supplement 1 pet212e spring 2011

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PE 212E Rock Properties Mustafa Onur Given Date: February 18, 2011 Subject: Porosity Computations Supplement No: 1 Solutions Problem 1:Consider both rhombohedral and hexagonal packing of spheres of uniform sizes, each with radius r. Rhombohedral packing is shown in Fig. 1a, while a hexagonal packing of spheres of uniform size is shown in Fig. 1b. Note that hexagonal packing is obtained by moving the top layer of spheres one radius to the right as shown in the figure below. As in the cubic packing that we discussed in the class, the bulk volumes for rhombohedral and hexagonal packings can be formed by connecting the centers of eight adjacent spheres (the four shown in front view and four behind them) so that the bulk volume contains one net sphere as the enclosed solid or grain volume. Find the porosity for the rhombohedral and hexagonal packings. Compare your results with that obtained for cubic packing, and discuss your results whether the porosities computed for cubic, rhombohedral and hexagonal packings make sense. Hexagonal packing Fig. 1a) Rhombohedral packing Fig. 1b) Hexagonal packing For the purpose of generality, I will start deriving the porosity of cubic packing of spherical grains with uniform size. Then, I will do the hexagonal and rhombohedral packing of spherical grains with uniform size. (a) Cubic packing: Recall from your lecture notes that cubic packing can be illustrated by the following figure:

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  • PE 212E Rock Properties Mustafa Onur

    Given Date: February 18, 2011 Subject: Porosity Computations

    Supplement No: 1

    Solutions Problem 1:Consider both rhombohedral and hexagonal packing of spheres of uniform sizes, each with radius r. Rhombohedral packing is shown in Fig. 1a, while a hexagonal packing of spheres of uniform size is shown in Fig. 1b. Note that hexagonal packing is obtained by moving the top layer of spheres one radius to the right as shown in the figure below. As in the cubic packing that we discussed in the class, the bulk volumes for rhombohedral and hexagonal packings can be formed by connecting the centers of eight adjacent spheres (the four shown in front view and four behind them) so that the bulk volume contains one net sphere as the enclosed solid or grain volume. Find the porosity for the rhombohedral and hexagonal packings. Compare your results with that obtained for cubic packing, and discuss your results whether the porosities computed for cubic, rhombohedral and hexagonal packings make sense.

    Hexagonal packing

    Fig. 1a) Rhombohedral packing Fig. 1b) Hexagonal packing For the purpose of generality, I will start deriving the porosity of cubic packing of spherical grains with uniform size. Then, I will do the hexagonal and rhombohedral packing of spherical grains with uniform size. (a) Cubic packing: Recall from your lecture notes that cubic packing can be illustrated by the following figure:

  • We can construct a cube by adjoining sphere centers so that the cube contains one sphere and pore space. The cross-sectional view is shown below:

    90o

    The bulk volume, Vb, is then the volume of the cube adjoining the centers of the spheres: 32 2 2 8bV r r r r= = (1) The grain volume is equal to the volume of one sphere because the cube adjoining the centers of sphere contains one sphere. Then, the grain volume, Vg is:

    343g

    V r= (2) Recall the definition of porosity, which is given by

    r

    = 90o

  • 3 3

    3

    4 48 83 3 0.4764

    8 8p b g

    b b

    r rV V VV V r

    = = = = = (or %47.64) (3) b) Hexagonal packing: As in the cubic packing we adjoining the centers of the spheres but the resulting volume in this case is not a cube, but is a parallelepiped with the angle 60o (see Figure 1b). This parallelepiped contains one sphere and pore space. The cross-sectional view is shown below:

    Note that the angle in this case is 60o. The bulk volume of the parallelepiped is in general: 3 32 2 sin 2 8 sin 8 sin 60bV r r r r r = = = (4) (Note that we actually multiply the area of a parallelogram by 2r. Note that the area of a parallelogram is equal to the multiplication of length of its chosen side by the height perpendicular to that side) And the grain volume is the same as the cubic packing because the parallelepiped formed contains one sphere, i.e.,

    343g

    V r= (5) Using the definition of porosity, and Eqs. 4 and 5, it follows that the porosity of hexagonal packing is:

    3 3

    3

    4 48 sin 8sin 603 3 0.3954

    8 sin 8 sin 60p b g

    b b

    r rV V VV V r

    = = = = = (or %39.54) (6)

    = 60o

  • c) Rhombohedral packing: As in the cubic packing we adjoining the centers of the spheres but the resulting volume in this case is not a cube, but is a parallelepiped with the angle 45o (see Figure 1b). This parallelepiped contains one sphere and pore space. The cross-sectional view is shown below:

    Note that the angle in this case is 45o. The bulk volume of the parallelepiped is in general: 3 32 2 sin 2 8 sin 45 8 sin 45bV r r r r r= = = (7) And the grain volume is the same as the cubic packing because the parallelepiped formed contains one sphere, i.e.,

    343g

    V r= (8) Using the definition of porosity, and Eqs. 7 and 8, it follows that the porosity of hexagonal packing is:

    3 3

    3

    4 48 sin 8sin 453 3 0.2595

    8 sin 8 sin 45p b g

    b b

    r rV V VV V r

    = = = = = (or %25.95) (9)

    According to the results, the porosities found from the cubic, rhombohedral and hexagonal packings make

    sense. Rhombohedral packing is the tightest packing because the angle formed between the grain is 45o and

    hence the bulk volume is the smallest and it gives the lowest porosity of 25.95. Then it comes hexagonal

    packing with the angle of 60o and it gives the porosity value of 39.54 which is greater than rhombohedral

    packing as it expected because the orientation of grain is the loosest (90o). The cubic packing has the

    highest porosity which is 47.64.

    = 45o