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 = 1.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1 m wide sits in the wall of a tank of glycerin, with its top edge 2.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2.5 mH = 1.5 mb = 1 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
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 3.5 m above the orifice and the discharge coefficient is 0.63. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 3.5 mCd = 0.63g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 25 mm while rising 9 m. At section 1 the pressure is 380 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 25 mmp1 = 380 kPav1 = 0.75 m/sΔz = 9 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 5 L/s through a pipe 50 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
Q = 5 L/sL = 50 mD = 75 mmf = 0.018g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 5) 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 100 L/s?
Q = 100 L/sL = 300 mD = 250 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 74 m while its shaft takes 27.9 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 74 mPshaft = 27.9 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 15 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 100 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. Find the head H the pump must add.
Q = 15 L/sΔz = 5 mL = 100 mD = 100 mmf = 0.015ΣK = 8
H = m
9.Open-channel flow (Manning)Medium
A rectangular brickwork channel is 4.5 m wide. Water flows uniformly 2.2 m deep. Manning n = 0.015 and the bed slope is S = 0.005. Find the flow rate Q in m³/s.
b = 4.5 my = 2.2 mn = 0.015S = 0.005
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal clean earth channel with a bottom width of 4.5 m and side slopes of 1:1 (horizontal:vertical) must carry 9.4 m³/s in uniform flow at a depth of 0.9 m. Manning n = 0.022. Find the required bed slope S (in m/m).
Q = 9.4 m³/sb = 4.5 my = 0.9 mz = 1n = 0.022
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 18.39 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 + 1.5/2 = 1.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.25·1.5 = 18.39 kN
Answer
F = 18.39 kN
2.Hydrostatic force
Answer: ycp = 3.308 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 = 2.5 + 1.5/2 = 3.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3.25·1.5 = 60.26 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·1.5³/12 = 0.2813 m⁴
ycp = yc + Ic/(yc·A) = 3.25 + 0.2813/(3.25·1.5) = 3.308 m
Answer
ycp = 3.308 m
3.Continuity & Bernoulli
Answer: Q = 10.25 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.
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) = 0.63·0.001963·8.287 = 0.01025 m³/s = 10.25 L/s
Answer
Q = 10.25 L/s
4.Continuity & Bernoulli
Answer: p2 = 220 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.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 0.75·(100/25)² = 12 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 12²)/1000 = −71.72 kPa
ρgΔz = 1000·9.81·9/1000 = 88.29 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 380 + (−71.72) − 88.29 = 220 kPa
Answer
p2 = 220 kPa
5.Pipe friction & head loss
Answer: hf = 0.7834 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·0.075²/4 = 0.004418 m²
V = Q/A = 0.005/0.004418 = 1.132 m/s
Velocity head
V²/(2g) = 1.132²/(2·9.81) = 0.06529 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(50/0.075)·1.132²/(2·9.81) = 0.7834 m
Answer
hf = 0.7834 m
6.Pipe friction & head loss
Answer: Δz = 6.071 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Loss coefficients
f·L/D = 0.018·300/0.25 = 21.6
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Total head loss
hf = 21.6·0.2115 = 4.569 m
hm = 7.1·0.2115 = 1.502 m
Δz = hf + hm = 6.071 m
Answer
Δz = 6.071 m
7.Pumps & pump power
Answer: η = 65.05 %
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 = 1000·9.81·0.025·74 = 18.15 kW
Efficiency
η = Ph/Pshaft = 18.15/27.9 = 65.05%
Answer
η = 65.05%
8.Pumps & pump power
Answer: H = 9.276 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (15 + 8)·0.1859 = 4.276 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 5 + 4.276 = 9.276 m
Answer
H = 9.276 m
9.Open-channel flow (Manning)
Answer: Q = 50.1 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 = 4.5·2.2 = 9.9 m²
P = b + 2y = 4.5 + 2·2.2 = 8.9 m
R = A/P = 1.112 m
Velocity
V = (1/0.015)·1.1122/3·0.0051/2 = 5.061 m/s
Flow rate
Q = A·V = 9.9·5.061 = 50.1 m³/s
Answer
Q = 50.1 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.002971 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 = (4.5 + 1·0.9)·0.9 = 4.86 m²
P = b + 2y·√(1 + z²) = 4.5 + 2·0.9·√(1 + 1²) = 7.046 m
R = A/P = 0.6898 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.022·9.4/(4.86·0.7807) = 0.0545
S = 0.0545² = 0.002971
Answer
S = 0.002971
That is a fall of 2.971 m per kilometre.
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