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A 2.5 -in.-diameter flow nozzle meter is installed in a 3.8 -in.- diameter pipe that carries water at \(160^{\circ} \mathrm{F}\). If the air-water manometer used te measure the pressure difference across the meter indicates a reading of \(3.1 \mathrm{ft}\), determine the flowrate

Short Answer

Expert verified
Following the above steps will provide the solution for the flowrate. Bernoulli's equation and continuity equation are integral concepts utilized in solving this problem. Be conscious of the units and include consideration of pressure loss in the situation.

Step by step solution

01

Compute the cross-sectional area of the pipe and meter

First, the cross-sectional area \(A\) of the pipe and the meter should be calculated. This can be found by using the formula for the area of a circle: \(A=\frac{\pi d^{2}}{4}\), where \(d\) is the diameter of the circle. Here, remember to convert the diameter from inches to feet since we are dealing with imperial units. This will provide the correct area in square feet.
02

Calculate flow velocity

Let's denote the flow velocity across the pipe as \(V_{1}\) and across the flow nozzle as \(V_{2}\). Use the Bernoulli's equation modified to real-world phenomena which is given as: \(\Delta h = \frac{(V_{2}^2 - V_{1}^2)}{2g}\). In this problem, the initial velocity \(V_{1}\) in the pipe can be ignored as the meter will have channelised the flow. Given that the pressure difference is represented in units of feet as \(\Delta h\) and \(g\) taken as 32.2 ft/s² (gravity), we solve for \(V_{2}\).
03

Compute the flow rate

Flow rate \(Q\) is given by the product of the flow velocity and the cross-sectional area of the nozzle: \(Q = A_2 \times V_{2}\), where \(A_{2}\) is the area of the flow nozzle we computed in step 1 and \(V_{2}\) is the flow velocity obtained in step 2. Solving this equation will give the value of the flow rate.

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Key Concepts

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

Flow Nozzle Meter
A flow nozzle meter is a device used to measure the flow rate of a fluid in a pipe. It operates on the principle of introducing a constriction in the flow, which causes a drop in pressure. This pressure drop is then related to the flow rate. Flow nozzle meters are widely used because they offer a balance between accuracy and pressure loss compared to other measuring devices such as orifice plates or venture meters.

Some reasons for their popularity include:
  • Ruggedness and durability for long-term use
  • Relatively low cost of installation and maintenance
  • High accuracy in measuring flow rates
Understanding how a flow nozzle meter works involves recognizing its role in reducing cross-sectional area, which in turn increases fluid velocity. This change in dynamics is key to determining flow rates in various applications.
Bernoulli's Equation
Bernoulli's Equation is a fundamental principle in fluid dynamics that relates different properties of a flowing fluid. The equation is essentially a statement of the conservation of energy for flowing fluids. It is expressed as:\[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \]where:
  • \(P\) is the pressure energy per unit volume,
  • \(\frac{1}{2} \rho v^2\) is the kinetic energy per unit volume,
  • \(\rho g h\) is the potential energy per unit volume, with \(\rho\) being the fluid density, \(v\) the velocity, \(g\) the gravitational acceleration, and \(h\) the height above a reference point.
This equation helps predict the behavior of fluids in motion and is particularly useful when assessing the pressure changes in flow nozzle meters. Modified versions of Bernoulli's equation, as used in practical applications, account for real-world factors such as friction and other losses, to provide more accurate predictions of fluid behavior.
Cross-Sectional Area Calculations
To determine flow-related properties, calculating the cross-sectional area of the pipe or nozzle is crucial. The area of a circle is found using the formula:\[ A = \frac{\pi d^2}{4} \]where \(d\) is the diameter of the pipe or nozzle. It's important to ensure that all measurements are consistent units, such as converting diameters from inches to feet when calculating areas in square feet. This formula allows us to calculate the area through which the fluid flows, influencing the flow rate significantly.

Understanding cross-sectional areas:
  • Bigger diameters mean larger cross-sectional areas, resulting in a broader channel for fluid flow.
  • Conversely, smaller diameters constrict the flow, increasing fluid velocity.
Hence, these calculations are essential in designing and assessing fluid systems to ensure that the desired flow rates and characteristics meet specific application needs.
Manometry in Fluid Mechanics
Manometry is a technique used in fluid mechanics to measure pressure differences between two points in a fluid system. A common type of manometer is the U-tube manometer, which uses a liquid column to indicate pressure differences. For example, an air-water manometer efficiently demonstrates how the pressure differential across a flow nozzle meter can be measured.

Basic operation involves:
  • Two fluid columns (such as water and another liquid)
  • The difference in height (\(\Delta h\)) between fluid levels directly relates to the pressure difference
  • Calibration and understanding of the fluid properties involved
This pressure difference allows for calculation of relative pressures across flow nozzles, providing critical data for understanding flow rates and dynamics.

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Most popular questions from this chapter

Assume a car's exhaust system can be approximated as \(14 \mathrm{ft}\) of 0.125 -ft-diameter cast-iron pipe with the equivalent of six \(90^{\circ}\) flanged elbows and a muffler. (See Video V8.14.) The muffler acts as a resistor with a loss coefficient of \(K_{t}=8.5 .\) Determine the pressure at the beginning of the exhaust system if the flowrate is \(0.10 \mathrm{cfs},\) the temperature is \(250^{\circ} \mathrm{F}\), and the exhaust has the same properties as air.

A 3 -in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.) carries 210 'F SAE 40 crankcase oil at the rate of 6.0 gal/min. The oil specific gravity is \(0.89,\) and the absolute viscosity is \(6,6 \times 10^{-7} 1 \mathrm{b} \cdot\) sec/ft \(^{2}\). For the same pressure drop, what must the acw pipe size be to carry the oil at 10.7 gal/min?

Water flows downward through a vertical 10 -mm-diameter galvanized iron pipe with an average velocity of \(5.0 \mathrm{m} / \mathrm{s}\) and exits as a free jet. There is a small hole in the pipe \(4 \mathrm{m}\) above the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe through the hole? Repeat the problem if the average velocity is \(0.5 \mathrm{m} / \mathrm{s}\)

For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by \(u(r)=2\left(1-r^{2} / R^{2}\right)\) in \(\mathrm{m} / \mathrm{s},\) where \(R\) is the inner radius of the pipe. Assuming that the pipe diameter is \(4 \mathrm{cm},\) find the maximum and average velocities in the pipe as well as the volume flow rate.

A thief siphoned 15 gal of gasoline from a gas tank in the middle of the night. The gas tank is 12 in. wide, 24 in. long, and 18 in. high and was full when the thief started. The siphoning plastic tube has an inside diameter of 0.5 in. and a length of \(4.0 \mathrm{ft}\). Assume that at any instant of time, the steady-state mechanical energy equation is adequate to predict the gasoline flow rate through the tube. As 15 gal is 3465 in \(^{3}\), the gasoline level in the tank will drop 12.0 in. You may use the gasoline level after it has dropped 6.0 in. to estimate the average gasoline flow rate. Use this flow rate to estimate the time needed to siphon the 15 gal of gasoline. Compare your answer with the answer of 190 sec found in problem 3.107 using Bernoulli's equation. The siphon cischarges at the level of the bottom of the gasoline tank. You may find it useful to use the Blasius equation for smooth pipes found in problem 8.45

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