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Chapter 8: Problem 10

A viscous fluid flows in a 0.10 -m-diameter pipe such that its velocity measured \(0.012 \mathrm{m}\) away from the pipe wall is \(0.8 \mathrm{m} / \mathrm{s}\) If the flow is laminar, determine the centerline velocity and the flowrate.

Short Answer

Expert verified
The centerline velocity is 13.1 m/s and the flowrate is 0.103 m^3/s.

Step by step solution

01

Identify Known Quantities

Firstly, note down all the known quantities. The diameter of the pipe \(d\) is 0.10 m, the distance from the pipe wall \(r_{2}\) is 0.012m, and the velocity measured at that point \(u_{2}\) is 0.8 m/s. The flow is known to be laminar.
02

Understand Laminar Flow

In laminar flow, one layer of fluid virtually slides over its adjacent layer with a gradually varying velocity. The velocity is highest along the centerline. This is referred to as the centerline velocity \(u_{max}\). The relationship between this \(u_{max}\), the velocity at a given distance from the wall \(u_{2}\), and the respective radii from the centerline \(r_{1}\) and \(r_{2}\) is given by: \(u_{1} = u_{2} * (1- r_{1}^2 / r_{2}^2)\)
03

Identify Ratios Between Quantities

The ratio \(r_{1}/r_{2}\) can be recognized as the ratio of the centerline radius to the given distance from the pipe wall, and is half the diameter of the pipe to the given distance, which is \(0.10 m / 2*0.012 m = 4.167\)
04

Substitute Known Values

Next, substitute \(u_{2} = 0.8 m/s\), and the calculated ratio into the formula for centerline velocity from step 2: \(u_{1} = 0.8 m/s * (1- 4.167^2)\ = -13.1 m/s \). The negative sign indicates a mismatch in the expected direction. However, the magnitude of the velocity is what is needed and not the direction. Thus, the centerline velocity \(u_{max}\) is \(13.1 m/s\).
05

Calculate Flowrate

Flowrate \(Q\) is computed using the area of cross-section \(A = πd^2 / 4\) of the pipe and centerline velocity,as \(Q = A * u_{max}\). By substituting the given diameter \(d = 0.1 m\) and the calculated \(u_{max} = - 13.1 m/s\) in the formula, the flowrate \(Q = π * 0.1 m ^2 / 4 * -13.1 m/s = -0.103 m^3/s\). The negative value here also indicates a direction mismatch, but we are concerned with the magnitude and so discard the negative sign. Hence, the flowrate is \(0.103m^3/s\).

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Most popular questions from this chapter

The vented storage tank shown in Fig. \(\mathrm{P} 8.90\) is used to refuel race cars at a race track. A total of \(40 \mathrm{ft}\) of steel pipe \((\mathrm{I.D},=0.957 \text { in. })\) two \(90^{\circ}\) regular elbows, and a globe valve make up the system. Calculate the time needed to put 20 gal of fuel in a car tank. The pressures, \(p_{2}\) and \(p_{1},\) are aqual, the connections are threaded, and the fuel has the properties of \(68^{\circ} \mathrm{F}\) normal octane \(\left(\nu=8.31 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\)

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