Fluid Statics Calculator | Hydrostatic Pressure & Manometry

📝 Configuration

Load Example Case:

Surface Pressure ($P_0$) [kPa]:

Depth of Interest ($z$) [m]:

Fluid Layers (Top to Bottom)

Density $\rho$ [kg/m³]

Thickness $h$ [m]

Density $\rho$ [kg/m³]

Thickness $h$ [m]

Surface Shape:

Dimension 1:

Dimension 2:

Dimension 3:

Inclination Angle ($\theta$) [degrees]:

Centroid Depth ($h_c$) [m]:

Fluid Density ($\rho$) [kg/m³]:

Atmospheric Pressure ($P_{atm}$) [kPa]:

Number of fluids in Left leg (0-3):

Number of fluids in Right leg (0-3):

Manometer Fluid

Manometer Fluid Density ($\rho_{man}$) [kg/m³]:

Liquid Level Deflection ($\Delta h$) [m] — Right surface height minus Left surface:

📊 Simulation Results & Diagrams

📊 Results Summary
📥 Download

Depth of Interest ($z$) 121.92 m

Pressure at Depth ($P_z$) 121.9190 kPa 20.5940 kPa (Gauge)

Surface Pressure ($P_0$) 101.33 kPa Absolute head reference

📈 Pressure Profile vs Depth

🖨️ Raw Fortran Output

 --- RESULTS MODE A ---
Surface Pressure (P0)      =     101.3250 kPa
Target Depth               =       2.5000 m
Pressure at Target Depth   =     121.9190 kPa

 --- INTERFACE PRESSURE DISTRIBUTION ---
    Boundary      Depth [m]    Pressure [kPa]
           0          0.000          101.3250
           1          2.000          117.0156
           2          5.000          146.4356

📘 Calculation Methodology

Mathematical Model & Theory

Fluid statics analyses fluids at rest where shear stresses are zero. Pressures and resultant forces are modeled as follows:

1. Hydrostatic pressure in layers

Hydrostatic pressure increases linearly with depth in a constant density liquid according to:

$$P(z) = P_0 + \int_{0}^{z} \rho(z) g dz = P_0 + \sum_{i=1}^{k} \rho_i g h_i$$

2. Forces on Immersed Plane Surfaces

The resultant force $F_R$ acts normal to the surface at the center of pressure ($y_{cp}$). By taking moments of the pressure force about the centroidal axis, we obtain:

$$F_R = P_{c} A = (\rho g h_c + P_0) A, \quad y_{cp} = y_c + \frac{I_{xx,c}}{y_c A}$$

3. Manometry

A manometer works on the principle that pressure at equivalent levels in a continuous fluid is equal. Thus, we balance heads in the Left and Right legs:

$$P_{left} + \sum \rho_{left,i} g h_{left,i} = P_{right} + \sum \rho_{right,j} g h_{right,j} + \rho_m g \Delta h$$

Academic References

White, Frank M.: Fluid Mechanics, McGraw-Hill.
Munson, B. R., Young, D. F., & Okiishi, T. H.: Fundamentals of Fluid Mechanics, Wiley.

Worked Engineering Example

Problem Statement:
A vertical rectangular gate of width $b = 2.0\text{ m}$ and height $a = 3.0\text{ m}$ is submerged in water ($\rho = 1000\text{ kg/m³}$). The centroid of the gate is at depth $h_c = 5.0\text{ m}$. Calculate the resultant gauge force $F_R$ and the center of pressure depth $h_{cp}$.

Step-by-step Solution:
1. Calculate the surface area $A$:
$$A = b \cdot a = 2.0 \times 3.0 = 6.0\text{ m²}$$ 2. Calculate the moment of inertia about the centroidal axis $I_{xx,c}$:
$$I_{xx,c} = \frac{1}{12} b a^3 = \frac{1}{12} \times 2.0 \times 3.0^3 = 4.5\text{ m⁴}$$ 3. Calculate the resultant gauge force $F_R$:
$$F_R = \rho g h_c A = 1000 \times 9.80665 \times 5.0 \times 6.0 = 294,200\text{ N} = 294.2\text{ kN}$$ 4. Since the gate is vertical ($\theta = 90^\circ$), the inclined coordinate $y$ is equal to depth $h$ ($y_c = h_c = 5.0\text{ m}$). Calculate center of pressure $y_{cp}$:
$$y_{cp} = y_c + \frac{I_{xx,c}}{y_c A} = 5.0 + \frac{4.5}{5.0 \times 6.0} = 5.0 + 0.15 = 5.15\text{ m}$$ 5. Calculate center of pressure depth $h_{cp}$:
$$h_{cp} = y_{cp} \sin(90^\circ) = 5.15\text{ m}$$ Final Results:
• Gate resultant force: 294.2 kN
• Center of pressure depth: 5.15 m (15 cm below the centroid)

📐 Fluid Statics Calculator