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Chapter 8: Problem 128

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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Most popular questions from this chapter

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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