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 2 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 = 2 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 glycerin, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2 mb = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 5 m above the orifice and the discharge coefficient is 0.6. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 5 mCd = 0.6g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 50 mm siphons water from an open tank. The highest point C of the hose is 0.5 m above the water surface, and the outlet (open to the air) is 2 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
d = 50 mmzC = +0.5 mz2 = −2 mg = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2 m/s through a horizontal pipe, 500 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2 m/sL = 500 mD = 100 mmf = 0.022ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.025. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 150 L/s?
Q = 150 L/sL = 100 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 56 m while its shaft takes 73.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 80 L/sH = 56 mPshaft = 73.2 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 14 m higher. The pipe is 100 m long with an inside diameter of 250 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 78%. Find the shaft power Pshaft in kW.
Q = 120 L/sΔz = 14 mL = 100 mD = 250 mmf = 0.015ΣK = 8η = 78%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A rectangular clean earth channel is 4 m wide. Water flows uniformly 1.4 m deep. Manning n = 0.022 and the bed slope is S = 0.0005. Find the flow rate Q in m³/s.
b = 4 my = 1.4 mn = 0.022S = 0.0005
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 375 mm (Manning n = 0.013) must reach a self-cleansing velocity of 1 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 375 mmn = 0.013V = 1 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 19.62 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 = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1·2 = 19.62 kN
Answer
F = 19.62 kN
2.Hydrostatic force
Answer: F = 74.16 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 = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·2·3 = 74.16 kN
Answer
F = 74.16 kN
3.Continuity & Bernoulli
Answer: Q = 11.67 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.
Ideal jet velocity
√(2gh) = √(2·9.81·5) = 9.905 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.6·0.001963·9.905 = 0.01167 m³/s = 11.67 L/s
Answer
Q = 11.67 L/s
4.Continuity & Bernoulli
Answer: Q = 12.3 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·2) = √(2·9.81·2) = 6.264 m/s
Flow rate
Q = A·v = 0.001963·6.264 = 0.0123 m³/s = 12.3 L/s
Answer
Q = 12.3 L/s
5.Pipe friction & head loss
Answer: Δp = 220 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) = 0.022·(500/0.1)·2²/(2·9.81) = 22.43 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 = 1000·9.81·22.43 = 220000 Pa = 220 kPa
Answer
Δp = 220 kPa
6.Pipe friction & head loss
Answer: Δz = 2.991 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 = 0.15/0.07069 = 2.122 m/s
V²/(2g) = 0.2295 m
Loss coefficients
f·L/D = 0.025·100/0.3 = 8.333
ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7
Total head loss
hf = 8.333·0.2295 = 1.913 m
hm = 4.7·0.2295 = 1.079 m
Δz = hf + hm = 2.991 m
Answer
Δz = 2.991 m
7.Pumps & pump power
Answer: η = 60.04 %
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 = 1000·9.81·0.08·56 = 43.95 kW
Efficiency
η = Ph/Pshaft = 43.95/73.2 = 60.04%
Answer
η = 60.04%
8.Pumps & pump power
Answer: Pshaft = 27.57 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (6 + 8)·0.3046 = 4.264 m
Pump head
H = Δz + hL = 14 + 4.264 = 18.26 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.12·18.26/0.78 = 27.57 kW
Answer
Pshaft = 27.57 kW
9.Open-channel flow (Manning)
Answer: Q = 5.001 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 = 4·1.4 = 5.6 m²
P = b + 2y = 4 + 2·1.4 = 6.8 m
R = A/P = 0.8235 m
Velocity
V = (1/0.022)·0.82352/3·0.00051/2 = 0.893 m/s
Flow rate
Q = A·V = 5.6·0.893 = 5.001 m³/s
Answer
Q = 5.001 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003968 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 = 0.375/4 = 0.09375 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1/0.2064)² = 0.003968
Answer
S = 0.003968
That is a fall of 3.968 m per kilometre.
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