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

d = 0.5 mH = 2 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1 m wide sits in the wall of a tank of water, 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 = 2 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water drains from a large open tank through a sharp-edged orifice with a diameter of 50 mm. The water level is kept 6 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².

d = 50 mmh = 6 mC_d = 0.62g = 9.81 m/s²

Q = L/s

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 75 mm to 40 mm while rising 7 m. At section 1 the pressure is 390 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 75 mmd₂ = 40 mmp₁ = 390 kPav₁ = 1 m/sΔz = 7 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1.9 m/s through a horizontal pipe, 300 m long, with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.9 m/sL = 300 mD = 150 mmf = 0.018ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 3 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 60 L/s?

Q = 60 L/sL = 300 mD = 250 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 10K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 58 m. The pump efficiency is 75% and the electric motor driving it has an efficiency of 90%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 60 L/sH = 58 mη_pump = 75%η_motor = 90%g = 9.81 m/s²

P_el = kW

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 28 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 80 L/sΔz = 28 mL = 150 mD = 200 mmf = 0.022ΣK = 6

H = m

3Solutions

1.Hydrostatic force

Answer: F = 88.29 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 = 0.5 + 2/2 = 1.5 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.5·6 = 88.29 kN

Answer

F = 88.29 kN

2.Hydrostatic force

Answer: y_cp = 3.111 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 + 2/2 = 3 m

A = H·b = 2·1 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3·2 = 58.86 kN

Centre of pressure

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

I_c = b·H³/12 = 1·2³/12 = 0.6667 m⁴

y_cp = y_c + I_c/(y_c·A) = 3 + 0.6667/(3·2) = 3.111 m

Answer

y_cp = 3.111 m

3.Continuity & Bernoulli

Answer: Q = 13.21 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.

Ideal jet velocity

√(2gh) = √(2·9.81·6) = 10.85 m/s

Actual flow rate

The discharge coefficient accounts for the contraction of the jet and friction at the edge.

Q = C_d·A·√(2gh) = 0.62·0.001963·10.85 = 0.01321 m³/s = 13.21 L/s

Answer

Q = 13.21 L/s

4.Continuity & Bernoulli

Answer: p₂ = 315.7 kPa

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v₂ = v₁·(d₁/d₂)² = 1·(75/40)² = 3.516 m/s

½ρ(v₁² − v₂²) = ½·1000·(1² − 3.516²)/1000 = −5.68 kPa

ρgΔz = 1000·9.81·7/1000 = 68.67 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 390 + (−5.68) − 68.67 = 315.7 kPa

Answer

p₂ = 315.7 kPa

5.Pipe friction & head loss

Answer: Δp = 64.98 kPa

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.

Head loss

h_f = f·(L/D)·V²/(2g) = 0.018·(300/0.15)·1.9²/(2·9.81) = 6.624 m

Pressure drop

The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·h_f.

Δp = ρ·g·h_f = 1000·9.81·6.624 = 64980 Pa = 64.98 kPa

Answer

Δp = 64.98 kPa

6.Pipe friction & head loss

Answer: Δz = 2.315 m

Velocity

V = Q/A = 0.06/0.04909 = 1.222 m/s

V²/(2g) = 0.07615 m

Loss coefficients

f·L/D = 0.015·300/0.25 = 18

ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4

Total head loss

h_f = 18·0.07615 = 1.371 m

h_m = 12.4·0.07615 = 0.9442 m

Δz = h_f + h_m = 2.315 m

Answer

Δz = 2.315 m

7.Pumps & pump power

Answer: P_el = 50.58 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 = 1000·9.81·0.06·58 = 34.14 kW

Shaft power

P_shaft = 34.14/0.75 = 45.52 kW

Electrical power

P_el = 45.52/0.9 = 50.58 kW

Answer

P_el = 50.58 kW

8.Pumps & pump power

Answer: H = 35.44 m

Velocity

V = Q/A = 0.08/0.03142 = 2.546 m/s

V²/(2g) = 0.3305 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (16.5 + 6)·0.3305 = 7.436 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 28 + 7.436 = 35.44 m

Answer

H = 35.44 m

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