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Model setting without an explicit use of $ \color{white} \vec{g}$ in the momentum equation

Let us take a look at the equation () as if its right hand side were known. Then we can introduce an alternative quantity, denoted by $ \color{white} p_{-\varrho gh}$, to the static pressure $ \color{white} p$

$\displaystyle \color{white} p_{-\varrho gh} = {}'' p - \varrho g h {}'' = p - \varrho \vec{g} \cdot \vec{r} = p + \varphi \quad .$(3.36)

By means of ([*]) we can formally substitute equation ([*]) by the following one

$\displaystyle \color{white} \vec{U} \cdot \nabla \vec{U} - \nu \Delta \vec{U} = - \tfrac{1}{\varrho} \nabla p_{-\varrho g h} \quad ,$(3.37)

for the right hand side takes the same values. This is the consequence of a fact that field ([*]) has potential and thus a change in potential energy of an arbitrary element of a fluid is not dependent on its path, but on its initial and final position only.

By introducing $ \color{white} p_{-\varrho gh}$ we formally drop img146 1 out from the momentum equation. However, in order to obtain the same solution as in the previous settings, it is necessary to alter the presrciption ([*]) by adding $ \color{white} \varrho \vec{g} \cdot (\vec{r}_{Outlet} - \vec{r}_{Inlet})$ to its right hand side3.7. By doing this we obtain a new prescription, but this time for the quantity $ \color{white} p_{-\varrho gh}$

$\displaystyle \color{white} p_{-\varrho g h, tot, In} (\vec{r}) = \varrho \vec{g} \cdot (\vec{r} - \vec{r}_{0,H} + \vec{r}_{0,L} - \vec{r}_{Inlet}) \quad .$(3.38)

If we calculate the mean value of ([*]), we obtain

$\displaystyle \color{white} \overline{p_{-\varrho g h, tot, In}}$$\displaystyle \color{white}= \tfrac{1}{S_{Inlet}} \iint_{Inlet} \varrho \vec{g} \cdot (\vec{r} - \vec{r}_{0,H} + \vec{r}_{0,L} - \vec{r}_{Inlet}) \mathrm{d} S$(3.39)
$\displaystyle \color{white}$$\displaystyle \color{white}= \tfrac{1}{S_{Inlet}} \iint_{Inlet} \varrho \vec{g} \cdot \vec{r} \mathrm{d} S + \varrho \vec{g} \cdot (- \vec{r}_{0,H} + \vec{r}_{0,L} - \vec{r}_{Inlet})$(3.40)
$\displaystyle \color{white}$$\displaystyle \color{white}= \varrho \vec{g} \cdot (\vec{r}_{0,L} - \vec{r}_{0,H})$(3.41)
$\displaystyle \color{white}$$\displaystyle \color{white}= \varrho g h \quad .$(3.42)

Prescription at the outlet surface remains formally the same as ([*]), but this time for the quantity $ \color{white} p_{-\varrho gh}$

$\displaystyle \color{white} p_{-\varrho gh, Out} (\vec{r}) = \varrho \vec{g} \cdot (\vec{r} - \vec{r}_{Outlet}) \quad ,$(3.43)

and hence its mean value is zero.

We can see that in this setting there is no need to know a position of a turbine with respect to the water level and there is also no need to even take the measurement of $ \color{white} h_{IO}$. It only suffice to know $ \color{white} h$, the head.