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 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 1.5 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 oil, with its top edge 2 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2 mH = 1.5 mb = 1.5 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows at 6 L/s through a horizontal pipe that narrows from 150 mm to 75 mm. The pressure in the narrow section is 260 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 6 L/sd₁ = 150 mmd₂ = 75 mmp₂ = 260 kPaρ = 1000 kg/m³

p₁ = kPa

4.Continuity & BernoulliHard

A hose with an inner diameter of 32 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 4 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure p_C at point C. The atmospheric pressure is 101.3 kPa.

d = 32 mmz_C = +0.5 mz₂ = −4 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 1.6 m/s through a pipe 100 m long with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the friction head loss h_f.

V = 1.6 m/sL = 100 mD = 75 mmf = 0.02g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 17 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 = 17 mL = 150 mD = 250 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 82% delivers 40 L/s of water (ρ = 1000 kg/m³) against a head of 48 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 40 L/sH = 48 mη = 82%g = 9.81 m/s²

P_shaft = kW

8.Pumps & pump powerHard

A pump moves 120 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 12 m higher. The pipe is 100 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 72%. Find the shaft power P_shaft in kW.

Q = 120 L/sΔz = 12 mL = 100 mD = 250 mmf = 0.018ΣK = 5η = 72%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 99.33 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 = 1.5 + 1.5/2 = 2.25 m

A = H·b = 1.5·3 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.25·4.5 = 99.33 kN

Answer

F = 99.33 kN

2.Hydrostatic force

Answer: y_cp = 2.818 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 = 2 + 1.5/2 = 2.75 m

A = H·b = 1.5·1.5 = 2.25 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·2.75·2.25 = 53.42 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·1.5³/12 = 0.4219 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.75 + 0.4219/(2.75·2.25) = 2.818 m

Answer

y_cp = 2.818 m

3.Continuity & Bernoulli

Answer: p₁ = 260.9 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 = 0.006/0.01767 = 0.3395 m/s

v₂ = Q/A₂ = 0.006/0.004418 = 1.358 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 260 + ½·1000·(1.358² − 0.3395²)/1000 = 260 + 0.8646 = 260.9 kPa

Answer

p₁ = 260.9 kPa

4.Continuity & Bernoulli

Answer: p_C = 57.15 kPa

Velocity in the hose

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

v = √(2g·4) = √(2·9.81·4) = 8.859 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(8.859²/2 + 9.81·0.5)/1000 = 101.3 − 44.15 = 57.15 kPa

Answer

p_C = 57.15 kPa

If this pressure falls to the vapour pressure of the water, the siphon stops working.

5.Pipe friction & head loss

Answer: h_f = 3.479 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.

Velocity head

V²/(2g) = 1.6²/(2·9.81) = 0.1305 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.02·(100/0.075)·1.6²/(2·9.81) = 3.479 m

Answer

h_f = 3.479 m

6.Pipe friction & head loss

Answer: Q = 213.7 L/s

Loss coefficients

f·L/D = 0.022·150/0.25 = 13.2

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.

17 = (13.2 + 4.4)·V²/(2·9.81)

V = √(2·9.81·17/17.6) = 4.353 m/s

Flow rate

Q = V·π·D²/4 = 4.353·0.04909 = 0.2137 m³/s = 213.7 L/s

Answer

Q = 213.7 L/s

7.Pumps & pump power

Answer: P_shaft = 22.97 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 = 1000·9.81·0.04·48 = 18.84 kW

Shaft power

P_shaft = P_h/η = 18.84/0.82 = 22.97 kW

Answer

P_shaft = 22.97 kW

8.Pumps & pump power

Answer: P_shaft = 25.7 kW

Velocity

V = Q/A = 0.12/0.04909 = 2.445 m/s

V²/(2g) = 0.3046 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (7.2 + 5)·0.3046 = 3.716 m

Pump head

H = Δz + h_L = 12 + 3.716 = 15.72 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.12·15.72/0.72 = 25.7 kW

Answer

P_shaft = 25.7 kW

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