/*! 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 10 A liquid of density \(900 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A liquid of density \(900 \mathrm{~kg} / \mathrm{m}^{3}\) flows through a horizontal pipe that has a cross-sectional area of \(1.80 \times 10^{-2} \mathrm{~m}^{2}\) in region \(A\) and a cross-sectional area of \(9.50 \times 10^{-2} \mathrm{~m}^{2}\) in region \(B\). The pressure difference between the two regions is \(7.20 \times 10^{3} \mathrm{~Pa}\). What are (a) the volume flow rate and (b) the mass flow rate?

Short Answer

Expert verified
Volume flow rate: \(3.24 \times 10^{-2} \, \mathrm{m}^3/\mathrm{s}\); mass flow rate: \(29.16 \, \mathrm{kg/s}\).

Step by step solution

01

Calculate the volume flow rate

To find the volume flow rate (Q), we need to use the principle of continuity, stating that the volume flow rate must be constant in steady flow. Therefore, the equation is \( Av_A = A_2v_2 \). Rearranging for velocity in the smaller section (\( v_A \)), we use Bernoulli's equation: \( P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2 \) where \( \Delta P = P_A - P_B = 7.20 \times 10^3 \, \text{Pa} \). Solving for \( v_A \), we get \( v_A = \sqrt{\frac{2\Delta P}{\rho(\frac{A_B^2}{A_A^2} - 1)}} \). After calculation, \( v_A = 1.80 \, \text{m/s} \). Substitute back to get \( Q = v_A \cdot A_A = 1.80 \, \text{m/s} \cdot 1.80 \times 10^{-2} \, \text{m}^2 = 3.24 \times 10^{-2} \, \text{m}^3/\text{s} \).
02

Calculate the mass flow rate

The mass flow rate \( \dot{m} \) is given by \( \dot{m} = \rho \cdot Q \). The volume flow rate \( Q \) was calculated to be \( 3.24 \times 10^{-2} \, \text{m}^3/\text{s} \) and we know \( \rho = 900 \, \text{kg/m}^3 \). So, the mass flow rate is \( \dot{m} = 900 \, \text{kg/m}^3 \times 3.24 \times 10^{-2} \, \text{m}^3/\text{s} = 29.16 \, \text{kg/s} \).

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
The Continuity Equation is a fundamental concept in fluid dynamics. It is derived from the conservation of mass, signifying that the mass of fluid entering a control volume must equal the mass leaving it in steady flow. In simple terms, this means that in any streamlined flow of an incompressible fluid, the quantity of fluid per unit time is constant across any section of the pipe.
Given by the equation \(A_1 v_1 = A_2 v_2\), where \(A\) is the cross-sectional area and \(v\) is the velocity, it signifies that the product of the area and velocity at any two points along the streamline is equal. This is why if a pipe narrows, the fluid's velocity must increase, and conversely, if it widens, the velocity decreases accordingly. This ensures that flow remains steady and consistent.
The concept is crucial for calculating flow rates and understanding fluid motion within constricted spaces such as pipes or channels. It's particularly useful in this exercise to calculate the fluid's velocity in different sections of the pipe, allowing us to derive other relevant quantities.
Bernoulli's Principle
Bernoulli's Principle is another cornerstone of fluid dynamics, relating the pressure of a fluid to its velocity and height. Formulated as \(P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}\) along a streamline, it describes how fluid velocity changes affect pressure and height.
In simpler terms, it expresses that an increase in the speed of the fluid occurs simultaneously with a decrease in the fluid's potential energy or pressure. In horizontal pipes, where height changes can be ignored, the principle simplifies to show a relationship between velocity and pressure.
  • When fluid speeds up in a narrow pipe, pressure drops.
  • Conversely, in wider sections, velocity decreases and pressure increases.
In our exercise, Bernoulli's equation helps relate the velocities and pressures between two regions in the pipe, verifying the consistency of pressure changes due to velocity differences.
Mass Flow Rate
The concept of Mass Flow Rate is a measure of the mass of fluid passing through a section of a pipe per unit time. Represented as \(\dot{m} = \rho Q\), it combines the fluid's density \(\rho\) and volume flow rate \(Q\). It is crucial because it allows engineers and scientists to quantify the amount of fluid being transported over time, ensuring that systems are designed to handle such flows without failure.
In simple terms, if you know the volume of fluid flowing every second and the density of that fluid, you can calculate how much mass is moving per second. This is essential for evaluating processes involving fluids, such as water delivery systems or fuel lines.
  • Mass flow rate helps in sizing pumps and pipes.
  • It's used to ensure adequate delivery and avoid over-pressurization.
In our exercise, using the mass flow rate formula helped determine that the mass flow rate is 29.16 kg/s, providing vital information about the fluid's motion within the pipe system.

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

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density \(1.30 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The area of each base is \(4.25 \mathrm{~cm}^{2}\), but in one vessel the liquid height is \(0.854 \mathrm{~m}\) and in the other it is \(1.560 \mathrm{~m}\). Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

The maximum depth \(d_{\max }\) that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of \(0.050 \mathrm{~atm}\). What is the difference in \(d_{\max }\) for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of \(1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) )?

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are \(1.50 \mathrm{~L}\), \(2.6 \mathrm{~g} / \mathrm{cm}^{3} ; 0.75 \mathrm{~L}, 1.0 \mathrm{~g} / \mathrm{cm}^{3}\); and \(0.60 \mathrm{~L}, 0.80 \mathrm{~g} / \mathrm{cm}^{3}\). What is the force on the bottom of the container due to these liquids? One liter \(=1 \mathrm{~L}=1000 \mathrm{~cm}^{3}\). (Ignore the contribution due to the atmosphere.)

A hollow sphere of inner radius \(8.0 \mathrm{~cm}\) and outer radius \(9.0 \mathrm{~cm}\) floats half-submerged in a liquid of density \(820 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.

A cylindrical tank with a large diameter is filled with water to a depth \(D=0.30 \mathrm{~m}\). A hole of cross-sectional area \(A=6.2 \mathrm{~cm}^{2}\) in the bottom of the tank allows water to drain out. (a) What is the drainage rate in cubic meters per second? (b) At what distance below the bottom of the tank is the cross-sectional area of the stream equal to one-half the area of the hole?

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.