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 1.5 m wide sits in the wall of a tank of water, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 3 mb = 1.5 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 oil, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2.5 mb = 2 mρ = 880 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 2 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 = 2 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 200 kPa. Water leaves through an opening with a diameter of 50 mm, 1.5 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
pair = 200 kPa (gauge)h = 1.5 md = 50 mmρ = 1000 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.3 m/s through a pipe 500 m long with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.015 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.3 m/sL = 500 mD = 200 mmf = 0.015g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.025. 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. What difference in water level Δz between the reservoirs gives a flow rate of 120 L/s?
Q = 120 L/sL = 200 mD = 250 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 60% delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 68 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 20 L/sH = 68 mη = 60%g = 9.81 m/s²
Pshaft = kW
8.Pumps & pump powerHard
A pump moves 20 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 100 mm and a Darcy friction factor f = 0.025; 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 = 20 L/sΔz = 11 mL = 100 mD = 100 mmf = 0.025ΣK = 5η = 70%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 242.8 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 = 4 + 3/2 = 5.5 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·4.5 = 242.8 kN
Answer
F = 242.8 kN
2.Hydrostatic force
Answer: F = 161.9 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 + 2.5/2 = 3.75 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·3.75·5 = 161.9 kN
Answer
F = 161.9 kN
3.Continuity & Bernoulli
Answer: Q = 7.995 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) = √(2·9.81·2) = 6.264 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.001963·6.264 = 0.007995 m³/s = 7.995 L/s
Answer
Q = 7.995 L/s
4.Continuity & Bernoulli
Answer: Q = 40.69 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.
Bernoulli from the surface to the jet
v = √(2·(200000/1000 + 9.81·1.5)) = 20.72 m/s
Flow rate
Q = A·v = 0.001963·20.72 = 0.04069 m³/s = 40.69 L/s
Answer
Q = 40.69 L/s
5.Pipe friction & head loss
Answer: hf = 10.11 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) = 2.3²/(2·9.81) = 0.2696 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.015·(500/0.2)·2.3²/(2·9.81) = 10.11 m
Answer
hf = 10.11 m
6.Pipe friction & head loss
Answer: Δz = 7.432 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Loss coefficients
f·L/D = 0.025·200/0.25 = 20
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Total head loss
hf = 20·0.3046 = 6.092 m
hm = 4.4·0.3046 = 1.34 m
Δz = hf + hm = 7.432 m
Answer
Δz = 7.432 m
7.Pumps & pump power
Answer: Pshaft = 22.24 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 = 1000·9.81·0.02·68 = 13.34 kW
Shaft power
Pshaft = Ph/η = 13.34/0.6 = 22.24 kW
Answer
Pshaft = 22.24 kW
8.Pumps & pump power
Answer: Pshaft = 5.862 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 = 0.02/0.007854 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (25 + 5)·0.3305 = 9.915 m
Pump head
H = Δz + hL = 11 + 9.915 = 20.92 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.02·20.92/0.7 = 5.862 kW
Answer
Pshaft = 5.862 kW
ExamProblems · examproblems.com/courses/fluid-mechanics