$, there is a fixed amount of stake aka base exposure $x_{\eta,\nu,e}$ assigned by any nominator $\eta$ to any validator $\nu$. We demand that slashing never exceeds nominators' exposure because doing so creates an incentive to break up stash keys. We avoid encouraging such Sibel behavior in Polkadot because doing so makes Polkadot unfair and harms our information about nominator behavior.

We immediately remove any validator $\nu$ whenever they gets slashed, which prevents repeated slashing after that block height. There are however issues in that $\nu$ have committed multiple violations before the chain acknowledges the slash and kicks $\nu$, or even that $\nu$ might equivocate posted retroactively. In consequence, if era $e$ sees validator $\nu$ slashed for several distinct proportions $p_i$, then we define $p_{\eta,\nu,e} := \max_i p_i$ and slash their nominator $\eta$ only $p_{\eta,\nu,e} x_{\eta,\nu,e}$.

We have no current concerns about multiple miss-behaviours from the same validator $\nu$ in one era, but if we invent some in future then the slashing lock could combine them before producing these $p_i$. We know this would complicate cross era logic, but such issues should be addressed by considering the specific miss-behaviour. In essence, this $p_{\eta,\nu,e} := \max_i p_i$ definition provides default mechanism for combining slashes within one era that is simple, fair, and commutative, but alternative logic remains possible so long as we slash the same regardless of the order in which offenses are detected.

We do however worry about miss-behaviours from different validators $\nu \ne \nu'$ both because nomination must restrict Sibels and also because correlated slashing need not necessarily involve the same validators. We therefore let $N_{\eta,e}$ denote the validators nominated by $\eta$ in era $e$ and slash $\sum_{\nu \in N_e} p_{\eta,\nu,e} x_{\eta,\nu,e}$ from $\eta$ when multiple validators $\nu \in N_{\eta,e}$ get slashed.

We cannot assume that all events that warrant slashing a particular stash account get detected early or occur within the same era. If $e$ and $e'$ are distinct eras then we expect $x_{\eta,\nu_j,e} \ne x_{\eta,\nu_j,e'}$ so the above arguments fail. Indeed, we cannot even sum slashes applied to different validators because doing so could quickly exceeds nominators exposure $x_{\eta,\nu,e}$.

We might assume $\min { x_{\eta,\nu_j,e}, x_{\eta,\nu_j,e'} }$ to be the "same" stake, but this does not obviously buy us much. We therefore suggest the slashing $\eta$ the amount $\max_e \sum_{\nu \in N_e} p_{\eta,\nu,e} x_{\eta,\nu,e}$ where again $N_e$ is the validators nominated by $\eta$ in era $e$

In particular, there is an extortion attack in which someone runs many poorly staked validators, receives nominations, and then threatens their nominators with being slashed. We cannot prevent such attacks entirely, but this outer $\max_e$ reduces the damage over formula that add slashing from different eras.

We cannot slash for anything beyond the unbonding period and must expire slashing records when they go past the unbonding period. We implement this by recording all slash events along with some value $s_{\eta,\nu,e}$ recording the amount actually slashed at that time. If $e'$ is later than $e$ then we record the initial slash $s_{\eta,\nu,e} := p_{\eta,\nu,e} x_{\eta,\nu_j,e}$ at $e$ and record a lesser slash $s_{\eta,\nu,e'} := p_{\eta,\nu,e'} x_{\eta,\nu_j,e'} - p_{\eta,\nu,e} x_{\eta,\nu_j,e}$ at the later $e'$. These $s_{\eta,\nu,e}$ values permit slashes to expire without unfairly increasing other slashes.

We take several additional actions whenever some validator $\nu$ causes the slashing of some nominator $\eta$:

First, we post a slashing transaction to the chain, which drops $\nu$ from the active validator list by invalidating their session keys, which makes everyone ignore $\nu$ for the remainder of the era, and also invalidates any future blocks that do not ignore $\nu$. We also remove all nomination approval votes by any nominator for $\nu$, even those who currently allocate $\nu$ zero stake.

Second, we remove all $\eta$'s nomination approval votes for future eras. We do not remove $\eta$'s current nominations for the current era or reduce the stake currently backing other validators. Also we permit $\eta$ to add new nomination approval votes for future eras during the current era. We also notify $\eta$ that $\nu$ cause them to be slashed.

We treat any future nominations by $\eta$ separately from any that happen in the current era or before. in other words, we partition the eras into slashing periods for $\eta$ which are maximal contiguous sequence of eras $\bar{e} = \left[ e_1, \ldots, e_n \right]$ such that $e_n$ is the least era in which $\eta$ gets slashed for actions in one of the $e_i$. We let $\bar{e}$ range over the slashing periods for $\eta$ then we have slashed $\eta$ in total

We ask that slashing be monotonic increasing for all parties so that validators cannot reduce any nominator's slash by additional miss-behavior. In other words, the amount any nominator gets slashed can only increase with more slashings events, even ones involving the same validator but not the same nominator.

We think fairness imposes this condition because otherwise validators can reduce the slash of their favoured nominators, normally by making other nominators be slashed more. We know trusted computing environments (TEE) avoid this issue, but we do not currently foresee requiring that all validators use them.

There are no meaningful limits on the diversity of nominators who nominated a particular validator within the unbonding period. In consequence, almost every validator can be slashed simultaneously, thanks to by monotonicity and the validator adding past equivocations, which enables an array of "rage quit attacks". In other words, we cannot bound the total stake destroyed by a combined slashing event much below the slash applied to the total stake of the network.


text:
# Deprecation

### tf.contrib.framework.deprecated(date, instructions)

Decorator for marking functions or methods deprecated.

This decorator logs a deprecation warning whenever the decorated function is called. It has the following format:

(from ) is deprecated and will be removed after . Instructions for updating:

will include the class name if it is a method.

It also edits the docstring of the function: ' (deprecated)' is appended to the first line of the docstring and a deprecation notice is prepended to the rest of the docstring.

##### Args:
• date: String. The date the function is scheduled to be removed. Must be ISO 8601 (YYYY-MM-DD).
• instructions: String. Instructions on how to update code using the deprecated function.
##### Returns:

Decorated function or method.

##### Raises:
• ValueError: If date is not in ISO 8601 format, or instructions are empty.

### tf.contrib.framework.deprecated_args(date, instructions, *deprecated_arg_names)

Decorator for marking specific function arguments as deprecated.

This decorator logs a deprecation warning whenever the decorated function is called with the deprecated argument. It has the following format:

Calling (from ) with is deprecated and will be removed after . Instructions for updating:

will include the class name if it is a method.

It also edits the docstring of the function: ' (deprecated arguments)' is appended to the first line of the docstring and a deprecation notice is prepended to the rest of the docstring.

##### Args:
• date: String. The date the function is scheduled to be removed. Must be ISO 8601 (YYYY-MM-DD).
• instructions: String. Instructions on how to update code using the deprecated function.
• *deprecated_arg_names: String. The deprecated arguments.
##### Returns:

Decorated function or method.

##### Raises:
• ValueError: If date is not in ISO 8601 format, instructions are empty, or the deprecated arguments are not present in the function signature.

### tf.contrib.framework.deprecated_arg_values(date, instructions, **deprecated_kwargs)

Decorator for marking specific function argument values as deprecated.

This decorator logs a deprecation warning whenever the decorated function is called with the deprecated argument values. It has the following format:

Calling (from ) with = is deprecated and will be removed after . Instructions for updating:

will include the class name if it is a method.

It also edits the docstring of the function: ' (deprecated arguments)' is appended to the first line of the docstring and a deprecation notice is prepended to the rest of the docstring.

##### Args:
• date: String. The date the function is scheduled to be removed. Must be ISO 8601 (YYYY-MM-DD).
• instructions: String. Instructions on how to update code using the deprecated function.
• **deprecated_kwargs: The deprecated argument values.
##### Returns:

Decorated function or method.

##### Raises:
• ValueError: If date is not in ISO 8601 format, or instructions are empty.


text:
# A pure imperative language

% Remco Bloemen % 2010-02-16, last updated 2014-03-04

In the previous post I developed a virtual machine for a silly language (Brainfuck) in a silly language (C++). This post is documents my attempt to create a language that is not silly. I’ll start with three use-cases, show their C++ implementation and then start mutulating the C++ syntax.

## Three challenging functions

The following three functions are my favorites when trying out a new language. You’ll be supprised at how often one or more of these examples are simply impossible in a language, yet they represent very common mathematical operations.

The first challenge is recursion:

int factorial(int n)
{
if(n == 1)
return 1;
return n * fac(n - 1);
}

I could make it more challening by requiring guaranteed tail call optimization, but for now let’s just leave it at supporting recursive functions.

The second challenge is multiple return values:

void divide(int& quotient, int& remainder,  int dividend, int divisor)
{
quotient = 0;
remainder = dividend;
while(remainder >= divisor) {
remainder -= divisor;
++quotient;
}
}

The C++ solution is functional, but syntacticly it mixes inputs and outputs.

The third challenge is first class functions:

typedef double (*function)(double);

function makeDerivative(function f, double dx)
{
return delegate double (double x) {
return (f(x + dx) - f(x)) / dx;
};
}

Note that this last example is not valid C! In fact, I doubt there is a way to express this function in C without calling an external compiler and linking in the result.

## Developing a solution

Since this is supposed to become a practical solution, I’ll start with a pragmatic language: C++. In a few steps I will completely disfigure this language, hopefully for the better.

### Step zero: removing the sugar

Templates in C++ are little more than statically typed macros. Remove them. Object oriented programming is little more than implicit passing of a this pointer to member functions. Type inheritance and virtuals functions are a bit more complex, but these can be simultated with first class functions. Oh yes and remove the whole pre-processor.

We are now basically left with C again, but without a few of the problems that C has. We can always add a new sugar coating later, but first we’ll redesign the cake.

### Step one: multiple return values

The second challenge is probably the easiest, so I’ll start with this one. Let’s extend C++ in the following way:

(int quotient, int remainder)divide(int dividend, int divisor)
{
quotient = 0;
remainder = dividend;
while(remainder >= divisor) {
remainder -= divisor;
++quotient;
}
}

// Function call syntax
(q, r) = divide(13, 4);

In summary: return values are named, return values are assigned as local variables, return no longer takes an argument, function calls take the form () = ();. Nesting of function calls like sqrt(sin(x)+cos(x)) is no longer supported (altough this could easily be reintroduced with some syntactic sugar).

### Step two: pure functions

For the sake of clarity and tractability, let’s make the all functions pure. That is, all inputs are constant or unchangable.

Pure functions make it difficult to interact with stateful objects. These not only include objects from object-oriented programming, but also thing such as IO operations. In fact, functions like currentTime() and randomNumber() are by their nature impure.

One solution is to introduce an explicit state variable:

(Object this, int returnValue) ObjectMemberFunction (Object this);

(pipe state, string input) readInput (pipe state);

(clockType state, timeType time) currentTime (clockType state);

(RNG state, int number) randomNumber (RNG state, int min, int max);

This will all be hidden under a sugar coating later. I have not yet tought about how to incorporate multi-threaded memory access into this.

### Step three: lambda expressions

The last challenge requires functions as values and thus the creation of anonymous functions.

I’ll introduce the following expression to create anonymous functions

( <outputs> ) λ ( <inputs> )
{
<statements>
}

Now challenge three can be implemented as:

typedef (double)*function(double);

(function derivative) makeDerivative (function f, double dx)
{
derivative = (double x) λ (double dfdx)
{
dfdx = (f(x + dx) - f(x)) / dx;
}
}

### Step four: only lambda expressions

We are now in the situation where the following two pieces of code are equivalent:

(double y) addFive (double x)
{
y = x + 5;
}

addFive = (double y) λ (double x)
{
y = x + 5;
}

Redundancy in notation is never good, it reflects a discrepancy between syntax and semantics. I’ll remove the redundancy by elliminating the first form. All functions are to be declared via the λ expression.

### Step five: re-introducing recursion

The last step introduced a problem, we can no longer define recursive functions! Look what happens:

factorial = (int f)λ(int n)
{
if(n == 0)
f = 1;
else
f = n * factorial(n - 1);
}

The problem is, factorial is not yet defined when inside the function. This is the problem of recursive anonymous lambda expressions. The common way to solve this is using a fixed point combinator (see also this article in lisp and the F♯ version).

A practical way to solve this is by introducing the self keyword for self referencing:

factorial = (int f)λ(int n)
{
if(n == 0)
f = 1;
else
f = n * self(n - 1);
}

Note that I am breaking my own function calling rule for the sake of clarity. To enforce the rule one would need to explicitly introduce temporary variables. The correct version would be:

factorial = (int f)λ(int n)
{
bool isZero;
(isZero) = equals(n, 0);
if(isZero) {
(f) = identity(1);
} else {
int prevN;
(prevN) = subtract(n, 1);
(prevF) = self(prevN);
(f) = multiply(n, prevF);
}
}

### Step six: syntax change

Since this language has become an odd one in the curly bracket family we might as well change all of the syntax:

Design principles:

• Very simple grammar

• Strong emphasize on following common mathematical syntax

• Symetric infix operators have symetric symbols

• Assymetric infix operators have assymetric symbols, of which the mirror image works as espected

Functionality Old notation New notation

Assignment x = y x ↤ y Anonymous tuple type (type, type, …) (type, type, …) Anonymous tuple (value, value, …) (value, value, …) Named tuples (type name, type name, …) (name: type, name: type Function type +outputs *inputs + → Function definition  { } +name ↤ λ inputsoutputs { statements | } Function call =  ↤

Where inputs and outputs are anonymous tuples types when in a function type, named tuples in a function definition and anonymous tuples in a function call.

The factorial example now becomes:

factorial ↤ λ (n: int) → (f: int)
{
isZero ↤ equals(n, 0)
if(isZero) {
f ↤ 1
} else {
f ↤ n × self(n - 1)
}
}

### Step seven: the if meta-function

In the above example still contains silly syntaxt like if (bool) {statements } else {statements }. This can be replaced by a simple meta-function:

if ↤ λ (condition: bool, then: T, else: T) → (output: T)

Depending on condition this function returns either then or else. Now, the naïve way to employ this functions is:

factorial ↤ λ (n: int) → (f: int)
{
f ↤ n × self(n - 1)
isZero ↤ equals(n, 0)
f ↤ if(isZero, 1, f)
}

But this will lead to infinite recursion. The correct way is to make use of the first-class functions:

factorial ↤ λ (n: int) → (f: int)
{
then ↤ λ (n: int) → (f: int)
{
f ↤ 1
}
else ↤ λ (n: int) → (f: int)
{
f ↤ n × self(n - 1)
}
isZero ↤ equals(n, 0)
ff ↤ if(isZero, then, else)
f ↤ ff(n)
}

### Step eight: the while meta-function

Another way to implement the factorial function is using a loop:

factorial ↤ λ (n: int) → (f: int)
{
f ↤ 1
do {
f ↤ n × f
n ↤ n - 1
repeat ↤ ni > 1
} while (repeat)
}

But this do { statements } while (condition) polutes the language. It can be replaced by the following meta function:

while ↤ λ (loopState: T, loop: (loopState: T) → (loopState: T, repeat: bool)) → (output: T)

This function starts with loopState and then itterates the function loop on it untill this function sets repeat to false. The factorial can now be implemented as:

factorial ↤ λ (n: int) → (f: int)
{
loop ↤ λ ((ni: int, fi: int)) → ((no: int, fo: int), repeat: bool)
{
fo ↤ ni × fi
no ↤ ni - 1
repeat ↤ ni > 1
}
f ↤ while((n, 1), loop)
}

All the other control flow constructions such as switch and for can be trivially rewritten using if and while.

## to be continued…

That’s enough for now. Some things for the future:

• Formal grammar and type system
• Parser and compiler (using Quex and LLVM)
• Explicit internal and external variables to functions
• Generics
• Homoiconicity
• Ability to extend parser and compiler in language and at runtime
• A large coating of syntactic sugar developed in language
• Arrays, list, containers, etc…
• Embedded proofs and a verifying compiler
• Self-hosting
• Self-awareness

Remco Bloemen
Math & Engineering
https://2π.com


text:
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text:
# A water filtration unit is made of uniform-size sand particles of 0.4 mm diameter with shape factor of 0.84 and specific gravity of 2.55. The depth of the filter bed is 0.70 m and the porosity is 0.35. The filter bed is to be expanded to a porosity of 0.65 by hydraulic backwash. If the terminal settling velocity of sand particles during during backwash is 4.5 cm/s, the required backwash velocity is

This question was previously asked in
GATE CE 2021 Official Paper: Shift 2
View all GATE CE Papers >
1. 0.69 cm/s
2. 6.35 × 10-3 m/s
3. 5.79 × 10-3 m/s
4. 0.75 cm/s

Option 2 : 6.35 × 10-3 m/s
Free
CT 1: Ratio and Proportion
2672
10 Questions 16 Marks 30 Mins

## Detailed Solution

Concept:

The head loss during the backwash (He) is given by

Hb = z × ( 1 – n ) × (S – 1) = ze × ( 1 – ne ) × ( S – 1 )

Where,

ze is the thickness of the expanded sand bed
ne is the porosity of the expanded sand bed
s is the specific gravity of sand
z is the thickness of sand bed and

n is the porosity of the sand bed before expansion

The rate of backwash Vis given by

VB = Vs × ne4.5,
where Vs is the settling velocity of sand

Given by

$${V_s} = \;\frac{{{\rm{g}}\left( {{\rm{s}} - 1} \right){{\rm{d}}^2}}}{{18{\rm{v}}}}$$

for laminar condition
where,

d is the size of sand particles and, v is the kinematic viscosity of water

Calculation

Given,

Shape factor = 0.85 , particle diameter (d) = 0.4 mm, Specific Gravity (S) = 2.55

Depth of filter Bed (z) = 0.75 m , Porosity (η ) = 0.35

The porosity of expanded bed  ηe = 0.65

Terminal velocity Vs = 4.5 cm/sec = 4.5 × 10-2 m/sec

The rate of backwash Vis:

VB = 4.5 × 10-2 × (0.65)4.5

VB = 6.47 × 10-3 m/sec


text:
# Trace of SU(3) generators

1. Jan 6, 2016

### Safinaz

Hi all,

The trace of two SU(3) generators can be calculated by:
$T_{ij} T_{ji} = \frac{1}{2}$, now how to calculate the trace of SU(3) generators:
$T_{il} T_{lk} T_{kj} T_{ji}$ ?

2. Jan 6, 2016

### Staff: Mentor

What's wrong about $∑ T_{ii}$?

3. Jan 6, 2016

### Safinaz

I didn't understand, can you be more clear please ?

4. Jan 6, 2016

### Staff: Mentor

The trace of any matrix is defined as the sum of its diagonal entries.

5. Jan 6, 2016

### ChrisVer

Tr(AB) = Tr(BA)
then use the comm. relations ?

6. Jan 6, 2016

### samalkhaiat

Are you asking about a generator or the generators? I don’t see any group index on $T$! Any way, for a single $3 \times 3$ traceless matrix, you have
$$\mbox{Tr}\left(T^{4}\right) = \frac{1}{2} \left( \mbox{Tr}\left( T^{2} \right) \right)^{2} .$$
For $SU(N)$, all the relevant traces can be obtained from the following relation
$$T^{a}T^{b} = \frac{1}{2} \Big \{ \frac{1}{N} \delta^{ab} \mathbb{I}_{N} + (d^{abe} + i f^{abe})T^{e} \Big \} , \ \ \ \ (1)$$
where $f^{abc}$ is the totally anti-symmetric structure constant:
$$[ T^{a} , T^{b} ] = if^{abc}T^{c}, \ \ a,b,c = 1,2, \cdots , N^{2}-1 ,$$
and $d^{abc}$ are totally symmetric constants satisfying
$$\big \{ T^{a} , T^{b} \big \} = \frac{1}{N}\delta^{ab} \mathbb{I}_{N} + d^{abc} T^{c} .$$
So, if you multiply (1) by $T^{c}$ and trace over, you get
$$\mbox{Tr}\left( T^{a}T^{b}T^{c} \right) = \frac{1}{4} (d^{abc} + i f^{abc}) . \ \ \ \ \ (2)$$
Then, you can keep going, multiplying (1) by $T^{c}T^{d}$, taking the trace and using (2) and the Jacobi Identities you obtain the following ugly relation
$$\begin{equation*} \begin{split} \mbox{Tr}\left( T^{a}T^{b}T^{c}T^{d} \right) =& \frac{1}{4N}\left( \delta^{ab}\delta^{cd} + \delta^{ad}\delta^{bc} - \delta^{ac}\delta^{bd} \right) \\ & + \frac{1}{8} \left( d^{abe}d^{cde} + d^{ade}d^{bce} - d^{ace}d^{bde} \right) \\ & + \frac{i}{8} \left( d^{abe}f^{cde} + d^{cde}f^{abe} \right) . \end{split} \end{equation*}$$
Another very useful identity for calculating traces in $SU(N)$ is
$$(T^{a})_{ij}(T^{a})_{kl} = \frac{1}{2} \left( \delta_{il}\delta_{jk} - \frac{1}{N} \delta_{ij}\delta_{kl} \right) , \ \ i,j = 1, 2, \cdots , N$$
Note that the left hand side is summed over the group index $a = 1, \cdots , N^{2}-1$. This identity can be used, for example, to calculate
$$\mbox{Tr}\left( T^{a}T^{b}T^{a}T^{c} \right) = - \frac{1}{4N} \delta^{bc} .$$

7. Jan 7, 2016

### vanhees71

Where is the relation in the second equation from below from? Is it really correct? Take the SU(2) adjoint representation, where
$$(T^a)_{ij} =-\mathrm{i} \epsilon_{aij}$$
and (Einstein summation relation implied)
$$(T^a)_{ij} (T^a)_{kl} =-\epsilon_{aij} \epsilon_{akl}=-(\delta_{ik} \delta_{jl}-\delta_{il} \delta_{jk}).$$

8. Jan 7, 2016

### Safinaz

Thanks guys for these inclusive answers.

9. Jan 7, 2016

### samalkhaiat

Okay, the question “where does it come from?” can be answered by deriving the identity. And since that identity is very important in QCD calculations, I will derive it in here.
Consider the algebra of $SU(n)$
$$[ \frac{\tau^{a}}{2} , \frac{\tau^{b}}{2} ]_{ij} = i f^{abc} \left(\frac{\tau^{c}}{2}\right)_{ij} , \ \ a = 1, 2, … , n^{2}-1, \ \ i = 1, … , n$$
Please observe the range of the of the group index $(a)$ and that of the fundamental matrix representation $(i)$.
We can always consider the normalization
$$\mbox{Tr}\left(\tau^{a}\tau^{b}\right) = 2 \delta^{ab} . \ \ \ \ \ \ \ (1)$$
Since the $n \times n$ hermitian traceless matrices $\tau^{a}$, together with the $n \times n$ identity matrix $\mathbb{I}_{n}$ form a complete set of $n \times n$ hermitian matrices, we can expand any arbitrary $n \times n$ hermitian matrix $M$ in terms of them (repeated indices are summed) :
$$M = C \ \mathbb{I}_{n} + C_{a} \tau^{a} . \ \ \ \ \ (2)$$
Using (1) and the tracelessness of the $\tau^{a}$’s, the coefficients $C$ and $C_{a}$ are calculated as follows
$$C = \frac{1}{n} \mbox{Tr}\left( M \right) = \frac{1}{n} \delta_{jk}M_{kj} , \ \ \ \ \ (3)$$
$$C_{a} = \frac{1}{2} \mbox{Tr}\left( M \tau^{a} \right) = \frac{1}{2} (\tau^{a})_{jk}M_{kj} . \ \ \ \ \ (4)$$
Substituting (3) and (4) back in (2), we find
$$M = \frac{1}{n} \delta_{jk} \ M_{kj} \ \mathbb{I}_{n} + \frac{1}{2} \tau^{a} \ (\tau^{a})_{jk} \ M_{kj} .$$
Okay, take the $il$ matrix element of the above
$$M_{il} = \frac{1}{n} \delta_{jk} \ M_{kj} \ \delta_{il} + \frac{1}{2} (\tau^{a})_{il} \ (\tau^{a})_{jk} \ M_{kj} .$$
Now, one left-hand-side substitute
$$M_{il} = \delta_{ik} \delta_{lj} M_{kj} .$$
Then, from the fact that $M_{kj}$ is arbitrary, we obtain
$$\delta_{ik} \ \delta_{lj} = \frac{1}{n} \delta_{jk} \ \delta_{il} + \frac{1}{2} \left(\tau^{a}\right)_{il} \ \left(\tau^{a}\right)_{jk} ,$$
which is just the identity in my previous post, with $T = \tau /2$. Warning: Please notice the range of the indices in the identity. $(i,j,k,l)$ run from $1$ to $n$, while $a$ is summed over from $1$ to $n^{2}-1$.
I can not believe you make such mistake! Read the warning regarding the indices. For $SU(2)$ the indices $(i,j,k,l)$ take values in $\{1,2\}$ while $a = 1,2,3$. The correct form of the generators in the adjoint representation is
$$\left( \frac{\tau^{a}}{2}\right)_{bc} = - i \epsilon_{abc} .$$

Last edited: Jan 7, 2016
10. Jan 8, 2016

### vanhees71

I don't get it. Of course, the adjoint representation is for spin 1 and thus 3-dimensional, but for spin 1/2 (the fundamental representation) 2D. So, how is the discrepancy between the two formulas resolved? As far as I see from your derivation, it's only valid in the fundamental representation(s) of SU(N).

11. Jan 9, 2016

### samalkhaiat

For $SU(2)$, the matrix $M$ in the previous post is $2 \times 2$, and the identity is just the following relation for Pauli matrices:
$$\frac{1}{2} \sigma^{a}_{ij}\sigma^{a}_{kl} = \delta_{il} \delta_{jk} - \frac{1}{2} \delta_{ij} \delta_{kl} , \ \ \forall (i,j,k,l) \in \{1,2\} .$$
From this, as you can check, we obtain the following correct relations
\begin{align*} \left( \sigma^{a}\sigma^{a} \right)_{il} &= \delta_{jk} \sigma^{a}_{ij}\sigma^{a}_{kl} \\ &= 2(2\delta_{il} - \frac{1}{2}\delta_{il}) \\ &= 3 \delta_{il}, \\ \mbox{Tr}(\sigma^{a}\sigma^{a}) &= 6 . \end{align*}
Yes, the adjoint representation of $SU(2)$ is 3-dimensional. This means that the generators matrices in the adjoint representation are $3 \times 3$, and there are 3 of them. So, your objects $(T^{a})_{ij} = - i \epsilon_{aij}$ with $a \in \{1,2,3 \}$ and $(i,j) \in \{1,2 \}$ are not correct because: (1) $\epsilon_{aij}$ is a $2 \times 2$ matrix, and (2) $T^{1} = T^{2} = 0$ and $T^{3} = \sigma^{2}$. So, $(T^{a})_{ij} = - i \epsilon_{aij}$ is not a representation, let alone adjoint representation.
You created the discrepancy by wanting the identity to apply to the single $2\times2$ matrix $-i \epsilon_{aij}$ which is not even a representation.
Yes, and I was stressing this fact all along, by specifying the range of different indices. However, this does not mean that the identity is good only for the fundamental representation (quarks). In fact, it is very useful in calculating the colour factors in gluon loop.
The sum over colour indices in gluon loop can be represented by either the trace factor $t_{2}(A)$:
$$\mbox{Tr}(A^{b}A^{c}) = t_{2}(A) \delta^{bc} ,$$
or by the eigenvalue of the quadratic Casimir operator in the adjoint representation $C_{2}(A)$
$$(A^{a}A^{a})_{bc} = C_{2}(A) \delta_{bc} ,$$
where $A^{a}$ is the generator in the adjoint representation, as required by the gluon field
$$(A^{a})_{bc} = - i f_{abc} = - \frac{1}{4} \mbox{Tr}\left( \tau_{c} [ \tau_{a} , \tau_{b} ] \right) .$$
Since the structure constant $f_{abc}$ of $SU(n)$ is totally antisymmetric, one can easily show that
$$\mbox{Tr}(A^{a}A^{b}) = \left( A^{c} A^{c} \right)_{ba} = f_{acd}f_{bcd} .$$
So, in order to calculate the value of the factor $t_{2}(A) = C_{2}(A)$, we need to compute
\begin{align*} f_{acd}f_{bcd} &= - \frac{1}{16} \mbox{Tr}\left( \tau_{d}[ \tau_{b} , \tau_{c} ] \right) \mbox{Tr}\left( \tau_{d}[ \tau_{a} , \tau_{c} ] \right) \\ &= - \frac{1}{16} \tau^{d}_{ij} \ [ \tau^{b} , \tau^{c} ]_{ji} \ \tau^{d}_{kl} \ [ \tau_{a} , \tau_{c} ]_{lk} . \end{align*}
Now, using our identity
$$\tau^{d}_{ij} \ \tau^{d}_{kl} = 2 \left( \delta_{il}\delta_{jk} - \frac{1}{n} \delta_{ij}\delta_{kl} \right) ,$$
we obtain
$$f_{acd}f_{bcd} = \frac{-1}{8} \mbox{Tr} \left( [ \tau^{b}, \tau^{c}] [ \tau^{a}, \tau^{c}] \right) - \frac{1}{n} \mbox{Tr}\left( [ \tau^{b} , \tau^{c}] \right) \mbox{Tr}\left( [ \tau^{a} , \tau^{c}] \right) .$$
The second term is zero because the trace of $[ A , B ]$ is zero for any two matrices $A$ and $B$. The first term, when expanded, consists of 4 terms each of which has a factor of the form $\tau^{e}_{ij}\tau^{e}_{kl}$. So, we can apply our identity to each term. I leave it for you as exercise to show that
\begin{align*} \mbox{1st term} &= \mbox{Tr}(\tau^{b}\tau^{c}\tau^{a}\tau^{c}) = - \frac{4}{n} \delta^{ab} \\ \mbox{2nd term} &= \mbox{Tr}(\tau^{b}\tau^{a}\tau^{c}\tau^{c}) = 4( n - \frac{1}{n}) \delta^{ab} \\ \mbox{3rd term} &= 4(n - \frac{1}{n}) \delta^{ab} \\ \mbox{4th term} &= - \frac{4}{n} \delta^{ab} . \end{align*}
This gives us
$$f_{acd}f_{bcd} = \frac{-2}{8} \Big \{ - \frac{4}{n} - 4 \left( n - \frac{1}{n} \right) \Big \} \delta^{ab} = n \delta^{ab} .$$
Which means that $t_{2}(A) = C_{2}(A) = n$.

12. Jan 9, 2016

### vanhees71

I think it's simply a misunderstanding. You talk about the fundamental representation and not about a general representation, and I thought it should be valid in any representation.

Of course the adjoint representation of su(2) (Lie algebra) is so(3) and you can use $(T^a)_{ij}=-\mathrm{i} \epsilon_{aij}$ as a basis of the Lie algebra in this representation.

13. Jan 9, 2016

### samalkhaiat

If and only if, the index $a$ and the pair $(i,j)$ take values in the same set $\{1,2,3\}$.

14. Jan 10, 2016

### vanhees71

Sure, that's the adjoint representation. You can "naturally" define it on the Lie algebra itself.