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 = 2.5 mb = 1.5 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 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1 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 150 mm and a throat diameter of 75 mm. The pressure at the inlet is 37 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 150 mmd2 = 75 mmp1 − p2 = 37 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 20 mm siphons water from an open tank. The highest point C of the hose is 2 m above the water surface, and the outlet (open to the air) is 2.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure pC at point C. The atmospheric pressure is 101.3 kPa.
d = 20 mmzC = +2 mz2 = −2.5 mg = 9.81 m/s²patm = 101.3 kPa
pC = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2 m/s through a pipe 150 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss hf.
V = 2 m/sL = 150 mD = 75 mmf = 0.02g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 400 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.015) 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 120 L/s?
Q = 120 L/sL = 400 mD = 250 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 42 m while its shaft takes 10.3 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 15 L/sH = 42 mPshaft = 10.3 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 40 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 85%. Find the shaft power Pshaft in kW.
Q = 40 L/sΔz = 5 mL = 100 mD = 200 mmf = 0.022ΣK = 6η = 85%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal unfinished concrete channel has a bottom width of 3 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 1.7 m deep. Manning n = 0.014 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.
b = 3 my = 1.7 mz = 1.5n = 0.014S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 300 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 = 300 mmn = 0.013V = 0.9 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 82.77 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 + 2.5/2 = 2.25 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·3.75 = 82.77 kN
Answer
F = 82.77 kN
2.Hydrostatic force
Answer: F = 38.85 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 + 1/2 = 1.5 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·1.5·3 = 38.85 kN
Answer
F = 38.85 kN
3.Continuity & Bernoulli
Answer: Q = 39.25 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·(75/150)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 = √(2·37000/(1000·(1 − 0.0625))) = 8.884 m/s
v1 = 2.221 m/s
Flow rate
Q = A2·v2 = 0.004418·8.884 = 0.03925 m³/s = 39.25 L/s
Answer
Q = 39.25 L/s
4.Continuity & Bernoulli
Answer: pC = 57.15 kPa
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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·2.5) = √(2·9.81·2.5) = 7.004 m/s
Bernoulli from the surface to C
patm = pC + ½ρv² + ρgzC
pC = 101.3 − 1000·(7.004²/2 + 9.81·2)/1000 = 101.3 − 44.15 = 57.15 kPa
Answer
pC = 57.15 kPa
If this pressure falls to the vapour pressure of the water, the siphon stops working.
5.Pipe friction & head loss
Answer: hf = 8.155 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) = 2²/(2·9.81) = 0.2039 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.02·(150/0.075)·2²/(2·9.81) = 8.155 m
Answer
hf = 8.155 m
6.Pipe friction & head loss
Answer: H = 8.559 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Loss coefficients
f·L/D = 0.015·400/0.25 = 24
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 + 24 + 3.1)·0.3046 = 8.559 m
Answer
H = 8.559 m
7.Pumps & pump power
Answer: η = 60 %
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 = 1000·9.81·0.015·42 = 6.18 kW
Efficiency
η = Ph/Pshaft = 6.18/10.3 = 60%
Answer
η = 60%
8.Pumps & pump power
Answer: Pshaft = 2.957 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 = 0.04/0.03142 = 1.273 m/s
V²/(2g) = 0.08263 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (11 + 6)·0.08263 = 1.405 m
Pump head
H = Δz + hL = 5 + 1.405 = 6.405 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.04·6.405/0.85 = 2.957 kW
Answer
Pshaft = 2.957 kW
9.Open-channel flow (Manning)
Answer: Q = 43.57 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 = (3 + 1.5·1.7)·1.7 = 9.435 m²
P = b + 2y·√(1 + z²) = 3 + 2·1.7·√(1 + 1.5²) = 9.129 m
R = A/P = 1.033 m
Velocity
V = (1/0.014)·1.0332/3·0.0041/2 = 4.618 m/s
Flow rate
Q = A·V = 9.435·4.618 = 43.57 m³/s
Answer
Q = 43.57 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.004328 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 = 0.3/4 = 0.075 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.9/0.1778)² = 0.004328
Answer
S = 0.004328
That is a fall of 4.328 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering