Box 8  Deriving the Tangent Law of Refraction

William W. Woessner; Eileen P. Poeter

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.

Figure showing geometry of flow lines at an interface between materials of differing hydraulic conductivity — Figure Box 8-1 – Geometry of flow lines at an interface between materials of differing hydraulic conductivity showing the angle of refraction, the width of the flow tube d₁ and d₂, and the distance between equipotential lines l₁ and l₂ on different sides of the interface.

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.

Q_n = – K_n i_n A_n

(Box 8-1)

where:

Q_n	=	volumetric flow through a unit width into the image for region n (L³/T)
K_n	=	hydraulic conductivity of region n (L/T)
i_n	=	hydraulic gradient in region n (dimensionless) = $-\frac{head difference}{distance}=-\frac{hdiff_{n}}{l_{n}}$
A_n	=	area perpendicular to flow in tube of region n (L²), A_n = d_n w where w is a unit width into the image

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

$\displaystyle -K_1\frac{{hdiff}_1}{l_1}\ d_1w={-K}_2\frac{{hdiff}_2}{l_2}\ d_2w$

(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.

$\displaystyle K_1\frac{d_1}{l_1}=K_2\frac{d_2}{l_2}$

(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:

θ_n

=

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.

$\displaystyle K_1\frac{b\ \cos{\theta_1}}{l_1}=K_2\frac{b\ \cos{\theta_2}}{l_2}$

(Box 8-6)

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

	$\displaystyle \frac{b\ }{l_1}=\frac{1}{\sin{\theta_1}}$	(Box 8-7)
	$\displaystyle \frac{b\ }{l_2}=\frac{1}{\sin{\theta_2}}$	(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.

$\displaystyle K_1\frac{\cos{\theta_1}}{\sin{\theta_1}}=K_2\frac{\cos{\theta_2}}{\sin{\theta_2}}$

(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.

$\displaystyle K_1\frac{\sin{\theta_2}}{\cos{\theta_2}}=K_2\frac{\sin{\theta_1}}{\cos{\theta_1}}$

(Box 8-10)

Using the Trigonometric identity of Equation Box 8-11.

$\displaystyle \tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}$

(Box 8-11)

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

$\displaystyle K_1\tan{\theta_2}=K_2\tan{\theta_1}$

(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.

$\displaystyle \frac{K_1}{K_2}=\frac{\tan{\theta_1}}{\tan{\theta_2}}$

(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)

Return to where text links to Box 8

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