cancel
Showing results for 
Search instead for 
Did you mean: 
cancel
Showing results for 
Search instead for 
Did you mean: 

Community Tip - If community subscription notifications are filling up your inbox you can set up a daily digest and get all your notifications in a single email. X

Preliminary design hydroelectric station problem.mcdx.pdf

KarlBogha
12-Amethyst

Preliminary design hydroelectric station problem.mcdx.pdf

Hi PTC Community,

This is a long example problem from a textbook. It goes thru the preliminary steps of the design of a hydro electric power station.

You may find it interesting enough to pick up the book yourself. If you need the Prime file please request.

N/A.
6 REPLIES 6

Looks like some questionable use of units early on - divide by 75 (why?) in the first equation, but with no mention of gravity; then the result has units of kg.m/s but is stated as"'metric hp"!

kg.m/s doesn't even have units of power - do you mean kgf.m/s?  In that case, it's volumetric weight of water, not density... but also 735.5 [W / metric hp] ÷ 9.81 [gravity] does (very nearly) equal 75, so at least it explains that.

Wouldn't it have been easier to calculate directly in watts in the first place, and let MathCAD convert to other units if you wish?

Hi John,

The units do not work out in the common terms. I was not going to work thru the Prime/Mathcad units to make it work because it was rather challenging. I left it as it were, and at times presented the final unit. You may go to Google Books and run thru the pages of the book. There were tables and charts I did not scan and insert them.

There were two chapters on hydro electric station, and at the end was this lengthy example.

N/A.

Hi John,

75 in the equation represents:

75 kg m/sec = 1 metric h.p. = 0.736 kW

Now here it is mass x speed, so mass would be the kg and velocity/speed m/sec, so its mv for momentum.

If its weight (m) then is mass times acceleration for the kg.

The calculations are from a mix of mechanical and electrical disciplines, and in this example the professor choose to explain his working in such a style possibly for the tie to the design criteria or process of identifying the equipment sizing as the solution progresses.

I had some difficulty with the units interpreted by Prime/Mathcad and said let it go and continued with the flow of the solution.

N/A.

The likely issue with units is that the 75 isn't unitless: it's [power / power / gravity], so ultimately s^2 / m (also kg / N).

As you rightly point out, what that first step calculates is not power but momentum...

I think the confusion primarily stems from trying to use kg as a force unit rather than a mass unit.  It also seems very strange to take a complete set of SI inputs and then immediately calculate a power in imperial units, only to convert it back to kW in the next step!

Hello John,

On another point with regards to your last statement above, in the metric engineering world for oil and gas, piping, and energy, both units are used in countries that are English speaking and once were Imperial (English) units. The new production of pipes and a lot of material/equipment produced were interconnected to existing plant. The manufacture of new plant related closely to the system in existing, plus the capacity handled was usually the same.

I used the English units and had to migrate to the metric units, so for me its quite okay using both.

Cheers.

N/A.

I can identify with the use of mixed units - as an automotive transmissions guy in the UK, I tend to think in bhp (746 W), rpm, and mph, as well as often needing to use km/h for the benefit of my European colleagues; but generally I try to keep these as input units and convert everything to m/s and W or kW for the purposes of calculation.  Once the calculation produces outputs I may then convert them back to 'familiar' units for sanity checking, but internally I try to keep the units consistent.

Top Tags