/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 You are trying to pump storm wat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 30

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. \(/ \mathrm{hr}\). What is the flow rate ( gpm) from the storm into the basement? The basement is \(30 \mathrm{ft} \times 20 \mathrm{ft}\).

Short Answer

Expert verified
The flow rate of storm water into the basement is equal to the amount of water pumped out per minute plus 27.5 gpm.

Step by step solution

01

Determine the Volume of Water Reduced

First let's find out how much water is being lowered in the basement every hour. Since the water level is dropping 4 inches per hour, the volume lowered by the pump per hour is \(4 \frac{\text{in}}{\text{hr}} \times 30 \text{ft} \times 20 \text{ft}\). But, since feet and inches aren't in the same units, we need to convert 4 inches to feet which result in \(\frac{4}{12}\) feet. Hence the volume of water pumped per hour is \(\frac{4}{12} \text{ft/hr} \times 30 \text{ft} \times 20 \text{ft}\).
02

Convert Volume to Gallons per Minute

The volume calculated is in cubic feet per hour, whereas we need the result in gallons per minute. Hence, we perform the following conversions: To change the cubic feet to gallons, we use the conversion 1 ft^3 = 7.48052 gallons. To change hours to minutes, we divide by 60. So, \( \text{Volume Pumped in Gallons/min} = \left(\frac{4}{12} \frac{\text{ft}}{\text{hr}} \times 30 \text{ft} \times 20 \text{ft}\right) \times 7.48052 \frac{\text{gallons}}{\text{ft}^3} \times \frac{1}{60} \frac{\text{hr}}{\text{min}} \).
03

Determine the Flow Rate

The flow rate of the storm water into the basement can be identified by adding the rate at which the water level is decreasing (calculated in Step 2) to the pump's extraction rate of 27.5 gpm. Therefore, \(\text{Flow Rate} = \text{Volume Pumped in Gallons/min} + 27.5 \text{gpm} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Water is discharged at a flow rate of \(0.3 \mathrm{m}^{3} / \mathrm{s}\) from a narrow slot in a 200 -mm-diameter pipe. The resulting horizontal twodimensional jet is \(1 \mathrm{m}\) long and \(20 \mathrm{mm}\) thick, but of nonuniform velocity, the velocity at location (2) is twice that at location (1). The pressure at the inlet section is \(50 \mathrm{kPa}\) (gage). Calculate (a) the velocity in the pipe and at locations (1) and (2) and (b) the forces required at the coupling to hold the spray pipe in place. Neglect the mass of the pipe and the water it contains.

Consider the steady adiabatic flow of air through a long straight pipe with \(0.05 \mathrm{m}^{2}\) cross-sectional area. At the inlet, the air is at \(200 \mathrm{kPa}\) (gage), \(60^{\circ} \mathrm{C}\), and has a velocity of \(150 \mathrm{m} / \mathrm{s}\). At the exit, the air is at \(80 \mathrm{kPa}\) and has a velocity of \(300 \mathrm{m} / \mathrm{s}\). Calculate the axial force of the air on the pipe. (Be sure to make the direction clear.)

A cylindrical tank, \(0.3 \mathrm{m}\) in diameter, drains through a hole in its bottom. At the instant when the water depth is \(0.6 \mathrm{m}\) the flow rate from the tank is observed to be \(4 \mathrm{kg} / \mathrm{s}\). Determine the rate of change of water level at this instant.

A block of copper of mass 5 kg is heated to \(90^{\circ} \mathrm{C}\) and then plunged into an insulated container containing 4 L of water at \(10^{\circ} \mathrm{C}\). Find the final temperature of the system. For copper, the specific heat is \(385 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K},\) and for water the specific heat is \(4186 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\).

A pipe branches symmetrically into two legs of length \(L,\) and the whole system rotates with angular speed \(\omega\) around its axis of symmetry. Each branch is inclined at angle \(\alpha\) to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate \(Q\). The pipe diameter, \(D,\) is much smaller than \(L\) Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration \(\dot{\omega} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Astrophysics

Read Explanation

Oscillations

Read Explanation

Space Physics

Read Explanation

Mechanics and Materials

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.