CELE Hydraulics & Fluid Mechanics Reviewer 2026

Answer: A. 78.48 kPa

Step 1 — Identify the formula for gauge pressure: p = γh. Step 2 — Substitute values: p = 9.81 kN/m³ × 8 m = 78.48 kPa. Step 3 — Why not 88.29 kPa? That corresponds to h = 9 m (9.81 × 9). Why not 98.10 kPa? That is p at h = 10 m. Why not 68.67 kPa? That is p at h = 7 m. The key point: gauge pressure ignores atmospheric pressure and increases linearly with depth.

Answer: A. 20 m

Step 1 — The pressure head formula is h = p/γ. Step 2 — Substitute: h = 196.2 kPa ÷ 9.81 kN/m³ = 20 m. Step 3 — Check units: kPa ÷ kN/m³ = (kN/m²) ÷ (kN/m³) = m. ✓ Step 4 — Distractors: 15 m → p would be 147.15 kPa; 25 m → p would be 245.25 kPa; 10 m → p would be 98.1 kPa. Pressure head is a useful way to express pressure as an equivalent column height of fluid.

Answer: A. Equally in all directions and undiminished

Step 1 — Pascal's Law: any pressure applied at one point of a confined fluid is transmitted equally and without diminishment to every point within the fluid and to the walls of the container. Step 2 — This is why hydraulic jacks and brakes work. Step 3 — 'Only downward' is wrong — that describes gravity-driven hydrostatic pressure increase, not Pascal's transmission. Step 4 — 'Proportionally to depth' describes how hydrostatic pressure varies, not how an applied pressure is transmitted. Step 5 — 'Only along the axis' is incorrect; fluids are isotropic in pressure transmission.

Answer: A. 29.43 kN

Step 1 — Area: A = 1.5 × 2 = 3.0 m². Step 2 — Centroid depth: h̄ = 2/2 = 1.0 m (half the height for a surface-piercing rectangle). Step 3 — Total force: F = γh̄A = 9.81 × 1.0 × 3.0 = 29.43 kN. Step 4 — 58.86 kN is double; that would occur if h̄ = 2 m (centroid at the bottom edge). Step 5 — 44.15 kN would require h̄ ≈ 1.5 m. Key reminder: use the centroid depth h̄, not the full depth of the plate.

Answer: A. Below the centroid

Step 1 — Center of pressure location: yp = ȳ + Ig/(ȳA). Step 2 — The term Ig/(ȳA) is always positive (since Ig > 0, ȳ > 0, A > 0), so yp > ȳ always. Step 3 — Since depth increases downward, yp being greater than ȳ means the center of pressure is deeper, i.e., below the centroid. Step 4 — 'Above the centroid' is never true for a static fluid. Step 5 — 'At the centroid' only if Ig = 0, which is impossible for a real surface. Board-exam trap: many students confuse where the force magnitude is computed (centroid) versus where it acts (center of pressure).

Answer: A. The vertical projection of the curved surface

Step 1 — For a curved surface, resolve the hydrostatic force into horizontal (FH) and vertical (FV) components. Step 2 — FH is the force on the vertical projection of the curved surface: FH = γh̄Avert. This is because horizontal pressure components are equivalent to those acting on the flat vertical plane that 'shadows' the curve. Step 3 — FV = weight of fluid (real or imaginary) directly above the curved surface up to the free surface. Step 4 — 'Horizontal projection' gives the vertical component weight, not the horizontal force. Step 5 — 'The curved surface itself' requires integration; using the projection is the standard simplification.

Answer: A. 4.50 m⁴

Step 1 — Formula for centroidal moment of inertia of a rectangle: Ig = bh³/12 where b = width and h = height. Step 2 — Substitute: Ig = (2)(3)³/12 = (2 × 27)/12 = 54/12 = 4.50 m⁴. Step 3 — 9.00 m⁴ is incorrect; this comes from using bh³/6, which is the moment about the base, not the centroid. Step 4 — 18.00 m⁴ = bh³/3, which is also the moment about the base (wrong axis). Step 5 — 2.25 m⁴ would be Ig for a 1 m × 3 m rectangle. Always use the centroidal axis formula Ig = bh³/12.

Answer: A. The weight of the fluid (real or imaginary) directly above the curved surface up to the free surface

Step 1 — FV = γV, where V is the volume of fluid directly above the curved surface up to the free surface. Step 2 — If the fluid is above the surface, FV acts downward. If the surface curves upward (no real fluid above), use the imaginary fluid column — FV acts upward. Step 3 — 'γh̄ × area of curved surface' is not a valid formula; the area in F = γh̄A must be a flat projected area. Step 4 — The weight of the structure itself is irrelevant for fluid force calculations. Step 5 — 'Pressure at centroid × projected horizontal area' correctly gives FV = γh̄ × Ahoriz, which is equivalent to the fluid weight above — this is an acceptable alternate form.

Answer: A. Add γh when moving downward through a fluid; subtract γh when moving upward

Step 1 — Manometry is based on the hydrostatic equation: pressure increases with depth. Step 2 — Moving downward means going to greater depth → pressure increases → add γh. Step 3 — Moving upward means going to lesser depth → pressure decreases → subtract γh. Step 4 — The process: start at the known pressure end, walk through the fluid columns applying these additions and subtractions, and set the expression equal to the pressure at the other end. Step 5 — 'Always add' ignores the direction, which would give incorrect results. All fluid columns in the manometer (not just the denser one) must be accounted for.

Answer: A. The center of curvature of the arc

Step 1 — On a circular-arc surface, every pressure force is directed radially (perpendicular to the curved surface), pointing toward (or away from) the center of the circle. Step 2 — Since all pressure force vectors are radial, they all pass through the center of curvature. Step 3 — Therefore, the resultant of all these forces also passes through the center of curvature. Step 4 — 'Centroid of the arc' is a geometric property but not where the resultant acts. Step 5 — 'Midpoint of the arc' is only the centroid for a semicircle, but is not generally the line of action. This property is very useful: for circular gates hinged at the center of curvature, the hydrostatic force produces zero moment about the hinge.

Answer: A. 196.2 N

Step 1: Recall Archimedes' principle: F_B = γ_fluid × V_displaced. Step 2: Since the object is fully submerged, V_displaced = volume of the object = 0.02 m³. Step 3: F_B = 9.81 kN/m³ × 0.02 m³ = 0.1962 kN = 196.2 N. Wrong options: 98.1 N would result if V = 0.01 m³; 392.4 N if V = 0.04 m³; 9.81 N confuses γ (kN/m³) with a force directly.

Answer: A. 0.28 m

Step 1: A homogeneous body with specific gravity s floats with a fraction s of its height submerged. Step 2: draft d = s × h = 0.7 × 0.4 m = 0.28 m. Step 3: Verify — Weight W = 0.7 × 9.81 × (0.5 × 0.5 × 0.4) = 0.686 kN; F_B = 9.81 × (0.5 × 0.5 × 0.28) = 0.686 kN ✓. Wrong options: 0.40 m means fully submerged (SG ≥ 1); 0.20 m corresponds to SG = 0.5; 0.35 m corresponds to SG = 0.875.