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
Fluid Mechanics: cram pack
Formula sheets, the key method steps for every topic and 8 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)
2Practice problems
1.Hydrostatic forceMedium
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 3 mb = 2 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 oil, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 mb = 1 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 25 mm. In the wide section the pressure is 310 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 25 mmp1 = 310 kPav1 = 0.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 80 mm, throat 40 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 50 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 = 80 mmd2 = 40 mmΔh = 50 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 1000 m long with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.7 m/sL = 1000 mD = 300 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015. 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 50 L/s?
Q = 50 L/sL = 150 mD = 200 mmf = 0.015Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 5Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 20 L/sH = 58 mηpump = 72%ηmotor = 94%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 40 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 500 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 40 L/sΔz = 8 mL = 500 mD = 150 mmf = 0.015ΣK = 8
H = m
3Solutions
1.Hydrostatic force
Answer: F = 235.4 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 = 2.5 + 3/2 = 4 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4·6 = 235.4 kN
Answer
F = 235.4 kN
2.Hydrostatic force
Answer: F = 22.66 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 + 1.5/2 = 1.75 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·1.75·1.5 = 22.66 kN
Answer
F = 22.66 kN
3.Continuity & Bernoulli
Answer: p2 = 278.1 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 = 0.5·(100/25)² = 8 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 8²)/1000 = −31.88 kPa
p2 = p1 + ½ρ(v1² − v2²) = 310 + (−31.88) = 278.1 kPa
Answer
p2 = 278.1 kPa
4.Continuity & Bernoulli
Answer: Q = 4.517 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)·9.81·0.05 = 6180 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 40/80 = 0.5
v2 = √(2·6180/(1000·(1 − 0.0625))) = 3.631 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.001257·3.631 = 0.004517 m³/s = 4.517 L/s
Answer
Q = 4.517 L/s
5.Pipe friction & head loss
Answer: hf = 7.365 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) = 0.015·(1000/0.3)·1.7²/(2·9.81) = 7.365 m
Answer
hf = 7.365 m
6.Pipe friction & head loss
Answer: Δz = 2.369 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 = 0.05/0.03142 = 1.592 m/s
V²/(2g) = 0.1291 m
Loss coefficients
f·L/D = 0.015·150/0.2 = 11.25
ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1
Total head loss
hf = 11.25·0.1291 = 1.452 m
hm = 7.1·0.1291 = 0.9166 m
Δz = hf + hm = 2.369 m
Answer
Δz = 2.369 m
7.Pumps & pump power
Answer: Pel = 16.81 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 = 1000·9.81·0.02·58 = 11.38 kW
Shaft power
Pshaft = 11.38/0.72 = 15.8 kW
Electrical power
Pel = 15.8/0.94 = 16.81 kW
Answer
Pel = 16.81 kW
8.Pumps & pump power
Answer: H = 23.15 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 = 0.04/0.01767 = 2.264 m/s
V²/(2g) = 0.2611 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (50 + 8)·0.2611 = 15.15 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 8 + 15.15 = 23.15 m
Answer
H = 23.15 m
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