A Compilation of Laboratory Reports

A Compilation of Laboratory Reports in

Chemical Engineering Laboratory I (ChE 402)

Elaine May L. Geneston BS Chemical Engineering -4

Engr. Rosario Dangin Laboratory Instructor

March 23,2015

Table of Contents

Experiment

Activity/Title

1.

Calibration of Venturi and Orifice Meters

2.

Reynolds Number

3.

Fluidization

4.

Heat Loss in Bare and Lagged Pipes

5.

Plate and Frame Filter Press

6.

Sedimentation

7.

Sieving

8.

Agitation

Calibration of Venturi and Orifice Meters

Elaine May L. Geneston BS Chemical Engineering – IV Group 5

Department of Chemical Engineering, College of Engineering and Architecture Cebu Institute of Technology – University N. Bacalso Ave. Cebu City, 6000 Philippines

The objective of this experiment is to gather data on pressure drop versus water flow rate and to plot the coefficient of discharge of a sharp – orifice versus Reynolds Number, and the coefficient of discharge of a venturi versus Reynolds Number. The two typical head meters that are used to measure flow rates are the orifice and venture meters. to fix the water flow rate , the pump was started and the main regulating flow valve was gradually opened. The tubes were connected from the venture or orifice pressure tapping points to the manometer (mouth or inlet tap point and throat tap point), care must be taken to ensure that there is no trapped air in the connecting lines. Ample time was allowed to stabilize the flow before taking any readings. The upstream and downstream of the manometer was read and recorded. For any reading of the manometer, the volume discharge at the outlet was collected and the time was measured to collect certain quantity of water either in the steeped measuring tank or using a graduated cylinder. Several trials were taken by adjusting the main flow regulating valve. The coefficient of discharge of the orifice apparatus and its Reynolds Number was obtained. Engr. Rosario G. Dangin Instructor

March 23, 2015

1 .Introduction The Orifice and venturi meter are two typical head meters that are used to measure flow rates. By their design, a pressure difference will occur between theupstream and downstream sides of the element. The cause is the constriction which changes the pressure head partly into velocity heads. This is the principle on which by a manometric apparatus attached to static bores located upstream and downstream or on the element. The orifice meter is the simplest and most inexpensive. It consists of a flat plate, perpendicularly placed to the stream of flow, attached to the conduit by flanges, with a concentric hole with respect to the conduit, on it. The hole is usually sharp – edge with respect to the upstream and resembles a squat frustum of a cone when seen in cross – section. Streamlines issuing from such a constriction attains its minimum cross sectional area at a mean distance of ½ pipe diameter downstream. That location is called the vena contracta. Thelocation is not constant but rather, the distance from the plate decreases as D (orifice)/D (pipe) increases with respect to a bore located 1 – 2 pipe diameters upstream. The maximum pressure different is noted when the downstream bore is at the vena contracta. The magnitude of the pressure differential is dependent on the location of tapping. There are 5 ways of tapping: corner taps, pipe taps, flange taps, vena contracta taps, and radius taps. The main disadvantage of the orifice meter is the permanent head loss is the always accounted for due to friction at the constriction. The venturi meter is composed of 2 frustums of cone and a throat. The converging cone is of a larger angle sine friction effects on the upstream side are usually negligible. The diverging cone is made of as smooth and as tapering as possible to eliminate drag and friction. The loss amounts to 10% to 20% of the initial head. Tapings are located at a part of the meter whose diameter is still equal to that of the pipe at the upstream and at the throat to assure maximum pressure difference. The equation that are derived for the flow rates are applicable for both the venturi and orifice meters and incompressible and compressible fluid flow, the only difference when the fluid is compressible the insertion of a coefficient of

expansion,ϒ, which is a function of D (constriction)/D (pipe), acoustic ratio and the pressure head. Theoretically for the variation of C with R e (which is also a measure of flow) opposite should be seen for the venturi and orifice flow meters. For the venturi, as NRe increases, C should be increase since friction effects decrease and flow rate approaches the theoretical. For the orifice meter, as N Re increases, C should decrease since friction increase and a greater head loss results.

2. Materials and Methods

2.1

Materials and Apparatus

     

Hydraulic bench apparatus Orifice meter, Venturi meter Stopwatch Manometer Water Caliper

2.2 Methods The venturi meter or orifice meter apparatus was set-up. The pump was started and the main regulating flow valve was gradually opened to fix the water flow rate.The tubes was connected from the venturi or orifice pressure tapping points to the manometer (mouth or inlet tap point and throat tap point), care must be taken to ensure that there is no trapped air in the connecting lines. Ample time was allowed to stabilize the flow before taking any readings. The upstream and downstream of the manometer was read and recorded. The diameter of the cylindrical cross-section of the tapping points of the venturi or orifice apparatus was recorded.The theoretical volumetric flow rate was computed.For any reading of manometer, the volume discharge at the outlet was collected and the time was measured to collect certain quantity of water either in the steeped measuring tank or using a graduated cylinder. The volume collected and the time was recorded. The actual volumetric flow rate from the volume collected divided by the time obtained was computed. Several trials were taken by adjusting the main flow regulating valve. The data was recorded and computed for the coefficient of discharge of the Venturi or Orifice apparatus and its Reynolds Number.

3. Results A. Venturi Flow Meter Gathered Data: Trial

Manometer Reading Upstream (cm) Downstream (cm)

Rm (m)

1 2 3

27.6 28.3 31.2

18.5 18.6 19.6

0.091 0.097 0.116

4

33.3

20.6

0.127

5

35.6

21.7

0.139

6 7 8 9 10

36 30.9 31.6 45.3 51.5

22.4 21 21.2 35.7 36.3

0.136 0.099 0.104 0.096 0.152

Volumetric Flowrate (Q actual) m3/s 0.00001 0.0000092 0.0000164 0.0000214 0.0000264 0.0000236 0.0000134 0.0000168 0.000016 0.00003

B. Orifice Flow Meter Gathered Data: Trial 1 2 3 4 5 6 7 8 9 10

Manometer Reading Upstream (cm) 23.8 32.5 23.2 29.7 38.8 47.7 68 60.3 63.5 74.5

Downstream (cm) 19.3 25.1 14.6 26.9 36.5 46.2 64.8 39.3 52.6 68.3

Volumetric Flowrate (Q actual) m 3/s

Rm (m) 0.045 0.074 0.086 0.028 0.023 0.015 0.032 0.21 0.109 0.062

0.00034 0.0003675 0.0003775 0.000275 0.00031 0.000295 0.00032 0.0004 0.00039 0.0003675

Theoretical Volumetric Flowrate A Qtheoretical =

√

√

2(P 1−P2) ρ

1−(

4

D2 ) D1

=

A √2 g (∆ H )

√

N ℜ=

4

1−(

D2 ) D1

D2 νρ D2 Qρ = μ Aμ

Where: D1= pipe diameter = 0.025m 0.015 m

D2= throat diameter =

μ = viscosity of water (H20) at 25 0C = 0.8937 x 10 -3 kg/msΔH= manometer reading difference (R)

ρ

= density of water (H20) at 25 0C = 997.08 kg / m

0.000176715 m2

A. Venturi Flow Meter

3

A= area of throat =

Computed Data: Volumetric Flowrate (Q theoretical ) m 3 /s 0.000253052 0.000261261 0.000285705 0.000298945 0.00031275 0.000309356 0.000263941 0.000270524 0.000259911 0.000327048

C (discharged coefficient) 0.039518 0.035214

NRe Actual

NRe Theoretical

9470.133278 8712.522616

239643.7444 247417.9862

0.057402

15531.01858

270566.7314

0.071585

20266.08521

283104.8184

0.084413 0.076287 0.050769 0.062102 0.061559 0.09173

25001.15185 22349.51454 12689.97859 15909.82391 15152.21324 28410.39983

296178.0031 292964.4035 249955.673 256189.939 246139.3317 309718.5337

B. Orifice Meter Flow

Computed Data:

Volumetric Flowrate (Q theoretical ) m 3 /s 0.000177949 0.000228194 0.000246002 0.000140368 0.000127219 0.000102739 0.00015006 0.000384414 0.000276951 0.000208874

C (Coefficient Discharge) 1.910659732 1.610468275 1.534540733 1.959135024 2.436735926 2.871356237 2.132483501 1.040545402 1.40819257 1.759430795

Graphs A. Venturi Flow Meter

NRe Actual

NRe Theoretical

321984.5315 348027.398 357497.5312 260428.6651 293574.1316 279368.9317 303044.2649 378805.3311 369335.1978 348027.398

168520.0803 216103.2311 232967.1175 132930.432 120478.4353 97295.11374 142108.6094 364044.981 262276.0592 197806.8128

C vs Actual NRe 0.1 0.08

f(x) = 0x + 0.01 R² = 0.98

0.06 Coefficient Discharge

0.04 0.02 0 5000

10000 15000 20000 25000 30000 Actual Reynolds Number

C vs Theoretical NRe 0.1 0.08

f(x) = 0x - 0.12 R² = 0.83

0.06 Coefficient Discharge

0.04 0.02 0 200000

250000

300000

Actual Reynolds Number

B. Orifice Flow Meter

350000

C vs Actual NRe 2.5 2 1.5 Coefficient Discharge

f(x) = - 0x + 5.41 R² = 0.69

1 0.5 0 300000

325000

350000

375000

400000

Actual Reynolds Number

C vs Theoretical NRe 3.5 3 2.5 2 Coefficient Discharge 1.5

f(x) = - 0x + 3.05 R² = 0.86

1 0.5 0 10000

110000

210000

310000

Theoretical Reynolds Number

4.Discussion As shown on the graph plotted above,

the used of Venturi meter is more

accurate than using orifice meters. As the fluid flows through the orifice plate the velocity increases, at the expense of pressure heat. The Orifice and venturi meter are two typical head meters that are used to measure flow rates. Pressure difference will occur between the upstream and downstream sides of the element. The cause is the constriction which changes the pressure head partly into velocity heads.For the orifice meter, as N Re increases, C should decrease since friction increase and a greater head loss results.

5. Conclusion As the orifice is passed, the pressure drops immediately. The drop continues until the vena contracta is reached. Then gradually increases until a maximum pressure point is reached that will be lower than the pressure upstream of the orifice. The pressure difference is obtained because the velocity is increased from V1 to V2. When the velocity decreases as the fluid leaves the orifice the pressure increases and tends to return to its original level. Because of some frictional losses, some of the difference P1 – P2 is not fully recovered. The permanent pressure loss is much higher because of eddies formed when the jet expands below the vena contracta. The pressure drop across the orifice increases when the rate of flow increases. When there us no differential there is no flow rate. To obtain the actual volumetric flow rate, the volume collected is divided by the time obtained. And for the theoretical volumetric flow rate given the pressure drop, the Bernoulli equation at a constant elevation and continuity equation was used.

6. Appendices A. Formulas

Bernoulli equation at a constant elevation from a point in the pipe in the pipe to the constriction. P1 U 21 P2 U 22 + = + ρ 2 ρ 2

U 22−U 21 =2

( P1 −P 2) ρ

From continuity equation, A1U1 = A2U2

U1 =

1 D¿ ¿ ¿ ¿ 2 ( D 2) U 2 ¿

Let B =

D2 D1

U1 = B

2

U2

U 22−U 21 =2

( P1 −P 2) ρ

U 22−(B 2−U 2 )2=2

U 22−B4 U 22 =2

U2 =

U2=

√

√

(P1−P 2) ρ

2( P1−P2 ) ρ

√ 1−( B )

4

2( P1−P2 ) ρ

√

D 1− 2 D1

4

( )

A2 Q(Theo)=

(P1 −P2) ρ

√

√

2( P 1−P2) ρ

D 1− 2 D1

( )

Q(actual)=Volume/Time

C=

NRe =

4

Q (actual) Q(theo) D ( U 2 ) ( ρ) µ

Where: P = P1 - P2 = pressure drop U = velocity A = area  = density D = diameter (orifice/venturi) C = discharged coefficient NRe = Reynolds number

7. References http://www.shambhaviimpex.com/venturimeter-orificemeter-calibration.html

30 Last

"Fundamentals of Orifice Metering." Afms.org. Smith Metering, Inc, n.d. Web. Sept. 2013. . accessed March 20, 2015.

Çengel, A., and J. M. Cimbala. Fluid Mechanics: Fundamentals and Applications. 2nd ed. Boston: McGraw-Hill Higher Education, 2010. Print. Last accessed March 20, 2015.

FLUIDIZATION

Elaine May L. Geneston BS Chemical Engineering – IV

Department of Chemical Engineering

College of Engineering and Architecture Cebu Institute of Technology – University 196 N. Bacalso Avenue Cebu City, 6000 Philippines

This experiment is about fluidization of a bed of solid by passing a fluid, usually a gas upwards through a bed of particles supported on a distributor. Fluidization or fluidizing, converts a bed of solid particles into an expanded mass that has many properties of a liquid. As a fluid is passed upward through a bed of particles, pressure loss due to frictional resistance increases as fluid flow increases. At a point, whereby the upward drag force exerted by the fluid on the particle equal to apparent weight of particles in the bed, fluidization occurs. The size of solid particle which can be fluidized varies greatly from less than1µm to 6cm. It is generally concluded that particles distributed in sizes between 150µm and 10µm are the best for smooth fluidization (least formation of large bubbles). Large particles cause instability and result in slugging or massive surges. Small particles (less than 20µm) frequently even though dry, act as if damp, forming agglomerates or fissures in the bed, or spouting. Adding finer sized particles to a coarse bed or coarse sized particles to a bed of fines usually results in better fluidization. The upward velocity of the gas is usually between 0.15m/s and 6m/s. This velocity is based upon the flow through the empty vessel as is referred to as the superficial velocity. As the velocity of flow increases, the particles rearrange themselves to offer less resistance to the fluid flow and the bed will tend to expand unless it is composed of large particles (mean diameter > 1mm). The expansion continues until a stage is reached where the drag force exerted on the particles will be sufficient to support the weight of the particles in the bed. The fluid/particle systems then begin to exhibit fluid like properties and it will flow under the influence of a hydrostatic head. This is the point of incipient fluidization and the gas velocity needed to achieve this is referred to as the minimum fluidization velocity, U mf. Beyond this velocity, the pressure drop across the bed will be approximately equal to the weight of the bed per unit area. The effective ∆P excludes the hydrostatic pressure drop across the bed which can be neglected in gas fluidized systems operating at atmospheric pressure. It is likely, however that this pressure drop will be exceeded just prior to fluidization with gas fluidized systems in order to overcome cohesive forces between the particles and break down the residual packing and interlocking of particles within the bed.

The behavior of fluidization is depends on the types of the particles composed in the vessel. Geldart (1973) classified powders into four groups according to their fluidization properties at ambient condition. There are 4 stages of particles that are (A) aerated, (B) bubble, (C) cohesive and (D) dense. In this experiment, we are considering with a coarse sand which is in group B, Ballotini which is in group A and Glutinous flour which is in group C. From this experiment, we can obtain the bed expansion, bed pressure drop and the flow rate of the fluid. By the equation given in the theory, superficial gas velocity, Umf and εmf for all cases can be calculated. Then only, we plotted two graphs which are bed pressure drop against superficial gas velocity and bed expansion against superficial gas velocity for all cases. The U mf predicted from the graph then is being compared with the calculated one. ______________________________________________________________________________ ___________________________

Engr. Rosario G. Dangin Instructor

March 23, 2015

1. INTRODUCTION Fluidized bed reactors are a relatively new tool in the chemical engineering field. Fritz Winkler developed the first fluidized bed gas generator in Germany in the 1920s. One of the first United States fluidized bed reactors used was the Catalytic Cracking Unit, created in Baton Rouge, LA in 1942 by the Standard Oil Company (Exxon Mobil). A fluidized bed is a packed bed through which fluid flows at such a high velocity that the particles in the bed are loosened such that the bed behaves as though it is a liquid. Fluidized beds provide a large surface area for contact between solids and a liquid or a gas that is conducive for heat and mass transfer. In this environment, nearly uniform temperatures can be maintained in the reactor even with highly exothermic reactions. This is important because a temperature gradient can form in a poorly mixed bed, leading to equipment failure, product degradation, and decreased efficacy of the reaction. A fluidized bed also provides uniform mixing, which is important for product quality and efficiency. Fluidized bed reactors are often a continuous process, meaning they are also very efficient compared to batch processes. However, there are some disadvantages to fluidized beds. One disadvantage is that the cost of a fluidized bed reactor is usually high because the vessels are typically larger than batch or other processes. Another disadvantage is that sometimes particles may become entrained, or blown along with the flow, which can be costly and problematic to repair. There is also an extra power input that is required for the pump to moderate the pressure drop. Finally, the fluid-like behavior of these fine particles may eventually cause erosion issues. Fluidized beds can be stimulated by either gas or liquid flows. In either case, the process of fluidization is a competition between the force of gravity pointing downwards and the upward pointing drag force caused by friction between the flowing fluid and the individual particles that make up the fluidized bed. As the flow rate of the working fluid is increased, it flows faster across the individual particles, increasing the magnitude of the drag force. Eventually, at a certain velocity (called the minimum fluidization velocity, V f), the drag and gravitational forces will be in balance, and the bed will begin to fluidize and bubble. As the velocity is further increased, the drag force becomes more and more dominant over gravity, and the bed bubbles more furiously. The individual particles are not carried away with the flow, because their settling velocities are far larger than the minimum fluidization velocity, perhaps 50-75 times larger. In this experiment, minimum fluidization velocity will be found for several different types of fluidized beds. The effects of pressure drop, bed height, bed type, grain size, and temperature will be investigated.

2. Materials and Methods: 2.1 Materials:     

Fluidization apparatus Tap water Tape Measure Graduated Cylinder Timer

2.2 Methods: 1. The weight of the bed and the average of the particle were noted. 2. The water container was filled with water sufficient enough to operate the apparatus. 3. Before operating, the height of the mercury was level by removing the clip and adjusting the pressure in the manometer slowly. 4. Before turning the pump the valve was surely closed to avoid mercury spillage. 5. The pump was turned on. 6. Initial height of the particle was recorded. 7. The pressure drop across the bed was measured starting with the lowest possible flow-rate of water. 8. By simultaneously collecting liquid sample in the exit tube using graduated cylinder the volumetric flow rate and the time of collection was recorded. 9. The volume of the liquid sample was measured and recorded. 10. The flow rate of water was increased and the corresponding pressure drop across the bed was recorded. 11. The new height of bed was measured if the bed expands. 12. Steps 10 and 11 was repeated until the maximum allowable flow-rate was reached. 3. Results: 3.1 Tables: Orifice Trial No.

Volume (mL)

Upstream

Downstrea m

1

850

21.8

4

2

992

34.5

24

Volumetric Rm (cm) flow rate, Q (m3/s) 8.5x10-4/ 17.8 5.13 9.92x10-4/ 10.5

3

850

49.5

37.6

4

540

66.1

52.7

5

550

63.7

50.5

6

500

81.2

66.3

7

500

41.1

29.8

8

530

53.7

40.7

9

620

89.7

28.8

10

680

17.7

7.8

Trial No.

Volume (mL)

Upstream

Throat

1

610

26.6

23.2

9.4 8.5x10-4/ 5.06 5.4x10-4/ 3.14 5.5x10-4/ 3.48 5x10-4/ 3.07 5.3x10-4/ 3.99 5.3x10-4/ 3.08 6.2x10-4/ 4.17 6.8x10-4/ 5.44

11.9 13.4 13.2 14.9 11.3 13.0 30.9 9.9

Table.1 Venturi

2

640

36.5

32.5

3

770

38.1

35.8

4

760

37.8

34.5

5

880

24.5

21.2

6

870

36.8

33.1

7

650

29.5

25.4

8

680

35.3

31.8

9

740

65.4

60.5

Volumetric flow rate, Q (m3/s) 6.1x10-4/ 3.20 6.4x10-4/ 4.61 7.7x10-4/ 5.54 7.6x10-4/ 5.38 8.8x104 /6.12 8.7x10-4/ 5.63 6.5x10-4/ 3.87 6.8x10-4/ 4.59 7.4x10-4/ 4.65

Rm (cm)

3.4 4 3.2 3.3 3.3 3.7 4.1 3.5 4.9

10

800

43

8x10-4/ 4.59

39

4

3. Results and Discussion: Tabulated Data and Results Pressure Time (sec) 2.37 2.76 2.35 2.32 2.04 1.63 2.56

Bed

(mmHg) Left Right 195 203 206 210 213 214 214

Porosity (ε)

height

191 185 182 177 173 174 174

ΔP in manometer (Pa) 533.2895 2399.8026 3199.7368 4399.6382 5332.8947 5332.8947 5332.8947

Volumetric ΔP in bed

Volume

(Pa)

(m3)

1.58655 8.95310 10.61965 15.07147 17.18321 24.16057 620.2174

Velocity v,

Superficial

Fanning

(m/s)

velocity v’,

friction

(m/s)

(f)

(m)

-5

8.3x10 2.77x10-4 2.60x10-4 3.12x10-4 2.95x10-4 2.84x10-4 2.42x10-4

flow rate (m3/s) 3.502x10-5 1.004x10-4 1.106x10-4 1.345x10-4 1.446x10-4 1.742x10-4 9.453x10-5

NRe, bed

NRe, particle

0.05

0.0278680305

0.0181627462

0

144.974

217.462

0.06

4 0.0798955946

7 0.0520712581

0

415.631

623.446

0.0880124777

7 0.0573613660

0

457.856

686.784

0.08

2 0.1070314489

7 0.0697568149

0

556.796

835.194

0.09

0.1150687548

5 0.0749950590

0

598.608

897.912

0.09

0.1386236313

9 0.0903467447

0

721.144

1081.72

0.7522440794

7 0.4902685293

0.256

3913.30

5869.96

0.07

0.09

0.651741293 5

4. Discussion: Fluidization is a process when a fl uid is passed upward trough a bed of particles the pressure loss in the fluid due to the frictional resistance with increases with increasing fl uid fl ow. A point is reached when the upward drag force exerted by the fl uid on the particles is equal to the apparent weight of particles in the bed. At this point the particles re l i f t e d b y t h e fl u i d , t h e s e p a r a t i o n o f t h e p a r t i c l e i n c re a s e s , a n d t h e b e d b e c o m e fluidized. The superficial fluid velocity at which the packed bed becomes a fluidized bed is known as the minimum fluidization velocity. This velocity increases with particle sizeand particle density and is affected by fluid properties.For the first types of particles which is coarse grade sand, the graph shows a little increasement in the pressure drop when the superfi cial velocity gas also increase. The fl u i d i z a t i o n s t a r t s w h e n i t re a c h e s m i n i m u m fl u i d i z a t i o n v e l o c i t y w h i c h i s a b o u t 0.0715m/s. The second type is fi ner sand or ballotini, from the graph we can saw that the pressure drop also increase as the superfi cial gas velocity increased. For this case, the minimum fl uidization velocity is 0.0682m/s. For these two types of particles, bubbles continue to grow, never achieving a maximum size. Lastly is the glutinous fl our, fl uidization did not occur in this case. The bed not expanding and resist aeration. This is because the flour is cohesive and the structure is so strong upon fl uidization. Besides that, it also because the interaction force between the particles is strong if compared to the hydronamic force by the fluiding gas. For Glutinous Flour, fluidization did not occur because group C particles exhibit cohesive tendencies. The structures are so strong which upon fluidization, cracks and rat hole is form and at a given pressure different, the bed not expanding and resist aeration.

5. Conclusion: Thus, t h e m i n i m u m fl u i d i z i n g v e l o c i t y , U mf for coarse grade sand is 0.0715m/s while Umf for Ballotini is 0.0682m/s.2 . T h e v o i d a g e a t m i n i m u m fl u i d i z i n g v e l o c i t y εmf for coarse grade sand is 0.321387while εmf for Ballotini is 0.383097.3 . C o n c l u d i n g t h a t

t h e b e d e x p a n s i o n a n d t h e p re s s u re d ro p o f t h e p a r t i c l e a re p ro p o r t i o n a l t o t h e superficial velocity of the gas supply.

References: [1] http://www.thermopedia.com/content/1241/ [2] http://www.slideshare.net/jeufier/calibration-of-orifice-venturi-meter [3] http://www.engineeringtoolbox.com/orifice-nozzle-venturi-d_590.html [4] Kim, S.: Calculations of Low Reynolds Number Rocket Nozzles. AIAA Paper 930888,Jan. 1993.

HEAT LOSS IN BARE AND LAGGED PIPES Geneston, Elaine May L. BS Chemical Engineering – IV

Group 5

Department of Chemical Engineering College of Engineering and Architecture Cebu Institute of Technology – University 196 N. Bacalso Avenue Cebu City, 6000 Philippines

The impact on heat transfer to or from the fluid will depend not only on the relative thermal resistances of the pipe wall, but also on the other thermal resistances in the system. For bare piping, the air surface coefficient normally represents the largest thermal resistance in the system. The wind speeds at the surface, along with the thermal emittance of the surface material, are dominant. As insulation is added to the system, the resistance of the insulation layer begins to dominate and other resistances become less important. As expected, the heat loss or gain depends on both the thickness of the insulation as well as the choice of the pipe material. However, the effect of insulation thickness is considerably more significant than the choice of pipe material.

Instructor Engr. Rosario Dangin

March 23, 2015 1. INTRODUCTION 2. Materials and Methods: 2.1 Materials:      

Boiler Test Pipes – bare, paint, silver chrome paint and 85% magnesia insulation Thermocouple Beakers Graduated Cylinder Stopwatch

2.2 Methods: Three runs with steam at approximately 30 psig for each run was made. 1 The drain cock was cracked under the header to remove the water from the steam line and header after adjusting the system to the desired pressure. 2 The four plug typ0e valve was opened to blow out any condensate from the pipes and was closed until only small amount of steam escapes along with the condensate. 3 The condensate from each pipe was measured and collected over a time interval of 15 – 30 minutes, when the system has reached the equilibrium, as determined by surface temperature measurements. 3. Results: 3.1 Tables:

PIPE NO. COVERING

1 Paint

OUTSIDE DIAMETER, in EMISSIVITY RUN NO. BAROMETRIC PRESSURE STEAM PRESSURE STEAM TEMPERATUR E ROOM TEMPERATUR E TIME/RUN

1.34

LENGTH OF PIPE 2 3 Bare Pipe Silver Chrome Paint 1.34 1.34

0.95

0.95

4 85% Magnesia Insulation 2.48%

0.35

0.95

1 29.58 in Hg 30 psig 134.38oC

25oC

15mins

Table.1 Pipe Trial 1

2

2 Silver Chrome 126oC 101oC 102oC

3 Paint 116oC 106oC 81oC

4 Bare Paint 139oC 112oC 61oC

D A B

62oC 68oC 62oC

112oC 117oC 122oC

117oC 119oC 101oC

121oC 113oC 107oC

C D A B C D

3

Average Volume of Condensate Table.2

2

A B C

1 Magnesia 61oC 65oC 64oC

65oC 66oC 66oC 63oC 70oC 64oC 64.667oC 138

Sample Calculations

109oC 115oC 119oC 116oC 102oC 107oC 112.33oC 205

117oC 118oC 110oC 113oC 121oC 113oC 111oC 219

105oC 113oC 112oC 124oC 100oC 118oC 110.417oC 240

1 atm 29.921 inHg

Atmospheric Pressure = 29.58 inHg x lb ¿2

2

x

14.696 lb /¿ 1 atm

(abs)

Pressure in the system = 30 psig + 14.696

HL1 = 242.922

HL2 = 180

Btu lb

HV2 = 1150.263

Btu lb

@ 44.696

@ 14.5285 Btu lb

lb ¿2

lb ¿2

lb ¿2

= 44.696

lb ¿2

(abs)

(abs)

@ 14.5285

lb ¿2

(abs)

Substitute values to the equation: HL1 = x (HL2) + (1 – x) (HV2) Paint:

Q θ

= 17.443

=

mL g ¿ ¿ 219 mL ¿ ¿

1.34∈¿ ¿ A= π¿ ¿

x

min hr 15 min

60

x 929.219

Btu lb

= 1918.200

(10ft) = 3.508 ft2

ΔT=111oC – 25oC = 86oC ≈ 186.8oF

(hc + hr) =

Q θ AΔT

Btu hr =2.9272 2 (3.508 ft )(186.8 F ) 1918.200

=

Btu hr ft 2 oF

Btu hr

(abs)

ΔT Hc=0.42 ( D ¿ 701.3 100 ¿ ¿ 537 100 Hr= ¿ ¿ ¿4 ¿ 0.173 ( 0.95 ) ¿ ¿

L.E. =

0.25

1.34∈¿ = 0.42( 186.8 F ¿

0.25

Btu = 1.443 hr ft 2 oF

= 1.40444

240−219 x100=8.75% 240

Silver Chrome:

Q θ

=

mL g ¿ ¿ 205 mL¿ ¿

1.34∈¿ ¿ A= π¿ ¿

min hr 15 min

60

x

x 929.219

Btu lb

= 1795.576

(10ft) = 3.508 ft2

ΔT=234.194oF – 77oF = 157.194oF

(hc + hr) =

Q θ AΔT

ΔT Hc=0.42 ( D ¿

Btu hr =3.2562 2 (3.508 ft )(157.194 F) 1795.576

=

0.25

1.34∈¿ = 0.42( 157.194 F ¿

0.25

Btu hr ft 2 oF

Btu = 1.3822 hr ft 2 oF

Btu hr

701.5 100 ¿ ¿ 537 100 Hr= ¿ ¿ ¿4 ¿ 0.173 ( 0.35 ) ¿ ¿

= 0.6125

237−234.194 L.E. = x100=1.183% 237

85% Magnesia Insulation

Q θ

=

mL g ¿ ¿ 138 mL ¿ ¿

2.48∈¿ ¿ A= π¿ ¿

min hr 15 min

60

x

x 929.219

Btu lb

= 1208.73

(10ft) = 6.493 ft2

ΔT=148.4006 - 77= 71.4006oF

(hc + hr) =

Q θ AΔT

ΔT Hc=0.42 ( D ¿

Btu hr =2.607 2 (6.493 ft )(71.4006 F ) 1208.73

=

0.25

2.48∈¿ 71.4006 F = 0.42( ¿

0.25

Btu hr ft 2 oF

Btu = 0.973 hr ft 2 oF

Btu hr

611.6 100 ¿ ¿ 537 100 Hr= ¿ ¿ ¿4 ¿ 0.173 ( 0.95 ) ¿ ¿

Btu = 1.30651 hr ft 2 oF

237−148.4006 L.E. = x100=37.384% 237 Bare Pipe

Q θ

=

mL g ¿ ¿ 240 mL¿ ¿

1.34∈¿ ¿ A= π¿ ¿

min hr 15 min

60

x

Btu lb

x 929.219

= 2102.138

(10ft) = 3.508 ft2

ΔT=10.417-25=85.417oC ≈185.7506oF

(hc + hr) =

Q θ AΔT

ΔT Hc=0.42 ( D ¿

Btu Btu hr 2 =3.226 hr ft oF (3.508 ft 2 )(185.7506 F ) 2102.138

=

0.25

1.34∈¿ = 0.42( 185.7506 F ¿

0.25

Btu = 1.441 hr ft 2 oF

Btu hr

698.9 100 ¿ ¿ 537 100 Hr= ¿ ¿ ¿4 ¿ 0.173 ( 0.95 ) ¿ ¿

Btu = 1.3753 hr ft 2 oF

4. Discussion: The impact on heat transfer to or from the fluid will depend not only on the relative thermal resistances of the pipe wall, but also on the other thermal resistances in the system. For bare piping, the air surface coefficient normally represents the largest thermal resistance in the system. The wind speeds at the surface, along with the thermal emittance of the surface material, are dominant. As insulation is added to the system, the resistance of the insulation layer begins to dominate and other resistances become less important. As expected, the heat loss or gain depends on both the thickness of the insulation as well as the choice of the pipe material. However, the effect of insulation thickness is considerably more significant than the choice of pipe material. 5. Conclusion: Therefore, when steam is first admitted to a pipe after a period of shutdown, the pipe is full of air. Further amounts of air and other noncondensable gases will enter with the steam, although the proportions of these gases are normally very small compared with the steam. When the steam condenses, these gases will accumulate in pipes and heat exchangers. Precautions should be taken to discharge them. The consequence of not removing air is a lengthy warming up period and a reduction in plant efficiency and process performance. Air in a steam system will also affect the system temperature. Air will exert its own pressure within the system, and will be added to the pressure of the steam to give a total pressure. Therefore, the actual steam pressure and temperature of the steam/air mixture will be lower than that suggested by a pressure gauge. References: [1] http://www.insulation.org/articles/article.cfm?id=IO120901

[2] https://www.scribd.com/doc/239142736/Bare-and-Lagged [3] https://www.scribd.com/doc/156984866/Heat-Loss-for-Bare-and-Lagged-Pipes

PLATE AND FRAME FILTER PRESS Geneston, Elaine May L. BS Chemical Engineering – IV Group 5

Department of Chemical Engineering College of Engineering and Architecture Cebu Institute of Technology – University 196 N. Bacalso Avenue Cebu City, 6000 Philippines

______________________________________________________________________________ Filtration is a widely used unit operation of Chemical Engineering Process which utilizes the separation of solid and liquid. It uses the theory of solid and liquid separation by flowing liquid with solid suspension through porous mediums or screens which retain the solid suspensions and liquid flows out as filtrate. It is generally used in various industry, such as wastewater treatment industry, food and beverages industry, pharmaceutical industry and chemical industry. Plate and frame filter press is one of the most used filtering machines around due to its simplicity in usage and efficiency it produced. The experiment`s objective was to study the operation of the filter press for filtration of calcium carbonate slurry and to determine the filter medium resistance and specific cake resistance from filtration data obtained. Calcium carbonate solution was used as the filtrate. The experiment was run, every 5 L of filtrate produced the time was taken. At three different times, three samples were taken to study the relationship of density and viscosity to volume of filtrate. The viscosity and density of the samples was decreasing as V is approaching the final volume. And using the equation given, the value of filter medium resistance and filter cake resistance was found. The viscosity and density values at the three different samples were taken and it is found that they are proportional in relationship. The value of filter medium resistance and specific cake resistance was determined from filtration data obtained. ______________________________________________________________________________ Engr. Rosario G. Dangin Instructor

December 20, 2014 1

INTRODUCTION

Many process operations produce SLURRY of solid particles suspended in a liquid which must be separated into the solid and liquid phases. The simplest method is to FILTER the slurry through a fine mesh FILTER CLOTH so that the solid FILTER CAKE is deposited on the cloth whilst the clear liquid FILTRATE flows through. Either the CAKE or the FILTRATE or both may be the useful products of this operation. As the cake builds up on the cloth the resistance to flow increases and a greater pressure would be required to force the liquid through the cake itself. In a vacuum filter the cake can be scrapped off the cloth continuously but this is slow and produces a very loose cake. High pressure filtration is faster and produces denser cakes but it is

necessary to contain the system within a PRESS which must then be opened to remove the cake. This method is therefore only suitable for batch operations. 2

Materials and Methods: 2.1 Materials:   

Plate and Frame Filter Press Equipment Water Calcium Carbonate

2.2 Methods:

1 2 3 4 5 6 7

18 L of slurry containing 10% by weight of CaCO3 and 90% by weight water was prepared. The feed tank was filled with the predetermined amount of slurry. The movable head was tightened to lock. The necessary valves (pressure and control valves) were opened. The agitator and the pump were switched on simultaneously. When the slurry from the feed tank was already consumed, the switch and the pump of both agitators were turned off. 2-6 procedure was repeated using 20% by weight of CaCO3.

3. Results: 3.1 Tables: Table 1. Test run data using 10% by weight CaCO3 in 18 L of slurry. Filtrate Volume, (Vf),L 1.2 2.3 3.4 4.5 5.6 6.7 7.8 Ρslurry=1.07 g/mL

Time, sec 12.30 18.10 22 26.40 30 35 39.2

Wt Slurry= 19.2kg

Vf/Time (Rate), 1/sec 0.16260 0.16575 0.18182 0.1894 0.2 0.2 0.2041 Wt. of water in slurry= 17.3 L Volume of water in slurry=17.3 kg

Solid in cake = 1.92 kg Table 2 Test data using 20% by weight CaCO3 in 18L of slurry. Filtrate Volume, (Vf),l 1.2 2.3 3.4 4.5 5.6 6.7 7.8 Ρslurry=1.144 g/mL Wt Slurry= 20.6 kg

Time, sec 10 17.5 24.4 30.44 36.8 43 48.72

Vf/Time (Rate), 1/sec 0.2 0.1714 0.1639 0.1643 0.1630 0.1628 0.1642 Wt. of water in slurry= 16.48 kg

Volume of water in slurry=16.48 L

Solid in cake = 4.12 kg

4. Sample Calculations From Perry’s Chemical Engineering Handbook g Specific GravityCaCO3 = 2.711 mL

M M H 2O

Ρslurry =

ρ H 2O

+

M CaCO 3 ρCaCO 3

@10% CaCO3 : 0.1 ) ( 0.91 )+( 2.711 ¿ 1g ¿

ρslurry =

@20%: 0.2 ) ( 0.81 )+( 2.711

g = 1.067 mL

¿ 1g ¿

ρslurry =

g 1.144 mL Solving for the weight of the slurry:

Solving for the weight of the slurry:

Ws = Vsρs

Ws = Vsρs

1000 mL 1L Ws = 18 L (1.067 ) g ¿¿ mL

1000 mL 1L Ws = 18 L (1.144 ) g ¿¿ mL

Ws = 19.2 kg

Ws = 20.6 kg

Solving for mass of CaCO3:

Solving for mass of CaCO3:

% by weight CaCO3 =

M CaCO 3 M Slurry

% by weight CaCO3 =

M CaCO3

= 0.10(19.2 kg)

M CaCO3

= 0.20(20.6 kg)

M CaCO3

= 1.92 kg

M CaCO3

= 4.12 kg

MH2O = Ws -

M CaCO3

= 19.2 kg – 9.2 kg MH2O = 17.28 kg

MH2O = Ws -

M CaCO3

= 20.6 kg – 4.12 kg MH2O = 16.48 kg

M CaCO 3 M Slurry

=

4. Discussion: A filter cake is formed by the substances that are retained on a filter. The filter cake grows in the course of filtration, becomes thicker as particulate matter is being retained. With increasing layer thickness the flow resistance of the filter cake increases. As time volume increase, the filter cake resistance is increasing until the cake was fully stuffed at the trays. The filter medium resistance changed only when the pressure drop changed. Cake resistance is based on the volume of filtrate stocked in the filter. With some filter cakes, the specific resistance varies with the pressure drop across it. This is because the cake becomes denser under the higher pressure and so provides fewer and smaller passages for flow. The effect is spoken of as the compressibility of the cake. The pressure drop of this experiment was constant at 40 psi. Filter cake builds up on the upstream side of the cloth that is the side away from the plate. In the early stages of the filtration c y c l e , t h e p r e s s u r e d r o p a c r o s s t h e c l o t h i s s m a l l a n d f i l t r a t i o n p r o c e e d s a t m o r e o r l e s s a constant rate. As the cake increases, the process becomes more and more a constant-pressure one and this is the case of this experiment. Calcium carbonate is directly affected by the pressure drop because of the size of its particle, this does not apply to materials with smaller particle like salt and sugar .The result of the experiment may not be correct because of the machine may not be functioning properly. This may cause error of the experiment which means it is not the same as the theory. The weight of the cake (dry and wet) was different for every tray. The difference of weight of wet cake and dry cake is because of the weight of moisture (water) in wet cake. 5. Conclusion: The filter medium resistance and specific cake resistance had been determined by using formulas and the data obtained. We explained the factor that affects the filter medium resistance and specific cake resistance value which are pressure drop, filter medium, the size of particle. The result of the experiment may not be correct as the machine may not be functioning properly. Proper maintenance should be done for better reliability of the experiment

6. References:

[1] https://www.scribd.com/doc/138327400/Lab-Experiment-12-Filter-Press [2] https://www.scribd.com/doc/58703428/Plate-and-Frame-Filtration [3] Coulson JM and Richardson JF, Chemical Engineering Vol2, New York: Pergamon Press, 1990 [4] https://www.scribd.com/doc/51658822/Plate-and-Filter-Frame-Press

SEDIMENTATION GENESTON, MAFEL MELODY L. BS Chemical Engineering – IV Group 5

Department of Chemical Engineering College of Engineering and Architecture Cebu Institute of Technology – University 196 N. Bacalso Avenue Cebu City, 6000 Philippines

Sedimentation is the tendency for particles in suspension to settle out of the fluid in which they are entrained, and come to rest against a barrier. This is due to their motion through the fluid in response to the forces acting on them: these forces can be due to gravity, centrifugal acceleration or electromagnetism. In geology sedimentation is often used as the polar opposite of erosion, i.e., the terminal end of sediment transport. In that sense it includes the termination of transport by saltation or true bed load transport. Settling is the falling of suspended particles through the liquid, whereas sedimentation is the termination of the settling process. Sedimentation may pertain to objects of various sizes, ranging from large rocks in flowing water to suspensions of dust and pollen particles to cellular suspensions to solutions of single molecules such as proteins and peptides. Even small molecules supply a sufficiently strong force to produce significant sedimentation. It is accomplished by decreasing the velocity of the water being treated to a point below

which the particles will no longer remain in suspension. When the velocity no longer supports the transport of the particles, gravity will remove them from the flow.

Engr. Rosario G. Dangin Instructor March 23, 2015 1. INTRODUCTION Sedimentation is a physical process whereby solid particles, of greater density than their suspending medium, will tend to separate into regions of higher concentration under the influence of gravity. As a solid/liquids separation technique it therefore possesses the great advantage of utilizing a natural and therefore countless, driving force. The sedimentation process is traditionally divided into settling within four regimes which are schematically depicted. 2. Materials and Methods: 2.1 Materials:       

Sedimentation Apparatus Mesh Stop Watch Beaker Spatula Stirring rod Powder of Calcium Carbonate

2.2 Methods Part A 1. The powdered calcium carbonate was sieved to achieve a uniform size of a particle using a mesh. 2. 1L of 2%, 4%, 6%, 8%, and 10% by weight calcium carbonate suspension in water was prepared. 3. The slurry every prepared sample was placed in sedimentation tubes at the same height. The light was turned on for better readings and at convenient time intervals the readings were noted. 4. The rise of the sludge interphase at the base of the cylinder was recorded. 5. A 24 hours period was allowed to elapse for final compaction readings.

Part B The above procedure was repeated except that the slurry is of the same concentration (4%) but with different in8itial heights. 3. Results: 3.1 Tables:

Part A Height Interface w/100 ml at differing concentrations 2% 4% 6% 8% 10% 480 490 480 490 480 92 398 449 471 472 78 308 421 453 461 64 243 395 436 450 53 200 370 418 440 49 172 347 402 430 47 153 325 386 421 44 120 285 355 420 43 105 266 338 402 42 96 252 326 394 41 88 237 313 386

Time Interval 0 5 10 15 20 25 30 35 40 45 50 AFTER 24 hours

36

61

90

111

140

Height Interface of Varying Concentration at time t 600 500 400 Height, mm

2%

300

4%

200

6%

100

8% 10%

0 0

5

10

15

20

25

Time, min

30

35

40

45

50

Table A1. Data for Limiting Concentration Vs Limiting Velocity at 2% Time Interval (min) 0 5 10 15 20 25 30 35 40 45 50

Zi (mm)

C (g/L)

Vs (mm/min)

480 92 78 64 53 49 47 44 43 42 41

20.26 105.7043478 124.6769231 151.95 183.4867925 198.4653061 206.9106383 221.0181818 226.1581395 231.5428571 237.1902439

77.6 40.2 27.73 21.35 17.24 14.43 12.46 10.93 9.73 8.78

Graph A1. Limiting Concentrations vs. Settling Velocity at 2%

Settling Velocity (V), mm/min

90 80 70 60 50 40 30 20 10 0 100

120

140

160

180

200

220

240

Concentration (C), g/L

Table B1. Data for Limiting Concentration Vs. Limiting Velocity at 4% Time Interval (min) 0 5 10 15 20 25 30

Zi (mm)

C (g/L)

Vs (mm/min)

490 398 308 243 200 172 153

41.03 50.51 65.28 82.74 100.52 116.89 131.4

18.4 18.2 16.47 14.5 12.72 11.23

35 40 45 50

120 105 96 88

167.54 191.47 209.42 228.46

10.57 9.63 8.76 8.04

Graph B1. Limiting Concentrations vs. Settling Velocity at 4% 20 15

Settling Velocity (Vs), mm/min

10 5 0 50

70

90

110 130 150 170 190 210 230 250

Concentration (C), g/L

Table C1. Data for Limiting Concentration Vs. Limiting Velocity at 6% Time Interval (min) Zi (mm) C (g/L) Vs (mm/min) 0 480 62.36 5 449 66.67 6.2 10 421 71.1 5.9 15 395 75.78 5.67 20 370 80.9 5.5 25 347 86.26 5.32 30 325 92.1 5.17 35 285 105.03 5.57 40 266 112.53 5.35 45 252 118.78 5.07 50 237 126.3 4.86 Graph C1. Limiting Concentrations vs. Settling Velocity at 6%

7 6 5 4

Settling Velocity (Vs), mm/min 3 2 1 0 60

70

80

90

100

110

120

130

Concentration (C), g/L

Table D1. Data for Limiting Concentration Vs. Limiting Velocity at 8% Time Interval (min) 0 5 10 15 20 25 30 35 40 45 50

Zi (mm)

C (g/L)

Vs (mm/min)

490 471 453 436 418 402 386 355 338 326 313

84.25 87.65 91.13 94.68 98.76 102.69 106.95 116.29 122.14 126.63 131.89

3.8 3.7 3.6 3.6 3.52 3.47 3.86 3.8 3.64 3.54

Graph D1. Limiting Concentrations vs. Settling Velocity at 8%

3.9 3.8 3.7 3.6

Settling Velocity (Vs), mm/min

3.5 3.4 3.3 3.2 80

90

100

110

120

130

140

Concentration (C), g/L

Table E1. Data for Limiting Concentration Vs. Limiting Velocity at 10%

Time Interval (min) Zi (mm) 0 480 5 472 10 461 15 450 20 440 25 430 30 421 35 480 40 402 45 394 50 386 Graph E1. Limiting Concentrations vs.

C (g/L) Vs (mm/min) 106.73 108.54 1.6 111.13 1.9 113.85 2 116.43 2 119.14 2 121.69 1.97 106.73 0 127.44 1.95 130.03 1.91 132.72 1.88 Settling Velocity at 10%

-

Settling Velocity (Vs)/ mm/min 105

110

115

120

125

130

135

Concentration (C), g/L

Table 2A. Data for initial Mass Settling Rate for each Concentration C (g/L) 2% 4% 6% 8% 10%

J 1572.18 754.95 386.63 320.15 170.77

Graph 2A. Concentration vs. Initial Mass Settling Rate 2000 1500

Initial Mass Settling Velocity Rate 1000 500 0 0%

2%

4%

6%

8%

10% 12%

Concentration

Part B Time Interval 0 5 10

Height of Interface w/ (4% Concentration but different initial heights) 1 w/ Hi = 2 w/ Hi = 3 w/ Hi = 4 w/ Hi = 5 w/ Hi = 500ml 600ml 700ml 800ml 900ml 236 290 339 396 439 96 147 192 220 290 28 38 54 55 138

15 20 25 30 35 40 45 50 AFTER 24 hours

23.5 17 21 20 19 18 18 17

27 25 23 22 21 20.5 20 19.5

32 30 28 27 26 25 24 23.5

37 36 31 30 29 28 27 26

51 42 40 38 37 35.5 35 33.5

17

18

21

23

28

Height of Interface of Varying Volumes at time t 500 450 400 350 300 Height, mm

1 w/ Hi = 500ml

250

2 w/ Hi = 600ml

200

3 w/ Hi = 700ml

150

4 w/ Hi = 800ml

100

5 w/ Hi = 900ml

50 0 0

5 10 15 20 25 30 35 40 45 50 Time, t

Table 3A. Data for Limiting Concentration vs. Settling Velocity for 4% 500 ml solution

Time Interval (min) 0 5 10 15 20 25 30 35 40 45 50

Zi (mm)

C (g/L)

Vs (mm/min)

236 96 28 23.5 17 21 20 19 18 18 17

41.03 100.87 345.82 412.05 569.59 461.1 484.154 509.64 537.95 537.95 569.69

28 20.8 14.17 10.95 8.6 7.2 6.2 5.45 4.84 4.38

Graph 3A. Limiting Concentrations vs. Settling Velocity for 4% 500 ml solution 30 25 20

Initial Settling Velocity (Vs), mm/min 15 10 5 0 100

200

300

400

500

600

Concentration (C), g/L

Table 3B. Data for Limiting Concentration vs. Settling Velocity for 4% 600 ml solution Time Interval C (g/L) Vs (mm/min) Zi (mm) (min) 0 290 41.03 5 147 80.94 28.6 10 38 313.12 25.2 15 27 440.69 17.53 20 25 475.95 13.25 25 23 517.33 10.68 30 22 540.85 8.93 35 21 566.6 7.69 40 20.5 580.42 6.74 45 20 594.94 6 50 19.5 610.19 5.41 Graph 3B. Limiting Concentrations vs. Settling Velocity for 4% 600 ml solution

35 30 25 20 Initial Settling Velocity (Vs), mm/min

15 10 5 0 80

180

280

380

480

580

680

Concentration (C), G/L

Table 3C. Data for Limiting Concentration vs. Settling Velocity for 4% 700 ml solution Time Interval (min) 0 5 10 15 20 25 30 35 40 45 50

Zi (mm) 339 192 54 32 30 28 27 26 25 24 23.5

C (g/L) 41.03 72.44 257.58 434.66 463.64 496.76 515.15 534.97 556.37 579.55 591.88

Vs (mm/min) 29.4 28.5 20.47 15.45 12.44 10.4 8.94 7.85 7 6.31

Graph 3C. Limiting Concentrations vs. Settling Velocity for 4% 700 ml solution

35 30 25 20

Initial Settling Velocity (Vs), mm/min

15 10 5 0 70

170

270

370

470

570

670

Concentration (C), g/L

Table 3D. Data for Limiting Concentration vs. Settling Velocity for 4% 800 ml solution Time Interval (min) 0 5 10 15 20 25 30 35 40 45 50

Zi (mm) 396 220 55 37 36 31 30 29 28 27 26

C (g/L) 41.03 73.85 295.42 439.13 451.33 524.13 541.5 560.27 580.28 601.77 624.92

Vs (mm/min) 35.2 34.1 23.93 18 14.6 12.2 10.49 9.2 8.2 7.4

Graph 3D. Limiting Concentrations vs. Settling Velocity for 4% 800 ml solution

Initial Settling Vvelocity (Vs), mm/min

40 35 30 25 20 15 10 5 0 70

170 270 370 470 570 670

Concentration (C), g/L

Table 3E. Data for Limiting Concentration vs. Settling Velocity for 4% 900 ml solution Time Interval (min) 0 5 10 15 20 25 30 35 40 45 50

Zi (mm)

C (g/L)

Vs (mm/min)

439 290 138 51 42 40 38 37 35.5 35 33.5

41.03 62.11 130.52 353.18 428.86 450.3 474 486.82 507.39 514.63 537.67

29.8 30.1 25.87 19.85 15.96 13.37 11.49 10.09 8.97 8.11

Graph 3E. Limiting Concentrations vs. Settling Velocity for 4% 900 ml solution 35 30 25 20

Initial Settling Velocity (vs), mm/min 15 10 5 0 60

160

260

360

460

Concentration (C), g/L

560

660

Table 4A. Data for initial Mass Settling Rate for each Concentration Weight of CaCO3 in slurry

J 574.5 6 704.1 3 844.6 6 1155. 62 1100. 51

20.52 24.62 28.73 32.83 36.93

Graph 4A. Concentration vs. Initial Mass Settling Rate 1400 1200 1000 800

Initial Mass Settling Rate (J)

600 400 200 0 20 22 24 26 28 30 32 34 36 38

Weight of CaCO3 in slurry

Equations C=

CoZo Z

V ¿

(Zo−Z) t

J= Co (Vo) Where: C

=

limiting concentration

Co

=

initial concentration

Vo

=

initial settling velocity

V

=

settling velocity

Zo

=

initial height

Z

=

height at a certain time

t

=

time needed to reach current height

=

initial mass settling rate

J

4. Discussion: Sedimentation is a treatment process where water is slowly flowing through a reservoir. Because of the low flow velocities particles are able to deposit on the bottom of the reservoir. Distinction should be made between discrete and flocculent settling. The cumulative frequency distribution of settling velocities cannot be calculated in a theoretical way because the size, shape and mass density of the particles are unknown and even variable with flocculent settling. The frequency distribution is found experimentally by plotting the percentage remaining SS (p) against the settling velocity versus that can be calculated from the depth of the sampling port under the water surface (H) divided by the elapsed time t.

5. Conclusion: Sedimentation practices are designed to be effective at retaining suspended solids that typically adsorb to solids. In every after the experiment, the retained solids must need to removed. Such apparatus is not that effective at retaining the dissolved pollutants. The panel must be translucent enough. At short period of time, particles are well distributed in a water phase at its flocculated suspension initial state based on the nature of the CaCO3. The formed flocs will cause an increase in sedimentation rate due to increase in size of sedimenting particles. References: [1] http://www.slideshare.net/jeufier/sedimentation-finalrepz1 [2] http://www.slideshare.net/GerardBHawkins/sedimentation [3] http://www.slideshare.net/jshrikant/l-10-sedimentation [4] http://www.slideshare.net/Redrika95/sedimentation-44097040

AGITATION GENESTON, ELAINE MAY L. BS Chemical Engineering – IV Group 5

Department of Chemical Engineering College of Engineering and Architecture Cebu Institute of Technology – University 196 N. Bacalso Avenue Cebu City, 6000 Philippines

______________________________________________________________________________ ______________________ This experiment only utilizes the agitator equipment. Before the start of the experiment a couple of measurement is made – the diameter of the vessel and impeller, which are vital in the computations later on. Due to its similarities with the properties of water, assumptions are made for fluids densities and viscosities. The fluid is liquid water. The data are obtained with and without the baffled vessel. Efficient mixing is obtained when correct answers are available to such questions as (1) size of motor required to drive the mixing equipment; (2) speed of the unit for quality of mixing desired; & (3) type of impeller and used of baffles. The paddle and the impeller are fixed on the shaft at near the bottom of the tank. The faster the rate of revolution of the shaft, the more the vortex becomes distinctive. With an increase of the rotating velocity, much more air is sucked into water. Doubling the impeller diameter will quadruple Reynold’s Number. This follows, as the impeller will sweep an area four times larger when the diameter is

doubled. Temperatures and pressures are accounted for in Reynold’s Number as they affect both density and viscosity. These factors are useful for sizing and selections of tanks, impellers, and the associated driving equipment. It is clear that major variables in the problem will be the size of the impeller and its speed. The presence of the baffles will affect the energy consumption materially.

Engr. Rosario G. Dangin Instructor

March 23, 2015

PROBLEM Components for a liquid detergent (µ=10 centipoises) are blended in the pilot plant in a 10 gal, baffled, flat-bottomed tank 10 in. in diameter. A

double-turbine agitator with blades 6 in. in diameter is used. A

1 2

hp motor

turns the agitator at 500 rpm for 30 min to attain complete dispersion. In the plant a geometrically similar unit is planned to blend 200 gal batches of this solution. Determine consistent values of agitator and tank diameter, revolutions per minute, power requirement, and batch time for the plant unit. Base the design on (a) constant peripheral speed, (c) constant rpm. Given: µ=10 cp V 1 = 10 gal

Nℜ

, (b) constant agitator

D T 1 = 10 in. (tanks diameter) D A1

= 6 in. (agitator diameter) T = 30 min (batch time)

Calculate: DT 2

,

DA2

N, P, t at

V2

= 200 gal

a Constant time b constant agitator peripheral speed c Constant rpm

Solution: a Constant time

Nℜ

R=

DT 2

V 2 13 V1

= 27.144 in.

=

=

DA2 N P µ DT 2 DT 1

;

200 13 10

=

DT 2 10

Nℜ

=

DA2

= R D A1

D A 2=

200 3 (6 in) 10

DA2

= 16.287 in.

DA2 N1 P µ

1

N2

=

N 2=¿

N2

N1

D A 12 D A 22

62 (500)( 16.287 2 )

= 67.86 or 68 rpm

=

D A2 N2 P µ

;

N2 N1

2

=

DA 1 2 DA 2