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Chapter 5: Problem 24

Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is \(1500 \mathrm{ft}^{2} .\) What capacity (gal/min) pump would you rent to (a) keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, and (b) reduce the water accumulation in your basement at a rate of 3 in./hr even while the backup problem exists?

Short Answer

Expert verified
(a) Rent a pump with at least 16 gal/min capacity. (b) Rent a pump with 63 gal/min capacity.

Step by step solution

01

Calculate the volume of water entering the basement per hour

Since the water depth increases by 1 inch per hour, we first calculate the volume of water entering the basement per hour in cubic feet. The volume of water entering the basement every hour is given by the floor area multiplied by the rise in water depth:\[\text{Volume per hour} = 1500 \, \text{ft}^2 \times \frac{1}{12} \, \text{ft} = 125 \, \text{ft}^3\]
02

Convert the volume from cubic feet to gallons

To find the volume in gallons, we use the conversion factor where 1 cubic foot is equivalent to approximately 7.48 gallons:\[\text{Volume in gallons} = 125 \, \text{ft}^3 \times 7.48 \, \text{gallons/ft}^3 = 935 \, \text{gallons} \]
03

Determine the pump capacity needed to keep the water level constant

To keep the water level constant, the pump needs to remove water at the same rate it is entering. Therefore, the required pump capacity is:\[\text{Capacity needed in gal/h} = 935 \, \text{gallons/hour}\]Convert this to gallons per minute (gal/min):\[\text{Capacity in gal/min} = \frac{935 \, \text{gallons/hour}}{60 \, \text{min/hour}} \approx 15.58 \, \text{gal/min}\]
04

Calculate the pump capacity to reduce water by 3 in/hr

The basement floods 1 in/hr, but the goal is to remove enough water to make the water depth drop by 3 in/hr, so the total removal rate should handle a 4 in/hr change.Find the additional volume of water to be removed:\[\text{Additional volume per hour} = 1500 \, \text{ft}^2 \times \frac{3}{12} \, \text{ft} = 375 \, \text{ft}^3\]Convert the additional volume to gallons:\[\text{Additional volume in gallons} = 375 \, \text{ft}^3 \times 7.48 \, \text{gallons/ft}^3 = 2805 \, \text{gallons}\]The total pumping capacity needed:\[\text{Total capacity in gal/h} = 935 + 2805 = 3740 \, \text{gallons/hour}\]Convert to gallons per minute:\[\text{Total capacity in gal/min} = \frac{3740 \, \text{gallons/hour}}{60 \, \text{min/hour}} \approx 62.33 \, \text{gal/min}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Storm Water Management
Storm water management involves controlling and using rainwater wisely to prevent flooding and pollution. When heavy storms occur, excess water can overwhelm sewer systems, leading to flooding, much like a situation where a basement starts to fill up with water. Proper storm water management is essential:
  • It prevents flooding and water damage by directing excess rainwater away from buildings.
  • Helps maintain clean water by preventing pollutants from washing into rivers and lakes.

In the given problem of a storm sewer backup flooding a basement, managing water inflow is essential. While the immediate focus is on selecting the right pump capacity to control water levels, it demonstrates the broader principle of controlling water flow to mitigate damage. Understanding the rate of water inflow and having appropriate mechanisms, like pumps, is crucial in managing storm water effectively.
Unit Conversion in Fluid Mechanics
Unit conversion is an essential skill in fluid mechanics because calculations often involve various units of measure. Converting units accurately ensures that engineers make correct calculations regarding fluid flow, volume, and pressure. In our basement flooding example, unit conversion plays a pivotal role.
Initially, the volume of water is calculated in cubic feet. To be more useful in practical applications like choosing a pump, this volume needs to be converted to gallons. The conversion factor used is that 1 cubic foot equals approximately 7.48 gallons. This conversion allows for determining the appropriate pump capacity in gallons per minute, a more practical unit for this context.
Accurate unit conversions let you:
  • Ensure compatibility with equipment specifications, like pumps that rate capacity in gallons per minute.
  • Compare different scenarios and solutions efficiently.
Pump Capacity Calculation
Calculating pump capacity is crucial when dealing with fluid dynamics, ensuring that a pump can handle the amount of water that needs to be moved. In the problem at hand, two scenarios are considered for calculating pump capacity: maintaining the current water level and reducing it despite ongoing water accumulation.
To find the pump capacity:
  • Determine the inflow rate of water that needs to be managed. In this exercise, it's calculated using the basement's surface area and the water depth change per hour.
  • Translate these measurements into pump requirements, considering how many gallons need to be pumped per hour and then converting this to gallons per minute for practical use.
For the scenario where extra pumping capacity is needed to reduce the water level by another 3 inches per hour, understanding these calculations ensures that a suitable pump is selected to meet demands. This means calculating the total pumping requirement so that it exceeds just balancing the inflow, effectively managing and reducing water levels even during continuous entry.

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Most popular questions from this chapter

Water enters an axial-flow turbine rotor with an absolute velocity tangential component, \(V_{\theta}\), of \(15 \mathrm{ft} / \mathrm{s}\). The corresponding blade velocity, \(U,\) is \(50 \mathrm{ft} / \mathrm{s}\). The water leaves the rotor blade row with no angular momentum. If the stagnation pressure drop across the turbine is 12 psi, determine the hydraulic efficiency of the turbine.

The exit plane of a 0.20 -m-diameter pipe is partially blocked by a plate with a hole in it that produces a 0.10 -m-diameter stream as shown in Fig. PS.53. The water velocity in the pipe is \(5 \mathrm{m} / \mathrm{s}\). Gravity and viscous effects are negligible. Determine the force needed to hold the plate against the pipe.

A small fan moves air at a mass flow rate of \(0.004 \mathrm{lbm} / \mathrm{s}\) Upstream of the fan, the pipe diameter is 2.5 in., the flow is laminar, the velocity distribution is parabolic, and the kinetic energy coefficient, \(\alpha_{1},\) is equal to \(2.0 .\) Downstream of the fan, the pipe diameter is 1 in., the flow is turbulent, the velocity profile is quite flat, and the kinetic energy coefficient, \(\alpha_{2}\), is equal to 1.08 . If the rise in static pressure across the fan is 0.015 psi and the fan shaft draws 0.00024 hp, compare the value of loss calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions.

A static thrust stand is to be designed for testing a specific jet engine. Knowing the following conditions for a typical test, \\[\begin{aligned}\text { intake air velocity } &=700 \mathrm{ft} / \mathrm{s} \\\\\text { exhaust gas velocity } &=1640 \mathrm{ft} / \mathrm{s} \\\\\text { intake cross section area } &=10 \mathrm{ft}^{2} \\\\\text { intake static pressure } &=11.4 \mathrm{psia} \\\\\text { intake static temperature } &=480^{\circ} \mathrm{R} \\ \text { exhaust gas pressure } &=0 \mathrm{psi}\end{aligned}\\] estimate a nominal thrust to design for.

A fire hose nozzle is designed to deliver water that will rise \(40 \mathrm{m}\) vertically. Calculate the stagnation pressure required at the nozzie inlet if (a) no loss is assumed, (b) a loss of \(30 \mathrm{N} \cdot \mathrm{m} / \mathrm{kg}\) is assumed.
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