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 1 m wide sits in the wall of a tank of water, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 1 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 50 mm. The pressure at the inlet is 23 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 100 mmd2 = 50 mmp1 − p2 = 23 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 75 mm and a gauge pressure of 20 kPa. The nozzle exit diameter is 40 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
d1 = 75 mmp1 = 20 kPa (gauge)d2 = 40 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.5 m/s through a pipe 50 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.5 m/sL = 50 mD = 200 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.025. The water surface of the upper reservoir is 31 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 = 0.2) 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 = 31 mL = 100 mD = 150 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 0.2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 10 m while its shaft takes 1.3 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 10 L/sH = 10 mPshaft = 1.3 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 25 m higher. The pipe is 400 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 78%. Find the shaft power Pshaft in kW.
Q = 50 L/sΔz = 25 mL = 400 mD = 150 mmf = 0.018ΣK = 8η = 78%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal brickwork channel has a bottom width of 1.5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.7 m deep. Manning n = 0.015 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 1.5 my = 1.7 mz = 1n = 0.015S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 250 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 = 250 mmn = 0.013V = 0.9 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 9.81 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 = 0.5 + 1/2 = 1 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1·1 = 9.81 kN
Answer
F = 9.81 kN
2.Hydrostatic force
Answer: F = 181.3 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 = 2.5 + 2/2 = 3.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3.5·6 = 181.3 kN
Answer
F = 181.3 kN
3.Continuity & Bernoulli
Answer: Q = 13.75 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·(50/100)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·23000/(1000·(1 − 0.0625))) = 7.005 m/s
v1 = 1.751 m/s
Flow rate
Q = A2·v2 = 0.001963·7.005 = 0.01375 m³/s = 13.75 L/s
Answer
Q = 13.75 L/s
4.Continuity & Bernoulli
Answer: H = 2.218 m
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.
Exit velocity
β = 40/75 = 0.5333
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·20000/(1000·0.9191)) = 6.597 m/s
Height of the jet
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
H = v2²/(2g) = 6.597²/(2·9.81) = 2.218 m
Answer
H = 2.218 m
5.Pipe friction & head loss
Answer: hf = 0.5161 m
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.
Velocity head
V²/(2g) = 1.5²/(2·9.81) = 0.1147 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(50/0.2)·1.5²/(2·9.81) = 0.5161 m
Answer
hf = 0.5161 m
6.Pipe friction & head loss
Answer: Q = 99.29 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 = 0.025·100/0.15 = 16.67
ΣK = 0.5 + 3·0.3 + 0.2 + 1.0 = 2.6
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
31 = (16.67 + 2.6)·V²/(2·9.81)
V = √(2·9.81·31/19.27) = 5.619 m/s
Flow rate
Q = V·π·D²/4 = 5.619·0.01767 = 0.09929 m³/s = 99.29 L/s
Answer
Q = 99.29 L/s
7.Pumps & pump power
Answer: η = 75.46 %
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 = 1000·9.81·0.01·10 = 0.981 kW
Efficiency
η = Ph/Pshaft = 0.981/1.3 = 75.46%
Answer
η = 75.46%
8.Pumps & pump power
Answer: Pshaft = 30.09 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (48 + 8)·0.408 = 22.85 m
Pump head
H = Δz + hL = 25 + 22.85 = 47.85 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.05·47.85/0.78 = 30.09 kW
Answer
Pshaft = 30.09 kW
9.Open-channel flow (Manning)
Answer: Q = 7.347 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 = (b + z·y)·y = (1.5 + 1·1.7)·1.7 = 5.44 m²
P = b + 2y·√(1 + z²) = 1.5 + 2·1.7·√(1 + 1²) = 6.308 m
R = A/P = 0.8624 m
Velocity
V = (1/0.015)·0.86242/3·0.00051/2 = 1.351 m/s
Flow rate
Q = A·V = 5.44·1.351 = 7.347 m³/s
Answer
Q = 7.347 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.005519 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 = 0.25/4 = 0.0625 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.9/0.1575)² = 0.005519
Answer
S = 0.005519
That is a fall of 5.519 m per kilometre.
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