Hydrostatic force
- Pressure at depth
- p = ρgh
- Force on a vertical wall
- F = ρg·hc·A
- Centre of pressure
- yp = yc + Ic/(ycA)
- Rectangle
- Ic = bH³/12
Hydraulic Engineering: cram pack
Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
1Formulas and methods
Continuity & Bernoulli
- Continuity
- A1v1 = A2v2
- Bernoulli
- p1 + ½ρv1² + ρgz1 = p2 + ½ρv2² + ρgz2
- Flow rate
- Q = A·v
- Circle area
- A = πd²/4
Pipe friction & head loss
- Reynolds number
- Re = V·D/ν = ρ·V·D/μ
- Flow regime
- laminar below Re ≈ 2000, turbulent above Re ≈ 4000
- Mean velocity
- V = Q/A, A = π·D²/4
- Darcy–Weisbach
- hf = f·(L/D)·V²/(2g)
- Minor losses
- hm = ΣK·V²/(2g)
- Typical K values
- entrance 0.5, exit 1.0, 90° bend ≈ 0.3
- Pressure drop
- Δp = ρ·g·hf
- Two open reservoirs
- Δz = (f·L/D + ΣK)·V²/(2g)
- Free outlet
- H = (1 + f·L/D + ΣK)·V²/(2g)
Pumps & pump power
- Hydraulic power
- Ph = ρ·g·Q·H = Q·Δp
- Shaft power
- Pshaft = Ph/ηpump
- Electrical power
- Pel = Pshaft/ηmotor
- Pump head
- H = Δz + Δp/(ρ·g) + hL
- System losses
- hL = (f·L/D + ΣK)·V²/(2g)
Open-channel flow (Manning)
- Manning equation
- V = (1/n)·R2/3·S1/2
- Flow rate
- Q = A·V
- Hydraulic radius
- R = A/P (P = wetted perimeter, free surface excluded)
- Rectangle
- A = b·y, P = b + 2y
- Trapezoid (side slope z:1)
- A = (b + z·y)·y, P = b + 2y·√(1 + z²)
- Circular pipe, full or half full
- R = D/4
- Required slope
- S = (n·Q/(A·R2/3))²
2Practice problems
1.Hydrostatic forceMedium
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of seawater, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 2.5 mb = 1.5 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 150 mm and a throat diameter of 100 mm. The pressure at the inlet is 30 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 150 mmd2 = 100 mmp1 − p2 = 30 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 60 kPa. Water leaves through an opening with a diameter of 50 mm, 5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
pair = 60 kPa (gauge)h = 5 md = 50 mmρ = 1000 kg/m³g = 9.81 m/s²
v = m/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.1 m/s through a horizontal pipe, 150 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.1 m/sL = 150 mD = 100 mmf = 0.02ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 400 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 39 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.
Δz = 39 mL = 400 mD = 300 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 80% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 22 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 22 mη = 80%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 60 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 13 m higher. The pipe is 600 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 72%. Find the shaft power Pshaft in kW.
Q = 60 L/sΔz = 13 mL = 600 mD = 200 mmf = 0.015ΣK = 8η = 72%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 300 mm flows half full. Manning n = 0.013 and the slope is S = 0.0005. Find the flow rate Q in m³/s.
D = 300 mmn = 0.013S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 200 mm (Manning n = 0.013) must reach a self-cleansing velocity of 0.9 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 200 mmn = 0.013V = 0.9 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 117.7 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1 + 2/2 = 2 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·6 = 117.7 kN
Answer
F = 117.7 kN
2.Hydrostatic force
Answer: F = 160.3 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 3 + 2.5/2 = 4.25 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4.25·3.75 = 160.3 kN
Answer
F = 160.3 kN
3.Continuity & Bernoulli
Answer: Q = 67.91 L/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Continuity
v1 = v2·(100/150)² = 0.4444·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·30000/(1000·(1 − 0.1975))) = 8.647 m/s
v1 = 3.843 m/s
Flow rate
Q = A2·v2 = 0.007854·8.647 = 0.06791 m³/s = 67.91 L/s
Answer
Q = 67.91 L/s
4.Continuity & Bernoulli
Answer: v = 14.77 m/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Bernoulli from the surface to the jet
v = √(2·(60000/1000 + 9.81·5)) = 14.77 m/s
Answer
v = 14.77 m/s
5.Pipe friction & head loss
Answer: Δp = 66.15 kPa
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Head loss
hf = f·(L/D)·V²/(2g) = 0.02·(150/0.1)·2.1²/(2·9.81) = 6.743 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·6.743 = 66150 Pa = 66.15 kPa
Answer
Δp = 66.15 kPa
6.Pipe friction & head loss
Answer: Q = 312.8 L/s
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Loss coefficients
f·L/D = 0.02·400/0.3 = 26.67
ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
39 = (26.67 + 12.4)·V²/(2·9.81)
V = √(2·9.81·39/39.07) = 4.426 m/s
Flow rate
Q = V·π·D²/4 = 4.426·0.07069 = 0.3128 m³/s = 312.8 L/s
Answer
Q = 312.8 L/s
7.Pumps & pump power
Answer: Pshaft = 5.396 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Hydraulic power
Ph = ρ·g·Q·H = 1000·9.81·0.02·22 = 4.316 kW
Shaft power
Pshaft = Ph/η = 4.316/0.8 = 5.396 kW
Answer
Pshaft = 5.396 kW
8.Pumps & pump power
Answer: Pshaft = 18.68 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Velocity
V = Q/A = 0.06/0.03142 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (45 + 8)·0.1859 = 9.853 m
Pump head
H = Δz + hL = 13 + 9.853 = 22.85 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.06·22.85/0.72 = 18.68 kW
Answer
Pshaft = 18.68 kW
9.Open-channel flow (Manning)
Answer: Q = 0.01081 m³/s
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Geometry
A = 0.03534 m², R = D/4 = 0.075 m
Velocity
V = (1/0.013)·0.0752/3·0.00051/2 = 0.3059 m/s
Flow rate
Q = A·V = 0.03534·0.3059 = 0.01081 m³/s
Answer
Q = 0.01081 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.007432 m/m
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Hydraulic radius
R = D/4 = 0.2/4 = 0.05 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.9/0.1357)² = 0.007432
Answer
S = 0.007432
That is a fall of 7.432 m per kilometre.
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