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 = 3 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1 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 = 3 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water drains from a large open tank through a sharp-edged orifice with a diameter of 20 mm. The water level is kept 1.5 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 20 mmh = 1.5 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 200 mm to 100 mm while rising 5 m. At section 1 the pressure is 300 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 200 mmd2 = 100 mmp1 = 300 kPav1 = 0.75 m/sΔz = 5 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 3 m/s through a horizontal pipe, 50 m long, with an inside diameter of 250 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 = 3 m/sL = 50 mD = 250 mmf = 0.02ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 500 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 15 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 = 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 = 15 mL = 500 mD = 300 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 56 m. The pump efficiency is 80% 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 = 25 L/sH = 56 mηpump = 80%ηmotor = 95%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 22 m higher. The pipe is 400 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022; 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 300 kPa. Find the head H the pump must add.
Q = 50 L/sΔz = 22 mL = 400 mD = 150 mmf = 0.022ΣK = 8ptank = 300 kPa
H = m
9.Open-channel flow (Manning)Medium
A rectangular unfinished concrete channel is 1 m wide. Water flows uniformly 2.1 m deep. Manning n = 0.014 and the bed slope is S = 0.0015. Find the flow rate Q in m³/s.
b = 1 my = 2.1 mn = 0.014S = 0.0015
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal unfinished concrete channel with a bottom width of 2 m and side slopes of 1.5:1 (horizontal:vertical) must carry 8 m³/s in uniform flow at a depth of 1.4 m. Manning n = 0.014. Find the required bed slope S (in m/m).
Q = 8 m³/sb = 2 my = 1.4 mz = 1.5n = 0.014
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 397.3 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 + 3/2 = 4.5 m
A = H·b = 3·3 = 9 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.5·9 = 397.3 kN
Answer
F = 397.3 kN
2.Hydrostatic force
Answer: ycp = 4.667 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 + 3/2 = 4.5 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.5·3 = 132.4 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·3³/12 = 2.25 m⁴
ycp = yc + Ic/(yc·A) = 4.5 + 2.25/(4.5·3) = 4.667 m
Answer
ycp = 4.667 m
3.Continuity & Bernoulli
Answer: Q = 1.108 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) = √(2·9.81·1.5) = 5.425 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.0003142·5.425 = 0.001108 m³/s = 1.108 L/s
Answer
Q = 1.108 L/s
4.Continuity & Bernoulli
Answer: p2 = 246.7 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 = 0.75·(200/100)² = 3 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 3²)/1000 = −4.219 kPa
ρgΔz = 1000·9.81·5/1000 = 49.05 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 300 + (−4.219) − 49.05 = 246.7 kPa
Answer
p2 = 246.7 kPa
5.Pipe friction & head loss
Answer: Δp = 18 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·(50/0.25)·3²/(2·9.81) = 1.835 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 = 1000·9.81·1.835 = 18000 Pa = 18 kPa
Answer
Δp = 18 kPa
6.Pipe friction & head loss
Answer: Q = 205.9 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 = 0.018·500/0.3 = 30
ΣK = 0.5 + 4·0.3 + 2 + 1.0 = 4.7
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
15 = (30 + 4.7)·V²/(2·9.81)
V = √(2·9.81·15/34.7) = 2.912 m/s
Flow rate
Q = V·π·D²/4 = 2.912·0.07069 = 0.2059 m³/s = 205.9 L/s
Answer
Q = 205.9 L/s
7.Pumps & pump power
Answer: Pel = 18.07 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 = 1000·9.81·0.025·56 = 13.73 kW
Shaft power
Pshaft = 13.73/0.8 = 17.17 kW
Electrical power
Pel = 17.17/0.95 = 18.07 kW
Answer
Pel = 18.07 kW
8.Pumps & pump power
Answer: H = 79.78 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (58.67 + 8)·0.408 = 27.2 m
Pressure head
p/(ρ·g) = 300000/(1000·9.81) = 30.58 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 22 + 30.58 + 27.2 = 79.78 m
Answer
H = 79.78 m
9.Open-channel flow (Manning)
Answer: Q = 3.174 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 = 1·2.1 = 2.1 m²
P = b + 2y = 1 + 2·2.1 = 5.2 m
R = A/P = 0.4038 m
Velocity
V = (1/0.014)·0.40382/3·0.00151/2 = 1.511 m/s
Flow rate
Q = A·V = 2.1·1.511 = 3.174 m³/s
Answer
Q = 3.174 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0005006 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 = (2 + 1.5·1.4)·1.4 = 5.74 m²
P = b + 2y·√(1 + z²) = 2 + 2·1.4·√(1 + 1.5²) = 7.048 m
R = A/P = 0.8144 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.014·8/(5.74·0.8721) = 0.02237
S = 0.02237² = 0.0005006
Answer
S = 0.0005006
That is a fall of 0.5006 m per kilometre.
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