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 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 2 mb = 1.5 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 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 0.5 mH = 1 mb = 1 mρ = 880 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
A pitot-static tube in a water pipe reads a difference of 10 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.
p0 − p = 10 kPaρ = 1000 kg/m³
v = m/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 1 m. At section 1 the pressure is 250 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 100 mmp1 = 250 kPav1 = 0.5 m/sΔz = 1 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at 5 L/s through a pipe 50 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss hf.
Q = 5 L/sL = 50 mD = 50 mmf = 0.022g = 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 = 0.025. 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. What difference in water level Δz between the reservoirs gives a flow rate of 50 L/s?
Q = 50 L/sL = 100 mD = 200 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 28 m while its shaft takes 11.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 30 L/sH = 28 mPshaft = 11.4 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 100 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 600 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 78%. Find the shaft power Pshaft in kW.
Q = 100 L/sΔz = 8 mL = 600 mD = 250 mmf = 0.018ΣK = 6η = 78%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 103 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 + 2/2 = 3.5 m
A = H·b = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.5·3 = 103 kN
Answer
F = 103 kN
2.Hydrostatic force
Answer: ycp = 1.083 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 = 0.5 + 1/2 = 1 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·1·1 = 8.633 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 1·1³/12 = 0.08333 m⁴
ycp = yc + Ic/(yc·A) = 1 + 0.08333/(1·1) = 1.083 m
Answer
ycp = 1.083 m
3.Continuity & Bernoulli
Answer: v = 4.472 m/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 to the stagnation point
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
p0 − p = ½ρv²
v = √(2·10000/1000) = 4.472 m/s
Answer
v = 4.472 m/s
4.Continuity & Bernoulli
Answer: p2 = 239.7 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 = 0.5·(150/100)² = 1.125 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 1.125²)/1000 = −0.5078 kPa
ρgΔz = 1000·9.81·1/1000 = 9.81 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 250 + (−0.5078) − 9.81 = 239.7 kPa
Answer
p2 = 239.7 kPa
5.Pipe friction & head loss
Answer: hf = 7.271 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·0.05²/4 = 0.001963 m²
V = Q/A = 0.005/0.001963 = 2.546 m/s
Velocity head
V²/(2g) = 2.546²/(2·9.81) = 0.3305 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.022·(50/0.05)·2.546²/(2·9.81) = 7.271 m
Answer
hf = 7.271 m
6.Pipe friction & head loss
Answer: Δz = 3.253 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 = 0.025·100/0.2 = 12.5
ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7
Total head loss
hf = 12.5·0.1291 = 1.614 m
hm = 12.7·0.1291 = 1.64 m
Δz = hf + hm = 3.253 m
Answer
Δz = 3.253 m
7.Pumps & pump power
Answer: η = 72.28 %
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 = 1000·9.81·0.03·28 = 8.24 kW
Efficiency
η = Ph/Pshaft = 8.24/11.4 = 72.28%
Answer
η = 72.28%
8.Pumps & pump power
Answer: Pshaft = 23.15 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (43.2 + 6)·0.2115 = 10.41 m
Pump head
H = Δz + hL = 8 + 10.41 = 18.41 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.1·18.41/0.78 = 23.15 kW
Answer
Pshaft = 23.15 kW
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