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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2 mb = 2 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
A pitot-static tube in a wind tunnel reads a difference of 100 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
p0 − p = 100 Paρ = 1.2 kg/m³
v = m/s
4.Continuity & BernoulliHard
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 120 kPa. The nozzle exit diameter is 20 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 = 50 mmp1 = 120 kPa (gauge)d2 = 20 mmρ = 1000 kg/m³g = 9.81 m/s²
H = m
5.Pipe friction & head lossMedium
Water flows at 80 L/s through a pipe 400 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss hf.
Q = 80 L/sL = 400 mD = 200 mmf = 0.02g = 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 200 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 26 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 0.2) 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 = 26 mL = 200 mD = 200 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 0.2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 30 m while its shaft takes 37.7 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 100 L/sH = 30 mPshaft = 37.7 kWg = 9.81 m/s²
η = %
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 20 m higher. The pipe is 100 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 = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 82%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 20 mL = 100 mD = 200 mmf = 0.02ΣK = 6η = 82%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A rectangular brickwork channel is 2 m wide. Water flows uniformly 1.3 m deep. Manning n = 0.015 and the bed slope is S = 0.0015. Find the flow rate Q in m³/s.
b = 2 my = 1.3 mn = 0.015S = 0.0015
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 600 mm (Manning n = 0.013) must reach a self-cleansing velocity of 1.2 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 600 mmn = 0.013V = 1.2 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 103 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 + 1/2 = 3.5 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.5·3 = 103 kN
Answer
F = 103 kN
2.Hydrostatic force
Answer: F = 103.6 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 + 2/2 = 3 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3·4 = 103.6 kN
Answer
F = 103.6 kN
3.Continuity & Bernoulli
Answer: v = 12.91 m/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.
Bernoulli to the stagnation point
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
p0 − p = ½ρv²
v = √(2·100/1.2) = 12.91 m/s
Answer
v = 12.91 m/s
4.Continuity & Bernoulli
Answer: H = 12.55 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 = 20/50 = 0.4
v2 = √(2p1/(ρ(1 − β⁴))) = √(2·120000/(1000·0.9744)) = 15.69 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) = 15.69²/(2·9.81) = 12.55 m
Answer
H = 12.55 m
5.Pipe friction & head loss
Answer: hf = 13.22 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·0.2²/4 = 0.03142 m²
V = Q/A = 0.08/0.03142 = 2.546 m/s
Velocity head
V²/(2g) = 2.546²/(2·9.81) = 0.3305 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.02·(400/0.2)·2.546²/(2·9.81) = 13.22 m
Answer
hf = 13.22 m
6.Pipe friction & head loss
Answer: Q = 148.3 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 = 0.02·200/0.2 = 20
ΣK = 0.5 + 4·0.3 + 0.2 + 1.0 = 2.9
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
26 = (20 + 2.9)·V²/(2·9.81)
V = √(2·9.81·26/22.9) = 4.72 m/s
Flow rate
Q = V·π·D²/4 = 4.72·0.03142 = 0.1483 m³/s = 148.3 L/s
Answer
Q = 148.3 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 = 1000·9.81·0.1·30 = 29.43 kW
Efficiency
η = Ph/Pshaft = 29.43/37.7 = 78.06%
Answer
η = 78.06%
8.Pumps & pump power
Answer: Pshaft = 24.2 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (10 + 6)·0.3305 = 5.288 m
Pump head
H = Δz + hL = 20 + 5.288 = 25.29 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·25.29/0.82 = 24.2 kW
Answer
Pshaft = 24.2 kW
9.Open-channel flow (Manning)
Answer: Q = 4.589 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 = 2·1.3 = 2.6 m²
P = b + 2y = 2 + 2·1.3 = 4.6 m
R = A/P = 0.5652 m
Velocity
V = (1/0.015)·0.56522/3·0.00151/2 = 1.765 m/s
Flow rate
Q = A·V = 2.6·1.765 = 4.589 m³/s
Answer
Q = 4.589 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003053 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 = 0.6/4 = 0.15 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·1.2/0.2823)² = 0.003053
Answer
S = 0.003053
That is a fall of 3.053 m per kilometre.
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