Negative head in Darcy's flow equation

This question appeared in my midterms.

A pipe transports water from a storage tank A to storage tank B. Storage tank A is at an elevation of 20 m above B, the pipe length is 6.4 km. The present discharge rate is 75% of the required discharge rate. A new design is proposed which would lay a new 2.4 km pipe segment (alongside the older one) running from point D to B.

(Consider the two pipes from DB to be in parallel, nothing complex) Diameter of original pipe is 200mm. Calculate the diameter for this new pipe that would allow the required discharge rate to occur. Assume friction factor 0.02 (fanning friction factor) for all pipes. (Assume no minor losses due to bends and other stuff. Flow friction only).

Here's how I went about the solution.

First I found the flow from the original design using:

$$H=\frac{fl Q^2}{3d^5}$$

$$Q= 0.0122 m^3/s$$

Next

$$Q = \frac {3}{4} Q_r$$

$$Q_r = 0.01633 m^3/s$$

Now I go to the new design and take head loss equation in the first segment of original pipe running from AB-1.

$$H = \frac {fl_1Q_1^2}{3d^5} + \frac {fl_2Q_2^2}{3d^5}$$

Since they're part of the same original pipe, everything is same except length and discharge, so:

$$H =\frac {f}{3d^5} (4000Q_1^2 +2400Q_2^2)$$

$$20 =\frac {0.02}{3(0.2)^5} (4000Q_1^2 +2400Q_2^2)$$

$$\frac {24}{25} = 4000Q_1^2 + 2400Q_2^2$$

$$\frac {24}{25} - 40000 (0.01633)^2 = 2400 Q_2^2$$

And here is where I hit a little snag:

$$0.96 - 1.0667 = 2400 Q_2^2$$

$$-0.1066 = 2400 Q_2^2$$

Negative. The square of the flow is negative.

How do I even proceed from here? The only thing that comes to mind is possibly calling the flow imaginary. But that makes no sense.

Is my solution correct? What did I do wrong? What does one do if this happens somewhere? Can you please look over my solution to see if I made a mistake anywhere?

• In case anyone's wondering what that $3$ in the denominator of each occurrence of the Darcy-Weisbach equation is all about, I think it's supposed to be an approximate value of $\mathrm{\pi}^2g/32$ in SI units. Aug 20 at 11:31
• Here's a thought experiment: Go back to the original scenario with a single straight pipe; then tilt the whole system so that $\mathsf{B}$ is above $\mathsf{A}$, i.e. the head is indeed negative. What do you think would happen? Hint: it's not an imaginary flow rate. Harder question: could you reformulate the Darcy-Weisbach equation in such a way that it covers the case where you don't know the sign of the head in advance? Aug 20 at 13:06

1 Answer

The substitution of 0.01633 for $$Q_1$$ may be incorrect. That is the total required discharge, and not the discharge from the original pipe outlet. Did you mean to substitute the original value of $$Q=0.0122$$ for $$Q_1$$? Doing that gives a positive value at the stage you are currently at. Follow that to completion and I imagine you will obtain a reasonable value.

• No, $Q_1$ is the flow in pipe $\mathsf{A}\mathsf{D}$ in the new situation, not the flow in the old situation. Aug 20 at 13:08