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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 1 mb = 1 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 seawater, with its top edge 3 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 3 mH = 1.5 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 200 mm to 75 mm. In the wide section the pressure is 160 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 200 mmd2 = 75 mmp1 = 160 kPav1 = 0.5 m/sρ = 1000 kg/m³
p2 = kPa
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 220 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 = 220 mmCd = 0.99ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.9 m/s through a pipe 1000 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.9 m/sL = 1000 mD = 50 mmf = 0.02g = 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 = 0.022. The water surface of the upper reservoir is 13 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 = 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 = 13 mL = 150 mD = 200 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
A pump with an efficiency of 70% delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 72 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 30 L/sH = 72 mη = 70%g = 9.81 m/s²
Pshaft = kW
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 11 m higher. The pipe is 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 2.5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 75%. Find the shaft power Pshaft in kW.
Q = 80 L/sΔz = 11 mL = 100 mD = 200 mmf = 0.025ΣK = 2.5η = 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 = 1.5 + 1/2 = 2 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2·1 = 19.62 kN
Answer
F = 19.62 kN
2.Hydrostatic force
Answer: F = 169.7 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 + 1.5/2 = 3.75 m
A = H·b = 1.5·3 = 4.5 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·3.75·4.5 = 169.7 kN
Answer
F = 169.7 kN
3.Continuity & Bernoulli
Answer: p2 = 153.8 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·(200/75)² = 3.556 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 3.556²)/1000 = −6.196 kPa
p2 = p1 + ½ρ(v1² − v2²) = 160 + (−6.196) = 153.8 kPa
Answer
p2 = 153.8 kPa
4.Continuity & Bernoulli
Answer: Q = 92.53 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)·9.81·0.22 = 27193 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 125/250 = 0.5
v2 = √(2·27193/(1000·(1 − 0.0625))) = 7.617 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.99·0.01227·7.617 = 0.09253 m³/s = 92.53 L/s
Answer
Q = 92.53 L/s
5.Pipe friction & head loss
Answer: hf = 73.6 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) = 1.9²/(2·9.81) = 0.184 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.02·(1000/0.05)·1.9²/(2·9.81) = 73.6 m
Answer
hf = 73.6 m
6.Pipe friction & head loss
Answer: Q = 110.5 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 = 0.022·150/0.2 = 16.5
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
13 = (16.5 + 4.1)·V²/(2·9.81)
V = √(2·9.81·13/20.6) = 3.519 m/s
Flow rate
Q = V·π·D²/4 = 3.519·0.03142 = 0.1105 m³/s = 110.5 L/s
Answer
Q = 110.5 L/s
7.Pumps & pump power
Answer: Pshaft = 30.27 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 = 1000·9.81·0.03·72 = 21.19 kW
Shaft power
Pshaft = Ph/η = 21.19/0.7 = 30.27 kW
Answer
Pshaft = 30.27 kW
8.Pumps & pump power
Answer: Pshaft = 16.7 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) = (12.5 + 2.5)·0.3305 = 4.958 m
Pump head
H = Δz + hL = 11 + 4.958 = 15.96 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.08·15.96/0.75 = 16.7 kW
Answer
Pshaft = 16.7 kW
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