HW 4 for CS 4220

You may (and probably should) talk about problems with the each other, with the TAs, and with me, providing attribution for any good ideas you might get. Your final write-up should be your own.

The Darcy friction coefficient $f$ for turbulent flow in a pipe is defined in terms of the Colebrook-White equation for large Reynolds number Re (greater than 4000 or so):

$$\frac{1}{\sqrt{f}}=-2{\log }_{10}(\frac{ϵ/D_h}{3.7}+\frac{2.51}{\mathrm{R}\mathrm{e}\sqrt{f}})$$

Here $ϵ$ is the height of the surface roughness and $D_h$ is the diameter of the pipe; we refer to $ϵ/D_h$ as the roughness. If we let $u=1/\sqrt{f}$, we can rewrite this as a fixed point equation

$$u_{\ast }=-\alpha \log (\beta +\gamma u_{\ast })$$

where $\alpha =2/\log (10)$, $\beta =(ϵ/D_h)/3.7$, and $\gamma =2.51/\mathrm{R}\mathrm{e}$. Alternately, we can write this as the problem of finding a zero of

$$g(u)=u+\alpha \log (\beta +\gamma u).$$

In this problem, we consider the computation of $u_{\ast }>0$ (and hence $f$) for different values of the Reynolds number. We note that $\alpha$, $\beta$, and $\gamma$, are all positive numbers that are less than one in this problem, and $u_{\ast }=1/\sqrt{f}$ must also be positive.

As a test case, we consider flow through a 10 cm pipe with roughness height of 0.1 mm (so the roughness factor is ${10}^{-3}$), and Reynolds numbers of ${10}^4$, ${10}^5$, and ${10}^6$.

1. Note that $0<\log (1+z)<z$ for $z>0$. Using properties of logs, show that for $u>0$, $g(u)=u+\alpha \log (\beta +\gamma u)$ satisfies

$$u+\alpha \log (\beta )\le g(u)\le (1+\alpha \gamma /\beta )u+\alpha \log (\beta )$$

2. Using the previous result, give an interval where $g$ changes sign (so $g(u)=0$ is guaranteed to have a solution in this interval). Verify the sign change for our test setting with Reynolds number of ${10}^4$.

3. Using the lower bound in the previous step as a starting guess, run the fixed point iteration $u_{k+1}=-\alpha \log (\beta +\gamma u_k)$ and plot $|g(u_k)|$ vs $k$ on a semilog scale. Run the calculation for 100 steps for Reynolds numbers ${10}^4$, ${10}^5$, and ${10}^6$. What do you observe about the convergence?

4. Subtract the fixed point equation from the fixed point iteration equation and use that $\log (1+z)=z+O(z^2)$ to get an error iteration of the form

$$e_{k+1}=-\frac{\alpha \gamma }{\beta +\gamma u_{\ast }}{e}_k+O(e_k^2).$$

5. Numerically verify that the rate of convergence from part 4 agrees with the rates of convergence seen in the part 3. You will want to use that if $e_{k+1}\approx \rho e_k$, then $\rho$ can be estimated from three successive iterates $u_{k+1}$, $u_k$, and $u_{k-1}$.

6. Write a Newton iteration for finding a zero of $g$ (with the lower bound from part 2 as an initial guess). Plot the residuals on a semilog scale for Reynolds number ${10}^4$ to illustrate quadratic convergence.

Answer: Fill in Newton iteration is given below, then add the requested plot.

7. Not a problem, but we can use the previous codes to show on a log-log plot the behavior of $u$ and its upper and lower bounds for Reynolds numbers between ${10}^4$ and ${10}^8$.