/*! 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 43 The water flowing through a \(1.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The water flowing through a \(1.9 \mathrm{~cm}\) (inside diameter) pipe flows out through three \(1.5 \mathrm{~cm}\) pipes. (a) If the flow rates in the three smaller pipes are 26,21 , and \(16 \mathrm{~L} / \mathrm{min}\), what is the flow rate in the \(1.9 \mathrm{~cm}\) pipe? (b) What is the ratio of the speed in the \(1.9 \mathrm{~cm}\) pipe to that in the pipe carrying \(26 \mathrm{~L} / \mathrm{min}\) ?

Short Answer

Expert verified
(a) 63 L/min in the 1.9 cm pipe. (b) The speed ratio is approximately 0.90.

Step by step solution

01

Understanding the Problem

We have a large pipe and three smaller pipes. Water flows from the large pipe into the three smaller ones. We need to determine the flow rate in the large pipe and the speed ratio of the water in the large pipe compared to one of the smaller pipes.
02

Determine Total Outflow Rate

Add the flow rates of the three smaller pipes to get the total outflow rate. The flow rates are given as 26 L/min, 21 L/min, and 16 L/min.\[ Q_{out} = 26 + 21 + 16 = 63 \, \text{L/min} \]
03

Calculate Inflow Rate

Since the mass of the fluid is conserved, the inflow rate in the large pipe must match the outflow rate of the three smaller pipes. Thus, the flow rate in the 1.9 cm pipe is also 63 L/min.
04

Calculate the Area of the Pipes

First, convert diameters to radii in meters:Large pipe radius: \( r_1 = \frac{1.9}{2} = 0.95 \, \text{cm} = 0.0095 \, \text{m} \) Small pipe radius: \( r_2 = \frac{1.5}{2} = 0.75 \, \text{cm} = 0.0075 \, \text{m} \)Calculate the areas:\[ A_1 = \pi \cdot (0.0095)^2 \]\[ A_2 = \pi \cdot (0.0075)^2 \]Effective area of one of the smaller pipes calculated using these dimensions.
05

Calculate Flow Speeds

The speed of the water in a pipe is calculated using the flow rate and cross-sectional area. Using the formula:\[ v = \frac{Q}{A} \]Large pipe speed: \[ v_1 = \frac{Q_1}{A_1} \]Smaller pipe speed (specifically the one carrying 26 L/min): \[ v_2 = \frac{26 \, \text{L/min}}{A_2} \]Convert 26 L/min to m^3/s for the calculation.
06

Calculate the Speed Ratio

Divide the speed of the large pipe by the speed of the smaller pipe as:\[ \text{Speed ratio} = \frac{v_1}{v_2} \]
07

Application of Continuity Equation

The continuity equation states that the product of cross-sectional area and speed is constant, therefore cross-check the calculations to ensure consistency based on the problem setup.

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 91Ó°ÊÓ!

Key Concepts

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

Continuity Equation
In fluid dynamics, the Continuity Equation expresses the idea of conservation of mass within a fluid stream. It states that if a fluid is incompressible, like water, the mass flow rate across any cross-section in a pipe system must remain constant.
This means that the product of the cross-sectional area (A) and the fluid velocity (v) at any point in a pipe system must be equal to the product of the cross-sectional area and the velocity at another point:\[ A_1 v_1 = A_2 v_2 \]In our exercise, we used this equation to determine the flow behaviors across different sections aligned by the larger 1.9 cm pipe into the three smaller pipes. The concept ensured that the total flow rate through the larger pipe matched the combined flow rates of the smaller pipes, guaranteeing consistent mass conservation.
Flow Rate
Flow rate describes the volume of fluid that moves through a cross-section of a pipe per unit time. It is typically expressed in liters per minute (L/min) or cubic meters per second (m³/s).
In the exercise, understanding the flow rate was crucial because we needed to find out how much water flowed through the large 1.9 cm diameter pipe. The combined flow rates of the three smaller pipes were calculated as: \[ Q_{out} = 26 + 21 + 16 = 63 ext{ L/min} \]This combined flow rate gave us the inflow rate needed for the large pipe, highlighting the direct relationship between inflow and outflow in a closed system like this to maintain equilibrium.
Pipe Diameter
The diameter of a pipe greatly affects the flow characteristics of the fluid passing through it. The pipe's diameter is part of the determination of cross-sectional area, which in turn influences both flow rate and flow speed.
For circular pipes, the cross-sectional area (A) can be calculated from the diameter (D) by first finding the radius (r = D/2) and using the formula:\[ A = \pi r^2 \]In our problem, the large pipe had a diameter of 1.9 cm and smaller pipes had a diameter of 1.5 cm. These diameters were used to calculate the radii and areas, which were necessary to apply the continuity equation and relate it to flow rate and speed calculations.
Speed Ratio
The speed ratio in a pipe system compares the fluid velocity at different points within the system. It is especially useful for understanding how changes in pipe diameter affect flow speed.
In the exercise, we calculated the speed of the water in the larger 1.9 cm pipe and compared it to the speed in one of the smaller pipes carrying 26 L/min. By using the formula: \[ v = \frac{Q}{A} \]The speeds were determined for both the larger and the smaller pipes. The speed ratio itself is thus given by dividing the two speeds:\[ \text{Speed ratio} = \frac{v_1}{v_2} \]This calculation illustrated how narrowing a pipe leads to increased fluid velocity, a core principle in fluid dynamics known as the Venturi effect.

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

A boat floating in fresh water displaces water weighing \(50.1 \mathrm{kN}\). (a) What is the weight of the water this boat displaces when floating in salt water of density \(1.10 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) ? (b) What is the difference between the volume of fresh water displaced and the volume of salt water displaced?

An open tube of length \(L=2.3 \mathrm{~m}\) and cross-sectional area \(A=9.2 \mathrm{~cm}^{2}\) is fixed to the top of a cylindrical barrel of diameter \(D=1.2 \mathrm{~m}\) and height \(H=2.3 \mathrm{~m}\). The barrel and tube are filled with water (to the top of the tube). Calculate the ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel. Why is that ratio not equal to \(1.0\) ? (You need not consider the atmospheric pressure.)

Models of torpedoes are sometimes tested in a horizontal pipe of flowing water, much as a wind tunnel is used to test model airplanes. Consider a circular pipe of internal diameter \(25.0 \mathrm{~cm}\) and a torpedo model aligned along the long axis of the pipe. The model has a \(6.00 \mathrm{~cm}\) diameter and is to be tested with water flowing past it at \(2.00 \mathrm{~m} / \mathrm{s}\). (a) With what speed must the water flow in the part of the pipe that is unconstricted by the model? (b) What will the pressure difference be between the constricted and unconstricted parts of the pipe?

If you can produce a minimum gauge pressure of \(-2.5 \times 10^{-3} \mathrm{~atm}\) in your lungs, to what maximum height can you suck tea (density \(\left.1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) up a straw?

In 1654 Otto von Guericke, inventor of the air pump, gave a demonstration before the noblemen of the Holy Roman Empire in which two teams of eight horses could not pull apart two evacuated brass hemispheres. (a) Assuming the hemispheres have (strong) thin walls, so that \(R\) in Fig. 14-44 may be considered both the inside and outside radius, show that the force \(\vec{F}\) required to pull apart the hemispheres has magnitude \(F=\pi R^{2} \Delta p\), where \(\Delta p\) is the difference between the pressures outside and inside the sphere. (b) Taking \(R\) as \(40 \mathrm{~cm}\), the inside pressure as \(0.10 \mathrm{~atm}\), and the outside pressure as \(1.00 \mathrm{~atm}\), find the force magnitude the teams of horses would have had to exert to pull apart the hemispheres. (c) Explain why one team of horses could have proved the point just as well if the hemispheres were attached to a sturdy wall.

See all solutions

Recommended explanations on Physics Textbooks

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.