Question

In the figure, the AB gate 3,464 long and 2m wide (perpendicular to the paper) separates 2 compartments. The piston weighs 20 KN and its area is 1 m2, determine F (in KN) so that the system is in equilibrium.

F 20 kPa I 15 GAS T 2.5 DR = 0.6 2.4 DR = 1.5 E 1.5 DR = 0.8 DR = 2 3 DR = 1 1.5 t 1.1 DR = 1.2

DR= Relative density

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Answer #1

F 20 kPa T 15 GAS T + DR = 0.6 2.5 DR = 1.5 2.4 DR = 0.8 1.5 DR = 2 3 Left hydrostatic force DR = 1 1.5 + DR = 1.2 1.1 Ľ Righ

As can be seen that the right and left hydrostatic forces act at the same point and thus no moments are created. We just have to balance the forces.

Since there are multiple fluids layers. We will CONVERT these different fluid layers to water.

On the left,

\frac{P_{left}}{\omega }+H_1S_1+H_2S_2+H_3S_3=H_wS_w

where

  • P(left) = 20 kPa = 20e3 Pa
  • \omega = Specific weight of water = 9810 N/m3
  • H1 = 1.5 m
  • H2 = 2.4 m
  • H3 = 3 m
  • S1 = 0 [We have already accounted this effect by considering the pressure]
  • S2 = 1.5
  • S3 = 2
  • Hw = Height of water in m = ?
  • Sw = SG of water = 1

\frac{P_{left}}{\omega }+H_1S_1+H_2S_2+H_3S_3=H_wS_w

\frac{20e3}{9810}+1.5\times 0+2.4\times 1.5+3\times 2=H_w\times 1=>H_{w-left}=11.638\:m

On the right,

\frac{\frac{W_{piston}+F}{Area_{piston}}}{\omega }+H_1S_1+H_2S_2+H_3S_3+H_4S_4=H_wS_w

where

  • W(piston) = Weight of piston = 20kN = 20e3 N
  • F = Additonal force exerted = ?
  • Area(piston = 1 m2
  • \omega = Specific weight of water = 9810 N/m3
  • H1 = 2.5 m
  • H2 = 1.5 m
  • H3 = 1.5 m
  • H4 = 1.1 m
  • S1 = 0.6
  • S2 = 0.8
  • S3 = 1
  • S4 = 1.2
  • Hw = Height of water in m = ?
  • Sw = SG of water = 1

\frac{\frac{W_{piston}+F}{Area_{piston}}}{\omega }+H_1S_1+H_2S_2+H_3S_3+H_4S_4=H_wS_w

\frac{\frac{20e3+F}{1}}{9810}+2.5\times 0.6+1.5\times 0.8+1.5\times 1+1.1\times 1.2=H_w\times 1

H_{w-right}=7.5587+\frac{F}{9810}

Since there has to be equilibrium, H(w-left) = H(w-right)

11.638=7.5587+\frac{F}{9810}=>\mathbf{Force = F = 40017.58\:N=40.018\:kN}

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