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 = 2 mb = 2 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 4 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 4 mH = 2 mb = 1.5 mρ = 880 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 25 mm. The water level is kept 2 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².
d = 25 mmh = 2 mCd = 0.65g = 9.81 m/s²
Q = L/s
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 90 kPa. Water leaves through an opening with a diameter of 25 mm, 2.5 m below the water surface, into the open air. Ignoring losses, find the jet velocity v. Use g = 9.81 m/s².
pair = 90 kPa (gauge)h = 2.5 md = 25 mmρ = 1000 kg/m³g = 9.81 m/s²
v = m/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 0.8 m/s through a pipe 50 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss hf.
V = 0.8 m/sL = 50 mD = 50 mmf = 0.018g = 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 300 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 = 0.2) 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 = 300 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 0.2Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 28 m while its shaft takes 9.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 25 L/sH = 28 mPshaft = 9.2 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 75 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 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 85%. Find the shaft power Pshaft in kW.
Q = 75 L/sΔz = 13 mL = 150 mD = 200 mmf = 0.02ΣK = 5η = 85%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 58.86 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 + 2/2 = 1.5 m
A = H·b = 2·2 = 4 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·4 = 58.86 kN
Answer
F = 58.86 kN
2.Hydrostatic force
Answer: F = 129.5 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 + 2/2 = 5 m
A = H·b = 2·1.5 = 3 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·5·3 = 129.5 kN
Answer
F = 129.5 kN
3.Continuity & Bernoulli
Answer: Q = 1.999 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) = √(2·9.81·2) = 6.264 m/s
Actual flow rate
The discharge coefficient accounts for the contraction of the jet and friction at the edge.
Q = Cd·A·√(2gh) = 0.65·0.0004909·6.264 = 0.001999 m³/s = 1.999 L/s
Answer
Q = 1.999 L/s
4.Continuity & Bernoulli
Answer: v = 15.13 m/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.
Bernoulli from the surface to the jet
v = √(2·(90000/1000 + 9.81·2.5)) = 15.13 m/s
Answer
v = 15.13 m/s
5.Pipe friction & head loss
Answer: hf = 0.5872 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) = 0.8²/(2·9.81) = 0.03262 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.018·(50/0.05)·0.8²/(2·9.81) = 0.5872 m
Answer
hf = 0.5872 m
6.Pipe friction & head loss
Answer: Q = 309.5 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 = 0.022·150/0.3 = 11
ΣK = 0.5 + 2·0.3 + 0.2 + 1.0 = 2.3
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
13 = (11 + 2.3)·V²/(2·9.81)
V = √(2·9.81·13/13.3) = 4.379 m/s
Flow rate
Q = V·π·D²/4 = 4.379·0.07069 = 0.3095 m³/s = 309.5 L/s
Answer
Q = 309.5 L/s
7.Pumps & pump power
Answer: η = 74.64 %
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 = 1000·9.81·0.025·28 = 6.867 kW
Efficiency
η = Ph/Pshaft = 6.867/9.2 = 74.64%
Answer
η = 74.64%
8.Pumps & pump power
Answer: Pshaft = 16.28 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (15 + 5)·0.2905 = 5.81 m
Pump head
H = Δz + hL = 13 + 5.81 = 18.81 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.075·18.81/0.85 = 16.28 kW
Answer
Pshaft = 16.28 kW
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