Question

Vector Calculus Suppose the hydraulic head in a vertically homogeneous aquifer is described by the equation h(x,y) = 10+Oy in the domain -10sxs10, -10sys10, as plotted in Fig 1. -10 -10 5 5 0 0 5 5 10 10 Figure 1 Figure 2 2a. What is the equation for the gradient of this surface (i.e., Vh; note that this is opposite the direction of flow)? (4 pts) 2b. What is the gradient of h at (x=2, y=4)? If the homogeneous, isotropic hydraulic conductivity is 0.55 m/day, what is the magnitude and direction of the specific discharge at this point? (6 pts) 2c. Now lets add a non-zero y component to the head field. The new equation for head is h(x,y)=15+ over the same domain (plotted in Figure 2). What is the equation for the gradient of this new surface? (4 pts) 2d. What is the gradient of h (i.e., Vh) at (x=2, y=4) using the new equation? If the homogeneous, isotropic hydraulic conductivity is 0.55 m/day, what is the magnitude and direction of the specific discharge at this point? (6 pts) 2e. Recall that the divergence of a vector field indicates the presence of sources and sinks. Since the flow of water will be parallel to the hydraulic gradient, the divergence of the hydraulic gradient will indicate sources/sinks of water. Find the divergence of -Vh. What does this mean? In what physical situation might this be happening? (10 pts)

Answer 1

Ans. (20 hlasy ) = 10-22 i and I are unit vectors alony x-axis and y-axis al (26) at X=2, y = 4 Th= -2; = -0.4i

DATE: 1 Specific discharge 2= kiloh) a 24X60X60 Ke Conductivity = 0:55m ay = 0·55 m/s A = area = 1m² Th= - 04F - 9= 0.55 x 0.4 i 24X60X60 q = 255x 10-6 ; 191 mointude = 2.55 X 106 direction is along positive X-axis. h(x, y) = 15+ y - x² - Į TO Th= Oh it dh Î Jx dy I -

DATE: 1 lado) Ohlaix=2, Y:4) = -0.4i + 0.51 10k = Jl0:43° +0-5) = 0.64 ñ ñ on = -0.4 i + 0.5 ? 0.64 0.64 | 7 = -0.625i + 0:28 | 9 = -K Jh A 0.55 x 0.64 24 x 60X60 191 - 4:07X /p-3 | mapintu de a direction (n = -0.60. -0.6251 +0.78