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 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1 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 glycerin, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 2.5 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 5.5 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 50 mmh = 5.5 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A hose with an inner diameter of 32 mm siphons water from an open tank. The highest point C of the hose is 0.5 m above the water surface, and the outlet (open to the air) is 2 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 = 32 mmzC = +0.5 mz2 = −2 mg = 9.81 m/s²patm = 101.3 kPa
pC = kPa
5.Pipe friction & head lossMedium
Water flows at 75 L/s through a pipe 150 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 75 L/sL = 150 mD = 200 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.022. 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), 3 bends (K = 0.3 each), a valve (K = 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 = 13 mL = 100 mD = 300 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 14 m. The pump efficiency is 78% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 20 L/sH = 14 mηpump = 78%ηmotor = 95%g = 9.81 m/s²
Pel = kW
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 16 m higher. The pipe is 500 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 120 L/sΔz = 16 mL = 500 mD = 250 mmf = 0.02ΣK = 5
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal gravel channel has a bottom width of 2.5 m and side slopes of 2:1 (horizontal:vertical). Water flows uniformly 1.3 m deep. Manning n = 0.025 and the bed slope is S = 0.005. Find the flow rate Q in m³/s.
b = 2.5 my = 1.3 mz = 2n = 0.025S = 0.005
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal brickwork channel with a bottom width of 5 m and side slopes of 2:1 (horizontal:vertical) must carry 5 m³/s in uniform flow at a depth of 0.5 m. Manning n = 0.015. Find the required bed slope S (in m/m).
Q = 5 m³/sb = 5 my = 0.5 mz = 2n = 0.015
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 44.15 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 + 1/2 = 1.5 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·3 = 44.15 kN
Answer
F = 44.15 kN
2.Hydrostatic force
Answer: F = 208.6 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 = 1 + 2.5/2 = 2.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·2.25·7.5 = 208.6 kN
Answer
F = 208.6 kN
3.Continuity & Bernoulli
Answer: Q = 13.26 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.
Ideal jet velocity
√(2gh) = √(2·9.81·5.5) = 10.39 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.001963·10.39 = 0.01326 m³/s = 13.26 L/s
Answer
Q = 13.26 L/s
4.Continuity & Bernoulli
Answer: pC = 76.78 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·2) = √(2·9.81·2) = 6.264 m/s
Bernoulli from the surface to C
patm = pC + ½ρv² + ρgzC
pC = 101.3 − 1000·(6.264²/2 + 9.81·0.5)/1000 = 101.3 − 24.53 = 76.78 kPa
Answer
pC = 76.78 kPa
If this pressure falls to the vapour pressure of the water, the siphon stops working.
5.Pipe friction & head loss
Answer: hf = 6.536 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 = 0.075/0.03142 = 2.387 m/s
Velocity head
V²/(2g) = 2.387²/(2·9.81) = 0.2905 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.03·(150/0.2)·2.387²/(2·9.81) = 6.536 m
Answer
hf = 6.536 m
6.Pipe friction & head loss
Answer: Q = 329.6 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 = 0.022·100/0.3 = 7.333
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
13 = (7.333 + 4.4)·V²/(2·9.81)
V = √(2·9.81·13/11.73) = 4.662 m/s
Flow rate
Q = V·π·D²/4 = 4.662·0.07069 = 0.3296 m³/s = 329.6 L/s
Answer
Q = 329.6 L/s
7.Pumps & pump power
Answer: Pel = 3.707 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.
Hydraulic power
Ph = ρ·g·Q·H = 1000·9.81·0.02·14 = 2.747 kW
Shaft power
Pshaft = 2.747/0.78 = 3.522 kW
Electrical power
Pel = 3.522/0.95 = 3.707 kW
Answer
Pel = 3.707 kW
8.Pumps & pump power
Answer: H = 29.71 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (40 + 5)·0.3046 = 13.71 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 16 + 13.71 = 29.71 m
Answer
H = 29.71 m
9.Open-channel flow (Manning)
Answer: Q = 16.13 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 + 2·1.3)·1.3 = 6.63 m²
P = b + 2y·√(1 + z²) = 2.5 + 2·1.3·√(1 + 2²) = 8.314 m
R = A/P = 0.7975 m
Velocity
V = (1/0.025)·0.79752/3·0.0051/2 = 2.432 m/s
Flow rate
Q = A·V = 6.63·2.432 = 16.13 m³/s
Answer
Q = 16.13 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.002022 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 + z·y)·y = (5 + 2·0.5)·0.5 = 3 m²
P = b + 2y·√(1 + z²) = 5 + 2·0.5·√(1 + 2²) = 7.236 m
R = A/P = 0.4146 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.015·5/(3·0.556) = 0.04496
S = 0.04496² = 0.002022
Answer
S = 0.002022
That is a fall of 2.022 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering