Between the connecting flanges of two pipes \(A\) and \(B\) is bolted a plate containing a sharp-edged orifice \(C\) for which \(C_{\mathrm{c}}=0.62\). The pipes and the orifice are coaxial and the diameters of \(A\), \(B\) and \(C\) are respectively \(150 \mathrm{~mm}, 200 \mathrm{~mm}\) and \(100 \mathrm{~mm}\). Water flows from \(A\) into \(B\) at the rate of \(42.5 \mathrm{~L} \cdot \mathrm{s}^{-1}\). Neglecting shear stresses at boundaries, determine ( \(a\) ) the difference of static head between sections in \(A\) and \(B\) at which the velocity is uniform, (b) the power dissipated.

Short Answer

Expert verified
The power dissipated summed up equal two yields that overall prominent solution

Step by step solution

01

Calculate the area of each cross-section

First, convert the diameters from mm to meters. Thus, 150 mm = 0.15 m, 200 mm = 0.2 m, and 100 mm = 0.1 m. The area (A) of a circle is given by the equation \[ A = \frac{\text{π}d^2}{4} \] For pipe A, \[ A_A = \frac{\text{π}(0.15)^2}{4} = 0.0177 \text{ m}^2 \]For pipe B, \[ A_B = \frac{\text{π}(0.2)^2}{4} = 0.0314 \text{ m}^2 \]For the orifice C, \[ A_C = \frac{\text{π}(0.1)^2}{4} = 0.00785 \text{ m}^2 \]
02

Calculate velocities in each section (A, B, and the vena contracta)

The velocity (V) can be found using the continuity equation \[ Q = A \times V \] where Q is the volume flow rate. Given Q = 42.5 \text{ L} \times \text{s}^{-1}= 0.0425 \text{ m}^3 \times \text{s}^{-1}For pipe A, \[ V_A = \frac{Q}{A_A} = \frac{0.0425}{0.0177} = 2.40 \text{ m} \times \text{s}^{-1} \]For pipe B, \[ V_B = \frac{Q}{A_B} = \frac{0.0425}{0.0314} = 1.35 \text{ m} \times \text{s}^{-1} \]For the orifice C, we use the discharge coefficient C_c = 0.62, which is the ratio between the actual and theoretical flow areas at the vena contracta. The velocity (vena contracta) is: \[ V_{\text{vena}} = \frac{Q}{C_c \times A_C} = \frac{0.0425}{0.62 \times 0.00785} = 8.8 \text{ m} \times \text{s}^{-1} \]
03

Calculate Power Dissipation

The power dissipated is given by the equation \[ P_{\text{dissipated}} = Q \times (P_{A} - P_{B}) \]

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Key Concepts

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

Continuity Equation
The continuity equation is a fundamental principle in fluid dynamics. It states that the mass flow rate must remain constant from one cross-section of a pipe to another if there are no leaks. Essentially, if you know the flow rate at one section of the pipe, you can use it to find the flow rates at other sections using the equation:
equation: \( Q = A_1 V_1 = A_2 V_2 \)
where \( Q \) is the flow rate, \( A \) is the cross-sectional area, and \( V \) is the velocity of the fluid. This equation helps to understand how fluids speed up or slow down when moving through pipes of varying diameters. By maintaining the product of area and velocity constant, we can predict fluid behavior effectively.
For example, if fluid flows from a wider pipe into a narrower section, its velocity must increase, and vice-versa. This principle is crucial when analyzing fluid mechanics problems, especially when dealing with complex pipeline systems. In our exercise, we use the continuity equation to calculate velocities at different points using known flow rates and cross-sectional areas.
Flow Rate Calculation
Flow rate is the volume of fluid that passes through a cross-section of a pipe per unit time. It's usually expressed in liters per second (L/s) or cubic meters per second (m³/s). The formula for flow rate is:
equation: \( Q = A \times V \)
where \( Q \) is the flow rate, \( A \) is the cross-sectional area, and \( V \) is the velocity of the fluid. Calculating flow rate is important for designing and analyzing plumbing systems, ensuring that they can handle required volumes efficiently. In our exercise, the given flow rate was 42.5 L/s, which converts to 0.0425 m³/s. Using this value, we could then determine the fluid velocities in different sections of the pipe.
Cross-Sectional Area
The cross-sectional area of a pipe affects how quickly a fluid can flow through it. It can be calculated using the diameter of the pipe with the formula for the area of a circle:
equation: \( A = \frac{\text{π} d^2}{4} \)
where \( d \) is the diameter. In the given exercise, the diameters of the pipes are converted from millimeters to meters (0.15 m, 0.2 m, and 0.1 m for sections A, B, and the orifice C, respectively). The cross-sectional areas are then calculated as follows:
  • For pipe A: \( A_A = 0.0177 \text{ m}^2 \)
  • For pipe B: \( A_B = 0.0314 \text{ m}^2 \)
  • For orifice C: \( A_C = 0.00785 \text{ m}^2 \)
    • The variation in cross-sectional area at different points is essential for determining fluid velocity and behavior in each section.
Velocity in Pipes
With the cross-sectional areas known, we can calculate the velocity at different pipe sections using the flow rate. The formula is derived from the continuity equation:
equation: \( V = \frac{Q}{A} \)
For our exercise:
  • Velocity in pipe A: \( V_A = \frac{0.0425}{0.0177} = 2.40 \text{ m} \times \text{s}^{-1} \)
  • Velocity in pipe B: \( V_B = \frac{0.0425}{0.0314} = 1.35 \text{ m} \times \text{s}^{-1} \)
  • For the orifice C, we apply the discharge coefficient \( C_C = 0.62 \) to get the velocity at the vena contracta: \( V_{\text{vena}} = \frac{0.0425}{0.62 \times 0.00785} = 8.8 \text{ m} \times \text{s}^{-1} \)
Understanding these velocities is crucial for pipeline design and for identifying any potential issues that might arise.
Power Dissipation
Power dissipation in fluid systems refers to the loss of energy, often due to frictional forces and other resistances within the pipe. The power dissipated can be calculated using the formula:
equation: \( P_{\text{dissipated}} = Q \times (P_A - P_B) \)
where \( Q \) is the flow rate, and \( P_A \) and \( P_B \) are the pressure at sections A and B, respectively. This helps in determining how much energy is being lost in the system and is crucial for efficient system design. By understanding and minimizing power dissipation, engineers can design piping systems that conserve energy and operate more efficiently.

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Most popular questions from this chapter

A large tank with vertical sides is divided by a vertical partition into two sections \(A\) and \(B\), with plan areas of \(1.5 \mathrm{~m}^{2}\) and \(7.5 \mathrm{~m}^{2}\) respectively. The partition contains a \(25 \mathrm{~mm}\) diameter orifice \(\left(C_{d}=0.6\right)\) at a height of \(300 \mathrm{~mm}\) above the base. Initially section \(A\) contains water to a depth of \(2.15 \mathrm{~m}\) and section \(B\) contains water to a depth of \(950 \mathrm{~mm}\). Calculate the time required for the water levels to equalize after the orifice is opened.

A fluid of constant density \(\varrho\) enters a horizontal pipe of radius \(R\) with uniform velocity \(V\) and pressure \(p_{1}\). At a downstream section the pressure is \(p_{2}\) and the velocity varies with radius \(r\) according to the equation \(u=2 V\left\\{1-\left(r^{2} / R^{2}\right)\right\\}\). Show that the friction force at the pipe walls from the inlet to the section considered is given by \(\pi R^{\frac{1}{2}}\left(p_{1}-p_{2}-\varrho V^{2} / 3\right)\).

A single uniform pipe joins two reservoirs. Calculate the percentage increase of flow rate obtainable if, from the mid-point of this pipe, another of the same diameter is added in parallel to it. Neglect all losses except pipe friction and assume a constant and equal \(f\) for both pipes.

A reservoir \(A\) feeds two lower reservoirs \(B\) and \(C\) through a single pipe \(10 \mathrm{~km}\) long, \(750 \mathrm{~mm}\) diameter, having a downward slope of \(2.2 \times 10^{-3}\). This pipe then divides into two branch pipes, one \(5.5 \mathrm{~km}\) long laid with a downward slope of \(2.75 \times\) \(10^{-3}\) (going to \(\left.B\right)\), the other \(3 \mathrm{~km}\) long having a downward slope of \(3.2 \times 10^{-3}\) (going to \(\left.C\right)\). Calculate the necessary diameters of the branch pipes so that the steady flow rate in each shall be \(0.24 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) when the level in each reservoir is \(3 \mathrm{~m}\) above the end of the corresponding pipe. Neglect all losses except pipe friction and take \(f=0.006\) throughout.

A trailer pump is to supply a hose \(40 \mathrm{~m}\) long and fitted with a \(50 \mathrm{~mm}\) diameter nozzle capable of throwing a jet of water to a height \(40 \mathrm{~m}\) above the pump level. If the power lost in friction in the hose is not to exceed \(15 \%\) of the available hydraulic power, determine the diameter of hose required. Friction in the nozzle may be neglected and \(f\) for the hose assumed to be in the range \(0.007-0.01\).

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