Flow to Wells – 2 Steady flow to a well in an unconfined aquifer

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Presentation transcript:

Flow to Wells – 2 Steady flow to a well in an unconfined aquifer Groundwater Hydraulics Daene C. McKinney

Summary Notation Steady flow to a well in an unconfined aquifer Steady flow to a well in an unconfined aquifer with recharge

Flow to a Well in an Unconfined Aquifer

Groundwater Notation Unconfined aquifer 𝛻∙𝐾ℎ𝛻ℎ= 𝜕 𝜕𝑥 𝐾 𝑥 ℎ 𝜕ℎ 𝜕𝑥 + 𝜕 𝜕𝑦 𝐾 𝑦 ℎ 𝜕ℎ 𝜕𝑦 = 𝑆 𝑦 𝜕ℎ 𝜕𝑡 General equation, rectangular coordinates ℎ 𝜕ℎ 𝜕𝑥 = 1 2 𝜕 ℎ 2 𝜕𝑥 𝛻∙ 𝐾 2 𝛻 ℎ 2 = 𝜕 𝜕𝑥 𝐾 𝑥 2 𝜕 ℎ 2 𝜕𝑥 + 𝜕 𝜕𝑦 𝐾 𝑦 2 𝜕 ℎ 2 𝜕𝑦 = 𝑆 𝑦 𝜕ℎ 𝜕𝑡 Steady flow, homogeneous, isotropic aquifer 𝛻 2 ℎ 2 = 𝜕 2 ℎ 2 𝜕 𝑥 2 + 𝜕 2 ℎ 2 𝜕 𝑦 2 =0

Groundwater Notation Unconfined aquifer 𝛻∙ 𝐾 2 𝛻 ℎ 2 = 1 𝑟 𝜕 𝜕𝑟 𝐾 𝑟 2 𝑟 𝜕 ℎ 2 𝜕𝑟 + 1 𝑟 2 𝜕 𝜕𝜃 𝐾 𝜃 2 𝑟 𝜕 ℎ 2 𝜕𝜃 = 𝑆 𝑦 𝜕ℎ 𝜕𝑡 General equation, radial coordinates 𝛻 2 ℎ 2 = 𝑑 𝑑𝑟 𝑟 𝑑 ℎ 2 𝑑𝑟 =0 Steady flow, homogeneous, isotropic aquifer

Steady Flow to a Well in an unconfined Aquifer 2rw Ground surface Bedrock Unconfined aquifer Q h0 Pre-pumping Water level r1 r2 h2 h1 hw Observation wells Water Table Pumping well Governing equation in radial coordinates 𝛻 2 ℎ 2 = 1 𝑟 𝜕 𝜕𝑟 𝑟 𝜕 ℎ 2 𝜕𝑟 + 1 𝑟 2 𝜕 2 ℎ 2 𝜕 𝜃 2 = 2 𝑆 𝑦 𝐾 𝜕ℎ 𝜕𝑡 Assumptions: Steady state flow, radial symmetry (isotropic medium), 𝑑 𝑑𝑟 𝑟 𝑑 ℎ 2 𝑑𝑟 =0

Steady Flow to a Well in an Unconfined Aquifer 𝑑 𝑑𝑟 𝑟 𝑑 ℎ 2 𝑑𝑟 =0 2rw Ground surface Bedrock Unconfined aquifer Q h0 Pre-pumping Water level r1 r2 h2 h1 hw Observation wells Water Table Pumping well 𝑟 𝑑 ℎ 2 𝑑𝑟 = 𝐶 1 Darcy’s law 𝑄=𝐴𝑞= 2𝜋𝑟ℎ 𝐾 𝑑ℎ 𝑑𝑟 =𝜋𝐾𝑟 𝑑 ℎ 2 𝑑𝑟 𝑟 𝑑 ℎ 2 𝑑𝑟 = 𝑄 𝜋𝐾 =𝐶 1 𝑑 ℎ 2 = 𝑄 𝜋𝐾 𝑑𝑟 𝑟 ℎ 2 = 𝑄 𝜋𝐾 𝑙𝑛 𝑟 + 𝐶 2

Steady Flow to a Well in an Unconfined Aquifer 2rw Ground surface Bedrock Unconfined aquifer Q h0 Pre-pumping Water level r1 r2 h2 h1 hw Observation wells Water Table Pumping well ℎ 2 = 𝑄 𝜋𝐾 𝑙𝑛 𝑟 + 𝐶 2 Boundary condition: ℎ= ℎ 1 @ 𝑟= 𝑟 1 ℎ 1 2 = 𝑄 𝜋𝐾 𝑙𝑛 𝑟 1 + 𝐶 2 𝐶 2 = ℎ 1 2 − 𝑄 𝜋𝐾 𝑙𝑛 𝑟 1 ℎ 2 = 𝑄 𝜋𝐾 𝑙𝑛 𝑟 + ℎ 1 2 − 𝑄 𝜋𝐾 𝑙𝑛 𝑟 1 ℎ 2 = ℎ 1 2 + 𝑄 𝜋𝐾 𝑙𝑛 𝑟 𝑟 1

Example – Steady Flow to a Well in an Unconfined Aquifer Given: Q = 300 m3/hr Two observation wells, 1. h1 = 40 m @ r1 = 50 m 2. h2 = 43 m @ r2 = 100 m Find: K 2rw Ground surface Bedrock Unconfined aquifer Q h0 Prepumping Water level r1 r2 h2 h1 hw Observation wells Water Table Pumping well ℎ 2 = ℎ 1 2 + 𝑄 𝜋𝐾 𝑙𝑛 𝑟 𝑟 1 ℎ 2 2 = ℎ 1 2 + 𝑄 𝜋𝐾 𝑙𝑛 𝑟 2 𝑟 1

Steady flow to a Well in an Unconfined Aquifer with Recharge 1 𝑟 𝜕 𝜕𝑟 𝐾 2 𝑟 𝜕 ℎ 2 𝜕𝑟 +𝑊=0 𝑑 𝑑𝑟 𝑟 𝑑 ℎ 2 𝑑𝑟 + 2𝑊 𝐾 𝑟=0 𝑟 𝑑 ℎ 2 𝑑𝑟 + 𝑊 𝐾 𝑟 2 = 𝐶 1 𝑑 ℎ 2 𝑑𝑟 = 𝐶 1 𝑟 − 𝑊 𝐾 𝑟 radius of influence

Steady flow to a Well in an Unconfined Aquifer with Recharge 𝑑 ℎ 2 𝑑𝑟 = 𝐶 1 𝑟 − 𝑊 𝐾 𝑟 Boundary condition 1: No flow across radius of influence 𝑞=−𝐾ℎ 𝑑ℎ 𝑑𝑟 =− 𝐾 2 𝑑 ℎ 2 𝑑𝑟 =0; 𝑑 ℎ 2 𝑑𝑟 𝑟= 𝑟 0 = 𝐶 1 𝑟 0 − 𝑊 𝐾 𝑟 0 =0 radius of influence 𝐶 1 = 𝑊 𝐾 𝑟 0 2

Steady flow to a Well in an Unconfined Aquifer with Recharge 𝑑 ℎ 2 𝑑𝑟 = 𝐶 1 𝑟 − 𝑊 𝐾 𝑟 𝐶 1 = 𝑊 𝐾 𝑟 0 2 Boundary condition 2: Recharge within radius of influence is captured by the well 𝑄=𝐴𝑊= 𝜋 𝑟 0 2 𝑊 𝑟 0 2 = 𝑄 𝜋𝑊 radius of influence 𝐶 1 = 𝑄 𝜋𝐾 𝑑 ℎ 2 𝑑𝑟 = 𝑄 𝜋𝐾 1 𝑟 − 𝑊 𝐾 𝑟

Steady flow to a Well in an Unconfined Aquifer with Recharge 𝑑 ℎ 2 𝑑𝑟 = 𝑄 𝜋𝐾 1 𝑟 − 𝑊 𝐾 𝑟 𝑑 ℎ 2 = 𝑄 𝜋𝐾 𝑑𝑟 𝑟 − 𝑊 𝐾 𝑟𝑑𝑟 ℎ 2 = 𝑄 𝜋𝐾 ln 𝑟 − 𝑊 2𝐾 𝑟 2 + 𝐶 2 Boundary condition 3: Recharge within radius of influence is captured by the well radius of influence ℎ= ℎ 0 @ 𝑟= 𝑟 0 ℎ 0 2 = 𝑄 𝜋𝐾 ln 𝑟 0 − 𝑊 2𝐾 𝑟 0 2 + 𝐶 2 ℎ 2 = ℎ 0 2 + 𝑄 𝜋𝐾 𝑙𝑛 𝑟 𝑟 0 − 𝑊 2𝐾 𝑟 2 − 𝑟 0 2

Example - Steady flow to a Well in an Unconfined Aquifer with Recharge 25 cm diameter well to maintain lowered water table K = 1x10-5 m/s (0.864 m/d) Bottom of aquifer is horizontal at 20 m bgs Water table is 1 m bgs Unconfined aquifer is recharged at W = 0.06 m/d Water table must be lowered 3 m over site r Drawdown at corners must be at least 3 m r=35.35 m ℎ 2 = ℎ 0 2 + 𝑄 𝑤 𝜋𝐾 𝑙𝑛 𝑟 𝑟 0 − 𝑊 2𝐾 𝑟 2 − 𝑟 0 2 16 2 = 19 2 + 0.06 𝜋 𝑟 0 2 𝜋𝐾 𝑙𝑛 35.35 𝑟 0 − 0.06 2(0.864) 35.35 2 − 𝑟 0 2 Solve by iteration: r0 = 70 m 𝑄 𝑤 =𝜋 𝑟 0 2 𝑊=𝜋 70 2 0.06 =923.6 𝑚 3 /𝑑

Summary Notation Steady flow to a well in an unconfined aquifer Steady flow to a well in an unconfined aquifer with recharge