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GEO-SCI 587 HYDROGEOLOGY Problem Set #1 Answers

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GEO-SCI 587 HYDROGEOLOGY Problem Set #1 Answers GEO-SCI 587 HYDROGEOLOGY Problem Set #1 Answers 1) Total Points: 10 Darcy’s Law (in a porous medium under linear conditions) relates the flow rate of water to the physical properties of the system, including area of flow, head, length of flow, and characteristics of the medium through which the flow is occurring. Q = - K A dh/dl The flow rate is directly proportional to the hydraulic conductivity of the medium, the cross-sectional area of flow, and head loss. It is inversely proportional to the length of flow. 2) Total Points: 20 Given: dL = 60 cm D = 10 cm R = D / 2 = 10 / 2 = 5 cm A = π * R2 = π * 25 = 78.5 cm2 K = 600 cm/d n = 0.20 hptop = 30 cm hpbottom = 80 cm Solution: Use the bottom end of the column as the datum. > Find the direction of flow httop = hptop + hztop = 30 cm + 60 cm = 90 cm htbottom = hpbottom + hzbottom = 80 cm + 0 cm = 80 cm Flow will be from high to low total head, or downward. hp = 30 cm hz = 60 cm ht = 30 + 60 = 90 cm dL = 60 cm hp = 80 cm hz = 0 cm ht = 80 + 0 = 80 cm > Calculate Q, q, and v Q = - K * A * dh/dL = - (600 cm/d) * (78.5 cm2) * (80 – 90 cm) / 60 cm Q = 7850 cm3/d Convert to cm3/min… 7850 cm3/d * (1 d / 1440 min) = 5.45 cm3/min q = Q / A = [5.45 cm3/min] / [78.5 cm2] = 0.069 cm/min v = q / n = [0.069 cm / min] / [0.20] = 0.35 cm/min > Is “Darcy velocity” (q) or “Average Linear Pore Water Velocity” (v) always larger in a porous medium? In the case where n<1 (a porous medium)… v > q since v = q / n (garden hose analogy) 3) Total Points: 15 a. Basin as a whole Inputs Outputs Precip 35 in/yr Evaporation 23 in/yr Streamflow (Total) 9 in/yr Subsea Outflow 3 in/yr b. Streams Inputs Outputs Stormflow 3 in/yr Streamflow (Total) 9 in/yr Baseflow 6 in/yr c. groundwater Inputs Outputs Recharge 9 in/yr Baseflow 6 in/yr Subsea Outflow 3 in/yr d. Average streamflow from the basin in cfs ~130 ft3/s A= 200 mi2 = 5.6 E9 ft2 Streamflow is calculated by taking length per time unit and multiplying by catchment area. e. Groundwater recharge First convert the 9 in/yr into ft3/day by using same technique used above, next convert to gallons per day, then divide answer by 200 mi2 to give you 419,000 gallons of water per sq. mile of surface area. Now divide by 1e6 to give you .419 millions of gallons water per day per sq. mile of surface area. 4) Total Points: 20 Given: A B C Depth to water (m) 15 40 40 Depth of Piezometer 140 70 100 (m) Use the bedrock surface as the datum for elevation head. A B C depth elevation above the datum 0m hz = 150 m water level 50 m 100 m shale 150 m hz = 0m > Part a: Calculate pressure, elevation, and hydraulic head hp = depth of piezometer – depth to water This is like the depth of water over your point of measurement. hz = depth to bedrock – depth of piezometer This is the elevation of your point of measure above some datum. ht = hp + hz A B C Pressure head (m) 140-15= 125 70-40= 30 100-40= 60 Elevation head (m) 150-140= 10 150-70= 80 150-100= 50 Hydraulic head (m) 125+10= 135 80+30= 110 60+50= 110 > Part b: Calculate the average vertical gradients. What is the significance of the results? A Æ B : dh/dl = [110-135] / [140-70] = - 0.36 There is a gradient from A to B. However, since there is a confining layer between them, there will not be significant flow, unless the aquitard is leaky. If there is water transmitted, it will be from A to B. B Æ C : dh / dl = [110-110] / [100-70] = 0 There is no gradient in total head, so there will be no flow between B and C. > Part c: Did the company do unnecessary work? YES. Wells B and C are in the same, unconfined aquifer. (Well A is in a different aquifer that is confined.) > Part d: To minimize pumping costs, you would want to pump from well A. Here, the water is only 15 vertical meters from the surface, versus 40 meters in the case of wells B and C. 5) Total Points: 20 First, calculate the total head at each of the three points. Total head at each point will be equal to [(land surface elevation)-(depth to water)-(elevation of datum)]. If we choose sea level as our datum, total head is simply [(land surface elevation)-(depth to water)]. Point 1: 5438 m – 49 m = 5389 m Point 2: 5442 m – 52 m = 5390 m Point 3: 5446 m – 55 m = 5391 m Next, plot the points on graph paper and label the head at each point (Figure). Since we only have three points available to define our potentiometric surface, we must assume it is planar. If the surface is planar, the gradient in any direction will be linear. As a result, we can interpret head values between holes 1, 2, and 3 along the lines connecting them. For instance, because the total head at point 1 is 5389 m and total head at point 3 is 5391 m, we can assume that the head half way between the two points is 5390 m. Now, draw lines connecting points of equal head to create equipotential lines (we only have two sets of points with equal head values, but all the equipotentials are parallel and equally spaced, so we can draw in other lines based on points where one head value is known). Assuming a homogenous, isotropic medium, groundwater flow will be perpendicular to the equipotentials, from high total head to low total head. In the scenario presented, flow is to the southwest (260 degrees). The distance (∆l) from the 5391 m equipotential to the 5389 m equipotential (∆h = 2 m) is 5 inches, so the hydraulic gradient is 2 m/(5 in * 200 m/in) = 2 m/1000 m = 2 x 10-3. 6) Total Points: 15 First, determine the total head at point A and D. Point A: 5 m Point D: 4 m We know that the total head is linear between these points, so Point B: 4.67 m Point C: 4.33 m Determine elevation head at all points. Point A: 5 m Point B: 0 m Point C: 0 m Point D: 4 m Next determine pressure head at A and D. Point A: 0 m Point D: 0 m Calculate pressure head at B and C using the equation: hT = hE + hP. Point B: 4.67 – 0 = 4.67 m Point C: 4.33 – 0 = 4.33 m Total, elevation and pressure head 5 4 ) ht m 3 ( he 2 ead hp h 1 0 012345678910111213 length (m) .
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