Topic 1 of 10 — free theory
Fluid Properties & Hydrostatics
Density, specific weight, viscosity, pressure distribution, and forces on plane surfaces.
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The properties you will use everywhere
Almost every water-resources question starts from the same handful of properties. Learn them once and they pay off across hydrostatics, pipe flow, open channels, and treatment:
γ = ρg
| γ | specific weight — weight per unit volume |
| ρ | mass density — mass per unit volume |
| g | gravitational acceleration, 32.2 ft/s² (9.81 m/s²) |
SG = γsubstance / γwater
| SG | specific gravity, dimensionless |
ν = μ / ρ
| μ | dynamic (absolute) viscosity |
| ν | kinematic viscosity |
For water at 60°F (the usual exam reference): γ = 62.4 lb/ft³, ρ = 1.94 slugs/ft³, ν = 1.22×10−5 ft²/s. In SI: γ = 9.79 kN/m³, ρ = 1000 kg/m³.
Hydrostatic pressure
Pressure in a static fluid increases linearly with depth. Using gauge pressure (atmospheric pressure taken as zero — the exam almost always works in gauge):
p = γh
| p | gauge pressure at depth h |
| h | vertical depth below the free surface |
FR = γ hc A
| FR | magnitude of hydrostatic force on a plane surface |
| hc | vertical depth of the surface's centroid |
| A | area of the surface |
yp = yc + Ixx,cycA
| yp | distance from the free surface to the centre of pressure, measured along the incline |
| yc | distance from the free surface to the centroid, measured along the incline |
| Ixx,c | second moment of area about the centroidal axis (rectangle: bh³/12, with h along the incline) |
For an inclined surface, depth and inclined distance are related by h = y sin θ. The centre of pressure is always below the centroid — pressure grows with depth, so the resultant acts low.
Advanced extension: multi-fluid manometers
Do not memorise a different equation for every tube shape. Start at a point of known pressure and “walk” through each fluid column: moving downward adds γΔh; moving upward subtracts it. Pressure is continuous across an interface at the same elevation.
pB = pA + Σ(γΔh)down − Σ(γΔh)up
Check: in one connected, static fluid, points at the same elevation have the same pressure. That single test catches most sign errors before arithmetic begins.
Worked example Force on a vertical sluice gate
Given:
- Rectangular gate 4 ft wide × 6 ft tall, vertical.
- Top edge of the gate is 3 ft below the water surface.
- Fresh water, γ = 62.4 lb/ft³.
Solution:
- Centroid depth: hc = 3 + 6/2 = 6.0 ft. Area: A = 4 × 6 = 24 ft².
- Resultant force: FR = 62.4 × 6.0 × 24 = 8,986 lb (≈ 8.99 kips).
- Centre of pressure: Ixx,c = (4)(6)³/12 = 72 ft4. yp = 6.0 + 72/(6.0 × 24) = 6.0 + 0.50 = 6.50 ft below the surface.
- So the 8.99-kip resultant acts 0.50 ft below the centroid — that offset is what creates the moment on the gate hinges.
Answer: FR ≈ 8.99 kips acting 6.50 ft below the free surface.