/*! 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 39 Oil having a density of \(\rho_{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Oil having a density of \(\rho_{o}=880 \mathrm{~kg} / \mathrm{m}^{3}\) and a viscosity of \(\mu_{o}=0.0680 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) flows through the 20 -mm-diameter pipe at \(0.001 \mathrm{~m}^{3} / \mathrm{s}\). Determine the reading \(h\) of the mercury manometer. Take \(\rho_{\mathrm{Hg}}=13550 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
Calculate the fluid velocity using the given flow rate and pipe diameter. Apply Bernoulli's equation to calculate \(P_{2}\). Then, equate this value to \(\Delta P = \rho_{\mathrm{Hg}} g h\) to find the value of \(h\), the manometer reading.

Step by step solution

01

Understand the Problem and Define Notations

First, understand that we're working with a pressure measurement scenario where fluid is flowing through a pipe, and a mercury manometer is used to measure the pressure difference. Define the following notations: \(\rho_{o}=880 \mathrm{~kg} / \mathrm{m}^{3}\) (oil density), \(\mu_{o}=0.0680 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) (oil viscosity), oil flow rate (\(Q=0.001 \mathrm{~m}^{3} / \mathrm{s}\)), pipe diameter (\(D=20 \mathrm{~mm}\)), and \(\rho_{\mathrm{Hg}}=13550 \mathrm{~kg} / \mathrm{m}^{3}\) (mercury density). The aim is to find 'h', the manometer reading.
02

Calculate Fluid Velocity

The velocity (\(V\)) of the fluid (oil) can be calculated using the continuity equation which is derived by equating the volume flow rate before and after a certain point in a fluid flow. Mathematically, \(V = \frac{Q}{A}\), where \(A = \frac{\pi D^{2}}{4}\) is the cross-sectional area of the pipe. Substitute given values to get the fluid velocity.
03

Apply Bernoulli's Equation

At the oil-mercury interface in the manometer, the total energy (pressure + kinetic energy + potential energy) is equal due to the incompressible and steady flow of the fluid. Consequently, Bernoulli's principle allows us to equate the pressure at Point 1 (where oil enters the pipe - \(P_{1}\)) and pressure at Point 2 (oil-mercury interface - \(P_{2}\)). The Bernoulli's equation is \(P_{1} + \frac{1}{2} \rho_{o} V^{2} = P_{2}\) (since the pipe is horizontal, potential energy term is disregarded). Simplify for \(P_{2}\).
04

Calculate Differential Pressure

The pressure difference (\(P_{2} - P_{1}\)) can also be expressed in terms of the height of the mercury column in the manometer. This is given by \(\Delta P = \rho_{\mathrm{Hg}} g h\) where \(h\) is the manometer reading we aim to find and \(g = 9.81 m/s^{2}\) is the acceleration due to gravity. Setting \(\Delta P = P_{2} - P_{1}\), we can solve for \(h\).

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.

Bernoulli's Equation
In fluid mechanics, Bernoulli's Equation is fundamental for understanding fluid dynamics, especially for incompressible and steady flows. This equation states that the total mechanical energy along a streamline (a path followed by fluid particles) is constant. It takes into account:
  • Pressure energy
  • Kinetic energy
  • Potential energy
For our problem, wherein oil flows through a pipe, the relevant form of Bernoulli's Equation applied is:\[ P_1 + \frac{1}{2} \rho_o V^2 = P_2 \]This equation allows us to compare the changes in pressure due to fluid motion between two points. In this scenario:
  • At Point 1, oil enters the horizontal pipe with a pressure \(P_1\).
  • At Point 2, the oil meets the mercury surface, providing pressure \(P_2\).
Potential energy changes are ignored as the pipe is horizontal, simplifying our calculations to focus on pressure and kinetic energy. Bernoulli’s principle bridges these energy changes to compute the unknown pressure difference measured by the manometer.
Continuity Equation
The Continuity Equation is essential in expressing the conservation of mass in fluid dynamics. It implies that the mass flow rate must remain constant from one cross-section of the pipe to another, assuming incompressibility and no fluid loss. This is formulated as:\[ Q = A_1 V_1 = A_2 V_2 \]Where:
  • \(Q\) is the volumetric flow rate (constant along the pipe)
  • \(A\) is the cross-sectional area
  • \(V\) is the fluid velocity
For our exercise, the flow rate \(Q = 0.001 \, \text{m}^3/\text{s}\) remains unchanged as the oil flows through the pipe. By knowing the diameter of the pipe \(D = 20 \, \text{mm}\), we find the area:\[ A = \frac{\pi D^2}{4} \]Inserting this area into our Continuity Equation allows us to solve for the fluid velocity \(V\). This velocity is crucial when we later apply Bernoulli's principle to understand the pressure difference observed by the manometer.
Manometer
A manometer is an instrument used to measure the pressure difference between two points in a fluid. In this exercise, a mercury manometer detects the pressure difference caused by oil flow. Understanding how it works is vital for calculating necessary measurements.The principle behind the manometer relies on balance between columns of liquid (here, mercury) in different arms of a U-tube. The difference in height \(h\) of the liquid columns represents the differential pressure \(\Delta P\) between the two points of interest. This can be expressed as:\[ \Delta P = \rho_{\text{Hg}} g h \]Where:
  • \(\rho_{\text{Hg}}\) is the mercury density
  • \(g\) is the gravitational acceleration (\(9.81 \, \text{m/s}^2\))
  • \(h\) is the height difference in columns captured
The calculation of pressure difference through this equation allows us to connect the measurable manometer reading directly to our understanding of fluid dynamics in the system.
Fluid Density
Fluid Density is a key property that affects fluid behavior and pressure relationships. In this exercise, two fluids are considered: oil and mercury.
  • Oil has a density of \(\rho_o = 880 \, \text{kg/m}^3\).
  • Mercury has a much higher density of \(\rho_{\text{Hg}} = 13550 \, \text{kg/m}^3\).
Density (\(\rho\)) indicates how much mass is contained within a certain volume:\[\rho = \frac{m}{V}\]In applications like our exercise, understanding the different densities of the fluids involved allows us to anticipate how the pressure levels might vary. High-density mercury in the manometer results in a significant shift in equilibrium height \((h)\), directly representing the pressure difference created by oil flow. This relationship is remarkable for translating flow conditions into a measurable difference in manometer readings.

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

Determine the plate's terminal velocity as it slides along the surface of the belt, if the belt is moving at a constant rate of \(2 \mathrm{~mm} / \mathrm{s}\). The plate of mass \(2.4 \mathrm{~kg}\) is resting on an 0.4 -mm- thick film of oil between the belt and surface, whereas oil between the top of the plate and the belt is 0.6 \(\mathrm{mm}\) thick. Assume the velocity profile is linear. Take \(\rho_{o}=\) \(900 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\mu_{o}=0.0675 \mathrm{~N} . \mathrm{s} / \mathrm{m}^{2}\).

=Crude oil is flowing vertically upward through a 50-mm-diameter pipe. If the difference in pressure between two points \(3 \mathrm{~m}\) apart along the pipe is \(26.4 \mathrm{kPa}\), determine the volumetric flow. Take \(\rho_{o}=880 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\mu_{o}=30.2\left(10^{-3}\right) \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\).

The water and oil films have the same thickness \(a\) and are subjected to the movement of the top plate. Plot the velocity profile and the shear-stress distribution for each fluid. There is no pressure gradient between \(A\) and \(B\). The viscosities of water and oil are \(\mu_{w}\) and \(\mu_{o},\) respectively.

When you inhale, air flows through the turbinate bones of your nasal passages as shown. Assume that for a short length of \(15 \mathrm{~mm}\), the flow is passing through parallel plates, the plates having a mean total width of \(w=20 \mathrm{~mm}\) and spacing of \(a=1 \mathrm{~mm}\). If the lungs produce a pressure drop of \(\Delta p=50 \mathrm{~Pa},\) and the air has a temperature of \(20^{\circ} \mathrm{C}\), determine the power needed to inhale air.

A pipe of \(60 \mathrm{~mm}\) diameter is used to drain out an engine oil from a tank at the rate of \(0.012 \mathrm{~m}^{3} / \mathrm{s}\). Determine the horizontal force required by the tank to exert on the pipe to hold it in place. Assume fully developed flow occurs along the pipe. Take \(\rho_{o}=900 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\mu_{o}=0.37 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\).

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.