Hydraulics. ONLINE CALCULATION

The force of hydrostatic pressure on a flat rectangular surface

Task № 1.1.1

Determine the force of hydrostatic pressure of water on a flat rectangular lid in the lateral vertical face of the tank (Fig. 1.1.1.a). Atmospheric pressure 101325 Pa acts on the free surface of the liquid. Cover size: width b = 0.1 m, height h = 0.3 m.

The condition of the problem:b=0.1 [m],h=0.3 [m]

p a https://www.k123.org.ua/ h b C D

Fig. 1.1.1.a

Guidelines (Methodical recommendations): calculate the magnitude of the force vector, the depth of immersion of the pressure center, build a plot of hydrostatic pressure on the wetted surface of the flat cover. The vector of hydrostatic pressure force on a flat surface is directed normally (at an angle of 90 degrees to the surface) and passes through the center of gravity of the plot.

Calculation algorithm

  1. depth of immersion of the center of gravity of the surface \(h_{C}=\frac{h}{2} [m]\);
  2. hydrostatic pressure in the center of gravity of the surface \(p=\rho gh_{C} [Pa]\);
  3. wetted surface area \(w=b\cdot h [m^2]\);
  4. the force of hydrostatic pressure on the wetted surface \(P=p\cdot w [N]\);
  5. the moment of inertia of the surface relative to the horizontal axis passing through the center of gravity \(I_{0}=\frac{b \cdot h^{3}}{12} [m^4]\);
  6. depth of immersion of the pressure center \(h_{D}=h_{C}+\frac{I_{0}}{h_{C}\cdot w} [m]\);
  7. to build a diagram of hydrostatic pressure.

Calculation

1.Depth of immersion of the center of gravity:

\[h_{C}=\frac{h}{2} [m],\]

\[h_{C}=\frac{0.3}{2}=0.15 [m].\]

2.Hydrostatic pressure in the center of gravity of the surface:

\[p=\rho gh_{C} [Pa],\]

\[p=1000 \cdot 9.81 \cdot 0.15=1471.5 [Pa].\]

3.Wet surface area:

\[w=b\cdot h [m^2],\]

\[w=0.1\cdot 0.3=0.03 [m^2].\]

4.The force of hydrostatic pressure on the surface:

\[P=p\cdot w [N],\]

\[P=1471.5\cdot 0.03=44.145 [N].\]

5.The moment of inertia of the surface relative to the horizontal axis passing through the center of gravity:

\[I_{0}=\frac{b \cdot h^{3}}{12} [m^4],\]

\[I_{0}=\frac{0.1 \cdot0.3^{3}}{12}=0.000225 [m^4].\]

6.Depth of immersion of the pressure center:

\[h_{D}=h_{C}+\frac{I_{0}}{h_{C}\cdot w} [m],\]

\[h_{D}=0.15+\frac{0.000225}{0.15\cdot 0.03}=0.2 [m].\]

7.Diagram of hydrostatic pressure

p a h h b C C h D D C D p a P https://www.k123.org.ua/

Fig. 1.1.1.b

Reply:

  1. the depth of immersion of the center of gravity of the surface relative to the free surface of the liquid - \(h_{C}=0.15 [m]\),
  2. pressure in the center of gravity - \(p=1471.5 [Pa]\),
  3. wetted surface area \(w=0.03 [m^2]\),
  4. hydrostatic pressure force - \(P=44.145 [N]\),
  5. the moment of inertia about the horizontal axis passing through the center of gravity of the surface - \(I_{0}=0.000225 [m^4]\),
  6. the depth of immersion of the pressure center relative to the free surface of the liquid - \(h_{D}=0.2 [m]\)

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