6.2 Lift Station Hydraulics, Pumping Rate & Drawdown Math

6.2 Lift Station Hydraulics, Pumping Rate & Drawdown Math

Exam Focus: Operators must be capable of calculating wet well holding capacity, performing volumetric pump drawdown calibrations accounting for continuous sewer inflow, computing pump cycle times, and determining water horsepower ($\text{WHP}$), brake horsepower ($\text{BHP}$), and daily power costs.

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Wet Well Geometry & Volumetric Capacity

Lift station wet wells provide temporary storage to buffer incoming wastewater surges between pump cycles. Calculating the volume per foot of depth allows operators to determine storage capacity, set level control floats, and conduct drawdown testing.

                  WET WELL VOLUMETRIC CONFIGURATIONS

        Rectangular Wet Well                    Cylindrical Wet Well
       +--------------------+                  /--------------------\
      /                    /|                 |                      |
     +--------------------+ |                 |                      |
     |                    | | H (Depth)       |                      | H (Depth)
     |                    | |                 |                      |
     |      L (Length)    |/                  |      D (Diameter)    |
     +--------------------+                    \--------------------/
            W (Width)
      V = L × W × H × 7.48 gal/cu ft         V = 0.7854 × D² × H × 7.48 gal/cu ft

1. Rectangular Wet Wells

$\mathbf{V_{\text{cu ft}} = L \times W \times H}$ $\mathbf{V_{\text{gallons}} = L \times W \times H \times 7.48\text{ gal/cu ft}}$ $\text{Gallons per Foot of Depth} = L \times W \times 7.48$ $\text{Gallons per Inch of Depth} = \frac{L \times W \times 7.48}{12}$

2. Cylindrical (Round) Wet Wells

$\mathbf{V_{\text{cu ft}} = 0.7854 \times D^2 \times H}$ $\mathbf{V_{\text{gallons}} = 0.7854 \times D^2 \times H \times 7.48\text{ gal/cu ft}}$ $\text{Gallons per Foot of Depth} = 0.7854 \times D^2 \times 7.48$ $\text{Gallons per Inch of Depth} = \frac{0.7854 \times D^2 \times 7.48}{12}$

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The Pump Drawdown Test

A drawdown test is the standard field method used by operators to determine the actual pumping capacity of a wastewater pump under field operating conditions. Because raw sewage continuously enters the wet well from the collection network during testing, operators must perform a two-step test to account for inflow.

                     PUMP DRAWDOWN TEST SEQUENCE

    STEP 1: INFLOW RATE TEST                   STEP 2: DRAWDOWN TEST
    (All Pumps OFF)                            (Lead Pump ON)
    
    +---------------------------+              +---------------------------+
    | Inflow (Qin) ---> [  ]    |              | Inflow (Qin) ---> [  ]    |
    |                           |              |                           |
    |   ↑ Liquid Rises (ΔH_in)  |              |   ↓ Liquid Drops (ΔH_out) |
    |~~~~~~~~~~~~~~~~~~~~~~~~~~~|              |~~~~~~~~~~~~~~~~~~~~~~~~~~~|
    |                           |              |   =====> [PUMP ON] ====>  |
    +---------------------------+              +---------------------------+
     Qin = Vol_rise / Time_in                   Qnet = Vol_drop / Time_pump
     
                 TOTAL PUMP DISCHARGE: Qpump = Qnet + Qin

Step-by-Step Drawdown Procedure

1. Step 1 — Measure Inflow Rate ($Q_{\text{in}}$):

   • Turn all pumps OFF.
   • Record the time ($t_{\text{in}}$ in minutes) it takes for incoming sewage to raise the wet well water level by a measured distance ($\Delta H_{\text{in}}$ in feet).
   • Calculate inflow volume: $V_{\text{in}} = \text{Area} \times \Delta H_{\text{in}} \times 7.48$.
   • Calculate inflow rate: $Q_{\text{in}} = V_{\text{in}} / t_{\text{in}}$ (in $\text{gpm}$).

2. Step 2 — Measure Net Pumping Rate ($Q_{\text{net}}$):

   • Turn the test pump ON (with inflow still occurring).
   • Record the time ($t_{\text{pump}}$ in minutes) required for the pump to draw down the water level by a measured distance ($\Delta H_{\text{drop}}$ in feet).
   • Calculate net volume pumped: $V_{\text{drop}} = \text{Area} \times \Delta H_{\text{drop}} \times 7.48$.
   • Calculate net drawdown rate: $Q_{\text{net}} = V_{\text{drop}} / t_{\text{pump}}$ (in $\text{gpm}$).

3. Step 3 — Compute Total Pumping Capacity ($Q_{\text{pump}}$): $\mathbf{Q_{\text{pump}} = Q_{\text{net}} + Q_{\text{in}} = \left(\frac{V_{\text{drop}}}{t_{\text{pump}}}\right) + Q_{\text{in}}}$

Class I Exam Trap: Never calculate pump output by simply dividing the drawdown volume by time. If you ignore the inflow entering during the test, you will significantly underestimate pump capacity.

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Pump Cycle Time & Starts per Hour

Frequent motor starts cause excessive heat buildup and premature motor winding failure. Designers typically limit lift station pump starts to $4\text{ to }6\text{ starts per hour}$.

• Fill Time ($t_{\text{fill}}$): Time required for incoming flow ($Q_{\text{in}}$) to fill the operating volume ($V_{\text{op}}$) between lead-off and lead-on float levels: $t_{\text{fill}} = \frac{V_{\text{op}}}{Q_{\text{in}}}$
• Pump Down Time ($t_{\text{pump}}$): Time required for the pump to evacuate the operating volume while inflow continues: $t_{\text{pump}} = \frac{V_{\text{op}}}{Q_{\text{pump}} - Q_{\text{in}}}$
• Total Operating Cycle Time ($T$): $T = t_{\text{fill}} + t_{\text{pump}}$
• Pump Starts per Hour: $\text{Starts per Hour} = \frac{60\text{ minutes/hour}}{T\text{ (minutes)}}$

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Horsepower, Efficiency & Energy Math

Pumping wastewater requires mechanical energy to overcome elevation differences (static head) and pipe friction (friction head), collectively known as Total Dynamic Head (TDH).

                 PUMP HORSEPOWER CONVERSION CASCADE

    Electrical Grid Input          Motor Output / Shaft          Liquid Work Done
    +--------------------+        +--------------------+        +--------------------+
    |  Motor Horsepower  | ------>|  Brake Horsepower  | ------>|  Water Horsepower  |
    |       (MHP)        | η_motor|       (BHP)        |  η_pump|       (WHP)        |
    +--------------------+        +--------------------+        +--------------------+
      MHP = BHP / η_motor           BHP = WHP / η_pump           WHP = (Q × TDH)/3960

1. Water Horsepower (WHP)

Water horsepower is the actual theoretical power imparted directly to the wastewater stream:

$\mathbf{WHP = \frac{Q_{\text{gpm}} \times \text{TDH}_{\text{ft}}}{3,960}}$

(The constant $3,960$ is derived from $33,000\text{ ft-lbs/min per HP} / 8.34\text{ lbs/gal}$.)

2. Brake Horsepower (BHP)

Brake horsepower is the actual mechanical horsepower delivered by the electric motor shaft to the pump impeller, accounting for hydraulic and friction losses inside the pump casing (pump efficiency $\eta_{\text{pump}}$):

$\mathbf{BHP = \frac{\text{WHP}}{\eta_{\text{pump}}} = \frac{Q_{\text{gpm}} \times \text{TDH}_{\text{ft}}}{3,960 \times \eta_{\text{pump}}}}$

3. Motor Horsepower (MHP) / Wire-to-Water Efficiency

Motor horsepower represents the total electrical power drawn from the electrical grid, accounting for motor electrical losses (motor efficiency $\eta_{\text{motor}}$):

$\mathbf{MHP = \frac{\text{BHP}}{\eta_{\text{motor}}} = \frac{\text{WHP}}{\eta_{\text{pump}} \times \eta_{\text{motor}}} = \frac{\text{WHP}}{\eta_{\text{wire-to-water}}}}$

Where: $\eta_{\text{wire-to-water}} = \eta_{\text{pump}} \times \eta_{\text{motor}}$

4. Electrical Power Consumption & Cost

• Kilowatt Demand ($\text{kW}$): $\text{kW} = \text{BHP} \times 0.746 \quad \text{or} \quad \text{kW} = \frac{\text{MHP} \times 0.746}{\eta_{\text{motor}}}$
• Daily Electrical Energy Consumption ($\text{kWh/day}$): $\text{kWh/day} = \text{kW} \times \text{Hours Operated per Day}$
• Daily Electrical Cost: $\text{Daily Cost} = \text{kWh/day} \times \text{Electrical Rate per kWh}$

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Step-by-Step Worked Lift Station Problems

Problem 1: Volumetric Drawdown Test

Question: A cylindrical wet well has a diameter of $10.0\text{ ft}$. With the lead pump operating for $5.0\text{ minutes}$, the water level drops exactly $2.50\text{ ft}$. A prior test established that raw sewage continuously flows into the wet well at $120\text{ gpm}$. Calculate the total pumping capacity of the lead pump in $\text{gpm}$.

• Step 1: Calculate wet well cross-sectional area: $A = 0.7854 \times D^2 = 0.7854 \times (10.0\text{ ft})^2 = 78.54\text{ sq ft}$
• Step 2: Calculate net volume drawn down: $V_{\text{drop}} = 78.54\text{ sq ft} \times 2.50\text{ ft} \times 7.48\text{ gal/cu ft} = 196.35\text{ cu ft} \times 7.48 = 1,468.7\text{ gallons}$
• Step 3: Calculate net pumping rate ($Q_{\text{net}}$): $Q_{\text{net}} = \frac{1,468.7\text{ gallons}}{5.0\text{ minutes}} = 293.7\text{ gpm}$
• Step 4: Calculate total pump capacity ($Q_{\text{pump}} = Q_{\text{net}} + Q_{\text{in}}$): $Q_{\text{pump}} = 293.7\text{ gpm} + 120.0\text{ gpm} = \mathbf{413.7\text{ gpm}} \approx \mathbf{414\text{ gpm}}$

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Problem 2: Water Horsepower Calculation

Question: A submersible lift station pump delivers $500\text{ gpm}$ against a total dynamic head (TDH) of $60.0\text{ feet}$. Calculate the Water Horsepower ($\text{WHP}$).

• Step 1: Apply WHP formula: $\text{WHP} = \frac{Q_{\text{gpm}} \times \text{TDH}_{\text{ft}}}{3,960} = \frac{500\text{ gpm} \times 60.0\text{ ft}}{3,960} = \frac{30,000}{3,960} = \mathbf{7.58\text{ WHP}} \approx \mathbf{7.6\text{ WHP}}$

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Problem 3: Lift Station Cycle Time

Question: A wet well has an operating volume of $3,000\text{ gallons}$ between the lead pump start float and stop float. If sewage enters at a constant rate of $150\text{ gpm}$ and the pump capacity is $450\text{ gpm}$, calculate the total cycle time in minutes.

• Step 1: Calculate fill time ($t_{\text{fill}}$): $t_{\text{fill}} = \frac{3,000\text{ gal}}{150\text{ gpm}} = 20.0\text{ minutes}$
• Step 2: Calculate pump drawdown time ($t_{\text{pump}}$): $t_{\text{pump}} = \frac{3,000\text{ gal}}{450\text{ gpm} - 150\text{ gpm}} = \frac{3,000\text{ gal}}{300\text{ gpm}} = 10.0\text{ minutes}$
• Step 3: Calculate total cycle time ($T$): $T = 20.0\text{ min} + 10.0\text{ min} = \mathbf{30.0\text{ minutes}}$
• Step 4: Check motor starts per hour: $\text{Starts per Hour} = \frac{60\text{ min/hr}}{30.0\text{ min/cycle}} = \mathbf{2\text{ starts/hour}} \quad (\text{Well within safe 4 to 6 limit})$