Specific Capacity - A Driller's Test

Last modified 8 February 2005 Specific capacity = Q/h_o-h = Q/s
    Obtained by dividing pump rate by drawdown in the pumping well.
Transmissivity can be estimated following Driscoll, 1986
   T=1500* Q/s (unconfined)
    T = Q/s * 2000 (confined)
    (T=gpd/ft; Q= gpm; s= ft)
Note: this calculation assumes t = 1 d r = 0.5 ft T = 30,000 gpd/ft; S = 0.003 for a confined aquifer and 0.075 for unconfined aquifer to find 1500 and 2000. While it may seem odd to use assumed T and S to calculate T, errors of less than 7% are reported by Driscoll.

Empirical Estimation from Single-Well Pump and Recovery
    Razack and Huntley, 1991 (Ground Water 29: 856-861)
        Metric Empirical (sq. m/day, cu. m/day, m)
                T=15.3(Q/s)^0.67
        English Empirical (sq. ft/day; cu. ft/d, ft)
             T=33.6(Q/s)^0.67
                (Can these empirical results be transferred elsewhere?)
    Custer and others 1991, Bozeman Fan
        log T = 1.36 log specific capcity + 3.53
        T = 3364 (spec. cap) ^1.36
                    Q=m³min.^-1 h_o-h = m T = m²d^-1
    Kauffman, 1999 Deer Lodge Valley and Flint Creek Valley, Montana
log T = 1.58 log (Q/(h_o-h)) + 2.53 with n of 7 in Deer Lodge Valley
log T = 1.53 log (Q/(h_o-h)) + 2.92 with n of 4 in Flint Creek Valley
    Q = (gpm); h_o-h = ft; T = gpd/ft.
Note:   to convert gpm/ft to m³/min/m=m²/min.   multiply by 0.0124
            to convert m³/min/m = m²/min to gpm/ft multiply by 80.5
            to convert m³/min/m=m²/min to gpd/ft multiply by 116000
           A web page that discusses log-log graphs is available

Pitfalls of specific capacity to transmissivity
   Specific capacity value depends on pumping time
    Specific capacity value depends on pumping rate
    Specific capacity value depends on well construction
    Transmissivity if calculated from specific capacity and recovery tests depends on the factors listed for specific capacity

Potential Well Yield Estimation
    Specific capacity can be used to estimate well yield potential. The pumped rate on the well log may not be the pump rate of the planned system. The question is, "What is the maximum yield expected from this well?" (NOTE: the calculation below should be verified with a real test, but this procedure helps determine whether the real test is likely to succeed or fail).
Example:
    Well yield from well log = 50 gpm
    Drawdown from well log = 20 ft
    Calculate specific capacity = 50/20=2.5 gpm/ft
    Water level from well log = 30 ft
    Total well depth from well log = 90 feet
    Calculate available drawdown = 90 ft-30 ft = 60 ft
    Subtract 10 feet because the pump cannot rest at the very bottom of the well and you do not want the water level to fall below the pump or the pump will burn out.
        60 ft - 10 ft = 50 ft
    Calculate the estimated potential yield.
          2.5 gpm/ft*50ft= 125 gpm
Another question that can be explored:
How much available head is needed to produce 200 gpm?
        200 gpm/ 2.5 gpm/ft = 80 feet
        but one needs at least 10 feet for pump (80+10=90 ft)
        This is clearly more than the head available (60 ft) so 200 gpm is not a likely expected yield. (Beware: Drilling deeper does not mean a longer water column will be available or that the same specific capacity will be found).

References:
Driscoll, F.G., 1986, Groundwater and Wells (2nd ed.): Johnson Division, St. Paul, Minnesota, p. 1021.

Custer, S.G., Donohue, D., Tanz, G., Nichols, T, Sill, W., Wideman, C., 1991, Groundwater potential in the Bozeman Fan Subarea, Gallatin County, Montana: Montana Department of Natural Resources, Helena, Montana, 336 p.

Kauffman, M. H., 1999, An investigation of ground water - surface water interaction in the Flint Creek Valley, Granite County, Montana: Master of Science Thesis, Montana State University, Bozeman, Montana, 196 p. (p. 31)

Ratzack, M., and Huntley, D., 1991, Assessing transmissivity from specific capacity data in large heterogeneous alluvial aquifers: Ground Water, v. 29, p. 856-861.