Eight streamlines are plotted in {numref}`Figure %s
A plot with both equipotentials and streamlines is called a flow net ({numref}`Figure %s
\n",
"
\n",
"\n",
"```"
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"```{exercise}\n",
":label: ex_66_redo65_stream_function\n",
"\n",
"For the example above, compute the stream function at $(x,y)=(0,100)$, $(x,y)=(0,\\varepsilon)$, $(x,y)=(0,-\\varepsilon)$, $(x,y)=(0,-100)$, where $\\varepsilon$ is a small increment, by determining the appropriate angles, and compare the answer to results with the `arctan2` function.\n",
"\n",
"Solution
\n", "\n", "$h_w = -3.68$ m\n", "\n", "\n",
"
\n",
"\n",
"```"
]
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" The method of images is an application of the superposition principle. Wells (or other singularities) are placed outside of the problem domain, and symmetry is used to satisfy conditions specified along boundaries. A well near a river with constant head is simulated by adding an image well on the other side of the river. The discharge of the image well must be equal, but opposite in sign to the discharge of the well, and the image well must be located at the “image location” of the well, that is, the same distance from the river, but on the other side. The solution for multiple wells near a river may be obtained in the same fashion, as long as a corresponding image well is added for each well.\n",
"Solution
\n", "\n", "$\\Psi = [-25, -50, 50, 25]$ m$^3$/d\n", "\n", "A well near an impermeable rock formation (an impermeable or no-flow boundary) may be simulated by adding an image well with the same discharge on the other side of the impermeable boundary. An example of a well near an impermeable boundary along $x = 0$ is shown in {numref}`Figure %s
\n",
"![](\"figures_chap6/sol_ex_6_7.png\")
\n",
"\n",
"
\n",
"\n",
"```"
]
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"```{exercise}\n",
":label: ex_68_discharge_well6.2\n",
"\n",
"A well is located a distance $d = 50$ m from a long and straight river with water level $h_r = 20$ m. The river is in direct contact with the aquifer, and the radius of the well is 0.3 m. The transmissivity of the aquifer may be approximated as constant and equal to $T = 200$ m$^2$/d. Compute the discharge of the well such that the head at the well is 2 m below the water level in the river.\n",
"\n",
"Solution
\n", "\n", "$h = -1.04$ m at extraction well\\\n", "$h = 1.18$ m at injection well\n", "\n", "\n",
"\n",
"\n",
"
\n",
"\n",
"```"
]
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"source": [
"```{exercise}\n",
":label: ex_69_total_recharge_compare\n",
"\n",
"Consider a well pumping with discharge $Q$ on a circular island with radius $R$; the radius of the well is $r_w$. The origin of the coordinate system is at the center of the island, and the well is located at $(x, y) = (d, 0)$. The head at the edge of the island is $h_0$. The transmissivity of the aquifer may be approximated as constant and equal to $T$. An expression for the head on the island may be obtained with the method of images for a circle ({cite:alp}`strack1989groundwater`).\n",
"\n",
"$$\n",
"h = \\frac{Q}{4\\pi T} \\ln\\left[\\frac{(x-d)^2 + y^2}{(x - R^2/d)^2 + y^2} \\frac{R^2}{d^2}\\right] + h_0\n",
"$$(eq_ex6.96.2)\n",
"\n",
"Compute and draw head contours, and compute the head at the well. Given: $T = 200$ m$^2$/d, $R = 500$, m, $d = 200$ m, $Q = 500$ m$^3$/d, $r_w = 0.3$ m, and $h_0 = 10$ m.\n",
"\n",
"Solution
\n", "\n", "$Q = 432.6$ m$^3$/d\n", "\n", "\n",
"![](\"figures_chap6/sol_ex_6_9.png\")
\n",
"\n",
"
\n",
"\n",
"```"
]
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Solution
\n", "\n", "$h_w = 7.12$ m\n", "\n", "\n",
"\n",
"