A piston type pump is operating at 2 gallons per stroke (7.57 liters per stroke) and 50 strokes per minute. If the influent flow is 1.0 MGD (3,785 LPD), the influent suspended solids is 200 mg/L and primary effluent suspended solids is 100 mg/L, how many minutes will the pump need to run each day to remove the 4% sludge that accumulates?

To solve this problem, we will determine how many minutes per day the piston pump needs to run to remove the 4% sludge that accumulates, given the influent and effluent conditions.

### Step-by-step breakdown:

1. **Influent Flow Rate**:
   - The influent flow rate is \( 1.0 \, \text{MGD} \) (million gallons per day), which is equivalent to:
     \[
     1.0 \, \text{MGD} = 1,000,000 \, \text{gallons per day} = 3,785,000 \, \text{liters per day}.
     \]

2. **Suspended Solids in the Influent and Effluent**:
   - Influent suspended solids concentration: \( 200 \, \text{mg/L} \).
   - Effluent suspended solids concentration: \( 100 \, \text{mg/L} \).

3. **Sludge Accumulation**:
   - The sludge consists of the difference in suspended solids between the influent and the effluent.
   - The amount of solids removed from the influent per liter of wastewater treated is:
     \[
     \text{Solids removed per liter} = (200 \, \text{mg/L} - 100 \, \text{mg/L}) = 100 \, \text{mg/L}.
     \]
   - The total amount of solids removed per day is:
     \[
     \text{Total solids removed per day} = 100 \, \text{mg/L} \times 3,785,000 \, \text{L/day} = 378,500,000 \, \text{mg/day} = 378.5 \, \text{kg/day}.
     \]

4. **4% Sludge Concentration**:
   - The removed solids are concentrated into 4% sludge. This means the sludge contains 4% solids by weight.
   - The total mass of solids removed per day is 378.5 kg.
   - The volume of sludge (with 4% solids) needed to hold this mass can be calculated as:
     \[
     \text{Volume of sludge per day} = \frac{\text{Mass of solids removed}}{0.04} = \frac{378.5 \, \text{kg/day}}{0.04} = 9462.5 \, \text{kg/day}.
     \]
   - Since \( 1 \, \text{kg} \) of water weighs \( 1 \, \text{liter} \), this is equivalent to \( 9462.5 \, \text{liters/day} \).

5. **Piston Pump Operation**:
   - The piston pump operates at 7.57 liters per stroke and 50 strokes per minute.
   - The flow rate of the pump is:
     \[
     \text{Flow rate of the pump} = 7.57 \, \text{L/stroke} \times 50 \, \text{strokes/minute} = 378.5 \, \text{L/minute}.
     \]
   - The time required to pump \( 9462.5 \, \text{liters/day} \) is:
     \[
     \text{Time per day} = \frac{9462.5 \, \text{liters}}{378.5 \, \text{L/min}} = 25 \, \text{minutes/day}.
     \]

### Conclusion:
The piston pump will need to run for approximately **25 minutes per day** to remove the 4% sludge that accumulates.

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Would you like further details on any step or have questions about this solution?

Here are 5 related questions for further understanding:
1. How does the concentration of solids in the sludge affect the required pump runtime?
2. What would happen if the pump operated at a lower stroke volume?
3. How would increasing the number of strokes per minute impact the overall daily runtime?
4. What factors could affect the suspended solids concentration in influent and effluent?
5. How does the efficiency of solid removal vary with changes in influent flow rate?

**Tip**: Regular maintenance of pumps ensures accurate flow rates and prevents inefficiencies in sludge removal operations.

Learn how to calculate the daily runtime of a piston pump to remove 4% sludge accumulation, given influent and effluent suspended solids concentrations, influent flow rate, and pump operation parameters. This step-by-step guide covers key concepts such as fluid dynamics, rate calculations, and proportional relationships. Suitable for advanced high school or college students.

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