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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 1.5 mb = 3 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 water, with its top edge 3 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 3 mH = 2.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 25 mm. In the wide section the pressure is 110 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 25 mmp1 = 110 kPav1 = 1.25 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 150 mm, throat 75 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 80 mm. The discharge coefficient is 0.95. 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 = 150 mmd2 = 75 mmΔh = 80 mmCd = 0.95ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.2 m/s through a horizontal pipe, 800 m long, with an inside diameter of 150 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 = 2.2 m/sL = 800 mD = 150 mmf = 0.02ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022. 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 = 100 mD = 200 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 72 m while its shaft takes 47.1 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 50 L/sH = 72 mPshaft = 47.1 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 7 m higher. The pipe is 400 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 8 (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 = 80 L/sΔz = 7 mL = 400 mD = 200 mmf = 0.018ΣK = 8ptank = 100 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular unfinished concrete channel is 3 m wide. Water flows uniformly 0.9 m deep. Manning n = 0.014 and the bed slope is S = 0.004. Find the flow rate Q in m³/s.
b = 3 my = 0.9 mn = 0.014S = 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 1 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 450 mmn = 0.013V = 1 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 99.33 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 = 1.5 + 1.5/2 = 2.25 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·4.5 = 99.33 kN
Answer
F = 99.33 kN
2.Hydrostatic force
Answer: ycp = 4.373 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 = 3 + 2.5/2 = 4.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.25·7.5 = 312.7 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2.5³/12 = 3.906 m⁴
ycp = yc + Ic/(yc·A) = 4.25 + 3.906/(4.25·7.5) = 4.373 m
Answer
ycp = 4.373 m
3.Continuity & Bernoulli
Answer: p2 = 47.5 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 = 1.25·(75/25)² = 11.25 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 11.25²)/1000 = −62.5 kPa
p2 = p1 + ½ρ(v1² − v2²) = 110 + (−62.5) = 47.5 kPa
Answer
p2 = 47.5 kPa
4.Continuity & Bernoulli
Answer: Q = 19.28 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)·9.81·0.08 = 9888 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 75/150 = 0.5
v2 = √(2·9888/(1000·(1 − 0.0625))) = 4.593 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.95·0.004418·4.593 = 0.01928 m³/s = 19.28 L/s
Answer
Q = 19.28 L/s
5.Pipe friction & head loss
Answer: Δp = 258.1 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·(800/0.15)·2.2²/(2·9.81) = 26.31 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 = 1000·9.81·26.31 = 258133 Pa = 258.1 kPa
Answer
Δp = 258.1 kPa
6.Pipe friction & head loss
Answer: Δz = 9.347 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 = 0.1/0.03142 = 3.183 m/s
V²/(2g) = 0.5164 m
Loss coefficients
f·L/D = 0.022·100/0.2 = 11
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Total head loss
hf = 11·0.5164 = 5.681 m
hm = 7.1·0.5164 = 3.667 m
Δz = hf + hm = 9.347 m
Answer
Δz = 9.347 m
7.Pumps & pump power
Answer: η = 74.98 %
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 = 1000·9.81·0.05·72 = 35.32 kW
Efficiency
η = Ph/Pshaft = 35.32/47.1 = 74.98%
Answer
η = 74.98%
8.Pumps & pump power
Answer: H = 31.74 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 8)·0.3305 = 14.54 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 + 14.54 = 31.74 m
Answer
H = 31.74 m
9.Open-channel flow (Manning)
Answer: Q = 8.312 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 = 3·0.9 = 2.7 m²
P = b + 2y = 3 + 2·0.9 = 4.8 m
R = A/P = 0.5625 m
Velocity
V = (1/0.014)·0.56252/3·0.0041/2 = 3.078 m/s
Flow rate
Q = A·V = 2.7·3.078 = 8.312 m³/s
Answer
Q = 8.312 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.003112 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 = (0.013·1/0.233)² = 0.003112
Answer
S = 0.003112
That is a fall of 3.112 m per kilometre.
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