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 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1 mb = 1.5 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 water, with its top edge 2.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
d = 2.5 mH = 2.5 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²
ycp = m
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 75 mm to 40 mm. In the wide section the pressure is 240 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 75 mmd2 = 40 mmp1 = 240 kPav1 = 2.75 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 300 mm, throat 150 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 370 mm. The discharge coefficient is 0.97. 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 = 300 mmd2 = 150 mmΔh = 370 mmCd = 0.97ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 30 L/s through a pipe 300 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Q = 30 L/sL = 300 mD = 150 mmf = 0.028g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.025. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 100 L/s?
Q = 100 L/sL = 300 mD = 300 mmf = 0.025Kentrance = 0.5Kbend = 0.3 (×4)Kvalve = 10Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump with an efficiency of 65% delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 80 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.
Q = 25 L/sH = 80 mη = 65%g = 9.81 m/s²
Pshaft = kW
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 19 m higher. The pipe is 150 m long with an inside diameter of 100 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 = 15 L/sΔz = 19 mL = 150 mD = 100 mmf = 0.022ΣK = 8ptank = 250 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 22.07 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 + 1/2 = 1.5 m
A = H·b = 1·1.5 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.5·1.5 = 22.07 kN
Answer
F = 22.07 kN
2.Hydrostatic force
Answer: ycp = 3.889 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 = 2.5 + 2.5/2 = 3.75 m
A = H·b = 2.5·2 = 5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·3.75·5 = 183.9 kN
Centre of pressure
Pressure grows with depth, so the resultant acts below the centroid.
Ic = b·H³/12 = 2·2.5³/12 = 2.604 m⁴
ycp = yc + Ic/(yc·A) = 3.75 + 2.604/(3.75·5) = 3.889 m
Answer
ycp = 3.889 m
3.Continuity & Bernoulli
Answer: p2 = 197 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) = 240 + (−42.95) = 197 kPa
Answer
p2 = 197 kPa
4.Continuity & Bernoulli
Answer: Q = 169.3 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)·9.81·0.37 = 45734 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 150/300 = 0.5
v2 = √(2·45734/(1000·(1 − 0.0625))) = 9.878 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.97·0.01767·9.878 = 0.1693 m³/s = 169.3 L/s
Answer
Q = 169.3 L/s
5.Pipe friction & head loss
Answer: hf = 8.226 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·0.15²/4 = 0.01767 m²
V = Q/A = 0.03/0.01767 = 1.698 m/s
Velocity head
V²/(2g) = 1.698²/(2·9.81) = 0.1469 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.028·(300/0.15)·1.698²/(2·9.81) = 8.226 m
Answer
hf = 8.226 m
6.Pipe friction & head loss
Answer: Δz = 3.846 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 = 0.1/0.07069 = 1.415 m/s
V²/(2g) = 0.102 m
Loss coefficients
f·L/D = 0.025·300/0.3 = 25
ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7
Total head loss
hf = 25·0.102 = 2.55 m
hm = 12.7·0.102 = 1.296 m
Δz = hf + hm = 3.846 m
Answer
Δz = 3.846 m
7.Pumps & pump power
Answer: Pshaft = 30.18 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 = 1000·9.81·0.025·80 = 19.62 kW
Shaft power
Pshaft = Ph/η = 19.62/0.65 = 30.18 kW
Answer
Pshaft = 30.18 kW
8.Pumps & pump power
Answer: H = 52.11 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 = 0.015/0.007854 = 1.91 m/s
V²/(2g) = 0.1859 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (33 + 8)·0.1859 = 7.622 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 = 19 + 25.48 + 7.622 = 52.11 m
Answer
H = 52.11 m
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