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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1 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 mb = 1.5 mρ = 1025 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 30 mm. The water level is kept 3.5 m above the orifice and the discharge coefficient is 0.64. Find the flow rate Q. Use g = 9.81 m/s².
d = 30 mmh = 3.5 mCd = 0.64g = 9.81 m/s²
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 110 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 = 110 kPa (gauge)d2 = 40 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at 75 L/s through a pipe 500 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss hf.
Q = 75 L/sL = 500 mD = 300 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 35 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 = 35 mL = 200 mD = 250 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 74 m while its shaft takes 18.6 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 20 L/sH = 74 mPshaft = 18.6 kWg = 9.81 m/s²
η = %
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 19 m higher. The pipe is 400 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 = 8 (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 = 60 L/sΔz = 19 mL = 400 mD = 200 mmf = 0.02ΣK = 8ptank = 350 kPa
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 900 mm flows half full. Manning n = 0.013 and the slope is S = 0.0005. Find the flow rate Q in m³/s.
D = 900 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 = 88.29 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 + 1/2 = 3 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·3 = 88.29 kN
Answer
F = 88.29 kN
2.Hydrostatic force
Answer: F = 120.7 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 = 3 + 2/2 = 4 m
A = H·b = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·4·3 = 120.7 kN
Answer
F = 120.7 kN
3.Continuity & Bernoulli
Answer: Q = 3.749 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) = √(2·9.81·3.5) = 8.287 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.64·0.0007069·8.287 = 0.003749 m³/s = 3.749 L/s
Answer
Q = 3.749 L/s
4.Continuity & Bernoulli
Answer: H = 12.2 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))) = √(2·110000/(1000·0.9191)) = 15.47 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) = 15.47²/(2·9.81) = 12.2 m
Answer
H = 12.2 m
5.Pipe friction & head loss
Answer: hf = 1.434 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.
Mean velocity
A = π·0.3²/4 = 0.07069 m²
V = Q/A = 0.075/0.07069 = 1.061 m/s
Velocity head
V²/(2g) = 1.061²/(2·9.81) = 0.05738 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.015·(500/0.3)·1.061²/(2·9.81) = 1.434 m
Answer
hf = 1.434 m
6.Pipe friction & head loss
Answer: Q = 241.4 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 = 0.02·200/0.25 = 16
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.
35 = (16 + 12.4)·V²/(2·9.81)
V = √(2·9.81·35/28.4) = 4.917 m/s
Flow rate
Q = V·π·D²/4 = 4.917·0.04909 = 0.2414 m³/s = 241.4 L/s
Answer
Q = 241.4 L/s
7.Pumps & pump power
Answer: η = 78.06 %
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 = 1000·9.81·0.02·74 = 14.52 kW
Efficiency
η = Ph/Pshaft = 14.52/18.6 = 78.06%
Answer
η = 78.06%
8.Pumps & pump power
Answer: H = 63.6 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 = 0.06/0.03142 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (40 + 8)·0.1859 = 8.924 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 = 19 + 35.68 + 8.924 = 63.6 m
Answer
H = 63.6 m
9.Open-channel flow (Manning)
Answer: Q = 0.2024 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 = 0.3181 m², R = D/4 = 0.225 m
Velocity
V = (1/0.013)·0.2252/3·0.00051/2 = 0.6363 m/s
Flow rate
Q = A·V = 0.3181·0.6363 = 0.2024 m³/s
Answer
Q = 0.2024 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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