Exercise 4 – Solution

1. Original system
2. The system can be stretched by a factor of 5 in the y direction to account for the anisotropy because 225^0.5/9^0.5 = 15 / 3 = 5
3. A flow net is sketched in the transformed image as if the system is isotropic
4. The system is transformed back to 1/5 the width to reveal the flow lines in the anisotropic system

Contour interval = (200 m) / (8 head drops) = 25 m

The aspect ratio is one, so a_r = 1

K_{equivalent for an anisotropic flow net} = (K_xK_y)^0.5 = ((225 m/d)(9 m/d))^0.5 = 45 m/d

Q_total = (45 m/d) (200 m) (100 m) (12 flow tubes) / (8 head drops) = 1,350,000 m³/d

Return to Exercise 4