/*! 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 15 What gaugc pressure must a machi... [FREE SOLUTION] | 91Ó°ÊÓ

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What gaugc 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 \(1.5 \mathrm{~m} ?\)

Short Answer

Expert verified
The gauge pressure needed is 26,487 Pa.

Step by step solution

01

Understand the problem

We need to find the gauge pressure required to lift mud with a density of \(1800 \text{ kg/m}^3\) up a height of \(1.5\text{ m}\). This involves calculating how much pressure is needed to counteract the weight of the mud column.
02

Use the hydrostatic pressure formula

The pressure required to lift a column of fluid can be calculated using the hydrostatic pressure formula: \[ P = \rho \cdot g \cdot h \] Where \( P \) is the pressure, \( \rho \) is the density of the fluid (\(1800 \text{ kg/m}^3\)), \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)), and \( h \) is the height (\(1.5 \text{ m}\)).
03

Substitute values into the formula

Substitute the known values into the hydrostatic pressure formula:\[ P = 1800 \cdot 9.81 \cdot 1.5 \]
04

Calculate the pressure

Perform the multiplication:\[ P = 1800 \times 9.81 \times 1.5 = 26,487 \text{ Pa} \]So, the gauge pressure required is \(26,487 \text{ Pa}\).

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Key Concepts

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

Gauge Pressure
Gauge pressure is a measure of pressure relative to the local atmospheric pressure. Unlike absolute pressure, which accounts for the pressure exerted by the atmosphere, gauge pressure only considers the pressure generated above this local atmospheric level.
For instance, when a machine needs to lift mud through a tube, it must produce sufficient gauge pressure to overcome the weight of the mud and the atmospheric pressure acting on it.
When calculating the gauge pressure required to lift a fluid, the objective is to find how much additional pressure needs to be applied by the machine. This is essential for any system operating under conditions where the external atmosphere also acts on the substance being moved.
  • Gauge pressure is convenient because many instruments measure pressure in this way, simplifying practical applications.
  • It's crucial in fields like fluid mechanics and engineering where it aids in designing systems capable of lifting or moving liquids efficiently.
  • In our problem, the gauge pressure of 26,487 Pa indicates how much extra pressure the machine must exert to elevate the mud 1.5 meters upward.
Fluid Density
Fluid density is an essential characteristic of a fluid, describing how much mass a fluid has in a given volume. It's expressed as \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. The density is typically expressed in units of kilograms per cubic meter (kg/m³).
This property significantly influences the behavior of the fluid under various pressures and forces, such as in our tubing scenario for mud lifting.
In the given exercise, the mud has a density of 1800 kg/m³.
  • High density usually means more mass in the same volume compared to a fluid with lower density.
  • Affects the gravitational pull on the fluid, as denser fluids require more pressure to lift the same height.
  • Understanding fluid density is crucial in applications where substances must be moved, such as in piping systems and pumps.
In our example, knowing the density helps compute how much pressure is needed to lift the mud, as denser substances naturally exert more gravitational force.
Pressure Calculation
Calculating pressure involves determining the force exerted over a particular area, and when dealing with fluids, one often uses the hydrostatic pressure formula: \[ P = \rho \cdot g \cdot h \] where \( \rho \) is fluid density, \( g \) is acceleration due to gravity, and \( h \) is height or depth.
In this problem, substituting the known values allows us to find the specific gauge pressure required.
When solving the problem, we calculate:
  • First, understand each component: \( \rho = 1800 \, \text{kg/m}^3 \), \( g = 9.81 \, \text{m/s}^2 \), \( h = 1.5 \, \text{m} \).
  • Substituting values gives \( P = 1800 \times 9.81 \times 1.5 \).
  • Calculate the product to get a final gauge pressure of 26,487 Pa.
This calculation is crucial for any such lifting system, providing insight into how much force must be generated by a machine or a pump to accomplish the task efficiently.

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Most popular questions from this chapter

(Come) back to the future. Suppose that a father is \(20.00 \mathrm{y}\) older than his daughter. He wants to travel outward from Earth for \(2.000 \mathrm{y}\) and then back for another \(2.000 \mathrm{y}\) (both intervals as he measures them ) such that he is then \(20.00 \mathrm{y}\) younger than his daughter. What constant speed parameter \(\beta\) (relative to Earth) is required?

For seawater of density \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\), find the weight of water on top of a submarine at a depth of \(255 \mathrm{~m}\) if the horizontal crosssectional hull area is \(2200.0 \mathrm{~m}^{2}\). (b) In atmospheres, what water pressure would a diver experience at this depth?

A pitot tube (Fig. \(14-48\) ) is used to determine the airspeed of an airplane. It consists of an outer tube with a number of small holes \(B\) (four are shown) that allow air into the tube; that tube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole \(A\) at the front end of the device, which points in the direction the plane is headed. At \(A\) the air becomes stagnant so that \(v_{A}=0 .\) At \(B,\) however, the speed of the air presumably equals the airspecd \(v\) of the plane. (a) Use Bernoulli's equation to show that $$ v=\sqrt{\frac{2 \rho g h}{\rho_{\text {ar }}}} $$ where \(\rho\) is the density of the liquid in the \(U\) -tube and \(h\) is the difference in the liquid levels in that tube. (b) Suppose that the tube contains alcohol and the level difference \(h\) is \(26.0 \mathrm{~cm}\). What is the plane's speed relative to the air? The density of the air is \(1.03 \mathrm{~kg} / \mathrm{m}^{3}\) and that of alcohol is \(810 \mathrm{~kg} / \mathrm{m}^{3}\)

The car-in-the-garage problem. Carman has just purchased the world's longest stretch limo, which has a proper length of \(L_{c}=30.5 \mathrm{~m} .\) In Fig. \(37-32 a,\) it is shown parked in front of a garage with a proper length of \(L_{g}=6.00 \mathrm{~m}\). The garage has a front door (shown open) and a back door (shown closed). The limo is obviously longer than the garage. Still, Garageman, who owns the garage and knows something about relativistic length contraction, makes a bet with Carman that the limo can fit in the garage with both doors closed. Carman, who dropped his physics course before reaching special relativity, says such a thing, even in principle, is impossible. To analyze Garageman's scheme, an \(x_{c}\) axis is attached to the limo, with \(x_{c}=0\) at the rear bumper, and an \(x_{k}\) axis is attached to the garage, with \(x_{g}=0\) at the (now open) front door. Then Carman is to drive the limo directly toward the front door at a velocity of \(0.9980 c\) (which is, of course, both technically and financially impossible). Carman is stationary in the \(x_{c}\) reference frame; Garageman is stationary in the \(x_{z}\) reference frame. There are two events to consider. Event 1 : When the rear bumper clears the front door, the front door is closed. Let the time of this event be zero to both Carman and Garageman: \(t_{\mathrm{s} 1}=t_{c 1}=0\). The event occurs at \(x_{c}=x_{g}=0 .\) Figure \(37-32 b\) shows event 1 according to the \(x_{g}\) reference frame. Event 2 : When the front bumper reaches the back door, that door opens. Figure \(37-32 c\) shows event 2 according to the \(x_{g}\) reference frame. According to Garageman, (a) what is the length of the limo, and what are the spacetime coordinates (b) \(x_{R 2}\) and (c) \(t_{g 2}\) of event \(2 ?\) (d) For how long is the limo temporarily "trapped" inside the garage with both doors shut? Now consider the situation from the \(x_{c}\) reference frame, in which the garage comes racing past the limo at a velocity of \(-0.9980 \mathrm{c}\). According to Carman, (e) what is the length of the passing garage, what are the spacetime coordinates (f) \(x_{c 2}\) and \((g) t_{c 2}\) of event \(2,(h)\) is the limo ever in the garage with both doors shut, and (i) which event occurs first? (j) Sketch events 1 and 2 as seen by Carman. (k) Are the events causally related; that is, does one of them cause the other? (I) Finally, who wins the bet?

Bullwinkle in reference frame \(S^ {\prime}\) passes you in reference frame \(S\) along the common direction of the \(x^ {\prime}\) and \(x\) axes, as in Fig. \(37-9 .\) He carrics three meter sticks: meter stick 1 is parallel to the \(x^{\prime}\) axis, meter stick 2 is parallel to the \(y^{\prime \prime}\) axis, and meter stick 3 is parallel to the \(z^ {\prime}\) axis. On his wristwatch he counts off \(15.0 \mathrm{~s},\) which takes \(30.0 \mathrm{~s}\) according to you. Two events occur during his passage. According to you, event 1 occurs at \(x_{1}=33.0 \mathrm {~m}\) and \(t_{1}=22.0 \mathrm{~ns},\) and event 2 occurs at \(x_{2}=53.0 \mathrm {~m}\) and \(t_{2}=62.0 \mathrm{~ns}\) According to your measurements, what is the length of (a) meter stick \(1,\) (b) meter stick \(2,\) and (c) meter stick 3 ? According to Bullwinkle, what are (d) the spatial separation and (e) the temporal separation between events 1 and \(2,\) and \((f)\) which event occurs first?

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