4.5.1.2: Multiply Layers

In the previous sections, the density was assumed to be constant. For non constant density the derivations aren't clean'' but are similar. Consider straight/flat body that is under liquid with a varying density. If density can be represented by average density, the force that is acting on the body is $F_{total} = \int_{A} g\rho h dA \sim \bar{\rho} \int_{A} g h dA \tag{164}$ In cases where average density cannot be represented reasonably, the integral has be carried out. In cases where density is non–continuous, but constant in segments, the following can be said $F_{total} = \int_{A} g \rho h dA = \int_{A_{1}} g \rho_{1} h dA + \int_{A_{2}} g \rho_{2} h dA + \cdot \cdot \cdot + \int_{A_{n}} g \rho_{n} h dA \tag{165}$ As before for single density, the following can be written $F_{total} = gsin \beta \left[\rho_{1} \int_{A_{1}} \xi dA + \rho_{2} \int_{A_{2}} \xi dA + \cdot \cdot \cdot + \rho_{n} \int_{A_{n}} \xi dA \right] \tag{166}$ Or in a compact form and in addition considering the atmospheric'' pressure can be written as

Total Static Force

$F_{total} = P_{atmos} A_{total} + g sin\beta \sum_{i=1}^{n} \rho_{i}x_{ci}A_{i}\tag{167}$

where the density, $$\rho_{i}$$ is the density of the layer $$i$$ and $$A_{i}$$ and $$x_{ci}$$ are geometrical properties of the area which is in contact with that layer. The atmospheric pressure can be entered into the calculation in the same way as before. Moreover, the atmospheric pressure can include all the layer(s) that do(es) not with the contact'' area. The moment around axis $$y$$, $$M_{y}$$ under the same considerations as before is $M_{y} = \int_{A} g \rho \xi ^{2} sin \beta dA \tag{168}$ After similar separation of the total integral, one can find that

Total Static Moment

$M_{y} = P_{atmos}x_{c}A_{total} + g sin\beta \sum_{i=1}^{n} \rho_{i} I_{x'x'i}\tag{170}$

In the same fashion one can obtain the moment for $$x$$ axis as

Total Static Moment

$M_{x} = P_{atmos}y_{c}A_{total} + g sin\beta \sum_{i=1}^{n} \rho_{i} I_{x'y'i}\tag{171}$

To illustrate how to work with these equations the following example is provided.

Example 4.16

Consider the hypothetical Figure 4.25 The last layer is made of water with density of $$1000 [kg/m^3]$$. The densities are $$\rho_1 = 500[kg/m^3]$$, $$\rho_2 = 800[kg/m^3]$$, $$\rho_3 = 850[kg/m^3]$$, and $$\rho_4 = 1000[kg/m^3]$$. Calculate the forces at points $$a_1$$ and $$b_1$$. Assume that the layers are stables without any movement between the liquids. Also neglect all mass transfer phenomena that may occur. The heights are: $$h_1 = 1[m]$$, $$h_2 = 2[m]$$, $$h_3 = 3[m]$$, and $$h_4 = 4[m]$$. The forces distances are $$a_1=1.5[m]$$, $$a_2=1.75[m]$$, and $$b_1=4.5[m]$$. The angle of inclination is is $$\beta= 45^\circ$$.

Fig. 4.25 The effects of multi layers density on static forces.

Solution 4.16

Since there are only two unknowns, only two equations are needed, which are (170) and (167). The solution method of this example is applied for cases with less layers (for example by setting the specific height difference to be zero). Equation (170) can be used by modifying it, as it can be noticed that instead of using the regular atmospheric pressure the new "atmospheric'' pressure can be used as
${P_{atmos}}^{'} = P_{atmos} + \rho_1\,g\,h_1 \tag{172}$
The distance for the center for each area is at the middle of each of the "small'' rectangular. The geometries of each areas are
\begin{array}{lcr}
{x_c}_1 = \dfrac{a_2 + \dfrac{h_2}{\sin\beta}}{2}
& A_1 = ll \left( \dfrac{h_2}{\sin\beta} -a_2 \right)
& {I_{x^{'}x^{'}}}_1 =
\dfrac{ll\left(\dfrac{h_2}{\sin\beta}-a_2\right)^{3}}{36} +
\left({x_c}_1\right)^2\, A_1    \\
{x_c}_2 = \dfrac{h_2 + h_3}{2\,\sin\beta}
& A_2 = \dfrac{ll}{\sin\beta} \left(h_3 - h_2\right)
& {I_{x^{'}x^{'}}}_2 =
\dfrac{ll\left({h_3}-h_2\right)^{3}}{36\,\sin\beta} +
\left({x_c}_2\right)^2\, A_2    \\
{x_c}_3 = \dfrac{h_3 + h_4}{2\,\sin\beta}
& A_3 = \dfrac{ll}{\sin\beta} \left(h_4 - h_3\right)
& {I_{x^{'}x^{'}}}_3 =
\dfrac{ll\left({h_4}-h_3\right)^{3}}{36\,\sin\beta} +
\left({x_c}_3\right)^2\, A_3     \tag{173}
\end{array}
After inserting the values, the following equations are obtained
Thus, the first equation is
$F_1 + F_2 = {P_{atmos}}^{'} \overbrace{ll (b_2-a_2)}^{A_{total}} + g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\, {x_c}_i\, A_i \tag{174}$
The second equation is (170) to be written for the moment around the point "O'' as
$F_1\,a_1 + F_2\,b_1 = {P_{atmos}}^{'}\, \overbrace{\dfrac{(b_2+a_2)}{2}{ll (b_2-a_2)}}^ ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Chemical_Engineering/Map:_Fluid_Mechanics_(Bar-Meir)/04:_Fluids_Statics/4.5:_Fluid_Forces_on_Surfaces/4.5.1:_Fluid_Forces_on_Straight_Surfaces/4.5.1.2:_Multiply_Layers), /content/body/div[5]/div/p[2]/span, line 1, column 4  + g\,\sin\beta\sum_{i=1}^{3}\rho_{i+1}\,{I_{x^{'}x^{'}}}_i \tag{175}$
The solution for the above equations is
$F1= \begin{array}{c} \dfrac{ 2\,b_1\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\, {x_c}_i\, A_i -2\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\,{I_{x^{'}x^{'}}}_i } {2\,b_1-2\,a_1} - \\ \qquad\dfrac{\left({b_2}^{2}-2\,b_1\,b_2+2\,a_2\,b_1-{a_2}^{2}\right) ll\,P_{atmos}} {2\,b_1-2\,a_1} \end{array}$
$F2= \begin{array}{c} \dfrac{ 2\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\,{I_{x^{'}x^{'}}}_i -2\,a_1\,g\,\sin\beta\,\sum_{i=1}^{3}\rho_{i+1}\, {x_c}_i\, A_i} {2\,b_1-2\,a_1} + \\ \dfrac{ \left( {b_2}^{2}+2\,a_1\,b_2+{a_2}^{2}-2\,a_1\,a_2\right) ll\,P_{atmos}} {2\,b_1-2\,a_1} \end{array}$
The solution provided isn't in the complete long form since it will makes things messy. It is simpler to compute the terms separately. A mini source code for the  calculations is provided in the the text source. The intermediate results in SI units ([m], [$$m^2$$], [$$m^4$$]) are:
$\begin{array}{lcr} x_{c1}=2.2892& x_{c2}=3.5355& x_{c3}=4.9497\\ A_1=2.696& A_2=3.535& A_3=3.535\\ {I_{x'x'}}_1=14.215& {I_{x'x'}}_2=44.292& {I_{x'x'}}_3=86.718 \end{array}$
$F_1=304809.79[N] \tag{176}$
$F_2=958923.92[N] \tag{177}$