Find: hmax [m] L 17.6 17.8 18.0 18.2 hmax h1 h2 L = 525 [m] w = 0.007 [m/day] hmax h1 h2 Find the maximum water height in meters. [pause] In this problem, L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L 17.6 17.8 18.0 18.2 hmax h1 h2 L = 525 [m] w = 0.007 [m/day] hmax h1 h2 two bodies of water, having different water surface elevations, are separated by a known distance. L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L 17.6 17.8 18.0 18.2 hmax h1 h2 L = 525 [m] w = 0.007 [m/day] hmax h1 h2 The dividing land mass has a known hydraulic conductivity and experiences a constant recharge. L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A hmax h1 h2 flowrate L = 525 [m] w = 0.007 [m/day] flowrate hmax h1 h2 Darcy’s law states that flowrate through a soil is equal to --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A hmax h1 h2 flowrate hydraulic w = 0.007 [m/day] flowrate hydraulic conductivity hmax h1 h2 the hydraulic conductivity, --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A hmax h1 h2 gradient flowrate hydraulic w = 0.007 [m/day] flowrate hydraulic conductivity hmax h1 h2 times the gradient, --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A hmax h1 h2 gradient total area flowrate w = 0.007 [m/day] flowrate hydraulic conductivity hmax h1 h2 times the total area. [pause] From a one-dimensional perspective, --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A Q hmax w= h1 h2 A gradient total area w = 0.007 [m/day] flowrate hydraulic conductivity Q hmax w= h1 h2 A the average recharge rate, w, is equal to the flowrate, divided by the area. L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A w=-K * i Q hmax w= h1 h2 A L = 525 [m] w = 0.007 [m/day] w=-K * i Q hmax w= h1 h2 A This term is then substituted into Darcy’s equation, --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] L Q=-K * i * A w=-K * i w - = i K hmax h1 h2 w = 0.007 [m/day] w=-K * i w - = i K hmax h1 h2 both sides are divided K, the hydraulic conductivity. [pause] After skipping a few steps --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 K w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h1 * h2 K 2 δx2 δy2 the gradient is rewritten as the sum of the partial gradients in the vertical and horizontal directions. L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 K w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h1 * h2 K 2 δx2 δy2 Assuming flow at the soil water interface is purely horizontal, --- L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 K w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h1 * h2 K 2 δx2 δy2 the partial differential equation is simplified by excluding the vertical component. L L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + - L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h1 * h2 K 2 δx2 δy2 The expression is then integrated with respect to x --- 2*w d2(h2) - L = K δx2 L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + - - L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h1 * h2 K 2 δx2 δy2 and the constants of integration are solved using known boundary conditions. 2*w d2(h2) - L = K δx2 w*x2 - h2= +c1*x + c2 K
Find: hmax [m] + - - L x Q=-K * i * A w=-K * i w - = i K w hmax - = h1 w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h1 * h2 K 2 δx2 δy2 If we set x equal to the horizontal distance from the left bank, in the direction of the right bank, --- 2*w d2(h2) - L = K δx2 x w*x2 - h2= +c1*x + c2 K
Find: hmax [m] + - - L x Q=-K * i * A w=-K * i w - = i K w hmax - = h w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h h1 * h2 K 2 δx2 δy2 and we set h, equal to the vertical distance up to the groundwater table, --- 2*w d2(h2) - L = K δx2 x w*x2 - h2= +c1*x + c2 K
Find: hmax [m] + - - L x Q=-K * i * A w=-K * i w - = i K w hmax - = h w = 0.007 [m/day] w=-K * i w - = i K w δ2h2 δ2h2 1 hmax - = + h h1 * h2 K 2 δx2 δy2 the first boundary condition will be x = 0, h = h1, and the second boundary condition will be --- 2*w d2(h2) - L = K δx2 x w*x2 BC#1: x=0, h=h1 - h2= +c1*x + c2 K
Find: hmax [m] + - - L x w - = i K w hmax - = h h1 h2 K = K w*x2 w = 0.007 [m/day] w - = i K w δ2h2 δ2h2 1 hmax - = + h h1 * h2 K 2 δx2 δy2 x = L, h = h2. Finally, the height of the groundwater table, 2*w d2(h2) - L = K δx2 x w*x2 BC#1: x=0, h=h1 - h2= +c1*x + c2 K BC#2: x=L, h=h2
Find: hmax [m] + - - + L x w - = i K w hmax - = h h1 h2 K = K w*x2 (h1 - h2 ) * x 2 2 *(L-x) * x h = h1- + [m/day] 2 K L w - = i K w δ2h2 δ2h2 1 hmax - = + h h1 * h2 K 2 δx2 δy2 as a function of x equals the square root of the quadratic, shown here. [pause] From the given data, --- 2*w d2(h2) - L = K δx2 x w*x2 BC#1: x=0, h=h1 - h2= +c1*x + c2 K BC#2: x=L, h=h2
Find: hmax [m] + - - + L x w - = i K w hmax - = h h1 h2 K = K w*x2 K w (h1 - h2 ) * x 2 2 *(L-x) * x h = h1- + [m/day] 2 K L w - = i K w δ2h2 δ2h2 1 hmax - = + h h1 * h2 K 2 δx2 δy2 we know h1, h2, the recharge w, the hydraulic conductivity K, and the length L, --- 2*w d2(h2) - L = K δx2 x L = 525 [m] w*x2 m day - h1=17.6 [m] h2= +c1*x + c2 K=14.5 K h2=15.3 [m]
Find: hmax [m] ? ? ? + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 *(L-x) * x h = h1- + [m/day] 2 K L ? ? ? hmax h h1 h2 but the value of x is unknown. [pause] Since the problem asks to find the maximum height of the groundwater table, L x L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] ? ? ? + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 *(L-x) * x h = h1- + [m/day] 2 K L ? ? ? hmax h h1 h2 we need to find the value of x which corresponds to this maximum height. L x L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] ? ? ? + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 *(L-x) * x h = h1- + [m/day] 2 K L ? ? ? dh =0 dx hmax h h1 h2 At the groundwater divide, the slope of the groundwater table is zero, therefore, we can --- L x L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d hmax h h1 h2 dx differentiate our equation for the height of the groundwater table, with respect to x, --- L x L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d =0 hmax h h1 h2 dx set that value equal to zero, and solve for x. Doing so will calculate --- L x L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d =0 hmax h h1 h2 dx the value of x, corresponding to the maximum groundwater height, h max. L K (h1 - h2 ) 2 2 - L xhmax= * w 2 2*L x L = 525 [m] m day h1=17.6 [m] K=14.5 h2=15.3 [m]
Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d =0 hmax h h1 h2 dx The known values of L, K, w, h1 and h2 are plugged in, and we compute the horizontal distance to from the left bank --- L K (h1 - h2 ) 2 2 - L xhmax= * w 2 2*L x L = 525 [m] m h1=17.6 [m] K=14.5 day h2=15.3 [m]
Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d =0 hmax h h1 h2 dx to the groundwater divide, to be 113.2 meters. [pause] Now that we know the appropriate value for x, --- L K (h1 - h2 ) 2 2 - L * w 2 2*L x L = 525 [m] m xhmax=113.2 [m] h1=17.6 [m] K=14.5 day h2=15.3 [m]
Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d =0 hmax h h1 h2 dx it can be plugged into the equation for the height of the groundwater table, --- L K (h1 - h2 ) 2 2 - L * w 2 2*L x L = 525 [m] m xhmax=113.2 [m] h1=17.6 [m] K=14.5 day h2=15.3 [m]
Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L *(L-x) * x h = h1- + [m/day] 2 K L dh =0 dx d =0 hmax h h1 h2 dx in addition to the other known variables, and the maximum groundwater height --- L K (h1 - h2 ) 2 2 - L * w 2 2*L x L = 525 [m] m xhmax=113.2 [m] h1=17.6 [m] K=14.5 day h2=15.3 [m]
Find: hmax [m] - + L x hmax=17.78 [m] h1 h2 w (h1 - h2 ) * x 2 2 *(L-x) * x h = h1- + [m/day] 2 K L d =0 hmax=17.78 [m] h1 h2 dx equals 17.78 meters. [pause] L K (h1 - h2 ) 2 2 - L * w 2 2*L x L = 525 [m] m xhmax=113.2 [m] h1=17.6 [m] K=14.5 day h2=15.3 [m]
Find: hmax [m] - + L x 17.6 17.8 18.0 hmax=17.78 [m] 18.2 h1 h2 w (h1 - h2 ) * x 2 2 *(L-x) * x h = h1- + [m/day] 2 K L 17.6 17.8 18.0 18.2 d =0 hmax=17.78 [m] h1 h2 dx Looking back at the possible solutions, L K (h1 - h2 ) 2 2 - L * w 2 2*L x L = 525 [m] m xhmax=113.2 [m] h1=17.6 [m] K=14.5 day h2=15.3 [m]
Find: hmax [m] - + AnswerB L x 17.6 17.8 18.0 hmax=17.78 [m] 18.2 h1 (h1 - h2 ) * x 2 2 *(L-x) * x h = h1- + [m/day] 2 K L 17.6 17.8 18.0 18.2 d =0 hmax=17.78 [m] h1 h2 dx AnswerB the answer is B. L K (h1 - h2 ) 2 2 - L * w 2 2*L x L = 525 [m] m xhmax=113.2 [m] h1=17.6 [m] K=14.5 day h2=15.3 [m]
? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4