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

d = 3 mH = 2 mb = 1.5 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 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 mb = 3 mρ = 1000 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 250 kPa and the velocity is 2.5 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 150 mmd₂ = 75 mmp₁ = 250 kPav₁ = 2.5 m/sρ = 1000 kg/m³

p₂ = 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 2.5 m above the water surface, and the outlet (open to the air) is 2.5 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 = 50 mmz_C = +2.5 mz₂ = −2.5 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1.4 m/s through a horizontal pipe, 50 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.03 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.4 m/sL = 50 mD = 200 mmf = 0.03ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 200 m long with an inside diameter of 150 mm (Darcy friction factor f = 0.015) 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 60 L/s?

Q = 60 L/sL = 200 mD = 150 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 2

H = m

7.Pumps & pump powerMedium

In a test, a pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 42 m while its shaft takes 5.7 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 10 L/sH = 42 mP_shaft = 5.7 kWg = 9.81 m/s²

η = %

8.Pumps & pump powerHard

A pump moves 20 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 32 m higher. The pipe is 100 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 8 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 20 L/sΔz = 32 mL = 100 mD = 100 mmf = 0.025ΣK = 8

H = m

3Solutions

1.Hydrostatic force

Answer: F = 117.7 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 = 3 + 2/2 = 4 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4·3 = 117.7 kN

Answer

F = 117.7 kN

2.Hydrostatic force

Answer: y_cp = 4.083 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 + 2/2 = 4 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4·6 = 235.4 kN

Centre of pressure

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

I_c = b·H³/12 = 3·2³/12 = 2 m⁴

y_cp = y_c + I_c/(y_c·A) = 4 + 2/(4·6) = 4.083 m

Answer

y_cp = 4.083 m

3.Continuity & Bernoulli

Answer: p₂ = 203.1 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 = 2.5·(150/75)² = 10 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.5² − 10²)/1000 = −46.88 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 250 + (−46.88) = 203.1 kPa

Answer

p₂ = 203.1 kPa

4.Continuity & Bernoulli

Answer: p_C = 52.25 kPa

Velocity in the hose

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

v = √(2g·2.5) = √(2·9.81·2.5) = 7.004 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(7.004²/2 + 9.81·2.5)/1000 = 101.3 − 49.05 = 52.25 kPa

Answer

p_C = 52.25 kPa

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

5.Pipe friction & head loss

Answer: Δp = 7.35 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) = 0.03·(50/0.2)·1.4²/(2·9.81) = 0.7492 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 = 1000·9.81·0.7492 = 7350 Pa = 7.35 kPa

Answer

Δp = 7.35 kPa

6.Pipe friction & head loss

Answer: H = 14.34 m

Velocity

V = Q/A = 0.06/0.01767 = 3.395 m/s

V²/(2g) = 0.5876 m

Loss coefficients

f·L/D = 0.015·200/0.15 = 20

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 + 20 + 3.4)·0.5876 = 14.34 m

Answer

H = 14.34 m

7.Pumps & pump power

Answer: η = 72.28 %

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 = 1000·9.81·0.01·42 = 4.12 kW

Efficiency

η = P_h/P_shaft = 4.12/5.7 = 72.28%

Answer

η = 72.28%

8.Pumps & pump power

Answer: H = 42.91 m

Velocity

V = Q/A = 0.02/0.007854 = 2.546 m/s

V²/(2g) = 0.3305 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (25 + 8)·0.3305 = 10.91 m

Pump head

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

H = Δz + h_L = 32 + 10.91 = 42.91 m

Answer

H = 42.91 m

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