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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 1 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 2 m wide sits in the wall of a tank of glycerin, with its top edge 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 0.5 mH = 1.5 mb = 2 mρ = 1260 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 40 mm. In the wide section the pressure is 110 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 40 mmp1 = 110 kPav1 = 2.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 100 mm to 40 mm while rising 10 m. At section 1 the pressure is 340 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 100 mmd2 = 40 mmp1 = 340 kPav1 = 0.75 m/sΔz = 10 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water flows at a mean velocity of 2.6 m/s through a pipe 200 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
V = 2.6 m/sL = 200 mD = 150 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
A pipe 100 m long with an inside diameter of 250 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 = 0.2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 100 L/s?
Q = 100 L/sL = 100 mD = 250 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 0.2
H = m
7.Pumps & pump powerMedium
A pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 54 m. The pump efficiency is 80% 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 = 50 L/sH = 54 mηpump = 80%ηmotor = 95%g = 9.81 m/s²
Pel = kW
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 5 m higher. The pipe is 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The upper tank is closed and the air above its water is at a gauge pressure of 250 kPa. Find the head H the pump must add.
Q = 75 L/sΔz = 5 mL = 300 mD = 250 mmf = 0.022ΣK = 8ptank = 250 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 9.81 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 + 1/2 = 1 m
A = H·b = 1·1 = 1 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1·1 = 9.81 kN
Answer
F = 9.81 kN
2.Hydrostatic force
Answer: F = 46.35 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 + 1.5/2 = 1.25 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 1260·9.81·1.25·3 = 46.35 kN
Answer
F = 46.35 kN
3.Continuity & Bernoulli
Answer: p2 = 67.05 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 = 2.75·(75/40)² = 9.668 m/s
½ρ(v1² − v2²) = ½·1000·(2.75² − 9.668²)/1000 = −42.95 kPa
p2 = p1 + ½ρ(v1² − v2²) = 110 + (−42.95) = 67.05 kPa
Answer
p2 = 67.05 kPa
4.Continuity & Bernoulli
Answer: p2 = 231.2 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·(100/40)² = 4.688 m/s
½ρ(v1² − v2²) = ½·1000·(0.75² − 4.688²)/1000 = −10.71 kPa
ρgΔz = 1000·9.81·10/1000 = 98.1 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 340 + (−10.71) − 98.1 = 231.2 kPa
Answer
p2 = 231.2 kPa
5.Pipe friction & head loss
Answer: hf = 11.48 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) = 2.6²/(2·9.81) = 0.3445 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(200/0.15)·2.6²/(2·9.81) = 11.48 m
Answer
hf = 11.48 m
6.Pipe friction & head loss
Answer: H = 2.665 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 = 0.1/0.04909 = 2.037 m/s
V²/(2g) = 0.2115 m
Loss coefficients
f·L/D = 0.025·100/0.25 = 10
ΣK = 0.5 + 3·0.3 + 0.2 = 1.6
Energy equation
The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.
H = (1 + 10 + 1.6)·0.2115 = 2.665 m
Answer
H = 2.665 m
7.Pumps & pump power
Answer: Pel = 34.85 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 = 1000·9.81·0.05·54 = 26.49 kW
Shaft power
Pshaft = 26.49/0.8 = 33.11 kW
Electrical power
Pel = 33.11/0.95 = 34.85 kW
Answer
Pel = 34.85 kW
8.Pumps & pump power
Answer: H = 34.58 m
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 = 0.075/0.04909 = 1.528 m/s
V²/(2g) = 0.119 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (26.4 + 8)·0.119 = 4.093 m
Pressure head
p/(ρ·g) = 250000/(1000·9.81) = 25.48 m
Pump head
Both surfaces are practically at rest, so there is no velocity head term.
H = Δz + p/(ρ·g) + hL = 5 + 25.48 + 4.093 = 34.58 m
Answer
H = 34.58 m
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