Fluid Mechanics: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Hydrostatic force

Pressure at depth: p = ρgh
Force on a vertical wall: F = ρg·h_c·A
Centre of pressure: y_p = y_c + I_c/(y_cA)
Rectangle: I_c = bH³/12

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: h_f = f·(L/D)·V²/(2g)
Minor losses: h_m = ΣK·V²/(2g)
Typical K values: entrance 0.5, exit 1.0, 90° bend ≈ 0.3
Pressure drop: Δp = ρ·g·h_f
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: P_h = ρ·g·Q·H = Q·Δp
Shaft power: P_shaft = P_h/η_pump
Electrical power: P_el = P_shaft/η_motor
Pump head: H = Δz + Δp/(ρ·g) + h_L
System losses: h_L = (f·L/D + ΣK)·V²/(2g)

2Practice problems

1.Hydrostatic forceMedium

A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 1.5 mb = 3 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 water, with its top edge 3 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 3 mH = 2.5 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows at 7 L/s through a horizontal pipe that narrows from 80 mm to 40 mm. The pressure in the narrow section is 130 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 7 L/sd₁ = 80 mmd₂ = 40 mmp₂ = 130 kPaρ = 1000 kg/m³

p₁ = kPa

4.Continuity & BernoulliHard

A hose with an inner diameter of 50 mm siphons water from an open tank. The highest point C of the hose is 0.5 m above the water surface, and the outlet (open to the air) is 1 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the flow rate Q.

d = 50 mmz_C = +0.5 mz₂ = −1 mg = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at 50 L/s through a pipe 150 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 50 L/sL = 150 mD = 250 mmf = 0.02g = 9.81 m/s²

h_f = 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 water surface of the upper reservoir is 23 m above that of the lower one. 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. Find the flow rate Q in L/s.

Δz = 23 mL = 100 mD = 200 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump with an efficiency of 72% delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 78 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 15 L/sH = 78 mη = 72%g = 9.81 m/s²

P_shaft = kW

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 26 m higher. The pipe is 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.015; 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 300 kPa. Find the head H the pump must add.

Q = 50 L/sΔz = 26 mL = 200 mD = 150 mmf = 0.015ΣK = 8p_tank = 300 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 143.5 kN

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 2.5 + 1.5/2 = 3.25 m

A = H·b = 1.5·3 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.25·4.5 = 143.5 kN

Answer

F = 143.5 kN

2.Hydrostatic force

Answer: y_cp = 4.373 m

Hydrostatic force on a plane surface: F = ρ·g·y_c·A. It acts at the centre of pressure, y_cp = y_c + I_c/(y_cA), which always lies below the centroid.

Centroid depth and area

y_c = d + H/2 = 3 + 2.5/2 = 4.25 m

A = H·b = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.25·3.75 = 156.3 kN

Centre of pressure

Pressure grows with depth, so the resultant acts below the centroid.

I_c = b·H³/12 = 1.5·2.5³/12 = 1.953 m⁴

y_cp = y_c + I_c/(y_c·A) = 4.25 + 1.953/(4.25·3.75) = 4.373 m

Answer

y_cp = 4.373 m

3.Continuity & Bernoulli

Answer: p₁ = 144.5 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.

Velocities

v₁ = Q/A₁ = 0.007/0.005027 = 1.393 m/s

v₂ = Q/A₂ = 0.007/0.001257 = 5.57 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 130 + ½·1000·(5.57² − 1.393²)/1000 = 130 + 14.55 = 144.5 kPa

Answer

p₁ = 144.5 kPa

4.Continuity & Bernoulli

Answer: Q = 8.697 L/s

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·1) = √(2·9.81·1) = 4.429 m/s

Flow rate

Q = A·v = 0.001963·4.429 = 0.008697 m³/s = 8.697 L/s

Answer

Q = 8.697 L/s

5.Pipe friction & head loss

Answer: h_f = 0.6346 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 h_f = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses h_m = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = h_f + h_m, with the exit loss K = 1 included.

Mean velocity

A = π·0.25²/4 = 0.04909 m²

V = Q/A = 0.05/0.04909 = 1.019 m/s

Velocity head

V²/(2g) = 1.019²/(2·9.81) = 0.05288 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(150/0.25)·1.019²/(2·9.81) = 0.6346 m

Answer

h_f = 0.6346 m

6.Pipe friction & head loss

Answer: Q = 170.1 L/s

Loss coefficients

f·L/D = 0.022·100/0.2 = 11

ΣK = 0.5 + 3·0.3 + 2 + 1.0 = 4.4

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

23 = (11 + 4.4)·V²/(2·9.81)

V = √(2·9.81·23/15.4) = 5.413 m/s

Flow rate

Q = V·π·D²/4 = 5.413·0.03142 = 0.1701 m³/s = 170.1 L/s

Answer

Q = 170.1 L/s

7.Pumps & pump power

Answer: P_shaft = 15.94 kW

Pump power: the power given to the water is the hydraulic power P_h = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: P_shaft = P_h/η_pump, and the motor draws P_el = P_shaft/η_motor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Hydraulic power

P_h = ρ·g·Q·H = 1000·9.81·0.015·78 = 11.48 kW

Shaft power

P_shaft = P_h/η = 11.48/0.72 = 15.94 kW

Answer

P_shaft = 15.94 kW

8.Pumps & pump power

Answer: H = 68.01 m

Velocity

V = Q/A = 0.05/0.01767 = 2.829 m/s

V²/(2g) = 0.408 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (20 + 8)·0.408 = 11.42 m

Pressure head

p/(ρ·g) = 300000/(1000·9.81) = 30.58 m

Pump head

Both surfaces are practically at rest, so there is no velocity head term.

H = Δz + p/(ρ·g) + h_L = 26 + 30.58 + 11.42 = 68.01 m

Answer

H = 68.01 m

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