Storm water pumping problem

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You are trying to pump storm water out of your basement during a storm. The pump can extract 27.5 gpm. The water level in the basement is now sinking by about 4 in./hr.What is the flow rate (gpm) from the storm into the basement? The basement is 30 ft 3 20 ft.

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This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. PURPOSE OF THE EXERCISE:
    Demonstrate the application of a control-volume material balance over a flooded space by relating the pump extraction rate, the rate of water-level change, and the storm water inflow rate.

  2. ENGINEERING CONTEXT:
    Basement flood control requires balancing the pump’s extraction capacity with unintended inflow through openings or cracks; solving this problem shows how an engineer can assess if the pump is sufficiently sized to overcome storm water infiltration.

  3. PHYSICAL CONTEXT:
    The dominant effect is the conservation of mass in the control volume defined by the basement, which has a uniform water level over its floor.

  4. ASSUMPTIONS:
    Assume steady, uniform water depth across the basement and constant inflow rate during the period considered; these assumptions are valid provided the basement geometry is simple and changes occur slowly relative to the measurement timescale.

  5. SOLUTION TECHNIQUE:
    Define the basement floor area as

A = 30~\mathrm{ft} \times 20~\mathrm{ft} = 600~\mathrm{ft^2}\,.

A water-level drop of 4 in/hr corresponds to

\frac{4~\mathrm{in}}{12~\mathrm{in/ft}} = 0.333~\mathrm{ft/hr}\,.

Thus, the rate of volume decrease is

\frac{dV}{dt} = 600~\mathrm{ft^2} \times 0.333~\mathrm{ft/hr} = 200~\mathrm{ft^3/hr}\,.

Convert this to gallons per minute (with 1~\mathrm{ft^3}=7.4805~\mathrm{gal} and 1~\mathrm{hr}=60~\mathrm{min}):

\frac{200~\mathrm{ft^3/hr}}{60} \times 7.4805 \approx 24.93~\mathrm{gpm}\,.

Let Q_{\mathrm{in}} be the storm inflow rate and Q_{\mathrm{pump}}=27.5~\mathrm{gpm} be the pump extraction; applying conservation of volume gives

\frac{dV}{dt} = Q_{\mathrm{in}} - Q_{\mathrm{pump}}\,.

Since the water volume is decreasing, dV/dt is negative when written as an outflow. In terms of absolute rates,

Q_{\mathrm{pump}} - Q_{\mathrm{in}} \approx 24.93~\mathrm{gpm}\,.

Thus, solving for the storm inflow rate yields

Q_{\mathrm{in}} \approx Q_{\mathrm{pump}} - 24.93 \approx 27.5 - 24.93 \approx 2.57~\mathrm{gpm}\,.
  1. REFLECTION:
    The analysis indicates that despite the pump’s capacity of 27.5 gpm, the inflow from the storm is only about 2.57 gpm. This outcome shows that the net water-level drop is primarily due to the pump extracting more water than is entering. The result guides the design and evaluation of basement flood mitigation systems by quantifying the effective storm water inflow.

In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 6 Fundamentals: Combining the language of fluid mechanics with the language of hydraulics en… (RELEVANT)
  2. Chapter 7.5: When to use Integral Conservation (RELEVANT)
  3. Chapter 7.2: Integral Conservation of Mass (RELEVANT)
  4. Chapter 2.1: Turbomachines Part 3 Part 1 (RELEVANT)
  5. Chapter 6.2: Modified Bernoulli equation Part 1 (RELEVANT)

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