# Box 8 Deriving the Tangent Law of Refraction

Eileen Poeter, William Woessner, and Paul Hsieh

Figure Box 8-1 shows flow lines and equipotential lines at an interface between materials of different hydraulic conductivity. One side of the interface, Region 1 has hydraulic conductivity K1; the other side of the interface, Region 2 has hydraulic conductivity K2. Under steady state conditions the discharge in a flow tube formed by two parallel flowlines must be the same on both sides of an interface (Q1 = Q2), and given that Darcy’s Law must be followed, the gradient and flow area (A) must differ on each side of the interface to accommodate the differing hydraulic conductivities. This causes the flow lines to refract at the interface as shown in Figure Box 8-1 because the gradient hdiff1/l1 in Region 1 and hdiff2/l2 in Region 2) and the flow area (indicated by the width of the flow tubes, d1 and d2, times a unit width into the image) must adjust to carry the same volumetric flow in materials of different hydraulic conductivity.

Darcy’s Law can be written for either Region 1, or 2, with the subscript n = 1 to represent region 1 or n = 2 to represent region 2, using Equation Box 8-1.

 Qn = – Kn in An (Box 8-1)

where:

 Qn = volumetric flow through a unit width into the image for region n (L3/T) Kn = hydraulic conductivity of region n (L/T) in = hydraulic gradient in region n (dimensionless) = An = area perpendicular to flow in tube of region n (L2), An = dn w where w is a unit width into the image

Setting Q1 = Q2 leads to Equation Box 8-2.

 (Box 8-2)

where:

 hdiff1 = head difference between equipotential lines region 1 (L) hdiff2 = head difference between equipotential lines region 2 (L) d1, d2, l1, l2 = distances defined in Figure Box 8-1 (L) w = a unit width into the image (L)

note: in this case hdiff1 equals hdiff2, hdiff = hdiff1 = hdiff2 = (129m – 130m = –1m).

Canceling the equal head differences, hdiff, and canceling the equal distances perpendicular to the flow direction, w, Equation Box 8-2 simplifies to Equation Box 8-3.

 (Box 8-3)

Given the distance b and angles of θ’s as defined in Figure Box 8-1, trigonometric relationships result in Equations Box 8-4 and Box 8-5.

 d1 = b cos θ1 (Box 8-4) d2 = b cos θ2 (Box 8-5)

where:

 θn = angles shown in Figure Box 8-1 (degrees)

Substituting Equations Box 8-4 and Box 8-5 for d1 and d2 in Equation Box 8-3 yields Equation Box 8-6.

 (Box 8-6)

Using rules of trigonometry, the Equation Box 8-7 and Equation Box 8-8 are true.

 (Box 8-7) (Box 8-8)

Substituting Equation Box 8-7 and Equation Box 8-8 into Equation Box 8-6 and canceling b, results in Equation Box 8-9.

 (Box 8-9)

Rearranging by multiplying both sides first by sin θ1 and then by sin θ2, and dividing both sides first by cos θ1 and then by cos θ2 results in Equation Box 8-10.

 (Box 8-10)

Using the Trigonometric identity of Equation Box 8-11.

 (Box 8-11)

Equation Box 8-10 can be expressed as Equation Box 8-12.

 (Box 8-12)

Then dividing both sides of Equation Box 8-12, first by K2 and then by tan θ2 the tangent law of Equation Box 8-13 is obtained.

 (Box 8-13)

where:

 K1 = hydraulic conductivity of layer 1 (L/T) K2 = hydraulic conductivity of layer 2 (L/T) θ1 = angle the flow line makes with a perpendicular to the boundary in stratum 1 as shown in Figure Box 8-1 (degrees) θ2 = angle the flow line makes with a perpendicular to the boundary in stratum 2 as shown in Figure Box 8-1 (degrees)