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A liquid \(\left(\rho=1.65 \mathrm{~g} / \mathrm{cm}^{3}\right)\) flows through two horizontal sections of tubing joined end to end. In the first section, the cross-sectional area is \(10.0 \mathrm{~cm}^{2}\), the flow speed is \(275 \mathrm{~cm} / \mathrm{s}\), and the pressure is \(1.20 \times 10^{5} \mathrm{~Pa} .\) In the second section, the cross-sectional area is \(2.50 \mathrm{~cm}^{2}\). Calculate the smaller section's (a) flow speed and (b) pressure.

Short Answer

Expert verified
The flow speed in the second section of tubing is 1100 cm/s. The pressure in the second section of tubing is approximately -0.23×10^5 Pa(gauge)

Step by step solution

01

Find Flow Speed

Using the principle of continuity, which states that the mass flow rate is conserved across sections in a flow for an incompressible fluid, we're able to calculate the flow speed in the second (smaller) section of tubing. This principle can be mathematically represented as \(A_1V_1 = A_2V_2\), where \(A_1\) and \(A_2\) are the cross-sectional areas and \(V_1\) and \(V_2\) are the fluid velocities in the first and second sections respectively. To find the flow speed \(V_2\) in the second section, we rearrange the equation as: \(V_2 = \frac{A_1V_1}{A_2} = \frac{10.0 cm^2 * 275 cm/s}{2.50 cm^2}\)
02

Calculate Flow Speed

After substituting the given values into the equation, we find that \(V_2 = 1100 cm/s\). So, the flow speed in the second section with smaller cross-sectional area is higher than the first section with larger cross-sectional area.
03

Find Pressure

Next, we will find the pressure in the second section. For that, we use Bernoulli’s equation, which represents the conservation of energy for flowing fluids. The simplified form of the equation, \(P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2\), is applicable since we are dealing with a horizontal flow, and there is no difference in gravitational potential energy. We can solve this equation for \(P_2\), the pressure in the second section. Thus, \(P_2 = P_1 + \frac{1}{2}\rho V_1^2 - \frac{1}{2}\rho V_2^2\)
04

Calculate Pressure

Substitute the given and calculated values into the equation to find the pressure in the second section. That will be \(P_2 = 1.20 ×10^5 Pa + \frac{1}{2}×1.65 g/cm^3×(275 cm/s)^2 - \frac{1}{2}×1.65 g/cm^3×(1100 cm/s)^2 \). After the necessary conversions and calculation, \(P_2 \) results as approximately \( -0.23 ×10^5 Pa\). A negative value signifies that the pressure at the second section is below ambient pressure; it represents a gauge pressure.

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Key Concepts

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

Bernoulli's equation
Bernoulli's equation is a fundamental principle in fluid dynamics that expresses the conservation of energy within a flowing fluid. It essentially states that for an incompressible, non-viscous fluid, the total mechanical energy remains constant along a streamline.

Mathematically, Bernoulli's equation can be written as:
\[ P + \frac{1}{2}\rho V^2 + \rho gh = \text{constant} \]
where \( P \) represents the pressure within the fluid, \( \rho \) is the fluid's density, \( V \) is the flow velocity, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point. In cases like our exercise, where we deal with horizontal flow with no change in height, the height term \( \rho gh \) can be dropped.

In practical applications, such as the exercise where we calculate pressure changes in different sections of a tube, it provides a way to account for differing speeds and pressures. When the fluid speeds up, pressure must decrease, and vice versa; this relationship is instrumental in various engineering and scientific fields.
Principle of continuity
The principle of continuity is grounded in the concept of the conservation of mass in fluid dynamics. For an incompressible fluid, this principle dictates that the mass flow rate—mass per unit time—should remain constant from one cross-sectional area of a pipeline to another.

The mass flow rate is represented mathematically as:
\[ A_1V_1 = A_2V_2 \]
where \( A_1 \) and \( A_2 \) are the cross-sectional areas and \( V_1 \) and \( V_2 \) are the fluid velocities at points 1 and 2, respectively. This equation implies that if the cross-sectional area decreases, the velocity of the fluid must accordingly increase to conserve mass flow rate, as was demonstrated in our exercise calculation.
Mass flow rate
Mass flow rate is a crucial concept in fluid dynamics and is essentially the amount of mass flowing through a given cross-section of a pipe or channel per unit time. It is a measure of how much fluid is being 'moved' in a system and is particularly significant when analyzing the dynamics of incompressible fluids, like water or most liquids in industrial applications.

Mathematically, the mass flow rate can be expressed as:
\[ \text{mass flow rate} = \rho A V \]
where \( \rho \) is the density of the fluid, \( A \) is the cross-sectional area, and \( V \) is the velocity of the fluid. In our exercise, we see how the mass flow rate remains unchanged even as the fluid moves from a wider to a narrower section of the tube, leading to an increase in the fluid's velocity.
Incompressible fluid
An incompressible fluid is one in which density remains approximately constant within the flow. This assumption simplifies the analysis of fluid flow because volume changes under pressure variations are negligible. Most liquids, especially water, are typically treated as incompressible due to their very low compressibility factor.

In our textbook exercise, the liquid with a density of \(1.65 \mathrm{~g} / \mathrm{cm}^{3}\) can be presumed to be incompressible, which allows us to use both the principle of continuity and Bernoulli’s equation without accounting for changes in density. This consideration is critical for the conservation equations to hold true and for the calculated flow speeds and pressures to be accurate.
Pressure calculation
Pressure calculation in fluid dynamics is related to determining the force exerted by the fluid per unit area within a flow system. Understanding pressure is necessary for analyzing force balance, predicting fluid behavior, and designing various hydraulic systems.

In the context of our exercise, Bernoulli's equation is used for the pressure calculation at different points along a streamline by considering velocity, density, and pressure changes within the fluid. It reveals the relationship between dynamic pressure (due to fluid velocity) and static pressure (the actual fluid pressure within the system) in a flowing fluid. Negative pressure, as indicated in our exercise, suggests the presence of a suction effect or lower pressure compared to the surrounding atmosphere, commonly referred to as gauge pressure. Calculating precise pressures is vital for engineering applications to ensure the safety and functionality of fluid systems.

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Most popular questions from this chapter

\(A\) straight horizontal pipe with a diameter of \(1.0 \mathrm{~cm}\) and â length of \(50 \mathrm{~m}\) carries oil with a coefficient of viscosity of \(0.12 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). At the output of the pipe, the flow rate is \(8.6 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\) and the pressure is \(1.0 \mathrm{~atm}\). Find the gauge pressure at the pipe input.

The block of ice (temperature \(0^{\circ} \mathrm{C}\) ) shown in Figure \(\mathrm{P} 9.63\) is drawn over a level surface lubricated by a layer of water \(0.10 \mathrm{~mm}\) thick. Determine the magnitude of the force \(\vec{F}\) needed to pull the block with a constant speed of \(0.50 \mathrm{~m} / \mathrm{s}\). At \(0^{\circ} \mathrm{C}\), the viscosity of water has the value \(\eta=1.79 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\)

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0-\mathrm{L}\) bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

The human brain and spinal cord are immersed in the cerebrospinal fluid. The fluid is normally continuous between the cranial and spinal cavities and exerts a pressure of 100 to \(200 \mathrm{~mm}\) of \(\mathrm{H}_{2} \mathrm{O}\) above the prevailing atmospheric pressure. In medical work, pressures are often measured in units of \(\mathrm{mm}\) of \(\mathrm{H}_{2} \mathrm{O}\) because body fhuids, inchuding the cerebrospinal fluid, typically have nearly the same density as water. The pressure of the cerebrospinal fluid can be measured by means of a spinal tap. A hollow tube is inserted into the spinal column, and the height to which the fluid rises is observed, as shown in Figure \(\mathrm{P} 9.83\). If the fluid rises to a height of \(160 \mathrm{~mm}\), we write its gauge pressure as \(160 \mathrm{~mm} \mathrm{H}_{2} \mathrm{O}\). (a) Express this pressure in pascals, in atmospheres, and in millimeters of mercury. (b) Sometimes it is necessary to determine whether an accident victim has suffered a crushed vertebra that is blocking the flow of cerebrospinal fluid in the spinal column. In other cases a physician may suspect that a tumor or other growth is blocking the spinal col\(\mathrm{umn}\) and inhibiting the flow of cerebrospinal fluid. Such conduons can be investigated by means of the Quecken. sted test. In this procedure the veins in the patient's neck are compressed, to make the blood pressure rise in the brain. The increase in pressure in the blood vessels is transmitted to the cerebrospinal fluid. What should be the normal effect on the height of the fluid in the spinal tap? (c) Suppose compressing the veins had no effect on the level of the fluid. What might account for this phenomenon?

Sucrose is allowed to diffuse along a \(10-\mathrm{cm}\) length of tubing filled with water. The tube is \(6.0 \mathrm{~cm}^{2}\) in crosssectional area. The diffusion coefficient is equal to \(5.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\), and \(8.0 \times 10^{-14} \mathrm{~kg}\) is transported along the tube in \(15 \mathrm{~s}\). What is the difference in the concentraton levels of sucrose at the two ends of the tube?

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