Certain solid substances, known as hydrated compounds, have well-defined molecular ratios of water to some other species. For example, calcium sulfate dihydrate (commonly known as gypsum, CaSO42H2O), has 2 moles of water per mole of calcium sulfate; alternatively, it may be said that 1 mole of gypsum consists of 1 mole of calcium sulfate and 2 moles of water. The water in such substances is called water of hydration. (More information about hydrated salts is given in Chapter 6.)
In order to eliminate the discharge of sulfuric acid into the environment, a process has been developed in which the acid is reacted with aragonite (CaCO3) to produce calcium sulfate. The calcium sulfate then comes out of solution in a crystallizer to form a slurry (a suspension of solid particles in a liquid) of solid gypsum particles suspended in an aqueous CaSO4 solution. The slurry flows front tlte crystallizer to a filter in which the particles arc collected as a filter cake. The filter cake, which is 95.0 wt% solid gypsum and the remainder CaSO4 solution, is fed to a dryer in which all water (including the water of hydration in the crystals) is driven off to yield anhydrous (water-free) CaSO4 as product A flowchart and relevant process data arc given below.
Solids content of slurry leaving crystallizer 0.35 kg CaSO4-2H2O/L slurry CaSO4 content of slurry liquid: 0.209g CaSO4/100g H2O Specific gravities: CaSO4-2H2O(s), 2.32; liquid solutions, 1.05
- Briefly explain in your own words the functions of the three units (crystallizer, filter, and dryer).
- Take a basis of one liter of solution leaving the crystallizer and calculate tlte mass (kg) and volume (L) of solid gypsum, the mass of CaSO4 in the gypsum, and the mass of CaSO4 in the liquid solution.
- Calculate the percentage recovery of CaSO4—that is, the percentage of the total CaSO4(precipitated plus dissolved) leaving the crystallizer recovered as solid anhydrous CaSO4.
- List five potential negative consequences of discharging H2SO4 into the river passing the plant.
Want to see the full answer?
Check out a sample textbook solutionChapter 3 Solutions
EBK ELEMENTARY PRINCIPLES OF CHEMICAL P
Additional Science Textbook Solutions
Database Concepts (8th Edition)
Starting Out with Java: From Control Structures through Data Structures (4th Edition) (What's New in Computer Science)
Starting Out with C++ from Control Structures to Objects (9th Edition)
Starting Out with Java: From Control Structures through Objects (7th Edition) (What's New in Computer Science)
Thinking Like an Engineer: An Active Learning Approach (4th Edition)
Starting Out with Programming Logic and Design (5th Edition) (What's New in Computer Science)
- #1 Chloral is consumed at a rate of 10 mol/L-s when reacting with chlorobenzene to form DDT and water in the reaction given below. Determine: i) the rate of disappearance of chlorobenzene. ii) the rate of formation of DDT. CCI CHO (Chloral) + 2C6H5Cl (Chlorobenzene) → (C6H4Cl)2CHCCI 3 (DDT) + H2Oarrow_forward#5 The irreversible liquid phase second order reaction, 2A → B, is carried out in a CSTR. The entering concentration of A, CAD is 2 mol/L, and the exit concentration of A, CA is 0.1 mol/L. The volumetric flow rate, vo, is at 3 L/s and is constant (v=vo). The reaction rate constant, k is 0.03 L/mol's. What is the corresponding reactor volume?arrow_forwardProblem 9.11 An 80 mm long line MN has its end M 15 mm in front of the V.P. The distance between the ends projector is 50 mm. The front view is parallel to and 20 mm above reference line. Draw the projections of the line and determine its inclination with the V.P. Also, locate the traces. Interpretation Front view of a line is parallel to xy, therefore, 1. The line is parallel to the H.P. 2. The top view of the line has true length. 3. The front view has projected length equal to the distance be- tween the projectors. Construction Refer to Fig. 9.11. 1. Draw a reference line xy. Mark point m' 20 mm above xy and point m 15 mm below xy. 2. Draw a 50 mm long line m'n' parallel to xy. 3. Draw an arc with centre m and radius 80 mm to meet projec- tor from point n' at point n. Join mn to represent the top view. Determine its inclination with xy as the inclination of line MN with the V.P. Here = 51°. 4. Traces Extend line mn to meet xy at point v. Project point v to meet m'n' produced at…arrow_forward
- oh 30 20 D и D P 60 60 80arrow_forward⑤ b Δε m ab C 40arrow_forwardProblem 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_forwardProblem 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_forward
- □ 40 a' 12 o' a O d'arrow_forwardTL = 85 D. Problem 9.12 The top view of a line measures 60 mm. The line is parallel to the V.P. and inclined at 45° to the H.P. One end of the line is 25 mm in front of the V.P. and lies on the H.P. Draw its projections and determine the true length. Interpretation Let the line be PQ parallel to the V.P. The front view has true length and the top view is parallel to xy. Construction Refer to Fig. 9.12. 1. Draw a reference line xy. Mark point p' on xy and point p 25 mm below xy. 2. Draw a 60 mm long line pq parallel to xy. This repre- sents the top view. 3. Draw line from point p', inclined at 45° to xy to meet the projector from point q at point q'. Join p'a' to represent the front view. Measure length of p'a' as true length of line PQ. Here T.L. = 85 mm. Result True length of line PQ is p'q' = 85 mm. 5 Fig. 9.12 p 60 σarrow_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_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Introduction to Chemical Engineering Thermodynami...Chemical EngineeringISBN:9781259696527Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark SwihartPublisher:McGraw-Hill EducationElementary Principles of Chemical Processes, Bind...Chemical EngineeringISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEYElements of Chemical Reaction Engineering (5th Ed...Chemical EngineeringISBN:9780133887518Author:H. Scott FoglerPublisher:Prentice Hall
- Industrial Plastics: Theory and ApplicationsChemical EngineeringISBN:9781285061238Author:Lokensgard, ErikPublisher:Delmar Cengage LearningUnit Operations of Chemical EngineeringChemical EngineeringISBN:9780072848236Author:Warren McCabe, Julian C. Smith, Peter HarriottPublisher:McGraw-Hill Companies, The