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 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 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 glycerin, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 2 mb = 1 mρ = 1260 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 30 mm. The water level is kept 3.5 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².
d = 30 mmh = 3.5 mCd = 0.62g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 250 mm, throat 125 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 90 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 = 250 mmd2 = 125 mmΔh = 90 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 2 m/s through a horizontal pipe, 150 m long, with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 2 m/sL = 150 mD = 75 mmf = 0.03ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.018) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each) and a valve (K = 10). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 200 L/s?
Q = 200 L/sL = 100 mD = 300 mmf = 0.018Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10
H = m
7.Pumps & pump powerMedium
A pump delivers 150 L/s of water (ρ = 1000 kg/m³) against a head of 34 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 150 L/sH = 34 mηpump = 72%ηmotor = 95%g = 9.81 m/s²
Pel = kW
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 13 m higher. The pipe is 400 m long with an inside diameter of 250 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 80%. Find the shaft power Pshaft in kW.
Q = 100 L/sΔz = 13 mL = 400 mD = 250 mmf = 0.025ΣK = 6η = 80%
Pshaft = kW
3Solutions
1.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 = 4 + 3/2 = 5.5 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·5.5·3 = 161.9 kN
Answer
F = 161.9 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 + 2/2 = 3 m
A = H·b = 2·1 = 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 = 3.632 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) = √(2·9.81·3.5) = 8.287 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.62·0.0007069·8.287 = 0.003632 m³/s = 3.632 L/s
Answer
Q = 3.632 L/s
4.Continuity & Bernoulli
Answer: Q = 59.19 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)·9.81·0.09 = 11125 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 125/250 = 0.5
v2 = √(2·11125/(1000·(1 − 0.0625))) = 4.872 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.01227·4.872 = 0.05919 m³/s = 59.19 L/s
Answer
Q = 59.19 L/s
5.Pipe friction & head loss
Answer: Δp = 120 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) = 0.03·(150/0.075)·2²/(2·9.81) = 12.23 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 = 1000·9.81·12.23 = 120000 Pa = 120 kPa
Answer
Δp = 120 kPa
6.Pipe friction & head loss
Answer: H = 7.63 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 = 0.2/0.07069 = 2.829 m/s
V²/(2g) = 0.408 m
Loss coefficients
f·L/D = 0.018·100/0.3 = 6
ΣK = 0.5 + 4·0.3 + 10 = 11.7
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 6 + 11.7)·0.408 = 7.63 m
Answer
H = 7.63 m
7.Pumps & pump power
Answer: Pel = 73.14 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 = 1000·9.81·0.15·34 = 50.03 kW
Shaft power
Pshaft = 50.03/0.72 = 69.49 kW
Electrical power
Pel = 69.49/0.95 = 73.14 kW
Answer
Pel = 73.14 kW
8.Pumps & pump power
Answer: Pshaft = 27.87 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) = (40 + 6)·0.2115 = 9.73 m
Pump head
H = Δz + hL = 13 + 9.73 = 22.73 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.1·22.73/0.8 = 27.87 kW
Answer
Pshaft = 27.87 kW
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