1. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- dHt = HtA - HtB = 0 m A 5 m -5 m 0 m B 8 m -8 m 0 m Hp Hz Ht dHt = HtA - HtB = 0 m A 5 m 10 m B 8 m 2 m 10 m Hp Hz Ht Ref is pool bottomRef is pool surfaceRef is 5 m below pool surface A 5 m 0 m 5 m B 8 m -3 m 5 m Hp Hz Ht dHt = HtA - HtB = 0 m Pressure and elevation head In certain systems water pressure is commonly expressed as height. This makes it convenient for estimation of flow potential as that of the effects of gravity upon flow can be easily accommodated as elevation head (Hz). Total pressure or total head is the summation of the two: Ht = Hp + Hz Comparison of the total water pressure between 2 different points in a flow system can be used to determine whether flow is occurring, in what direction and the magnitude of the driving force. Example 1: A tank of water 10 m deep has an angled pipe for emptying or filling attached to it. The pipe has two standpipes (vertical risers used to measure pressure by observing water height). The angled pipe is closed thus the heights of water in the two standpipes are the same and also the same water level as the surface of the pool. A B 5 m8 m 5 m2 m Water height above points of measurement (A, B) Distance above pool bottom Valve is closed Although in this visual example it is obvious that there is no flow occurring in many systems this is not known and thus reliance must be upon such pressure measurements. Determination of flow is by comparison of the total pressures between points of measurement. Hp is the height of water or pressure head, above the point of measurement and Hz is the elevation of the point of measurement relative to a reference plane. The three tables below show that it does not matter what the plane of reference is the difference between the two points, A and B, is always the same. 10 m 1 Hydraulic Gradients: Determination of direction of groundwater flow

2. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- Distance above pool bottom dHt = HtA - HtB = A B Hp Hz Ht Ref is as shown on diagram A B 2 m 5 m2 m Example 2: Determine whether flow is occurring and in what direction for the diagram below: dHt = HtA - HtB = A B Hp Hz Ht Ref is pool bottom dHt = HtA - HtB = A B Hp Hz Ht Ref is 20 m above poolbottom Water height above points of measurement (A, B) 10 m Example 3: Determine whether flow is occurring and in what direction for the diagram below: Distance above reference plane A B 3 m 7 m 25 m22 m Water height above points of measurement (A, B) 2

3. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- Liquid transport in porous materials (laminar flow) is the change in pressure over a distance: i = dHt/dx where dHt = Ht A - Ht B, the change in total pressure between points A and B, Ht = Hp + Hz, where Hp is the water pressure and Hz is the position of the point relative to the reference plane (z = 0) dx = Hx A - Hx B, the distance between points A and B in a horizontal plane, and 'H' represents pressure as a hydraulic head of water measured as height (m or cm). q = - K i Calculation of hydraulic gradient; horizontal flow Example 1; Piezometers in a groundwater aquifer. Both piezometers are the same depth. z Ref plane (z = 0) is 2 m below piezometer bottoms x ref plane (x = 0) is through point A z = 0 m 100 m AB 5 m AB Hz Hp Ht 2 m 5 m 7 m 2 m 5 m 7 m i = dHt dHx = 7 m - 7 m 0 m - 100 m = 0 m -100 m = 0 Ht A - Ht B Hx A - Hx B = x = 0 m No flow as total pressures are the same. Flux (q, m/s) V A t Hydraulic conductivity (K, m/s) is a function of the ability of the media to conduct (permeability) and the ability of the liquid to flow (fluidity) Hydraulic gradient (i, m/m) Volume of flow per cross-sectional area of media per unit time; q = V A -1 t -1 = m 3 m -2 s -1 = m/s Flux is a velocity term. Flow (q) is proportional (K) to the hydraulic gradient (i) and will occur in the direction of (-) decreasing potential 3

4. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- Hydraulic gradient Example 3 (same as 2, but z ref changes); Piezometers in a groundwater aquifer. Both piezometers are the same depth. z Ref plane (z = 0) is at ground surface x ref plane (x = 0) is through point A z = 0 m 100 m AB 8 m5 m AB Hz Hp Ht -15 m 8 m -7 m -15 m 5 m -10 m i = dHt dHx = -7 m - -10 m 0 m - 100 m = 3 m -100 m = -0.03 m/m Ht A - Ht B Hx A - Hx B = x = 0 m 15 m Flow occurs from high pressure to low pressure (A to B). The negative sign of the gradient indicates flow to the left in the x-axis. Note: although z-ref plane changed position and thus the total pressures changed the gradient remained constant in sign and value Example 2; Piezometers in a groundwater aquifer. Both piezometers are the same depth. z Ref plane (z = 0) is 2 m below piezometer bottoms x ref plane (x = 0) is through point A z = 0 m 100 m AB 8 m5 m AB Hz Hp Ht 2 m 8 m 10 m 2 m 5 m 7 m i = dHt dHx = 10 m - 7 m 0 m - 100 m = 3 m -100 m = -0.03 m/m Ht A - Ht B Hx A - Hx B = x = 0 m Flow is from high to low pressures (from A to B) or from left to right as indicated by the negative sign Calculation of hydraulic gradient; horizontal flow 4

5. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- Hydraulic gradient Example 1; Piezometers in a groundwater aquifer. Piezometers are at 2 different depths, one above the other. z Ref plane (z = 0) is 2 m below piezometer B A B 3 m 7 m Hz 7 m 2 m Hp 3 m 8 m ABAB i = dHt dHz = 10 m - 10 m 7 m - 2 m = 0 m 5 m = 0 m/m Ht A - Ht B Hz A - Hz B = z = 0 m There is no flow as the hydraulic gradient is '0'. Calculation of hydraulic gradient; vertical flow 8 m 2 m Ht 10 m Example 2; Piezometers in a groundwater aquifer. Piezometers are at 2 different depths, one above the other. z Ref plane (z = 0) is 2 m below piezometer B A B Hz 7 m 2 m Hp 3 m ABAB i = dHt dHz = 10 m - 5 m 7 m - 2 m = 5 m = 1 m/m Ht A - Ht B Hz A - Hz B = z = 0 m There is flow as the total pressure heads are different. Flow is from high to low (A to B). Flow is down as indicated by the postive gradient. Ht 10 m 5 m 3 m 7 m 3 m 2 m 5

6. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- Hydraulic gradient Example 3; upward flow Piezometers in a groundwater aquifer. Piezometers are at 2 different depths, one above the other. z Ref plane (z = 0) is 2 m below piezometer B Hz 7 m 2 m Hp 1 m 8 m ABAB i = dHt dHz = 8 m - 10 m 7 m - 2 m = -2 m 5 m = - 0.4 m/m Ht A - Ht B Hz A - Hz B = There is flow upwards as indicated by the lower total pressure at A and by the negative hydraulic gradient. Calculation of hydraulic gradient; vertical flow (continued) A B 1 m 7 m z = 0 m 8 m 2 m Ht 8 m 10 m Example 4; same as #3 but ref plane is moved. Piezometers in a groundwater aquifer. Piezometers are at 2 different depths, one above the other. z Ref plane (z = 0) is at groundsurface. Hz -5 m -10 m Hp 1 m 8 m ABAB i = dHt dHz = -4 m - -2 m -5 m - -10 m = -2 m 5 m = - 0.4 m/m Ht A - Ht B Hz A - Hz B = Flow is still upwards. Location of the reference plane has nothing to due with the sign nor the value of the hydraulic gradient. Ht -4 m -2 m A B z = 0 m 8 m 10 m 1 m 5 m 6

7. Gradients 309/m06gradients.ppt Dept of Agricultural & Bioresource Engineering -- Hydraulic Gradients; Groundwater flow in an hillslope. A B C D 200 m Change in elevation from A to C, 5 m x = 0 m z = 0 m E 200 m Piezometer A B C D E Depth beneath soil surface (m) 9.0 6.0 4.0 2.0 Water height (from bottom of piezometer, m) 4.0 1.2 3.5 3.0 0.5 3 m Find: ABCDEABCDE Hz (m) + Hp (m) = Ht (m) Hydraulic gradients i A-B i A-C i A-D i D-E 7 Flow Direction A and B A and C C and D D and E