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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 3 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 oil, 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 = 1.5 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 50 mm. In the wide section the pressure is 270 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 50 mmp1 = 270 kPav1 = 1.5 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A hose with an inner diameter of 25 mm siphons water from an open tank. The highest point C of the hose is 0.5 m above the water surface, and the outlet (open to the air) is 1 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.
d = 25 mmzC = +0.5 mz2 = −1 mg = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 0.9 m/s through a pipe 300 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
V = 0.9 m/sL = 300 mD = 50 mmf = 0.028g = 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 = 0.02. The water surface of the upper reservoir is 22 m above that of the lower one. 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. Find the flow rate Q in L/s.
Δz = 22 mL = 150 mD = 200 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 56 m while its shaft takes 45.8 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 60 L/sH = 56 mPshaft = 45.8 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 80 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 12 m higher. The pipe is 500 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 78%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 12 mL = 500 mD = 200 mmf = 0.015ΣK = 5η = 78%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 132.4 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 = 1.5 + 3/2 = 3 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3·4.5 = 132.4 kN
Answer
F = 132.4 kN
2.Hydrostatic force
Answer: F = 155.4 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 = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·4·4.5 = 155.4 kN
Answer
F = 155.4 kN
3.Continuity & Bernoulli
Answer: p2 = 265.4 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 = 1.5·(75/50)² = 3.375 m/s
½ρ(v1² − v2²) = ½·1000·(1.5² − 3.375²)/1000 = −4.57 kPa
p2 = p1 + ½ρ(v1² − v2²) = 270 + (−4.57) = 265.4 kPa
Answer
p2 = 265.4 kPa
4.Continuity & Bernoulli
Answer: Q = 2.174 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.
Velocity in the hose
Surface and outlet are both open to the air, so the pressure terms cancel.
v = √(2g·1) = √(2·9.81·1) = 4.429 m/s
Flow rate
Q = A·v = 0.0004909·4.429 = 0.002174 m³/s = 2.174 L/s
Answer
Q = 2.174 L/s
5.Pipe friction & head loss
Answer: hf = 6.936 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) = 0.9²/(2·9.81) = 0.04128 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(300/0.05)·0.9²/(2·9.81) = 6.936 m
Answer
hf = 6.936 m
6.Pipe friction & head loss
Answer: Q = 148.2 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 = 0.02·150/0.2 = 15
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
22 = (15 + 4.4)·V²/(2·9.81)
V = √(2·9.81·22/19.4) = 4.717 m/s
Flow rate
Q = V·π·D²/4 = 4.717·0.03142 = 0.1482 m³/s = 148.2 L/s
Answer
Q = 148.2 L/s
7.Pumps & pump power
Answer: η = 71.97 %
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 = 1000·9.81·0.06·56 = 32.96 kW
Efficiency
η = Ph/Pshaft = 32.96/45.8 = 71.97%
Answer
η = 71.97%
8.Pumps & pump power
Answer: Pshaft = 26.21 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 = 0.08/0.03142 = 2.546 m/s
V²/(2g) = 0.3305 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (37.5 + 5)·0.3305 = 14.05 m
Pump head
H = Δz + hL = 12 + 14.05 = 26.05 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·26.05/0.78 = 26.21 kW
Answer
Pshaft = 26.21 kW
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