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 = 3 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 oil, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1.5 mb = 1 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 50 mm. In the wide section the pressure is 230 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 50 mmp1 = 230 kPav1 = 0.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 100 mm, throat 50 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 110 mm. The discharge coefficient is 0.96. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
d1 = 100 mmd2 = 50 mmΔh = 110 mmCd = 0.96ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.6 m/s through a horizontal pipe, 300 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.6 m/sL = 300 mD = 100 mmf = 0.02ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 400 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018. The water surface of the upper reservoir is 39 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 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 = 39 mL = 400 mD = 250 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 72% delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 54 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 30 L/sH = 54 mη = 72%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 7 m higher. The pipe is 100 m long with an inside diameter of 250 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 pump efficiency is 85%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 7 mL = 100 mD = 250 mmf = 0.02ΣK = 8η = 85%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A rectangular gravel channel is 1.5 m wide. Water flows uniformly 2.2 m deep. Manning n = 0.025 and the bed slope is S = 0.0015. Find the flow rate Q in m³/s.
b = 1.5 my = 2.2 mn = 0.025S = 0.0015
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal gravel channel with a bottom width of 1 m and side slopes of 1.5:1 (horizontal:vertical) must carry 3.9 m³/s in uniform flow at a depth of 1.4 m. Manning n = 0.025. Find the required bed slope S (in m/m).
Q = 3.9 m³/sb = 1 my = 1.4 mz = 1.5n = 0.025
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 58.86 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 + 3/2 = 2 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·3 = 58.86 kN
Answer
F = 58.86 kN
2.Hydrostatic force
Answer: F = 48.56 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 + 1.5/2 = 3.75 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3.75·1.5 = 48.56 kN
Answer
F = 48.56 kN
3.Continuity & Bernoulli
Answer: p2 = 207.5 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·(150/50)² = 6.75 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 6.75²)/1000 = −22.5 kPa
p2 = p1 + ½ρ(v1² − v2²) = 230 + (−22.5) = 207.5 kPa
Answer
p2 = 207.5 kPa
4.Continuity & Bernoulli
Answer: Q = 10.15 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.
Pressure difference from the manometer
Δp = (13600 − 1000)·9.81·0.11 = 13597 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 50/100 = 0.5
v2 = √(2·13597/(1000·(1 − 0.0625))) = 5.386 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.96·0.001963·5.386 = 0.01015 m³/s = 10.15 L/s
Answer
Q = 10.15 L/s
5.Pipe friction & head loss
Answer: Δp = 202.8 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) = 0.02·(300/0.1)·2.6²/(2·9.81) = 20.67 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 = 1000·9.81·20.67 = 202800 Pa = 202.8 kPa
Answer
Δp = 202.8 kPa
6.Pipe friction & head loss
Answer: Q = 212.3 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 = 0.018·400/0.25 = 28.8
ΣK = 0.5 + 2·0.3 + 10 + 1.0 = 12.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
39 = (28.8 + 12.1)·V²/(2·9.81)
V = √(2·9.81·39/40.9) = 4.325 m/s
Flow rate
Q = V·π·D²/4 = 4.325·0.04909 = 0.2123 m³/s = 212.3 L/s
Answer
Q = 212.3 L/s
7.Pumps & pump power
Answer: Pshaft = 22.07 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 = 1000·9.81·0.03·54 = 15.89 kW
Shaft power
Pshaft = Ph/η = 15.89/0.72 = 22.07 kW
Answer
Pshaft = 22.07 kW
8.Pumps & pump power
Answer: Pshaft = 8.463 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (8 + 8)·0.1354 = 2.166 m
Pump head
H = Δz + hL = 7 + 2.166 = 9.166 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·9.166/0.85 = 8.463 kW
Answer
Pshaft = 8.463 kW
9.Open-channel flow (Manning)
Answer: Q = 3.471 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 = 1.5·2.2 = 3.3 m²
P = b + 2y = 1.5 + 2·2.2 = 5.9 m
R = A/P = 0.5593 m
Velocity
V = (1/0.025)·0.55932/3·0.00151/2 = 1.052 m/s
Flow rate
Q = A·V = 3.3·1.052 = 3.471 m³/s
Answer
Q = 3.471 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0007855 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 = (1 + 1.5·1.4)·1.4 = 4.34 m²
P = b + 2y·√(1 + z²) = 1 + 2·1.4·√(1 + 1.5²) = 6.048 m
R = A/P = 0.7176 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.025·3.9/(4.34·0.8015) = 0.02803
S = 0.02803² = 0.0007855
Answer
S = 0.0007855
That is a fall of 0.7855 m per kilometre.
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