Different equations for Hydropower

Question

I was doing some reading in hydropower and I came across these two power equations:

$$P = ρghQ $$

and

$$ P = QHg $$ Q is flow rate, ρ is water density, g is gravity, h is height, and H is head.

I do not understand why the second equation has no density term. To my understanding it just seems like the second equation is simply wrong. But I did not just want to assume that. I have no reference to the second equation as I came across it on some lecture slides, it appears to be form another text. I do not know its source but here is a picture

Answer 1

Unless the second version has Q in litres per second ie kg per second, which is a version hydraulic engineers use.

They will also use g as 10 instead of 9.81, because it is "close enough" when the actual power is estimated using an efficiency of 50%

This assumes the water density is 1000kg/m^3 - worked with hydraulic engineers on small hydro and it is a quick ball-park calculation for the estimated power output.

Edit: So replacing g with 10 using the example figures you just added will give 30kW, then at 50% efficiency that is 15kW - close enough to get a reality check.

Given that the river flow rate is estimated using pooh sticks and a measurement of the cross-sectional area it will be good enough.

Often a lot of effort is put into making a V-notch weir on the smaller rivers as the next step in improving the estimation.

A friend put a small turbine in on a river to produce 1kW, great care was taken to get all the flow and keep the water clean ie no sticks or leaves. As it was 500m from the house we also upped the voltage to 600V from the 230V AC and dropped it back at the house - reducing cable size and cost.

Answer 2

The second equation is wrong, while the result is correct. This just happens to work, as the density of water can be assumed to be $\rho=1'000kg/m^3$. So basically, the numerical result you get from the second equation (using SI-units $m^3/s$, $m$ and $m/s^2$) would have to be multiplied by $1000$ to lead to the result in Watts.

In other words, the second equation yields a result in Kilowatts, assuming a density of $\rho=1'000kg/m^3$.

Answer 3

Let's try plugging in the units:

$P = \rho ghQ = \dfrac{kg}{m^3}*\dfrac{m}{s^2}*m*\dfrac{m^3}{s} = \dfrac{kg-m^2}{s^3} = watt$

$P = QHg = \dfrac{m^3}{s}*m*\dfrac{m}{s^2} = \dfrac{m^5}{s^3} = ?$

Obviously, the second equation was wrong. (You are correct to say it missed the "mass density" in the equation)

Answer 4

It looks like it was a matter of units. First equation is usable with any units and the power results in the appropriate units.

The second equation however needs the numbers to be in specific units and the result is also restricted to resulting kW. Thanks to these unit requirements, second equation does come with an implicit density conversion of 1000kg/m3 as the density of water. It comes from an age when we weren't punching everything into excel spreadsheets and calculators/computers, and the efficient route was to treat something commonly approximated as 997kg/m3 as approximately 1000kg/m3.

Finally some comments (hopefully clarifying the why):

Mad props to poster for coming back and putting up the equation units.

At least a bit of shame on everyone such that came here and said the second equation is wrong. It may be an outdated way of doing things, but it works well enough for most. People who punch things into spreadsheets and calculators unaware of binary and floating point behavior can be just as wrong under certain conditions. Heck to assume you know the density of your water to this degree without even considering temperature and impurities can be just as wrong too. Engineering is always an approximation, and processes, education, and methods change over time to meet the needs of life. Ultimately we all answer to the same laws of physics, but all the different disciplines use their own approximations and equations to get their jobs done.