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Chapter 4: Problem 91

Water flows at a rate of \(0.025 \mathrm{~m}^{3} / \mathrm{s}\) in a horizontal pipe whose diameter increases from 6 to \(11 \mathrm{~cm}\) by an enlargement section. If the head loss across the enlargement section is \(0.45 \mathrm{~m}\) and the kinetic energy correction factor at both the inlet and the outlet is \(1.05,\) determine the pressure change.

Short Answer

Expert verified
The pressure decreases by approximately 2890.7 Pa across the enlargement.

Step by step solution

01

Identify Known Values and Convert Units

First, identify the known values from the problem. The flow rate \(Q\) is given as \(0.025 \, \text{m}^3/\text{s}\). The initial diameter \(D_1\) is \(6 \, \text{cm}\) or \(0.06 \, \text{m}\), and the final diameter \(D_2\) is \(11 \, \text{cm}\) or \(0.11 \, \text{m}\). The head loss \(h_f\) is \(0.45 \, \text{m}\), and the kinetic energy correction factor at both inlet and outlet is \( \alpha = 1.05 \).
02

Calculate Cross-sectional Areas

Calculate the cross-sectional areas using the formula for the area of a circle, \( A = \frac{\pi D^2}{4} \).For inlet area, \(A_1 = \frac{\pi \times (0.06)^2}{4} = 0.002827 \, \text{m}^2\).For outlet area, \(A_2 = \frac{\pi \times (0.11)^2}{4} = 0.009503 \, \text{m}^2\).
03

Calculate Flow Velocities

Use the continuity equation \(Q = A \times v\) to find velocities.Velocity at inlet, \(v_1 = \frac{Q}{A_1} = \frac{0.025}{0.002827} = 8.84 \, \text{m/s}\).Velocity at outlet, \(v_2 = \frac{Q}{A_2} = \frac{0.025}{0.009503} = 2.63 \, \text{m/s}\).
04

Apply Bernoulli's Equation

Bernoulli's equation including head loss is:\[P_1 + \alpha \frac{\rho v_1^2}{2} + \rho gh_1 = P_2 + \alpha \frac{\rho v_2^2}{2} + \rho gh_2 + \rho g h_f\]Since the pipe is horizontal, \(h_1 = h_2\), and \(\rho g h_1\) cancels out. Rearrange to find\[P_2 - P_1 = \alpha \frac{\rho (v_1^2 - v_2^2)}{2} - \rho gh_f\].
05

Calculate Pressure Change

Substitute values into the equation for pressure change:\[P_2 - P_1 = \frac{1.05 \times 1000 \times ((8.84)^2 - (2.63)^2)}{2} - 1000 \times 9.81 \times 0.45\]This calculation yields:\[P_2 - P_1 = \frac{1.05 \times 1000 \times (78.06 - 6.92)}{2} - 4414.5\]Simplify to find \(P_2 - P_1\) resulting in approximately \(-2890.7 \, \text{Pa}\).

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Key Concepts

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

Flow Rate
Flow rate refers to the volume of fluid that passes through a given cross-sectional area in a pipe or channel per unit time. In this exercise, the flow rate is given as \(0.025 \, ext{m}^3/ ext{s}\) which means 0.025 cubic meters of water passes through the pipe every second.

Flow rate is crucial in determining how fast a fluid moves through a system. It's calculated using the formula:
  • \(Q = A \times v\)
where:
  • \(Q\) is the flow rate,
  • \(A\) is the cross-sectional area,
  • \(v\) is the velocity of the fluid.
Understanding flow rate helps us anticipate changes in fluid behavior as it navigates different pipe dimensions and configurations.
Head Loss
Head loss signifies the decrease in total mechanical energy (or hydraulic energy) of the fluid as it moves through a pipe. This is due to friction and turbulence caused by irregular surfaces, bends, or expansions. In this problem, the head loss provided is \(0.45 \, ext{m}\).

Head loss is critical to consider as it affects fluid velocity and can signify energy inefficiencies in piping systems. The head loss can be included in Bernoulli's equation to determine changes in pressure, using
  • \( \rho g h_f\)
where:
  • \(\rho\) is the fluid density,
  • \(g\) is the acceleration due to gravity, approximately \(9.81 \, ext{m/s}^2\),
  • \(h_f\) is the head loss.
Recognizing the impact of head loss on fluid dynamics is vital for the design and maintenance of efficient piping systems.
Cross-sectional Area
The cross-sectional area refers to the size of the slice through a pipe where fluid flows. This area directly influences both the flow velocity and pressure of a fluid moving through the pipe.

In this exercise, the areas at the inlet and outlet are calculated using the formula for the area of a circle:
  • \( A = \frac{\pi D^2}{4} \)
For the given pipe diameters:
  • Inlet: \(D_1 = 0.06 \, ext{m}\) leading to \(A_1 = 0.002827 \, ext{m}^2\)
  • Outlet: \(D_2 = 0.11 \, ext{m}\) leading to \(A_2 = 0.009503 \, ext{m}^2\)
Changes in cross-sectional area, such as those from expansions or contractions in a pipe, necessitate adjustments in pressure and velocity, which is reflected through Bernoulli's principle and continuity equation.
Pressure Change
Pressure change indicates the difference in pressure between two points in a flow system. In the exercise, Bernoulli's equation incorporates both kinetic energy and head loss to determine this change between pipe sections.

Calculated pressure change is
  • \(P_2 - P_1 = \alpha \frac{\rho (v_1^2 - v_2^2)}{2} - \rho gh_f\)
where
  • \(\alpha\) is the kinetic energy correction factor,
  • \(v_1\) and \(v_2\) are the velocities at the inlet and outlet, respectively.

The impact of head loss and diameter changes results in a calculated pressure difference of approximately \(-2890.7 \, ext{Pa}\). Understanding how these parameters affect pressure allows for better system management and safety.
Continuity Equation
The continuity equation is a principle derived from the conservation of mass. It asserts that for a fluid flowing through a pipe, the mass flow rate must remain constant throughout the system. Thus, it can be expressed as:
  • \(Q = A_1v_1 = A_2v_2\)
In this problem, it ensures that since the flow rate \(Q\) is constant, any change in cross-sectional area \(A\) requires a compensatory change in fluid velocity \(v\).

This equation confirms that:
  • As cross-sectional area increases (from 0.06 m to 0.11 m), velocity decreases.
  • It's why \(v_1 = 8.84 \text{ m/s}\) and \(v_2 = 2.63 \text{ m/s}\).
Practically, the continuity equation helps engineers ensure system efficiency and safety by analyzing how changes in a given area affect flow dynamics.

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Most popular questions from this chapter

A valve connects two pipes, where the upstream pipe has a diameter of \(300 \mathrm{~mm}\) and the downstream pipe has a diameter of \(450 \mathrm{~mm}\). When water flows through the valve and the velocity in the upstream pipe is \(2.5 \mathrm{~m} / \mathrm{s}\), the difference in pressure between the inflow pipe and the outflow pipe is measured as \(60 \mathrm{kPa}\). If the inflow and outflow pipes are at the same elevation, what is the head loss through the valve? What is the power loss across the valve?

A room has dimensions \(10 \mathrm{~m} \times 10 \mathrm{~m} \times 3 \mathrm{~m},\) and air from the room is vented outside through a 100 -mm- diameter vent. The head loss in this type of vent can be estimated as \(h_{\ell}=0.1 V^{2} / 2 g\), where \(V\) is the flow velocity through the vent. If the pressure inside the room is \(2 \mathrm{kPa}\) higher than the outside pressure, estimate the volume flow rate of air through the vent and the time it takes to fully exchange the air in the room. Assume standard air.

Water at \(20^{\circ} \mathrm{C}\) is flowing in a 100 -mm-diameter pipe at an average velocity of \(2 \mathrm{~m} / \mathrm{s}\). If the diameter of the pipe is suddenly expanded to \(150 \mathrm{~mm}\), what is the velocity in the expanded pipe? What are the volume and mass flow rates in the pipe?

The performance of a rocket with a 220 -mm-diameter nozzle exit is tested on a stand as shown in Figure \(4.81 .\) Under the conditions of a particular test, atmospheric pressure is \(101 \mathrm{kPa}\) and the exhaust gas has an exit velocity of \(1550 \mathrm{~m} / \mathrm{s}\), a pressure of \(120 \mathrm{kPa}\) absolute, and a mass flow rate of \(15 \mathrm{~kg} / \mathrm{s}\). (a) Estimate the thrust generated by the rocket. (b) How is the mass flow rate of the exhaust gas related to the fuel consumption rate of the rocket?

In the United States, flow rates through showerheads are regulated to be no greater than \(9.5 \mathrm{~L} / \mathrm{min}\) under any water pressure condition likely to be encountered in a home. Water pressures in homes are typically less than \(550 \mathrm{kPa}\). A practical showerhead delivers water at a velocity of at least \(5 \mathrm{~m} / \mathrm{s}\). If nozzles in a showerhead can be manufactured with diameters of \(0.75 \mathrm{~mm}\), what is the maximum number of nozzles required to make a practical showerhead?
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