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 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2.5 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 oil, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1 mb = 1 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 200 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 50 mmp1 = 200 kPav1 = 1.75 m/sρ = 1000 kg/m³
p2 = kPa
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 2.5 m above the water surface, and the outlet (open to the air) is 1 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 = +2.5 mz2 = −1 mg = 9.81 m/s²patm = 101.3 kPa
pC = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.5 m/s through a pipe 50 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.5 m/sL = 50 mD = 300 mmf = 0.028g = 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 300 mm and a Darcy friction factor f = 0.025. The water surface of the upper reservoir is 22 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 = 5) 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 = 22 mL = 200 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 5Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 14 m while its shaft takes 1.7 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 10 L/sH = 14 mPshaft = 1.7 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 20 m higher. The pipe is 150 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 5 (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 100 kPa. Find the head H the pump must add.
Q = 50 L/sΔz = 20 mL = 150 mD = 150 mmf = 0.015ΣK = 5ptank = 100 kPa
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal finished concrete channel has a bottom width of 2.5 m and side slopes of 1:1 (horizontal:vertical). Water flows uniformly 1.6 m deep. Manning n = 0.012 and the bed slope is S = 0.0008. Find the flow rate Q in m³/s.
b = 2.5 my = 1.6 mz = 1n = 0.012S = 0.0008
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 250 mm (Manning n = 0.013) must reach a self-cleansing velocity of 0.75 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 250 mmn = 0.013V = 0.75 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 79.71 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 + 2.5/2 = 3.25 m
A = H·b = 2.5·1 = 2.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·2.5 = 79.71 kN
Answer
F = 79.71 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 = 4 + 1/2 = 4.5 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·4.5·1 = 38.85 kN
Answer
F = 38.85 kN
3.Continuity & Bernoulli
Answer: p2 = 193.8 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 = 1.75·(75/50)² = 3.938 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 3.938²)/1000 = −6.221 kPa
p2 = p1 + ½ρ(v1² − v2²) = 200 + (−6.221) = 193.8 kPa
Answer
p2 = 193.8 kPa
4.Continuity & Bernoulli
Answer: pC = 66.97 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·1) = √(2·9.81·1) = 4.429 m/s
Bernoulli from the surface to C
patm = pC + ½ρv² + ρgzC
pC = 101.3 − 1000·(4.429²/2 + 9.81·2.5)/1000 = 101.3 − 34.34 = 66.97 kPa
Answer
pC = 66.97 kPa
If this pressure falls to the vapour pressure of the water, the siphon stops working.
5.Pipe friction & head loss
Answer: hf = 1.487 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) = 2.5²/(2·9.81) = 0.3186 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(50/0.3)·2.5²/(2·9.81) = 1.487 m
Answer
hf = 1.487 m
6.Pipe friction & head loss
Answer: Q = 299.4 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 = 0.025·200/0.3 = 16.67
ΣK = 0.5 + 3·0.3 + 5 + 1.0 = 7.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
22 = (16.67 + 7.4)·V²/(2·9.81)
V = √(2·9.81·22/24.07) = 4.235 m/s
Flow rate
Q = V·π·D²/4 = 4.235·0.07069 = 0.2994 m³/s = 299.4 L/s
Answer
Q = 299.4 L/s
7.Pumps & pump power
Answer: η = 80.79 %
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 = 1000·9.81·0.01·14 = 1.373 kW
Efficiency
η = Ph/Pshaft = 1.373/1.7 = 80.79%
Answer
η = 80.79%
8.Pumps & pump power
Answer: H = 38.35 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (15 + 5)·0.408 = 8.161 m
Pressure head
p/(ρ·g) = 100000/(1000·9.81) = 10.19 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 20 + 10.19 + 8.161 = 38.35 m
Answer
H = 38.35 m
9.Open-channel flow (Manning)
Answer: Q = 14.77 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 = (2.5 + 1·1.6)·1.6 = 6.56 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·1.6·√(1 + 1²) = 7.025 m
R = A/P = 0.9337 m
Velocity
V = (1/0.012)·0.93372/3·0.00081/2 = 2.252 m/s
Flow rate
Q = A·V = 6.56·2.252 = 14.77 m³/s
Answer
Q = 14.77 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003833 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 = 0.25/4 = 0.0625 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.75/0.1575)² = 0.003833
Answer
S = 0.003833
That is a fall of 3.833 m per kilometre.
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