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  • 8/13/2019 Compression and Compaction-2011Lect5

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    Compression and compaction

    Dr Ali Nokhodchi, PharmD, PhD

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    References

    Pharmaceutics: the science of dosage

    form design. Edited by M.E. Aulton, 2nd

    Edition, Chap. 27, pp: 423-439

    Remington, the Science and Practice of

    Pharmacy, 21st Edition, Chap. 45, Oral

    solid dosage Forms, pp:894-896

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    3

    Basic Mechanical Unit of

    Compression

    Die

    lower punch

    upper punch

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    SINGLE TABLETING MACHINE

    Time

    Displaceme

    nt upper

    lower

    Contact time

    Dwell time

    force

    Time

    Ejection forceforce

    Time

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    ROTARY TABLETING MACHNIE

    Die

    Upper punch

    Lower punch

    Upper compression

    roll

    lower compression

    roll

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    Rotary tableting machine profile

    Displacem

    ent

    Time

    Upper punch

    lower punch

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    TABLETING PROCESS

    HARDNESS

    (bonding)

    DISSOLUTION

    (porosity)

    COMPACTIONincrease in mechanicalstrength (consolidation

    of particles)

    COMPRESSIONreduction in bulk volume

    (displacement of gaseous

    phase)

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    Stage 1: Particle rearrangement

    Size of the particles: Small particles

    enter the voids between larger particles

    Initial porosity of powder bed: The

    greater the porosity, the greater the

    particle rearrangement

    Shape of the particles: Spherical

    particles tend to assume closer packing

    arrangements than irregular particles

    Partic le rearrangemen t : A decrease in the relative

    volum e caused b y interparticulate sl ippage of powder

    leading to close packing

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    Stage 2: Deformation

    Elastic deformation: Particlesreturn to their original shape

    (deformation disappears completely)

    upon release of the stress

    Plastic deformation: Particles do

    not recover their original shape after

    release of the stress

    The force required to initiate a plastic

    deformation is known as the yield

    Deformat ion at point of con tact : Increasing the appl iedpressure causes deformation of th e part ic les = change of

    shape.

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    Stage 3: Fragmentation

    Fracture occurs when the stresses

    within the particles become great

    enough to propagate cracks.

    The fragments infiltrate theremaining voids to increase

    densification.

    Fragmentat ion : Ini t ia l part ic les are divid ed into a

    smaller discrete part ic les.

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    Stage 4: Bonding

    Bonding : Plast ic deformat ion at the

    part ic le interfaces and al ignment of

    part ic le su rfaces so that

    interpar ticu late bond ing can occu r.

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    Stage 5: Deformation of the solid

    body

    Deformat ion of the so l id body : If the

    app l ied pressu re is inc reased fu rther,

    the bonded so l id is conso l idatedtowards zero porosi ty

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    Stage 6: DecompressionDecompress ion : The interpart icu late bonds

    formed must be suf fic ient ly robust towiths tand release of the appl ied pressu re

    Removal of the upper punch relieves the axial

    pressure but a radial pressure from the die

    remains. This allows axial elastic recovery to take

    place.

    If elastic. If plastic.

    Axial elastic

    recovery

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    Stage 7: Ejection

    Eject ion: The pro cess b y which the tablet is

    removed from the die cavi ty .

    The force necessary to eject the tablet from the die

    (EF) must be greater than the quotient of the

    residual die wall force (RDWF) exerted from thetablet and the friction between the tablet and the

    die wall (w).RDWF

    EF

    RDWF = Residual Die Wall Force

    EF = RDWF *w

    w = Coeff. friction at die wall

    RDWFRDWF

    w

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    Example

    True density = 1.25 g/cc

    Weight = 500 mgDiameter = 10 mm

    Thickness after ejection = 7.5 mm

    Calculate % ER? Assume at max. pressure porosity is zero

    Porosity = 0 it means Volume of powder = volume of tabletVt = weight/density = 0.5/1.25 = 0.4 cm

    3

    Vt= 3.14 R2H 0.4 = 3.14 (0.5)2 H

    H = (0.4)/(3.14 0.25) H = 0.5 cm or 5 mm

    ER = [(7.5-5)/5] 100 ER = 50%

    100*0

    PH

    PHH

    ERHp: Tablet thickness after compression.

    H0: Tablet thickness upon removalof the Compression pressure (it

    could be 24 h after the ejection).

    Elastic Recovery

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    Effect of compression pressure

    on a bed of powder

    Examples

    Paracetamol

    Avicel, Starch

    Nacl

    Dicalcium Phosphate

    Lactose

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    COMPRESSION

    MECHANISMS REVERSIBLETIME

    DEPENDENT

    ELASTIC(rubber)

    YES NO

    PLASTIC

    (avicel)

    NO YES

    BRITTLE

    (emcompress)

    NO NO

    VISCO-ELASTIC

    (starch)

    PARTLY YES

    BRITTLE-PLASTIC

    (lactose)

    PARTLY YES

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    COMPACTIBILITY PROFILE

    0

    2

    4

    6

    8

    0 5 10 15 20

    Compaction Force (kN)

    Hardness

    (kP)

    starch

    avicellactose

    emcompress

    COMPRESSIBILITY PROFILE

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    25

    0

    20

    40

    60

    80

    100

    0 5 10 15 20

    Compaction Force (kN)

    Porosity(%)

    COMPRESSIBILITY PROFILE

    starch

    avicel lactose

    emcompress

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    COMPACTIBILITY PROFILE

    0

    2

    4

    6

    8

    0 1 2 3 4

    Compaction Force (kN)

    Hardness(kP) Avicel

    Highspeed

    Avicel

    Low

    speed

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    0

    20

    40

    60

    80

    100

    0 1 2 3

    Compaction Force (kN)

    Porosity(%)

    Avicel

    High speed

    AvicelLow speed

    COMPRESSIBILITY PROFILE

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    FACTORS IN TABLETING

    Press Force Press Speed

    Hardness Porosity

    Surface Area

    Dissolution

    Disintegration

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    Assessment of plastic deformation.

    Scanning electron microscopy:Size of particles after compression remains the same as before

    compression although a change in the particle shape could be observed.

    Percentage elastic recovery of the compressed tablets:Plastically deforming material exhibit no or small increase in tablet

    thickness after storage.

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    Assessment of plastic deformation

    (Cont inued)

    1. Force volume relationships: Heckel

    analysis

    ln [1/1-D] = kP + A

    k and A are constants obtained from the slope and intercept of the plot

    Ln(1-/1-D) versus P respectively, D is the relative density of a powderbed at the pressure P

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    Heckel plot

    -die filling

    -Particle

    rearrangement

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    ln [1/1-D] = kP + A

    k = (1/3)Y k gives a measure of the plasticity of a compressed

    material

    Greater slopes indicate a greater degree of plasticity of materials

    The slope was also related to the yield strength (Y) of the material

    The reciprocal of k to be the mean yield pressure (PY)

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    33True density =5 g/cm3 Weight = 300 mg Diameter =8 mm

    1 0.7 0.352 0.853 0.171 0.1871

    2 0.6 0.301 0.995 0.199 0.2219

    3 0.5 0.251 1.194 0.239 0.2729

    4 0.4 0.201 1.493 0.299 0.3546

    5 0.35 0.176 1.706 0.341 0.4174

    7.5 0.305 0.153 1.958 0.392 0.4969

    10 0.3 0.151 1.990 0.398 0.5076

    12.5 0.29 0.146 2.059 0.412 0.5307

    15 0.28 0.141 2.133 0.427 0.5560

    20 0.27 0.136 2.212 0.442 0.5840

    30 0.26 0.131 2.297 0.459 0.6150

    40 0.25 0.126 2.389 0.478 0.649550 0.24 0.121 2.488 0.498 0.6884

    60 0.23 0.116 2.596 0.519 0.7324

    80 0.2 0.100 2.986 0.597 0.9092

    100 0.15 0.075 3.981 0.796 1.5905

    Pressure Thickness App. Vol App. Density Relative ln(1/1-D)

    (Mpa) (cm) Cm3 (g/Cm3)

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    Slope = 0.0042 MPa-1

    Yield pressure = 238 MPa

    Q ? True density =4 g/cm3

    Weight = 300 mg

    Diameter = 8 mm

    Ln(1/1-Dr)

    y = 0.0042x + 0.4842

    R

    2

    =0.982

    0.0000

    0.2000

    0.4000

    0.6000

    0.8000

    1.0000

    1.2000

    1.4000

    1.6000

    1.8000

    0 20 40 60 80 100 120Compression pressure (MPa)

    Application of Heckel plot

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    Application of Heckel plot

    100*

    2

    )12

    (

    yP

    yP

    yP

    SRS

    Strain rate sensitivity

    (increase in compression speed increases the mean yield pressure).

    High speedLow speed

    Exp. Yield pressures for PEG

    at low and high speeds are 25

    and 50 Mpa respectively,

    Calculate SRS and determine

    the deformation mechanism?

    SRS= [(50-25)/50] x100

    SRS = (25/50)x100

    SRS = 50%

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    Highly fragmenting

    Highly plastic

    Strain rate sensitivity of some excipients and

    drugs

    materials SRS(%)

    Maize starch

    Mannitol

    NaCl

    Lactose b-anhydrous

    Spray dried lactose

    A-lactose monohydrate

    Paracetamol

    Paracetamol DC

    DCP

    97.12

    86.5

    66.3

    25.5

    23.8

    19.4

    11.9

    11.8

    0

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    Assessment of plastic deformation

    (Cont inued)

    Reduction in tablet tensile strength or tablethardness with increasing compression

    speed.

    Compression speed (mm/s)

    H

    ardness(N)

    AVICEL

    DCP

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    High Lubricant sensitivity (the same

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    High Lubricant sensitivity (the same

    compression force and speed)

    Lubricant concentration (%w/w)

    Hardness

    (N)

    PEG

    DCP

    compressioncompression

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    We require to make ibuprofen tablets, hardness about 100 N.

    The results showed that at these conditions the hardness of

    tablets is 70 N.

    Plastic deforming

    drug

    Use fragmenting

    Excipinets, DCP

    Increasecompression

    force

    Decreasecompression

    speed

    Conditions:

    500 mm/s

    10 kN

    Ibuprofen 400 mg

    Starch 20 mg

    Avicel 129 mg

    Lubricant 1 mg

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    We require to make carbamazepin tablets, hardness about

    100 N. The results showed that at these conditions the

    hardness of tablets is 70 N.

    Highly fragmenting

    drug

    Use plasticdeforming

    Excipinets, Avicel

    Increasecompression

    force

    Decreasecompression

    speed

    carbamazepin 400 mgStarch 20 mg

    DCP 129 mg

    Lubricant 1 mg

    Conditions:

    500 mm/s

    10 kN

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    Ibuprofen 400 mg

    Starch 20 mg

    DCP 129 mgLubricant 10 mg

    The hardness of ibuprofen tablets for above formulation is less than our target.

    Give your approaches to increase the hardness?

    1. Reduction in compression speed

    2. Increase DCP concentration

    3. Increase compression force4. Decrease lubricant concentration

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    Assessment of elastic deformation

    (Continued)

    Energy analysis: Force displacement

    profile: High amount of gross energy isspent in bonds disruption: Elastic energy.

    High percentage friability

    Low tensile strength or tablet hardness

    value

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    Assessment of Fragmentation

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    Assessment of Fragmentation(Continued)

    Force volume relationships: Heckelanalysis: High mean yield pressure value.

    Not Strain rate sensitive (increase in

    compression speed has no effect on themean yield pressure value).

    Stress relaxation: Little or no decrease in the

    maximum applied compression force withtime.

    No reduction in tablet tensile strength with