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 1.5 m wide sits in the wall of a tank of water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1.5 m wide sits in the wall of a tank of glycerin, 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 = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 150 mm and a throat diameter of 100 mm. The pressure at the inlet is 59 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 150 mmd2 = 100 mmp1 − p2 = 59 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 150 mm, throat 100 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 350 mm. The discharge coefficient is 0.99. 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 = 100 mmΔh = 350 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.7 m/s through a pipe 50 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.7 m/sL = 50 mD = 300 mmf = 0.028g = 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 200 mm and a Darcy friction factor f = 0.02. The water surface of the upper reservoir is 37 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 = 10) 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 = 37 mL = 100 mD = 200 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 44 m while its shaft takes 13.8 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 44 mPshaft = 13.8 kWg = 9.81 m/s²
η = %
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 11 m higher. The pipe is 100 m long with an inside diameter of 150 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. The upper tank is closed and the air above its water is at a gauge pressure of 350 kPa. Find the head H the pump must add.
Q = 50 L/sΔz = 11 mL = 100 mD = 150 mmf = 0.02ΣK = 5ptank = 350 kPa
H = m
9.Open-channel flow (Manning)Medium
A concrete storm sewer with an inside diameter of 750 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 = 750 mmn = 0.013S = 0.0015
Q = m³/s
10.Open-channel flow (Manning)Hard
A trapezoidal grass-lined earth channel with a bottom width of 4 m and side slopes of 2:1 (horizontal:vertical) must carry 16 m³/s in uniform flow at a depth of 1.9 m. Manning n = 0.03. Find the required bed slope S (in m/m).
Q = 16 m³/sb = 4 my = 1.9 mz = 2n = 0.03
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 119.6 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 = 2 + 2.5/2 = 3.25 m
A = H·b = 2.5·1.5 = 3.75 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.25·3.75 = 119.6 kN
Answer
F = 119.6 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 = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·4.5·4.5 = 250.3 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1.5·3³/12 = 3.375 m⁴
ycp = yc + Ic/(yc·A) = 4.5 + 3.375/(4.5·4.5) = 4.667 m
Answer
ycp = 4.667 m
3.Continuity & Bernoulli
Answer: Q = 95.24 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.
Continuity
v1 = v2·(100/150)² = 0.4444·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·59000/(1000·(1 − 0.1975))) = 12.13 m/s
v1 = 5.389 m/s
Flow rate
Q = A2·v2 = 0.007854·12.13 = 0.09524 m³/s = 95.24 L/s
Answer
Q = 95.24 L/s
4.Continuity & Bernoulli
Answer: Q = 80.74 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)·9.81·0.35 = 43262 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 100/150 = 0.6667
v2 = √(2·43262/(1000·(1 − 0.1975))) = 10.38 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.007854·10.38 = 0.08074 m³/s = 80.74 L/s
Answer
Q = 80.74 L/s
5.Pipe friction & head loss
Answer: hf = 0.6874 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) = 1.7²/(2·9.81) = 0.1473 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(50/0.3)·1.7²/(2·9.81) = 0.6874 m
Answer
hf = 0.6874 m
6.Pipe friction & head loss
Answer: Q = 177.7 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 = 0.02·100/0.2 = 10
ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
37 = (10 + 12.7)·V²/(2·9.81)
V = √(2·9.81·37/22.7) = 5.655 m/s
Flow rate
Q = V·π·D²/4 = 5.655·0.03142 = 0.1777 m³/s = 177.7 L/s
Answer
Q = 177.7 L/s
7.Pumps & pump power
Answer: η = 78.2 %
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 = 1000·9.81·0.025·44 = 10.79 kW
Efficiency
η = Ph/Pshaft = 10.79/13.8 = 78.2%
Answer
η = 78.2%
8.Pumps & pump power
Answer: H = 54.16 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) = (13.33 + 5)·0.408 = 7.481 m
Pressure head
p/(ρ·g) = 350000/(1000·9.81) = 35.68 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 11 + 35.68 + 7.481 = 54.16 m
Answer
H = 54.16 m
9.Open-channel flow (Manning)
Answer: Q = 0.4312 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 = 0.4418 m², R = D/4 = 0.1875 m
Velocity
V = (1/0.013)·0.18752/3·0.00151/2 = 0.976 m/s
Flow rate
Q = A·V = 0.4418·0.976 = 0.4312 m³/s
Answer
Q = 0.4312 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.0008357 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 = (4 + 2·1.9)·1.9 = 14.82 m²
P = b + 2y·√(1 + z²) = 4 + 2·1.9·√(1 + 2²) = 12.5 m
R = A/P = 1.186 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.03·16/(14.82·1.12) = 0.02891
S = 0.02891² = 0.0008357
Answer
S = 0.0008357
That is a fall of 0.8357 m per kilometre.
ExamProblems · examproblems.com/courses/hydraulic-engineering