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 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 3 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of glycerin, with its top edge 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 1 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows at 20 L/s through a horizontal pipe that narrows from 100 mm to 50 mm. The pressure in the narrow section is 280 kPa. Ignoring losses, find the pressure p1 in the wide section.
Q = 20 L/sd1 = 100 mmd2 = 50 mmp2 = 280 kPaρ = 1000 kg/m³
p1 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 100 mm, throat 60 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 150 mm. The discharge coefficient is 0.96. 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 = 100 mmd2 = 60 mmΔh = 150 mmCd = 0.96ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 5 L/s through a pipe 400 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
Q = 5 L/sL = 400 mD = 50 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.025) 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), 3 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 80 L/s?
Q = 80 L/sL = 200 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 78 m while its shaft takes 76.5 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 60 L/sH = 78 mPshaft = 76.5 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 15 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 10 m higher. The pipe is 300 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.015; 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 = 15 L/sΔz = 10 mL = 300 mD = 100 mmf = 0.015ΣK = 6η = 75%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 198.7 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 = 3 + 3/2 = 4.5 m
A = H·b = 3·1.5 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.5·4.5 = 198.7 kN
Answer
F = 198.7 kN
2.Hydrostatic force
Answer: F = 166.9 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 + 1/2 = 4.5 m
A = H·b = 1·3 = 3 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·4.5·3 = 166.9 kN
Answer
F = 166.9 kN
3.Continuity & Bernoulli
Answer: p1 = 328.6 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.
Velocities
v1 = Q/A1 = 0.02/0.007854 = 2.546 m/s
v2 = Q/A2 = 0.02/0.001963 = 10.19 m/s
Bernoulli
p1 = p2 + ½ρ(v2² − v1²) = 280 + ½·1000·(10.19² − 2.546²)/1000 = 280 + 48.63 = 328.6 kPa
Answer
p1 = 328.6 kPa
4.Continuity & Bernoulli
Answer: Q = 17.72 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)·9.81·0.15 = 18541 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 60/100 = 0.6
v2 = √(2·18541/(1000·(1 − 0.1296))) = 6.527 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.96·0.002827·6.527 = 0.01772 m³/s = 17.72 L/s
Answer
Q = 17.72 L/s
5.Pipe friction & head loss
Answer: hf = 79.32 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) = 0.03·(400/0.05)·2.546²/(2·9.81) = 79.32 m
Answer
hf = 79.32 m
6.Pipe friction & head loss
Answer: H = 1.898 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 = 0.08/0.07069 = 1.132 m/s
V²/(2g) = 0.06529 m
Loss coefficients
f·L/D = 0.025·200/0.3 = 16.67
ΣK = 0.5 + 3·0.3 + 10 = 11.4
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 16.67 + 11.4)·0.06529 = 1.898 m
Answer
H = 1.898 m
7.Pumps & pump power
Answer: η = 60.01 %
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 = 1000·9.81·0.06·78 = 45.91 kW
Efficiency
η = Ph/Pshaft = 45.91/76.5 = 60.01%
Answer
η = 60.01%
8.Pumps & pump power
Answer: Pshaft = 3.822 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (45 + 6)·0.1859 = 9.481 m
Pump head
H = Δz + hL = 10 + 9.481 = 19.48 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.015·19.48/0.75 = 3.822 kW
Answer
Pshaft = 3.822 kW
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