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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of seawater, with its top edge 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2 mH = 3 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
A pitot-static tube in a wind tunnel reads a difference of 450 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.
p0 − p = 450 Paρ = 1.2 kg/m³
v = m/s
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 60 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 = 60 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 2.4 m/s through a horizontal pipe, 800 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 pressure drop Δp over the pipe in kPa.
V = 2.4 m/sL = 800 mD = 75 mmf = 0.018ρ = 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 300 mm and a Darcy friction factor f = 0.018. The water surface of the upper reservoir is 13 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 = 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 = 13 mL = 400 mD = 300 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 58 m while its shaft takes 11.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 15 L/sH = 58 mPshaft = 11.4 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 75 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 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 6 (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 = 75 L/sΔz = 7 mL = 100 mD = 200 mmf = 0.025ΣK = 6ptank = 100 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular brickwork channel is 1 m wide. Water flows uniformly 1 m deep. Manning n = 0.015 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.
b = 1 my = 1 mn = 0.015S = 0.004
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 450 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 = 450 mmn = 0.013V = 0.9 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 68.67 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 + 2/2 = 3.5 m
A = H·b = 2·1 = 2 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.5·2 = 68.67 kN
Answer
F = 68.67 kN
2.Hydrostatic force
Answer: ycp = 3.714 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 + 3/2 = 3.5 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3.5·9 = 316.7 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·3³/12 = 6.75 m⁴
ycp = yc + Ic/(yc·A) = 3.5 + 6.75/(3.5·9) = 3.714 m
Answer
ycp = 3.714 m
3.Continuity & Bernoulli
Answer: v = 27.39 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 = √(2·450/1.2) = 27.39 m/s
Answer
v = 27.39 m/s
4.Continuity & Bernoulli
Answer: Q = 7.498 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)·9.81·0.06 = 7416 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 50/100 = 0.5
v2 = √(2·7416/(1000·(1 − 0.0625))) = 3.978 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.96·0.001963·3.978 = 0.007498 m³/s = 7.498 L/s
Answer
Q = 7.498 L/s
5.Pipe friction & head loss
Answer: Δp = 553 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) = 0.018·(800/0.075)·2.4²/(2·9.81) = 56.37 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 = 1000·9.81·56.37 = 552960 Pa = 553 kPa
Answer
Δp = 553 kPa
6.Pipe friction & head loss
Answer: Q = 202.4 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 = 0.018·400/0.3 = 24
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
13 = (24 + 7.1)·V²/(2·9.81)
V = √(2·9.81·13/31.1) = 2.864 m/s
Flow rate
Q = V·π·D²/4 = 2.864·0.07069 = 0.2024 m³/s = 202.4 L/s
Answer
Q = 202.4 L/s
7.Pumps & pump power
Answer: η = 74.87 %
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 = 1000·9.81·0.015·58 = 8.535 kW
Efficiency
η = Ph/Pshaft = 8.535/11.4 = 74.87%
Answer
η = 74.87%
8.Pumps & pump power
Answer: H = 22.57 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (12.5 + 6)·0.2905 = 5.374 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 = 7 + 10.19 + 5.374 = 22.57 m
Answer
H = 22.57 m
9.Open-channel flow (Manning)
Answer: Q = 2.027 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 = 1·1 = 1 m²
P = b + 2y = 1 + 2·1 = 3 m
R = A/P = 0.3333 m
Velocity
V = (1/0.015)·0.33332/3·0.0041/2 = 2.027 m/s
Flow rate
Q = A·V = 1·2.027 = 2.027 m³/s
Answer
Q = 2.027 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.002521 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 = 0.45/4 = 0.1125 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.9/0.233)² = 0.002521
Answer
S = 0.002521
That is a fall of 2.521 m per kilometre.
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