(a)
Interpretation:
The units of
Concept introduction:
Units of any quantity can be determined with the help of dimensional analysis. It is the analysis done for different physical quantities and their relationships with each other by their base quantities such as length, mass, time etc. It uses the rules of algebra for the conversion of units.
(b)
Interpretation:
The given equation (1) is to be transformed into the form
Concept introduction:
For the equation of type
And intercept
(c)
Interpretation:
An excel spreadsheet is to be prepared according to the given structure and from it, conclusion about the given equation (1) is to be stated.
Concept introduction:
For a given set of data to fit in a straight-line,
(d)
Interpretation:
The values of
(e)
Interpretation:
The difference in the calculated values of
Concept introduction:
Rate of a reaction is dependent on the initial concentration of the reaction, temperature, activation energy and nature of the reactant.
For a reaction whose temperature and final concentration is pre-defined, alterations in the initial concentration and activation energy gives the desired results.
(f)
Interpretation:
The given equation (2) is to be transformed into the form
Concept introduction:
It is a strong function of the absolute temperature in the reactor:
It is known as the Arrhenius equation. Here,
For the equation of type
And intercept
(g)
Interpretation:
The reason for the temperature control and well-mixing of the reactor contents are to be explained.
Concept introduction:
Yield of a reaction is defined as theoretical amount of the desired product formed in a chemical reaction.
Selectivity of a chemical reaction is the ratio of the desired product and the undesired product formed.
(h)
Interpretation:
The rows 1-3 and 12 of columns J and K of the excel spreadsheet are to be filled according to the structure given. The value of
Concept introduction:
Rate of a reaction is dependent on the initial concentration of the reaction, temperature, activation energy and nature of the reactant.
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Chapter 3 Solutions
ELEM.PRIN.OF CHEM.PROCESS-ACCESS
- Problem 10.16 An isosceles triangle of base 40 mm and altitude 54 mm has its base in the V.P. The surface of the plane is inclined at 50° to the V.P. and perpendicular to the H.P. Draw its projections. Construction Refer to Fig. 10.17. An isosceles triangle has its base in the V.P., so con- sider that initially the triangle ABC is placed in the V.P. with base AB perpendicular to the H.P. 1. First stage Draw a triangle a'b'c' keeping a'b' perpendicular to xy to represent the front view. Project the corners to xy and obtain ac as the top view. 2. Second stage Reproduce the top view of first stage keeping ab on xy and ac inclined at 50° to xy. Obtain new points a', b' and c' in the front view by joining the points of intersection of the vertical projectors from a, b and c of the second stage with the corresponding horizontal locus lines from a', b' and c' of the first stage. Join a'b'c' to represent the final front view. Here, the front view is an equilateral triangle of side 40 mm. X 54…arrow_forward%9..+ ۱:۱۹ X خطأ عذرا ، الرقم الذي أدخلته خاطئ. يرجى إدخال رقم بطاقة الشحن الصالحة والمحاولة مرة أخرى. رصيد هاتفك قم بمسح الرمز = رقم بطاقة التعبئة 7794839909080 رمز مكون من 13 او 14 رقماً طريقة إعادة التعبئة قم باعادة تعبئة الرصيد إعادة تعبئة الإنترنت إعادة تعبئة الرصيد O >arrow_forwardProblem 10.14 A hexagonal plane of side 30 mm has a corner in the V.P. The surface of the plane is inclined at 45° to the V.P. and perpendicular to the H.P. Draw its projections. Assume that the diagonal through the corner in the V.P. is parallel to the H.P. d' a 2 b b.f C' c.e b 'C' H.P. (a) V.P E HEX 30 e' O' d' a a' b' C' b' X y a b,f c,e d b,f (b) c,earrow_forward
- Problem 10.18 A 60° set-square has the shortest edge of 40 mm lying in the V.P. The surface is in- clined to the V.P. and perpendicular to the H.P. such that the front view appears as an isosceles triangle. Draw the projections of the set-square and determine its inclination with the V.P. Construction Refer to Fig. 10.18. A 60° set-square inclined to the V.P. and per- pendicular to the H.P. can appear as an isosceles triangle in the front view, when the shorter edge is in the V.P. 1. First stage Draw a right angled triangle 40 a' a' b' c' b' X 40 C' a,b C a,b a'b'c' keeping 40 mm long a'b' perpen- dicular to xy. Project the corners to xy and obtain ac as the top view. 2. Second stage Draw another right angled triangle a'b'c' on the horizontal locus line from points a', b' and c' of the first stage such that length of b'c' is equal to that of a'b'. Project a'b' to meet xy at ab. Draw an arc with centre a and radius equal to ac of the first stage to meet the vertical projector of c' at…arrow_forwardProblem 9.13 A 70 mm long line PQ does not have H.T. and V.T. One end of the line is 30 mm in front of the V.P. and 20 mm above the H.P. Draw its projections. Interpretation As the line PQ does not have H.T. and V.T., it is parallel to both H.P. and V.P. Construction Refer to Fig. 9.13. 1. Draw a reference line xy. Mark point p' 20 mm above xy and point p 30 mm below xy. 2. Draw a 70 mm long line p'a' parallel to xy to repre- sent the front view. X 20 p Fig. 9.13 3. Also, draw a 70 mm long line pq parallel to xy to represent the top view. 70 q yarrow_forwardProblem 10.19 A square lamina ABCD of side 40 mm is suspended from a point O such that its surface is inclined at 30° to the V.P. The point O lies on the side AB 12 mm away from A. Draw its projections. Construction Refer to Fig. 10.19. 1. First stage Draw a square a'b'c'd keeping a'd' parallel to xy. Mark a point o' on a'ď at a distance 12 mm from end a' as the point of suspension. Also, mark the centre of the square g' to represent the centre of gravity. 2. Second stage Reproduce the front view of first stage such that o'g' is perpendicular to xy. Project corners and obtain bd as the top view. 3. Third stage Reproduce the top view keeping bd inclined at 30° to xy. Obtain new points a', b', c' and ď' in the front view by joining the points of intersection of the vertical projectors drawn from points a, b, c and d of the third stage with the corresponding horizontal locus lines drawn from points a', b', c' and ď of the second stage. Join new a'b'c'd to represent the final front view.arrow_forwardProblem 10.15 A circular plane of diameter 50 mm is resting on a point of the circumference on the V.P. The plane is inclined at 30° to the V.P. and the centre is 35 mm above the H.P. Draw its projections.arrow_forwardYou are asked to manufacture 10 kg of polyester with a number-average molecular weight of 1000 by polymerizing butane-1,4-diol(HO(CH2)4OH) with adipic acid (HOOC-(CH2)4-COOH).a) What weight of diol and diacid do you need, respectively? To whatextent, p, should the reaction be carried out to? Assume a stoichiometricbalance.b) What are the number and weight fractions of dimer, trimer and tetramerat this point in the reaction?c) Because of the polymerization by dehydration to olefin, 3 mol% of thediol will be lost. What would be the number-average molecular weightwhen the reaction is carried out to the same extent? How could you offsetthis loss so that the desired molecular weightarrow_forward9.4. A PID temperature controller is at steady state with an output pressure of 9 psig. The set point and process temperature are initially the same. At time = 0, the set point is increased at the rate of 0.5°F/min. The motion of the set point is in the direction of lower temperatures. If the current settings are PART 3 LINEAR CLOSED-LOOP SYSTEMS Ke = 2 psig/°F Ti = 1.25 min TD = 0.4 min plot the output pressure versus time.arrow_forward9.6. A PI controller has the transfer function Determine the values of K, and T. 5s + 10 Ge Sarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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