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By
Ben Smith, The Newnan Times-Herald

Dec. 20, 2013Updated Dec. 22, 2013 at 10:16 AM

Lee, Ga.

I am trying to find the effective crack diameter (d) and K-value of the fractures. for the specific case of a rectangular prism with the base area of 20 cm and height of 8 cm.
It looks like the crack depth should be 8 cm, but I am uncertain if there is a crack width.

A general approach to calculating fracture permeability is to assume a constant crack length and a uniform crack spacing in the face plane of the prism. Then, you can calculate the fracture permeability as a function of the fracture length, the depth of the fracture, and the crack angle from the face plane. For a prism, the corresponding expression in terms of the crack length, the crack depth, and the crack angle is
$$phi=tan^{ -1}left(frac{d}{L}right)$$
$$frac{phi}{2}=sinleft(frac{phi}{2}right)$$
$$tanleft(frac{phi}{2}right)=frac{d}{L}$$
$$phi=2sin^{ -1}left(frac{d}{L}right)$$
$$frac{r_i}{r_o}=frac{L}{2}$$
$$frac{r_f}{r_o}=frac{L}{pi}$$
$$L=r_ftimespi$$
$$L_{cr}=2pi r_f$$
$$text{K}=frac{dot Q}{text{A} times L_{cr}^epsilon}$$
$$text{K}=frac{dot Q}{text{A}times 2pi r_f^epsilon}$$
In the last step, I took the geometrical average of the ratios of crack length and length to height. Since the crack length is the same as the cross-sectional area of the prism, and there are four parallel sides, the area to which the area ratio applies is
$$text{A}=frac{4}{sin(phi/2)}text{A}_f$$
$$text{A}=4text{A}_f=4times L_f$$
$$text{A}=4r_f^2$$
$$

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