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 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 seawater, with its top edge 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 0.5 mH = 2 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 50 mm. In the wide section the pressure is 150 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 50 mmp1 = 150 kPav1 = 1.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 200 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 230 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 = 200 mmd2 = 100 mmΔh = 230 mmCd = 0.96ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 0.8 m/s through a pipe 150 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
V = 0.8 m/sL = 150 mD = 50 mmf = 0.022g = 9.81 m/s²
hf = m
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 minor losses are a square-edged entrance (K = 0.5), 3 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. What difference in water level Δz between the reservoirs gives a flow rate of 60 L/s?
Q = 60 L/sL = 400 mD = 250 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 34 m while its shaft takes 39.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 100 L/sH = 34 mPshaft = 39.2 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 10 m higher. The pipe is 600 m long with an inside diameter of 250 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 pump efficiency is 70%. Find the shaft power Pshaft in kW.
Q = 120 L/sΔz = 10 mL = 600 mD = 250 mmf = 0.015ΣK = 5η = 70%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 900 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.0015. Find the flow rate Q in m³/s.
D = 900 mmn = 0.013S = 0.0015
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular unfinished concrete channel 6 m wide must carry 32 m³/s in uniform flow at a depth of 1.8 m. Manning n = 0.014. Find the required bed slope S (in m/m).
Q = 32 m³/sb = 6 my = 1.8 mn = 0.014
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 = 1.5 + 1/2 = 2 m
A = H·b = 1·3 = 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: ycp = 1.722 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 = 0.5 + 2/2 = 1.5 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·1.5·4 = 60.33 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·2³/12 = 1.333 m⁴
ycp = yc + Ic/(yc·A) = 1.5 + 1.333/(1.5·4) = 1.722 m
Answer
ycp = 1.722 m
3.Continuity & Bernoulli
Answer: p2 = 127 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·(100/50)² = 7 m/s
½ρ(v1² − v2²) = ½·1000·(1.75² − 7²)/1000 = −22.97 kPa
p2 = p1 + ½ρ(v1² − v2²) = 150 + (−22.97) = 127 kPa
Answer
p2 = 127 kPa
4.Continuity & Bernoulli
Answer: Q = 58.72 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)·9.81·0.23 = 28429 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 100/200 = 0.5
v2 = √(2·28429/(1000·(1 − 0.0625))) = 7.788 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.96·0.007854·7.788 = 0.05872 m³/s = 58.72 L/s
Answer
Q = 58.72 L/s
5.Pipe friction & head loss
Answer: hf = 2.153 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) = 0.8²/(2·9.81) = 0.03262 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(150/0.05)·0.8²/(2·9.81) = 2.153 m
Answer
hf = 2.153 m
6.Pipe friction & head loss
Answer: Δz = 3.137 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 = 0.06/0.04909 = 1.222 m/s
V²/(2g) = 0.07615 m
Loss coefficients
f·L/D = 0.018·400/0.25 = 28.8
ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4
Total head loss
hf = 28.8·0.07615 = 2.193 m
hm = 12.4·0.07615 = 0.9442 m
Δz = hf + hm = 3.137 m
Answer
Δz = 3.137 m
7.Pumps & pump power
Answer: η = 85.09 %
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 = 1000·9.81·0.1·34 = 33.35 kW
Efficiency
η = Ph/Pshaft = 33.35/39.2 = 85.09%
Answer
η = 85.09%
8.Pumps & pump power
Answer: Pshaft = 37.82 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (36 + 5)·0.3046 = 12.49 m
Pump head
H = Δz + hL = 10 + 12.49 = 22.49 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.12·22.49/0.7 = 37.82 kW
Answer
Pshaft = 37.82 kW
9.Open-channel flow (Manning)
Answer: Q = 0.7011 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 = 0.6362 m², R = D/4 = 0.225 m
Velocity
V = (1/0.013)·0.2252/3·0.00151/2 = 1.102 m/s
Flow rate
Q = A·V = 0.6362·1.102 = 0.7011 m³/s
Answer
Q = 0.7011 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001471 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·y = 6·1.8 = 10.8 m²
P = b + 2y = 6 + 2·1.8 = 9.6 m
R = A/P = 1.125 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.014·32/(10.8·1.082) = 0.03835
S = 0.03835² = 0.001471
Answer
S = 0.001471
That is a fall of 1.471 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering