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 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 1.5 mb = 2 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 seawater, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 3 mb = 2 mρ = 1025 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 100 mm to 40 mm. In the wide section the pressure is 140 kPa and the velocity is 1.25 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 100 mmd2 = 40 mmp1 = 140 kPav1 = 1.25 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 140 kPa. Water leaves through an opening with a diameter of 50 mm, 5 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².
pair = 140 kPa (gauge)h = 5 md = 50 mmρ = 1000 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 10 L/s through a pipe 150 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
Q = 10 L/sL = 150 mD = 100 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 3 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. What difference in water level Δz between the reservoirs gives a flow rate of 75 L/s?
Q = 75 L/sL = 100 mD = 200 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 68 m while its shaft takes 76.2 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 80 L/sH = 68 mPshaft = 76.2 kWg = 9.81 m/s²
η = %
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 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 = 5 (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 = 80 L/sΔz = 5 mL = 300 mD = 250 mmf = 0.022ΣK = 5ptank = 250 kPa
H = m
3Solutions
1.Hydrostatic force
Answer: F = 51.5 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 = 1 + 1.5/2 = 1.75 m
A = H·b = 1.5·2 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·1.75·3 = 51.5 kN
Answer
F = 51.5 kN
2.Hydrostatic force
Answer: F = 150.8 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 + 3/2 = 2.5 m
A = H·b = 3·2 = 6 m²
Resultant force
F = ρ·g·yc·A = 1025·9.81·2.5·6 = 150.8 kN
Answer
F = 150.8 kN
3.Continuity & Bernoulli
Answer: p2 = 110.3 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 = 1.25·(100/40)² = 7.813 m/s
½ρ(v1² − v2²) = ½·1000·(1.25² − 7.813²)/1000 = −29.74 kPa
p2 = p1 + ½ρ(v1² − v2²) = 140 + (−29.74) = 110.3 kPa
Answer
p2 = 110.3 kPa
4.Continuity & Bernoulli
Answer: Q = 38.18 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.
Bernoulli from the surface to the jet
v = √(2·(140000/1000 + 9.81·5)) = 19.44 m/s
Flow rate
Q = A·v = 0.001963·19.44 = 0.03818 m³/s = 38.18 L/s
Answer
Q = 38.18 L/s
5.Pipe friction & head loss
Answer: hf = 3.099 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·0.1²/4 = 0.007854 m²
V = Q/A = 0.01/0.007854 = 1.273 m/s
Velocity head
V²/(2g) = 1.273²/(2·9.81) = 0.08263 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(150/0.1)·1.273²/(2·9.81) = 3.099 m
Answer
hf = 3.099 m
6.Pipe friction & head loss
Answer: Δz = 4.473 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 = 0.075/0.03142 = 2.387 m/s
V²/(2g) = 0.2905 m
Loss coefficients
f·L/D = 0.022·100/0.2 = 11
ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4
Total head loss
hf = 11·0.2905 = 3.195 m
hm = 4.4·0.2905 = 1.278 m
Δz = hf + hm = 4.473 m
Answer
Δz = 4.473 m
7.Pumps & pump power
Answer: η = 70.03 %
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 = 1000·9.81·0.08·68 = 53.37 kW
Efficiency
η = Ph/Pshaft = 53.37/76.2 = 70.03%
Answer
η = 70.03%
8.Pumps & pump power
Answer: H = 34.74 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 = 0.08/0.04909 = 1.63 m/s
V²/(2g) = 0.1354 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (26.4 + 5)·0.1354 = 4.251 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 + 4.251 = 34.74 m
Answer
H = 34.74 m
ExamProblems · examproblems.com/courses/fluid-mechanics