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

d = 2 mH = 1 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 seawater, with its top edge 4 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 4 mH = 1.5 mb = 3 mρ = 1025 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 75 mm to 40 mm. In the wide section the pressure is 210 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 75 mmd₂ = 40 mmp₁ = 210 kPav₁ = 0.75 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 6 m. At section 1 the pressure is 230 kPa and the velocity is 1.75 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 150 mmd₂ = 75 mmp₁ = 230 kPav₁ = 1.75 m/sΔz = 6 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at 5 L/s through a pipe 100 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 5 L/sL = 100 mD = 50 mmf = 0.018g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018. The minor losses are a square-edged entrance (K = 0.5), 4 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 40 L/s?

Q = 40 L/sL = 300 mD = 200 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 10K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 22 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 25 L/sH = 22 mη_pump = 72%η_motor = 92%g = 9.81 m/s²

P_el = 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 8 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 6 (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 = 8 mL = 150 mD = 200 mmf = 0.018ΣK = 6p_tank = 300 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 36.79 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 + 1/2 = 2.5 m

A = H·b = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.5·1.5 = 36.79 kN

Answer

F = 36.79 kN

2.Hydrostatic force

Answer: y_cp = 4.789 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 = 4 + 1.5/2 = 4.75 m

A = H·b = 1.5·3 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1025·9.81·4.75·4.5 = 214.9 kN

Centre of pressure

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

I_c = b·H³/12 = 3·1.5³/12 = 0.8438 m⁴

y_cp = y_c + I_c/(y_c·A) = 4.75 + 0.8438/(4.75·4.5) = 4.789 m

Answer

y_cp = 4.789 m

3.Continuity & Bernoulli

Answer: p₂ = 206.8 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 = 0.75·(75/40)² = 2.637 m/s

½ρ(v₁² − v₂²) = ½·1000·(0.75² − 2.637²)/1000 = −3.195 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 210 + (−3.195) = 206.8 kPa

Answer

p₂ = 206.8 kPa

4.Continuity & Bernoulli

Answer: p₂ = 148.2 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.75·(150/75)² = 7 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.75² − 7²)/1000 = −22.97 kPa

ρgΔz = 1000·9.81·6/1000 = 58.86 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 230 + (−22.97) − 58.86 = 148.2 kPa

Answer

p₂ = 148.2 kPa

5.Pipe friction & head loss

Answer: h_f = 11.9 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·0.05²/4 = 0.001963 m²

V = Q/A = 0.005/0.001963 = 2.546 m/s

Velocity head

V²/(2g) = 2.546²/(2·9.81) = 0.3305 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.018·(100/0.05)·2.546²/(2·9.81) = 11.9 m

Answer

h_f = 11.9 m

6.Pipe friction & head loss

Answer: Δz = 3.28 m

Velocity

V = Q/A = 0.04/0.03142 = 1.273 m/s

V²/(2g) = 0.08263 m

Loss coefficients

f·L/D = 0.018·300/0.2 = 27

ΣK = 0.5 + 4·0.3 + 10 + 1.0 = 12.7

Total head loss

h_f = 27·0.08263 = 2.231 m

h_m = 12.7·0.08263 = 1.049 m

Δz = h_f + h_m = 3.28 m

Answer

Δz = 3.28 m

7.Pumps & pump power

Answer: P_el = 8.145 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 = 1000·9.81·0.025·22 = 5.396 kW

Shaft power

P_shaft = 5.396/0.72 = 7.494 kW

Electrical power

P_el = 7.494/0.92 = 8.145 kW

Answer

P_el = 8.145 kW

8.Pumps & pump power

Answer: H = 41.1 m

Velocity

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

V²/(2g) = 0.1291 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (13.5 + 6)·0.1291 = 2.518 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 = 8 + 30.58 + 2.518 = 41.1 m

Answer

H = 41.1 m

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