Showing posts with label General engineering data. Show all posts
Showing posts with label General engineering data. Show all posts

Thursday, August 30, 2012

1- Tubes
2- Tube Sheets
3- Shell and Shell-Side Nozzles
4- Tube Size Channels and Nozzles
5- Channel Covers
6- Pass Divider
7- Baffles

Components inside Heat Exchanger building

Thursday, September 29, 2011

Purpose
 To demonstrate how to measure the energy changes associated with various   neutralization reactions, and how to measure the heat capacity of a calorimeter.


Acids and Bases

nAcid:  donates H+ ions to the aqueous solution in which it is dissolved
nBase: donates OH- ions to the aqueous solution or accepts H+ ions from the aqueous solution in which it is dissolved
nStrong acid: dissociates completely in aqueous solution
nStrong base: dissociates completely in aqueous solution

Neutralization Reactions
nStrong acids and strong bases react to produce a salt and water; a solution with an overall pH of 7 under ideal conditions (neutralization reaction):
HCl(aq) + KOH(aq) ® KCl(aq) + H2O(l) + heat
nNeutralization reactions are exothermic, thus liberating heat to the surroundings.


Calorimetry
nCalorimeters measure the heat associated with a chemical reaction.
nWe have to account for the heat transferred from the solution to the calorimeter so we can add it back in. For this reason we perform a calibration on the calorimeter prior to running the experiment on the acid / base reaction.
nHeat (q) is the transfer of energy between two objects; always from hot to cold. The amount of heat transferred depends on the mass of the object.
nTemperature reflects the random motion of particles in an object and is independent of mass.
nYou will be working with both heat and temperature today.

Procedure
nCalibrate your IC probe.
nCalibrate your calorimeter to find CCAL.
nUse the sample calculations on pp. 212 and 213 to help you with your calculations.
nDetermine the energy transfer involved with your neutralization reaction.
nUse the sample calculations on pp. 214 and 215 to help you with your calculations.


Safety Concerns – HCl (0.2N)
nIngestion Hazards:
nMay cause gastrointestinal disturbances
nEye Contact:
nMay cause irritation, redness and pain

Safety Concerns – HNO3 (1.0N)
nEye Contact:
nCorrosive.  Irritating and damaging to eyes.  Burns and permanent eye damage.
nSkin Contact:
nCorrosive.  Redness, pain and severe skin burns.  Ulcers may form and skin stains.
nInhalation:
nCorrosive.  Breathing difficulties can lead to pneumonia and pulmonary edema, which may be fatal.  Coughing, choking, irritation of nose, throat, and respiratory tract.
nIngestion:
nCorrosive.  Pain and burns of the mouth, throat, esophagus and gastrointestinal tract.  

Safety Concerns – NaOH (0.2N)
nWhy do you not want to inhale NaOH?
nRespiratory irritant
nWhy do you not want to drink NaOH?
nNaOH is corrosive!  Will cause burns all the way down.
nWhat happens when you get NaOH on your skin?
nSevere burns and scarring may occur.
nWhat happens when you get NaOH in your eyes?
nYou could go blind…and it hurts!

Safety Concerns – Potassium Hydroxide
nEye Contact:
nIrritant and corrosive.  Tearing, redness, pain and impaired vision
nSkin Contact:
nIrritant and corrosive.  Soreness, redness, destruction of skin
nInhalation:
nIrritant.  May cause serious burns.
nIngestion:
nToxic.  Corrosive to mucous membranes.  May cause perforation of the esophagus and stomach.  Abdominal pain, nausea, vomiting, general gastro-intestinal upset.

Waste
nAcid and base waste must be added together prior to disposal down the sink.  Flush with a lot of water.
Neutralized acid / base solutions may go down the sink.


Calorimetry and the Enthalpies of Neutralization - Experiment


Acids are compounds that dissolve in water to produce solutions containing hydronium ions, H3O+ (aq), and bases dissolve in water to produce hydroxide ions, HO- (aq).  When an acidic solution and a basic solution are mixed, hydroxide ions react with hydronium ions to produce water molecules:
Abbreviating the hydronium ion as H+ (aq) makes it easier to count the water molecules produced.

     
This chemical process is called neutralization.  What happens to the other ions that were present in the original solutions?  Nothing!  They are still separately solvated and moving independently, unaffected by the reaction going on around them.  Let's look at the reaction between hydrochloric acid and sodium hydroxide:

Equations 1 and 2 represent the formation of the solutions of hydrochloric acid and sodium hydroxide, and equation 3 shows what happens when they are mixed.  Notice that sodium ions and chloride ions appear on both sides of equation 3, called the total ionic equation.  If we cancel out these spectator ions we are left with equation 4, the net ionic equation, which represents the chemistry of the neutralization reaction.  If we represent the sodium ions and chloride ions as NaCl (aq), and write formulas for hydrochloric acid and aqueous sodium hydroxide, equation 3 becomes equation 5, which summarizes the overall reaction.

Salts
     If we were to evaporate the water from the solution after this neutralization reaction, we would be left with solid sodium chloride, commonly known as table salt.  In general, the term salt refers to any ionic compound that could be produced by a neutralization reaction.  Salts contain metal cations, and anions that may be monatomic (like chloride) or polyatomic anions derived from oxyacids.  The formula of a salt can be predicted from the known charges of the cation and anion as we have seen previously for binary ionic compounds.  Sodium sulfate, for example, has the formula Na2SO4.  When a formula includes more than one polyatomic ion, it is enclosed in parentheses with a subscript to indicate the relative number of ions.  Magnesium nitrate is Mg(NO3)2.  There are two nitrate ions (charge 1- each) for every Mg2+ ion.  When necessary, we include the charge on the cation as part of the name.  Mercury(II) acetate is Hg(C2H3O2)2.

  Some salts form crystalline solids which include some water molecules as part of the structure.  Such salts are called hydrates, and we include the water molecules in the formula using a dot and a coefficient.  For example, copper(II) sulfate forms blue crystals with five water molecules per formula unit, so the formula is CuSO4.5H2O.  Hydrates can be dried by strong heating, which drives the water molecules out of the crystal.

Balancing Equations
     Equations for neutralization reactions are usually written in the form of equation 5 above, a summary using the formulas for the reactants and products.  To be useful, a chemical equation must be balanced, i.e. for each element there must be the same number of atoms on both sides of the equation.  This is because the law of conservation of mass requires that atoms are not created or destroyed during a chemical reaction.

     One way to balance the equation for a neutralization reaction is to remember that the underlying chemical process is the reaction of hydrogen ions and hydroxide ions in a one to one ratio.  We balance the overall equation by adding coefficients so that the numbers of hydrogen ions and hydroxide ions are equal.

Consider as an example the reaction of hydrogen bromide with barium hydroxide.  Hydrogen ions from the acid will react with hydroxide ions from the base, and barium and bromide ions will be left in solution.  The ionic compound made up of barium ions and bromide ions is barium bromide, which has the formula BaBr2 (remember that barium is a group IIA metal and forms Ba2+ cations).  We first write an overall equation for the process using the correct formulas for all of the compounds involved:
unbalanced
    We can then quickly balance the equation by recognizing that we have 2 moles of hydroxide ions from 1 mole of barium hydroxide, so we need 2 moles of hydrogen bromide and will get 2 moles of water:

     Inspection shows us that the other elements, barium and bromine are also balanced.  When balancing an equation, start with the correct formula for each compound and change the coefficients not the formulas.
     Most equations that we will meet can be balanced by direct inspection, but for more complicated situations an understanding of the chemical processes going on in the reaction can help.


http://www.chem.memphis.edu/bridson/FundChem/T16a1100.htm







Neutralization Reactions

Generally the theory based on that all indicators are either weak acids or weak bases in which the color of the ionized form is different from the color before dissociation.
They were first systematically employed in analytical chemistry by Robert Boyle, who used the aqueous extracts of the coloured principles present in red-cabbage, violets and cornflowers. The indicator most in use to-day is litmus, whose solution is turned red by an acid, and blue by an alkali. Several synthetic indicators are employed in acidimetry and alkalimetry. The choice is not altogether arbitrary, for experiments have shown that some are more suitable for acidimetry, while others are only applicable in alkalimetry; moreover, the strength of the acids and bases employed may exert a considerable influence on the behavior of the indicator.

litmus paper

Theory of Indicators 
 The ionic theory of solutions permitted the formulation of a logical conception of the action of indicators by W. Ostwald which for many years held its ground practically unchallenged; and even now the arguments originally advanced hold good, except for certain qualifications rendered necessary by more recent research. In the language of the ionic theory, an acid solution is one containing free hydrions, and an alkaline solution is one containing free hydroxidions. A neutral solution contains hydrions and hydroxidions in equal concentration; this is a consequence of the fact that pure water itself undergoes a certain dissociation, and several different methods show that in the purest water obtainable the concentration of the free hydrions and hydroxidions is 107 at 24°. Moreover, the law of mass-action (see Chemical Action) demands that the product of the concentrations of the hydrions and hydroxidions in any solution is constant at a given temperature, and we see from the above values that this constant is 1014. It follows, therefore, that the acidity or alkalinity of any solution can be expressed both in terms of hydrion or hydroxidion concentration. Many researches have been directed to classify acid and alkaline solutions according to the concentration of the hydrion. Conductivity determinations show that the maximum concentration of hydrion occurs in 5.8 - N nitric acid, where it has a value of about 2 -N, and the minimum occurs in 6.7N potassium hydroxide, where its value is 5 X I 015, that of the hydroxidion being about 2 - N. These figures apply to a temperature of 24 °. Bearing in mind the concentration of the ions in a neutral solution, it is seen that a scheme of seven grades of "neutrality," differing by successive powers of ten, may be formulated. The concentration of hydrion and hydrox idion in any solution may be determined by several independent methods, and it is therefore a simple matter to prepare solutions of definite ionic concentrations and to test these with the object of obtaining a list of indicators according to their sensitiveness. It is found that litmus responds to concentrations of io-- 6 H and 106 0H', a result which shows this dye to be the best indicator of true neutrality. Methyl orange responds to between - 4H and I 05 H - para-nitrophenol to between io - 5 H- and 106 Hand phenolphthalein to between r05 0H' and t06 0W. Salm (Zeit. Elektrochem., 1904, 10, p. 341) gives a list of twenty-seven indicators classified on this principle. Other papers bearing on this subject are Friedenthal, ibid., p. 113; Salessky, ibid., p. 204; Fels, ibid., p. 208; Scholtz, ibid., p. 549; M. Handa, Ber., 1909, 4 2, p. 3179.

The actual mechanism by which the indicator changes colour with varying concentrations of hydrion or hydroxidion is now to be considered. Ostwald formulated his ionization theory which assumes the change to be due to the transition of the non-dissociated indicator to the ionized condition, which are necessarily of different colours. On this theory, an indicator must be weakly basic or acid, for if it were a strong acid or base high dissociation would occur when it was in the free state, and there would be no change of colour when the solution was neutralized. Take the case of a weakly acid indicator such as phenolphthalein. The presence of an acid depresses the very slight dissociation of the indicator, and the colour of the solution is that of the non-dissociated molecule. The addition of an alkali, if it be strong, brings about the formation of a salt of phenolphthalein, which is readily ionized, and so reveals the intense red coloration of the anion; a weak base, however, fails to give free ions. An acid indicator of medium strength is methyl orange. When free this substance is ionized and the solution shows an orange colour, due to a mixing of the red of the non-dissociated molecule and the yellow of the ionized molecule. Addition of hydrions lessens the dissociation and the solution assumes the red colour, while a base increases the dissociation and so brings about the yellow colour. If the alkaline solution be titrated with a strong acid, the hydrions present in a very small amount of the acid suffices to reverse the colour; a weak acid, however, must be added in considerable excess of the quantity properly required to neutralize the solution, owing to its weak dissociation. This indicator is therefore only useful when strong acids are being dealt with, while it's strongly acid nature renders it serviceable for both strong and weak bases.


pH indicator 
 A pH indicator is a halochromic chemical compound that is added in small amounts to a solution so that the pH (acidity or alkali nity) of the solution can be determined easily. Hence a pH indicator is a chemical detector for hydronium ions (H3O+) (or Hydrogen ions (H+) in the Arrhenius model). Normally, the indicator causes the color of the solution to change depending on the pH. Solutions with a pH value above 7.0 are alkali, and solutions with a pH value below 7.0 are acidic. Solutions with a pH value of 7.0 are neutral 


Theory 
 pH indicators themselves are frequently weak acids or bases. When introduced into a solution, they may bind to H+ (Hydrogen ion) or OH- (hydroxide) ions. The different electron configurations of the bound indicator causes the indicator's color to change, which allows the pH to be determined by the different colors. 


Application 
pH indicators are frequently employed in titrations in analytic chemistry and biology experiments to determine the extent of a chemical reaction. Because of the subjective determination of color, pH indicators are susceptible to imprecise readings. For applications requiring precise measurement of pH, a pH meter is frequently used.

There are several common laboratory pH indicators. Indicators usually exhibit intermediate colors at pH values inside the listed transition range. For example, phenol red exhibits an orange color between pH 6.8 and pH 8.4. The transition range may shift slightly depending on the concentration of the indicator in the solution and on the temperature at which it is used.

Naturally occurring pH indicators 
 Many plants or plant parts contain chemicals from the naturally-colored anthocyanin family of compounds. They are red in acidic solutions and blue in basic. Extracting anthocyanins from red cabbage leaves or the skin of a lemon to form a crude acid-base indicator is a popular introductory chemistry demonstration.

Anthocyanins can be extracted from a multitude of colored plants or plant parts, including from leaves (red cabbage); flowers (geranium, poppy, or rose petals); berries (blueberries, blackcurrant); and stems (rhubarb). An exhaustive list would be beyond the scope of this article


Sources :
www.thefreedictionary.com
www.1911encyclopedia.org/Indicator
en.wikipedia.org/wiki/PH_indicator


Theory of Indicators

•Here we are deal with solutions of solid solutes dissolved in liquid solvents.
•Colligative properties:
1.Lowering of vapor pressure
2.Elevation of boiling point
3.Depression of freezing point
4.Osmotic pressure
•These properties depend on the number of the particles of the dissolved substance

and independent on their size or nature, therefore they have been grouped together as Colligative properties, these properties are important for determination the molecular weight of the unknown dissolved substance.

Lowering of vapor pressure, Raoult”s law:
Vapor pressure of a pure solvent is decreased when a non – volatile solute dissolved in it.
If P is the vapor pressure of the solvent,and

Ps that of the solution, then the lowering is:

P – Ps and P – Ps/P is called the relative lowering.

Raoult”s law states that” the relative lowering is equal to the mole fraction of the solute,

Thus P – Ps /P = n / n + N
Where n the number of moles of solute, and N the number of moles of solvent.


Derivation of Raoult”s law

•The vapor pressure of the solution is proportional to the mole fraction of the solvent, since the solute is non-volatile.

Thus Ps α N / n + N …………(1) 

In case of pure solvent n = 0 

Hence mol fraction of solvent = N / N + 0 = 1 

Thus from (1) P = k 

Ps = P . N /n + N 

Ps / P = N / n + N 

1 - Ps / P = 1 - N / n + N 

P – Ps / P = n / n + N 



Ideal solutions and deviation from Raoult”s law
•Suppose that γAB the attractive force between A and B molecules and γAA between A and A molecules. Then;

if γAB = γAA the solution is ideal

if γAB > γAA negative deviation is observed

if γAB < γAA positive deviation is observed

In very dilute solutions of non-electrolytes, the solvent and solute molecules are very much alike, such solutions approach the ideal behavior and obey Raoult”s law.


Determination of molecular weight from vapor pressure lowering.

if w gram of solute is dissolved in W gram of solvent, and m & M are molecular weight of solute and solvent respectively, then moles of solute (n) = w /m

moles of solvent (N) = W/M

substituting in Raoult”s law we get:

P- Ps/ P = n / n + N = (w/m) / W /M + w/m

Since for very dilute solution, n is very small it can be neglected in the denominator so

P – Ps / = w . M /mW



Measurement of lowering of vapor pressure by gas saturation method


• if w1 and w2 be the loss of weight in set of solution and in set of solvent respectively, 
then w1 α ps ………… (1) 

w2 α P – Ps …….(2) 

adding (1) and (2) we have 

w1 + w2 α ps+ P – Ps 

α P …. (3) 

Dividing (2) by (3) we get: 

P – Ps / P = w2 / w1 + w2 ………(4) 


Boiling point elevation
•As a result of vapor pressure lowering, the boiling point is elevated.
•If Tb is the boiling point of pure solvent, and T is that of the solution, the difference in boiling point (ΔT) is:
• ΔT = T - Tb 
•AB /AC = AD /AE 
•T1 – Tb / T2 – Tb = 
• P – P1/ P - P2 




•Hence the elevation of boiling point is directly proportional to the lowering of vapor pressure.

ΔT α P - Ps ………….(1) 

But since P is constant for the same solvent at fixed temperature, thus,

ΔT α P - Ps/ P ………..(2) 

But for Raoult”s law P - Ps/ P α w M /mW

M of solvent is constant, thus:

P - Ps/ P α w /mW……….(4) 

From (2) and (4) we get: 

ΔT α w /mW 

ΔT= kb . w/m . 1/W……….(5) 

•Kb is called molal elevation constant, if W of the solvent is expressed in kilograms, but if the mass of the solvent W is given in grames, it has to be converted to kilograms. Thus :
• ΔT = kb . w/m . 1/ W/1000………..(6)

From which we get: m = 1000 x kb x w / ΔT W

If the value of kb is given in k0 per 0.1 kg (100g) the last equation becomes:

m = 100 x kb x w / ΔT W

Thermodynamically kb can be calculated from

kb = RTb2 / 1000 Lv


•By Cottrell”s method:

Theory of dilute solutions

Unlike the internal energy and heat content ,the work obtainable when a system undergoes change depends not only on the initial and final states but also on how the change takes place .
The work performed by a system can vary all the way from zero, “in case of expansion into vacuum ,up to a maximum under specified conditions .
Any process in which the driving force is only infinitesimally greater than the opposing force ,and which canbe reversed by increasing the opposing force by an infinitesimal amount constitute reversible process .
All naturally occurring process are irreversible
Reversibility can be approached in galvanic cells . 
    


Reversibility and maximum work
•The amount of work a system has to perform to give certain change depends on the opposition the system experiences to the change, the greater that resistance is, the more work must be done by the system.
•Consider the expansion of an ideal gas against a pressure P through an infinitesimal volume change dv, the work done is zero , if P is zero, however as P increased more and more work has to be done as  the pressure approaches that of the gas ifself.

•It is evident that the work which may performed by a system is a maximum when the opposing pressure P differs only infinitesimally from the internal pressure of the gas itself. Thus maximum work is obtainable only from a system when the change taking place in it is entirely reversible.

since  w =  ∫ Pdv

for reversible process the external pressure is at all times only infinitesimally lower than the pressure of the gas itself, we may substitute  P = nRT/ v
So    Wm=  v1v2 nRTdv/v
   Wm     = nRTv1v2  dv/v  = nRT ln  v2/v1
      alternatively Wm = nRT  ln  p1/p2
Maximum work of adiabatic expansion.
The expression in question is arrived at as:
Differentiate the expression   pvγ = constant.
   γ pvγ-1 dv  + vγdp  =  0   dividing by vγ-1
   γ pdv  +  v dp    =  0
                  v dp   =   - γ pdv
  complete differentiation of  pv  = nRT gives:

            P dv  + v dp = nRT
On substitution of the expression for v dp
            P dv  - γ P dv  = nR dT
           P dv ( 1- γ) =  nR dT
           P dv    =   nRdT / ( 1- γ)
         Wm = ∫v1v2  p dv  =  T1T2 nR dT / ( 1- γ)
               =  nR ( T2 – T1) / ( 1- γ)
T2> T1   Wm  is negative and work is done on the gas, when work is done by the system
T2< T1   and   Wm is positive.

The maximum work function A
We have seen that the amount of work obtainable from a process depends on the manner in which the work has been performed. Under isothermal and reversible conditions a definite path of passing from the initial to the final state along which the maximum work done is definite and dependent only on the two states.

At constant temperature, when the maximum work being a function of the states as the internal energy E and heat content H.
Therefore , we may think of a system as possessing in each state a certain amount of maximum work content A, called (the Helmholtez free energy), when the system passes from one state to another, the change in the work function ∆ A is equal;
       ∆ A = A2 -  A1
∆ A > 0  means increase in the work content
∆ A < 0  means decrease in the work content




When the work is performed isothermally and reversibly, then the work Wm is done at the expense of the maximum work content of the system.
Thus Wm  =  - ∆A
From the first law;  ∆E = qr  - Wm
= qr + ∆A

The joule – Thomson effect

For an ideal gas, pv product is a constant , if this gas expands under adiabatic conditions into a vacuum, then;

q = 0       w  =  0      ∆E  = 0

i.e ∆E for the gas remains constant as vp and consequently the temperature. This is equivalent to say” At constant temperature the internal energy of an ideal gas is independent of the volume the gas occupies or ( ∂ E / ∂v)T = 0

With real gases the situation is different;


Joule – Thomson experiment

By applying pressure on the piston on the left slowly enough so as not to change the pressure p1, a volume of gas v1 was forced slowly through the pours plug and allowed then to expand to the pressure p2 and volume v2by moving the piston on the right.

      At left piston             At right piston
Work done is – p1v1      work done is p2v2
The net work   w = p2v2– p1v1
    q = 0   since the process is adiabatic
∆E  =  E2 – E1 =  - w  =  - (p2v2– p1v1 )
        E2+ p2v2 = E1 + p1v1
           H2  =  H1                       ∆H  =  0
i.e. the process was conducted at constant enthalpy. 

Under these conditions joule - Thomson observed near room temperatures that all gases show cooling except hydrogen actually became warmer, the extent of the temperature change was found to depend on the initial temperature and pressure of the gas.

•The magnitude of the observed effects controlled by the joule – Thomson coefficient µ which is defined as:
‘The number of degrees temperature change produced by atmosphere drop in pressure under constant enthalpy”.


µ = (∂T / ∂p)H 

µ positive                cooling 
 µ negative               heating



Hydrogen and helium can be cooled adiabatically if they are first brought to a sufficient low temperature, this temperature is called inversion temperature, namely, the temperature at which the gas exhibit neither cooling nor heating.

This effect is of great importance in liquefaction of gases.



The carnot cycle 

A question of great significance to how extent heat is convertible to other forms of energy which may utilize to do work, experience has shown that operating heat engines which absorb heat at temperature T2 and reject the waste heat at a lower temperature T1, can convert only a fraction of the absorbed heat into work.

Theoretical consideration show that even an ideal engine , operating under ideal conditions, would be able to convert only a certain fraction of the absorbed heat into work and that this fraction depend only on the operating tempe.

To establish the above deductions, let us consider the sequence of operation called a carnot cycle, which consists of four steps, two isothermal and two adiabatic.


W1 = nRT2 ln v1/v2 = q2 (a) 

W2 = - ∆E = - ncv ( T1 –T2) (b) 

W3 = nRT1 ln v4/v3 = - q1 ( c) 

W4 = - ∆E = -ncv (T2 – T1)= ncv (T1- T2) (d) 

Wm = w1 + w2 + w3 + w4 

= nRT2ln v2 /v1 + nRT1 ln v4/v3 = q2-q1 (A) 

Division of equation (A) by q2 yields:

Wm/q2 = q2 – q1 /q2 = 

nRT2 ln v1/v2 + nRT1 ln v4/v3 / nRT2 ln v1/v2(B) 

Equation (B) can be simplified,since the points (p1v1) and (p3v3) lie on the same adiabatic ,

then: v1T2cv/R = v3T1cv/R 

Similarly, the points (p2v2) and (p4v4) lie on the same adiabatic then:,

V2 T2cv/R = v4T1cv/R 

On dividing the first equation by the second we obtain: v1 /v2 = v3 /v4 consequently

ln v4 /v3 = - ln v1 /v2 

substituting this value in equation (B) we get

W/q2 =RT2 ln v1 /v2-RT1 ln v1 /v2/RT2ln v1 /v2 

= (T2 –T1) /T2 

and wm = q2 ( T2-T1 / T2) 

•Thus the maximum work that may be recovered from a system during a reversible cyclical process is equal to the heat absorbed multiplied by the ratio (T2-T1/T2 )
•The ratio of wm /q2 is called the thermodynamic efficiency of the process.
•Thermodynamic efficiency must be the same for all the process operating under the given temperature conditions.

The concept of the reversibility

 
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