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

d = 1 mH = 3 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of glycerin, with its top edge 1 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 1 mH = 3 mb = 3 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal venturi meter with an inlet diameter of 100 mm and a throat diameter of 60 mm. The pressure at the inlet is 18 kPa higher than at the throat. Ignoring losses, find the flow rate Q.

d₁ = 100 mmd₂ = 60 mmp₁ − p₂ = 18 kPaρ = 1000 kg/m³

Q = L/s

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 75 mm while rising 2 m. At section 1 the pressure is 370 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 75 mmp₁ = 370 kPav₁ = 0.5 m/sΔz = 2 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 75 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.9 m/sL = 300 mD = 75 mmf = 0.02ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 200 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.025) runs from an open reservoir and discharges as a free jet into the air. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each) and a valve (K = 2). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 50 L/s?

Q = 50 L/sL = 200 mD = 200 mmf = 0.025K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2

H = m

7.Pumps & pump powerMedium

In a test, a pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 78 m while its shaft takes 23.5 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 20 L/sH = 78 mP_shaft = 23.5 kWg = 9.81 m/s²

η = %

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 15 m higher. The pipe is 500 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 70%. Find the shaft power P_shaft in kW.

Q = 50 L/sΔz = 15 mL = 500 mD = 150 mmf = 0.02ΣK = 8η = 70%

P_shaft = kW

3Solutions

1.Hydrostatic force

Answer: F = 220.7 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 + 3/2 = 2.5 m

A = H·b = 3·3 = 9 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.5·9 = 220.7 kN

Answer

F = 220.7 kN

2.Hydrostatic force

Answer: y_cp = 2.8 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 = 1 + 3/2 = 2.5 m

A = H·b = 3·3 = 9 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·2.5·9 = 278.1 kN

Centre of pressure

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

I_c = b·H³/12 = 3·3³/12 = 6.75 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.5 + 6.75/(2.5·9) = 2.8 m

Answer

y_cp = 2.8 m

3.Continuity & Bernoulli

Answer: Q = 18.18 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.

Continuity

v₁ = v₂·(60/100)² = 0.36·v₂

Bernoulli between inlet and throat

The pipe is horizontal, so only pressure and velocity change.

p₁ − p₂ = ½ρv₂²(1 − β⁴)

v₂ = √(2·18000/(1000·(1 − 0.1296))) = 6.431 m/s

v₁ = 2.315 m/s

Flow rate

Q = A₂·v₂ = 0.002827·6.431 = 0.01818 m³/s = 18.18 L/s

Answer

Q = 18.18 L/s

4.Continuity & Bernoulli

Answer: p₂ = 350.1 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 0.5·(100/75)² = 0.8889 m/s

½ρ(v₁² − v₂²) = ½·1000·(0.5² − 0.8889²)/1000 = −0.2701 kPa

ρgΔz = 1000·9.81·2/1000 = 19.62 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 370 + (−0.2701) − 19.62 = 350.1 kPa

Answer

p₂ = 350.1 kPa

5.Pipe friction & head loss

Answer: Δp = 144.4 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) = 0.02·(300/0.075)·1.9²/(2·9.81) = 14.72 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 = 1000·9.81·14.72 = 144400 Pa = 144.4 kPa

Answer

Δp = 144.4 kPa

6.Pipe friction & head loss

Answer: H = 3.796 m

Velocity

V = Q/A = 0.05/0.03142 = 1.592 m/s

V²/(2g) = 0.1291 m

Loss coefficients

f·L/D = 0.025·200/0.2 = 25

ΣK = 0.5 + 3·0.3 + 2 = 3.4

Energy equation

The outlet jet still carries its velocity head; it is not a loss but it has to be supplied.

H = (1 + 25 + 3.4)·0.1291 = 3.796 m

Answer

H = 3.796 m

7.Pumps & pump power

Answer: η = 65.12 %

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 = 1000·9.81·0.02·78 = 15.3 kW

Efficiency

η = P_h/P_shaft = 15.3/23.5 = 65.12%

Answer

η = 65.12%

8.Pumps & pump power

Answer: P_shaft = 31.86 kW

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) = (66.67 + 8)·0.408 = 30.47 m

Pump head

H = Δz + h_L = 15 + 30.47 = 45.47 m

Shaft power

P_shaft = ρ·g·Q·H/η = 1000·9.81·0.05·45.47/0.7 = 31.86 kW

Answer

P_shaft = 31.86 kW

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