/*! 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 21 What gauge pressure must a machi... [FREE SOLUTION] | 91Ó°ÊÓ

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What gauge pressure must a machine produce in order to suck mud of density \(1800 \mathrm{~kg} / \mathrm{m}^{3}\) up a tube by a height of \(2.0 \mathrm{~m} ?\)

Short Answer

Expert verified
The required gauge pressure is 35316 Pa.

Step by step solution

01

Understanding the Problem

We need to find the gauge pressure required to lift mud of density \(1800 \, \text{kg/m}^3\) up \(2.0 \, \text{m}\) in a tube. Gauge pressure is the pressure relative to atmospheric pressure.
02

Identify the Formula

We use the formula for hydrostatic pressure, which is the pressure exerted by a fluid column: \[ p = \rho g h \]where \(p\) is the pressure, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the height of the fluid column.
03

Insert Known Values

We substitute the known values into the formula: \[ \rho = 1800 \, \text{kg/m}^3, \quad g = 9.81 \, \text{m/s}^2, \quad h = 2.0 \, \text{m} \]Thus, the pressure required is \[ p = 1800 \times 9.81 \times 2.0 \].
04

Calculate the Pressure

Compute the pressure using the substituted values:\[ p = 1800 \times 9.81 \times 2.0 = 35316 \, \text{Pa} \].
05

Interpret the Result

The gauge pressure needed to lift the mud up the tube by \(2.0 \, \text{m}\) is \(35316 \, \text{Pa}\). This is the pressure that must be applied to overcome the gravitational force on the column of mud.

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Key Concepts

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

Gauge Pressure
In many everyday situations, when we measure pressure, we often talk about gauge pressure. Think of gauge pressure as the difference between the actual pressure and the atmospheric pressure around us. This is particularly important because atmospheric pressure can vary with weather conditions and altitude, and ignoring it can lead to inaccurate calculations.

For instance, if you're checking tire pressure with a gauge, what you read is the gauge pressure. It reflects how much more pressure is inside the tire than the air outside. When applying this concept to lifting fluids, like mud in our exercise, the gauge pressure calculation focuses on how much added pressure is needed beyond what's already exerted by the atmosphere. The key idea is that gauge pressure helps you know exactly the extra pressure required to achieve the desired outcome, such as moving fluid through a tube.
Fluid Density
Fluid density is a fundamental concept when dealing with pressure in fluids. It refers to the mass of the fluid per unit volume and is typically measured in kilograms per cubic meter ( ext{kg/m}^3). The density of a fluid directly affects how much pressure it exerts. In our scenario, the mud we're addressing has a density of 1800 ext{kg/m}^3, which is relatively high compared to many other liquids like water (1000 ext{kg/m}^3).

This higher density means that the mud weighs more for the same amount of volume. Hence, it'll exert a greater force if the same volume is housed in a container. Understanding this concept is crucial for calculating the hydrostatic pressure because density plays a key role. It is the component that varies between different fluids, making each scenario unique based on the fluid involved.
Pressure Calculation
Calculating pressure, especially when considering hydrostatic pressure, involves understanding how different factors interact. In this case, the formula for pressure, \( p = \rho g h \), incorporates:
  • \( \rho \) - the density of the fluid
  • \( g \) - the acceleration due to gravity (approximately 9.81 m/s^2 on Earth)
  • \( h \) - the height of the fluid column
Together, these factors help calculate the total pressure exerted by a fluid at a certain height. When attempting to lift the mud up 2 meters, like in our task, each aspect is essential to compute the needed pressure properly.

Gravity pulls the fluid down, so to lift it, the machine must create a pressure that counters this gravitational force. By substituting the known values into the equation, the pressure calculation provides the needed gauge pressure to just overcome the gravitational pull and move the fluid as intended.

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Most popular questions from this chapter

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. 14-43); the cross- sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed \(V\) and then through a narrow "throat" of crosssectional area \(a\) with speed \(v\). A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta p\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta p\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. 14-43, show that $$ V=\sqrt{\frac{2 a^{2} \Delta p}{\rho\left(a^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(60 \mathrm{~cm}^{2}\) in the pipe and \(32 \mathrm{~cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and \(41 \mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?

Water is moving with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) through a pipe with a cross-sectional area of \(4.0 \mathrm{~cm}^{2}\). The water gradually descends \(12 \mathrm{~m}\) as the pipe cross-sectional area increases to \(8.0 \mathrm{~cm}^{2}\). (a) What is the speed at the lower level? (b) If the pressure at the upper level is \(1.5 \times 10^{5} \mathrm{~Pa}\), what is the pressure at the lower level?

What fraction of the volume of an iceberg (density \(917 \mathrm{~kg} / \mathrm{m}^{3}\) ) would be visible if the iceberg floats (a) in the ocean (salt water, density \(1024 \mathrm{~kg} / \mathrm{m}^{3}\) ) and (b) in a river (fresh water, density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) )? (When salt water freezes to form ice, the salt is excluded. So, an iceberg could provide fresh water to a community.)

What would be the height of the atmosphere if the air density (a) were uniform and (b) decreased linearly to zero with height? Assume that at sea level the air pressure is \(1.0 \mathrm{~atm}\) and the air density is \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\).

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.

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