## Fluids Question

## Fluids Question

(OP)

Hello All,

I am studying for my PE exam and during the review I have come across a problem that I cannot solve. I know the answer, it's $11.45 but I'm not sure how that is being calculated. It's been a few years since I've taken fluids so I'm a little rusty to say the least. Any help as to how they arrived at that answer would help. I've attached a copy of the question to this thread.

Thanks,

I am studying for my PE exam and during the review I have come across a problem that I cannot solve. I know the answer, it's $11.45 but I'm not sure how that is being calculated. It's been a few years since I've taken fluids so I'm a little rusty to say the least. Any help as to how they arrived at that answer would help. I've attached a copy of the question to this thread.

Thanks,

## RE: Fluids Question

Also, you are not given the schedule of the PVC tubing and you don't know the height from the centre line of the pump. Therefore you do not know the head at the outlet of the vertical annular tube.

Yes, there is a K value of 2. But without knowing the velocity in the tube, you can't calculate the head loss. Without knowing the mass flow rate, you cannot calculate the velocity.

Without this information, the power required cannot be calculated accurately. It is therefore not possible to calculate the cost to 2 dp.

How did you get the $11.45???

HPost CEng MIMechE

## RE: Fluids Question

Where Velocity = Q/A

Head loss = (K x V^2)/2g

Need to know what Q is

HPost CEng MIMechE

## RE: Fluids Question

It's bit of a strange question... I'd start off by assuming a volumetric flowrate such that the discharge height does reach 3m. From there, it'd be iterative in order to reach a decent answer.

## RE: Fluids Question

This is a diabolical problem that I just spent nearly 2 hours working out in MathCad with access to the Interwebz. In that room with just books and a calculator it would take a week, but it is solvable. I get $1.03.

The key is to realize that you can use Dynamics equations to get the exit velocity since you know how high the water is able to go. At the top of the 3 m arc, vertical velocity is zero, so:

v

_{0}=(6gh)^{0.5}/3=4.43 m/sWith the annular area and that velocity, volume flow rate is 0.174 m

^{3}/sI used the Petroleum Method (see FAQ378-1142: Flow in Annular Space) to get an effective flow diameter of 19.04 cm for a dP in the vertical pipe of 0.367 psid plus the gradient of the 180 cm vertical leg gives 2.93 psid in the annulus. That gives a pump discharge of 2.97 psid over atmospheric. Since they didn't give you an elevation difference between the surface of the water and the pump inlet I assumed that the pump suction was on the level of the bottom of the pond, so the pump head is 1.97 psi and the power required is 2.56 kW. Run that for 8 hours and you are at 20.5 kW-hr or $1.03.

I'm sure that this isn't the way the authors intended to solve this (I don't use the equation that has loss value), but it should be close.

David Simpson, PE

MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual.—Galileo Galilei, Italian Physicist## RE: Fluids Question

## RE: Fluids Question

David Simpson, PE

MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual.—Galileo Galilei, Italian Physicist## RE: Fluids Question

## RE: Fluids Question

Everything needed is presented in the problem statement, but it presumes that supplementary information is available such as friction factors, actual dimensions of the nominal pipe, etc. These tests presume the making of reasonable assumptions for missing information. The most important phase of solving these problems is to very carefully read and understand the actual problem. The problems are usually stated in a manner that provides needless or irrelevant information while leaving the happy test-taker to provide necessary missing information through the use of supplementary information from allowed texts or the exercise of reasonable engineering judgment.

Note: I hereby exercise my "white hair" or "old fogey" exemption from actually doing any calculations. (There are some benefits to advanced age, and besides, I already did all that pesky stuff when I took those exams.)

1- Always start by clearly listing your idealizations and assumptions. In this case, these would include assuming some reasonable temperature and altitude values. (800 degrees F and 75,000 feet would obviously not qualify as reasonable values while the assumption of pumping pure water and no leakage from the pump or piping system would qualify as being reasonable.)

2- Start with the maximum height of the spray above the plane of the annular discharge opening. That provides the basis for finding the required vertical velocity of the maximum velocity within the velocity profile of the annular flow.

3- The peak velocity within the profile can be used to determine the volumetric flow rate of the entire flow. (This stage can probably be adequately solved in a couple of iterations to get a sufficiently reasonable estimate of the total flow rate. The ratio of the maximum velocity to the average velocity is likely to be small, but it may prove to be significant. Only actually doing the calculations will provide that information.)

4- The total volumetric flow rate can then be used with the piping dimensions and information provided (or available in tables in allowed texts, etc.) to determine the head losses in the various portions of the piping system. Note that the diagram suggests short-radius bends in the pipes, so the losses associated with them can be taken into account by adding "equivalent lengths" to the stated pipe lengths. The loss factor for the transition from the pipe to the annulus is given. The loss factor for the transition from the pool to the pump inlet pipe must be assumed for the configuration shown. The pool depth is apparently sufficient that vortex formation is unlikely to be a concern, the calculated flow rate should be used to verify that vortex formation will not provide a troublesome additional loss. The diagram clearly suggests that the plane of the discharge opening of the annulus is effectively the same as the surface of the pool.

5- The wonderfully efficient pump and motor (92% combined--WOW!), suggests that any temperature rise due to fluid friction can be presumed to be trivial.

6- The total volumetric flow rate and the total combined head loss provides the basis for calculating the total hydraulic power provided by the pump.

7- The stated efficiency of the combined pump-motor system provides the basis for calculating the electric power draw of the motor.

8- The stated price per kwhr provides the basis for calculating the power cost for the stated operating period.

Valuable advice from a professor many years ago: First, design for graceful failure. Everything we build will eventually fail, so we must strive to avoid injuries or secondary damage when that failure occurs. Only then can practicality and economics be properly considered.

## RE: Fluids Question

However, I am still not sure that there is enough detail to be able to work this out to give a cost to 2 decimal places.

How much room are you allowed for error? Do you get marks for just knowing how to go about calculating the result?

Hpost

## RE: Fluids Question

David Simpson, PE

MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual.—Galileo Galilei, Italian Physicist