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Chapter 4: Problem 17

The velocity distribution for laminar flow in a long circular tube of radius \(R\) is given by the one-dimensional expression, \\[ \vec{V}=u \hat{i}=u_{\max }\left[1-\left(\frac{r}{R}\right)^{2}\right] \\] For this profile obtain expressions for the volume flow rate and the momentum flux through a section normal to the pipe axis.

Short Answer

Expert verified
The volume flow rate \(Q\) and momentum flux \(I\) for this velocity profile can be expressed as integrals over the cross-sectional area of the pipe: \[Q = 2\pi\int_{0}^{R} u_{max}\left[1-\left(\frac{r}{R}\right)^{2}\right] rdr, \] and \[I=2\pi\rho\int_{0}^{R} u_{max} \left[1- \left(\frac{r}{R}\right)^2\right]^2 rdr. \] The actual numerical values will depend on the specific values of \(u_{max}\), \(\rho\), and \(R\).

Step by step solution

01

Understanding the velocity profile

The velocity profile is given by \[\vec{V}=u \hat{i}=u_{max}\left[1-\left(\frac{r}{R}\right)^{2}\right]\], where \(\vec{V}\) is the velocity at a point at a distance \(r\) from the axis of the pipe, \(u_{max}\) is the maximum speed in the pipe (achieved at the centre of the pipe), and \(R\) is the radius of the pipe. Notice that the velocity varies from \(u_{max}\) at \(r = 0\) (centre of the pipe) to 0 at \(r = R\) (edge of the pipe).
02

Calculate the Volume Flow Rate

The volume flow rate \(Q\) can be calculated by integrating the velocity profile across the cross-sectional area of the pipe.\[Q = 2\pi\int_{0}^{R} u_{max}\left[1-\left(\frac{r}{R}\right)^{2}\right] rdr.\] Carry out the integration to obtain the flow rate.
03

Calculate the momentum flux

The momentum flux \(I\) can be calculated by integrating the product of the velocity profile and the radial component of the area over the cross-sectional area of the pipe. - the radial component of the area is simply \(2\pi r dr\).\[I=2\pi\rho\int_{0}^{R} u_{max} \left[1- \left(\frac{r}{R}\right)^2\right]^2 rdr\]Carry out this integration substituting the value of \(u_{max}\) from the volume flow rate to find the momentum flux.
04

Conclusion

By carrying out the above integrations and simplifications, we have found expressions for the volume flow rate and momentum flux. Note that both of these desired quantities depend on \(u_{max}\), the maximum velocity of the flow.

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Most popular questions from this chapter

A conical flask contains water to height \(H=36.8 \mathrm{mm}\) where the flask diameter is \(D=29.4 \mathrm{mm}\). Water drains out through a smoothly rounded hole of diameter \(d=7.35 \mathrm{mm}\) at the apex of the cone. The flow speed at the exit is approximately \(V=\sqrt{2 g y},\) where \(y\) is the height of the liquid free surface above the hole. A stream of water flows into the top of the flask at constant volume flow rate, \(Q=3.75 \times 10^{-7} \mathrm{m}^{3} / \mathrm{hr}\) Find the volume flow rate from the bottom of the flask. Evaluate the direction and rate of change of water surface level in the flask at this instant.

A cylindrical tank, \(0.3 \mathrm{m}\) in diameter, drains through a hole in its bottom. At the instant when the water depth is \(0.6 \mathrm{m}\) the flow rate from the tank is observed to be \(4 \mathrm{kg} / \mathrm{s}\). Determine the rate of change of water level at this instant.

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