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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 2.5 mb = 2 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 water, with its top edge 3 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 3 mH = 2.5 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 50 mm. In the wide section the pressure is 330 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 50 mmp1 = 330 kPav1 = 0.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A closed water tank has compressed air above the water at a gauge pressure of 40 kPa. Water leaves through an opening with a diameter of 20 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 = 40 kPa (gauge)h = 2.5 md = 20 mmρ = 1000 kg/m³g = 9.81 m/s²
v = m/s
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 1.9 m/s through a pipe 400 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the friction head loss hf.
V = 1.9 m/sL = 400 mD = 100 mmf = 0.03g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 200 m long with an inside diameter of 250 mm (Darcy friction factor f = 0.022) 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 120 L/s?
Q = 120 L/sL = 200 mD = 250 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10
H = m
7.Pumps & pump powerMedium
In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 42 m while its shaft takes 54.9 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 80 L/sH = 42 mPshaft = 54.9 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 50 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 33 m higher. The pipe is 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 82%. Find the shaft power Pshaft in kW.
Q = 50 L/sΔz = 33 mL = 200 mD = 150 mmf = 0.02ΣK = 8η = 82%
Pshaft = kW
3Solutions
1.Hydrostatic force
Answer: F = 134.9 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 = 1.5 + 2.5/2 = 2.75 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.75·5 = 134.9 kN
Answer
F = 134.9 kN
2.Hydrostatic force
Answer: ycp = 4.373 m
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 + 2.5/2 = 4.25 m
A = H·b = 2.5·3 = 7.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·4.25·7.5 = 312.7 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 3·2.5³/12 = 3.906 m⁴
ycp = yc + Ic/(yc·A) = 4.25 + 3.906/(4.25·7.5) = 4.373 m
Answer
ycp = 4.373 m
3.Continuity & Bernoulli
Answer: p2 = 325.8 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 = 0.75·(100/50)² = 3 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 3²)/1000 = −4.219 kPa
p2 = p1 + ½ρ(v1² − v2²) = 330 + (−4.219) = 325.8 kPa
Answer
p2 = 325.8 kPa
4.Continuity & Bernoulli
Answer: v = 11.36 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·(40000/1000 + 9.81·2.5)) = 11.36 m/s
Answer
v = 11.36 m/s
5.Pipe friction & head loss
Answer: hf = 22.08 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) = 0.03·(400/0.1)·1.9²/(2·9.81) = 22.08 m
Answer
hf = 22.08 m
6.Pipe friction & head loss
Answer: H = 9.138 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 = 0.12/0.04909 = 2.445 m/s
V²/(2g) = 0.3046 m
Loss coefficients
f·L/D = 0.022·200/0.25 = 17.6
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 + 17.6 + 11.4)·0.3046 = 9.138 m
Answer
H = 9.138 m
7.Pumps & pump power
Answer: η = 60.04 %
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 = 1000·9.81·0.08·42 = 32.96 kW
Efficiency
η = Ph/Pshaft = 32.96/54.9 = 60.04%
Answer
η = 60.04%
8.Pumps & pump power
Answer: Pshaft = 28.2 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 = 0.05/0.01767 = 2.829 m/s
V²/(2g) = 0.408 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (26.67 + 8)·0.408 = 14.15 m
Pump head
H = Δz + hL = 33 + 14.15 = 47.15 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.05·47.15/0.82 = 28.2 kW
Answer
Pshaft = 28.2 kW
ExamProblems · examproblems.com/courses/fluid-mechanics