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 1.5 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 = 3 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, with its top edge 1.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 1.5 mH = 2.5 mb = 2 mρ = 880 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 300 mm and a throat diameter of 150 mm. The pressure at the inlet is 27 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 300 mmd2 = 150 mmp1 − p2 = 27 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 100 mm and a gauge pressure of 100 kPa. The nozzle exit diameter is 30 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 = 100 mmp1 = 100 kPa (gauge)d2 = 30 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.8 m/s through a horizontal pipe, 50 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.8 m/sL = 50 mD = 200 mmf = 0.015ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.018) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each) and a valve (K = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 100 L/s?
Q = 100 L/sL = 200 mD = 200 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2
H = m
7.Pumps & pump powerMedium
A pump with an efficiency of 80% delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 12 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 50 L/sH = 12 mη = 80%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 80 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 3 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 350 kPa. Find the head H the pump must add.
Q = 80 L/sΔz = 8 mL = 300 mD = 200 mmf = 0.02ΣK = 3ptank = 350 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular grass-lined earth channel is 3 m wide. Water flows uniformly 2.3 m deep. Manning n = 0.03 and the bed slope is S = 0.002. Find the flow rate Q in m³/s.
b = 3 my = 2.3 mn = 0.03S = 0.002
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal finished concrete channel with a bottom width of 2 m and side slopes of 2:1 (horizontal:vertical) must carry 23 m³/s in uniform flow at a depth of 1.3 m. Manning n = 0.012. Find the required bed slope S (in m/m).
Q = 23 m³/sb = 2 my = 1.3 mz = 2n = 0.012
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 110.4 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 + 3/2 = 2.5 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·4.5 = 110.4 kN
Answer
F = 110.4 kN
2.Hydrostatic force
Answer: ycp = 2.939 m
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 = 1.5 + 2.5/2 = 2.75 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·2.75·5 = 118.7 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·2.5³/12 = 2.604 m⁴
ycp = yc + Ic/(yc·A) = 2.75 + 2.604/(2.75·5) = 2.939 m
Answer
ycp = 2.939 m
3.Continuity & Bernoulli
Answer: Q = 134.1 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·(150/300)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 = √(2·27000/(1000·(1 − 0.0625))) = 7.589 m/s
v1 = 1.897 m/s
Flow rate
Q = A2·v2 = 0.01767·7.589 = 0.1341 m³/s = 134.1 L/s
Answer
Q = 134.1 L/s
4.Continuity & Bernoulli
Answer: H = 10.28 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 = 30/100 = 0.3
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·100000/(1000·0.9919)) = 14.2 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) = 14.2²/(2·9.81) = 10.28 m
Answer
H = 10.28 m
5.Pipe friction & head loss
Answer: Δp = 14.7 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) = 0.015·(50/0.2)·2.8²/(2·9.81) = 1.498 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 = 1000·9.81·1.498 = 14700 Pa = 14.7 kPa
Answer
Δp = 14.7 kPa
6.Pipe friction & head loss
Answer: H = 11.41 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
V = Q/A = 0.1/0.03142 = 3.183 m/s
V²/(2g) = 0.5164 m
Loss coefficients
f·L/D = 0.018·200/0.2 = 18
ΣK = 0.5 + 2·0.3 + 2 = 3.1
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 18 + 3.1)·0.5164 = 11.41 m
Answer
H = 11.41 m
7.Pumps & pump power
Answer: Pshaft = 7.357 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 = 1000·9.81·0.05·12 = 5.886 kW
Shaft power
Pshaft = Ph/η = 5.886/0.8 = 7.357 kW
Answer
Pshaft = 7.357 kW
8.Pumps & pump power
Answer: H = 54.58 m
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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (30 + 3)·0.3305 = 10.91 m
Pressure head
p/(ρ·g) = 350000/(1000·9.81) = 35.68 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 8 + 35.68 + 10.91 = 54.58 m
Answer
H = 54.58 m
9.Open-channel flow (Manning)
Answer: Q = 9.644 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·y = 3·2.3 = 6.9 m²
P = b + 2y = 3 + 2·2.3 = 7.6 m
R = A/P = 0.9079 m
Velocity
V = (1/0.03)·0.90792/3·0.0021/2 = 1.398 m/s
Flow rate
Q = A·V = 6.9·1.398 = 9.644 m³/s
Answer
Q = 9.644 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003043 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.
Geometry
A = (b + z·y)·y = (2 + 2·1.3)·1.3 = 5.98 m²
P = b + 2y·√(1 + z²) = 2 + 2·1.3·√(1 + 2²) = 7.814 m
R = A/P = 0.7653 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.012·23/(5.98·0.8367) = 0.05516
S = 0.05516² = 0.003043
Answer
S = 0.003043
That is a fall of 3.043 m per kilometre.
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