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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 1 mb = 2 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 glycerin, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2.5 mH = 1 mb = 2 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal venturi meter with an inlet diameter of 150 mm and a throat diameter of 75 mm. The pressure at the inlet is 55 kPa higher than at the throat. Ignoring losses, find the flow rate Q.
d1 = 150 mmd2 = 75 mmp1 − p2 = 55 kPaρ = 1000 kg/m³
Q = L/s
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 75 mm to 50 mm while rising 2 m. At section 1 the pressure is 250 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 75 mmd2 = 50 mmp1 = 250 kPav1 = 2.25 m/sΔz = 2 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2.4 m/s through a horizontal pipe, 1000 m long, with an inside diameter of 300 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2.4 m/sL = 1000 mD = 300 mmf = 0.022ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018. The water surface of the upper reservoir is 23 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 2 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 = 23 mL = 300 mD = 200 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 42 m while its shaft takes 25.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 40 L/sH = 42 mPshaft = 25.4 kWg = 9.81 m/s²
η = %
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 9 m higher. The pipe is 300 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 75%. Find the shaft power Pshaft in kW.
Q = 40 L/sΔz = 9 mL = 300 mD = 150 mmf = 0.025ΣK = 6η = 75%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 19.62 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 = 0.5 + 1/2 = 1 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1·2 = 19.62 kN
Answer
F = 19.62 kN
2.Hydrostatic force
Answer: F = 74.16 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 + 1/2 = 3 m
A = H·b = 1·2 = 2 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·3·2 = 74.16 kN
Answer
F = 74.16 kN
3.Continuity & Bernoulli
Answer: Q = 47.85 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·(75/150)² = 0.25·v2
Bernoulli between inlet and throat
The pipe is horizontal, so only pressure and velocity change.
p1 − p2 = ½ρv2²(1 − β⁴)
v2 = √(2·55000/(1000·(1 − 0.0625))) = 10.83 m/s
v1 = 2.708 m/s
Flow rate
Q = A2·v2 = 0.004418·10.83 = 0.04785 m³/s = 47.85 L/s
Answer
Q = 47.85 L/s
4.Continuity & Bernoulli
Answer: p2 = 220.1 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 = 2.25·(75/50)² = 5.063 m/s
½ρ(v1² − v2²) = ½·1000·(2.25² − 5.063²)/1000 = −10.28 kPa
ρgΔz = 1000·9.81·2/1000 = 19.62 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 250 + (−10.28) − 19.62 = 220.1 kPa
Answer
p2 = 220.1 kPa
5.Pipe friction & head loss
Answer: Δp = 211.2 kPa
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.
Head loss
hf = f·(L/D)·V²/(2g) = 0.022·(1000/0.3)·2.4²/(2·9.81) = 21.53 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·21.53 = 211200 Pa = 211.2 kPa
Answer
Δp = 211.2 kPa
6.Pipe friction & head loss
Answer: Q = 106.7 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 = 0.018·300/0.2 = 27
ΣK = 0.5 + 2·0.3 + 10 + 1.0 = 12.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
23 = (27 + 12.1)·V²/(2·9.81)
V = √(2·9.81·23/39.1) = 3.397 m/s
Flow rate
Q = V·π·D²/4 = 3.397·0.03142 = 0.1067 m³/s = 106.7 L/s
Answer
Q = 106.7 L/s
7.Pumps & pump power
Answer: η = 64.89 %
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 = 1000·9.81·0.04·42 = 16.48 kW
Efficiency
η = Ph/Pshaft = 16.48/25.4 = 64.89%
Answer
η = 64.89%
8.Pumps & pump power
Answer: Pshaft = 12.36 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 = 0.04/0.01767 = 2.264 m/s
V²/(2g) = 0.2611 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (50 + 6)·0.2611 = 14.62 m
Pump head
H = Δz + hL = 9 + 14.62 = 23.62 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.04·23.62/0.75 = 12.36 kW
Answer
Pshaft = 12.36 kW
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