Gross Capacity & Farm Well Discharge Calculator

The pump-sizing question: given how much water the crop needs (net capacity), application inefficiency, downtime, and losses in getting water from the well to each field, how big does the well/pump actually need to be? Handles any number of fields sharing one supply.

1. Farm-Wide Capacity

C_g = C_n / [(E_LQ/100%)(1−D_t/100%)]

Net system capacity (C_n)

in/d

Application efficiency of the low quarter (E_LQ)

System downtime (D_t)

—

Gross system capacity (C_g), gpm/ac

2. Fields Served by This Supply

Each field's own conveyance (field canal/ditch) efficiency accounts for losses getting water from the main supply to that field.

Area

Conveyance efficiency

Discharge needed

3. Well / Main Supply Discharge

Main supply canal conveyance efficiency

Leave at 100% if the well/main line feeds the fields directly with no shared supply losses.

About this calculator

System capacity has to account for more than just crop demand: application inefficiency and downtime inflate the net capacity into a gross capacity, and then each field's own conveyance losses (and any shared main-canal losses) inflate that further into what the well actually has to discharge.

Q_{f, field} = C_g × Area / (E_{c, field}/100%) — Q_well = ΣQ_f / (E_{c, main}/100%)

Textbook Example 5.8: C_n=0.3 in/d, E_LQ=65%, D_t=10% → C_g≈9.7 gpm/ac. Two 80-ac furrow fields, field efficiencies 80% and 90%, main canal at 90% → Q_f1≈975 gpm, Q_f2≈867 gpm, well discharge ≈2,050 gpm. Click "Load textbook Example 5.8" above to see it worked through.