Lecture Packet #10: Superposition

1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet #10: Superposition Q Q t0 t0 t1 t1 x1 Q Q Q Q t0 t0 t1 t1 x1 x1 x1 Impermeable boundary Stream Piezometer ri rr Discharging image well x1 x1 x1 1.72, Groundwater Hydrology Prof. Charles Harvey Discharging real well Piezometer ri rr Recharging image well x1 x1 Discharging real well Lecture Packet 10 Page 1 of 6 Superposition Aquifer flow equation: ∂T∂H ∂T∂H ∂H + +Q = S 2 2 ∂t ∂y ∂x Subject to: ∂H ∂η H = A, and/or = 0 along some boundary H = A at t = 0 Suppose the system has constant head boundaries (H = A, where A is uniform) and no-flow boundaries, then we can convert the GW equation to be in terms of drawdown rather than head. d=A–H H=A–d ∂T∂d ∂T∂d ∂d + −Q = S 2 2 ∂t ∂y ∂x d = 0, and/or ∂d = 0 along some boundary ∂η Now we can investigate the effects on increasing the pumping rate, by multiplying the whole equation by a constant: ∂T∂ 2d ∂T∂2d ∂2d + − 2Q = S 2 2 ∂t ∂y ∂x 2d = 0, and/or ∂2d = 0 along some boundary ∂η Thus the solution is the same, except all the drawdowns have been increased by the increase in the pumping rate. We did not make any assumptions about the geometry of the domain, the spatial pattern of T, or the location of the well. Q=0 Q=1 Q=2 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 10 Page 2 of 6 Now we suppose we have two wells at different locations. We can add the two equations to get one equation: ∂T∂d1 ∂T∂d1 + = Q1 ∂x 2 ∂y 2 ∂T∂d 2 ∂T∂d 2 + = Q2 ∂x 2 ∂y 2 ∂T∂ (d1 + d 2 ) ∂T∂ (d1 + d 2 ) = Q1 + Q2 + ∂y 2 ∂x 2 A Q1 Q2 t0 (h0-h)B t1 r1 r2 B A’ Image Well Theory & Superposition Concept: To predict aquifer behavior in the presence of boundaries, introduce imaginary wells such that the response at the boundary is made true. Well Q -x1 Lake – constant head initial head Pumping water level Recharging Image Well Drawdown Image Well Q -x1 Q +x1 initial head Pumping water level Drawdown Real 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 10 Page 3 of 6 To maintain constant head at lake • Introduce an image well • It recharges (artificially) • It creates a cone of impression • Resultant cone is due to pumping well and recharge well Consider an impermeable boundary • Drawdown is enhanced due to reflection Well Q -x1 initial head Impermeable boundary drawdown Pumping water level Well Q -x1 +x1 Pumping Image Well Q initial head Real Drawdown Drawdown Image • A pumping image well creates the effect of the boundary Two things you can do with superposition or image wells. 1) Find the hidden boundary For a semi-log plot one might obtain: Impermeable boundary ti Impermeable boundary ∆H ∆Ha tr Ideal Unbounded ri rr Image Pumping well a Observation well a Pumping well ∆Ha Time (log scale) (applies to any linear system – can work with Theis equation as well) 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 10 Page 4 of 6 [h0 − h(r , t )]r 2.30Q [log10 ] 2.252 Tt 4πT r S [h0 − h(r , t )]i = 2.30Q [log10 ] 2.252Tt 4πT ri = Drawdown due to real well is set to drawdown due to image pumping well (at different times Å f(boundary)) [h0 − h(r , t )]r = [h0 − h(r , t )]i 2 which gives 2 ri t r = r Æ ri = rr i ti tr tr • • • • • Slope of line on semi-log plot depends on Q and T If an impermeable boundary is present, drawdown will double under the influence of the image well at some point in time Select an arbitrary drawdown ∆Ha and the time for this drawdown to occur under the real pumping well Find the same drawdown to be produced by the image well and its corresponding time Knowing drawdowns and times one can determine distance to boundary – uniqueness? multiple observation wells are needed 2) Analysis of water level recovery test data At end of pumping test, pump is stopped and water levels recover, this is called recovery. Water level below the original head during recover is called residual drawdown. Advantages of recovery test – inexpensive and “pumping” rate is nearly constant. What is drawdown from a pulse input? Pumping Schedule Q time Drawdown ∆H Residual drawdown recovery time 1.72, Groundwater Hydrology Prof. Charles Harvey Lecture Packet 10 Page 5 of 6 Superposed Schedules Superposed Solutions ∆H Q t time -Q t* − ∆H residual drawdown = ∆h* = u= Q Q W (u ) − W (u*) where 4πT 4πT r 2S r 2S and u* = 4Tt 4Tt * For small r and large t*, we get ∆h* = t 2.30Q log 4πT t* for a plot of ∆h vs. log t/t* over one log cycle, we get T as T= 1.72, Groundwater Hydrology Prof. Charles Harvey 2.30Q 4π∆h * Lecture Packet 10 Page 6 of 6