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 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 = 1 mb = 1 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 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 = 2.5 mb = 1.5 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 2 m above the orifice and the discharge coefficient is 0.65. Find the flow rate Q. Use g = 9.81 m/s².

d = 50 mmh = 2 mC_d = 0.65g = 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 2 m. At section 1 the pressure is 280 kPa and the velocity is 0.75 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 75 mmd₂ = 40 mmp₁ = 280 kPav₁ = 0.75 m/sΔz = 2 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 2.3 m/s through a horizontal pipe, 50 m long, with an inside diameter of 200 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 2.3 m/sL = 50 mD = 200 mmf = 0.022ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 300 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.018) 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 = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 75 L/s?

Q = 75 L/sL = 300 mD = 200 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 5

H = m

7.Pumps & pump powerMedium

In a test, a pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 46 m while its shaft takes 33.8 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 60 L/sH = 46 mP_shaft = 33.8 kWg = 9.81 m/s²

η = %

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 15 m higher. The pipe is 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.02; 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 150 kPa. Find the head H the pump must add.

Q = 80 L/sΔz = 15 mL = 150 mD = 200 mmf = 0.02ΣK = 6p_tank = 150 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 14.71 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 + 1/2 = 1.5 m

A = H·b = 1·1 = 1 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.5·1 = 14.71 kN

Answer

F = 14.71 kN

2.Hydrostatic force

Answer: y_cp = 4.373 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 + 2.5/2 = 4.25 m

A = H·b = 2.5·1.5 = 3.75 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.25·3.75 = 156.3 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·2.5³/12 = 1.953 m⁴

y_cp = y_c + I_c/(y_c·A) = 4.25 + 1.953/(4.25·3.75) = 4.373 m

Answer

y_cp = 4.373 m

3.Continuity & Bernoulli

Answer: Q = 7.995 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) = √(2·9.81·2) = 6.264 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) = 0.65·0.001963·6.264 = 0.007995 m³/s = 7.995 L/s

Answer

Q = 7.995 L/s

4.Continuity & Bernoulli

Answer: p₂ = 257.2 kPa

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

ρgΔz = 1000·9.81·2/1000 = 19.62 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 280 + (−3.195) − 19.62 = 257.2 kPa

Answer

p₂ = 257.2 kPa

5.Pipe friction & head loss

Answer: Δp = 14.55 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) = 0.022·(50/0.2)·2.3²/(2·9.81) = 1.483 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 = 1000·9.81·1.483 = 14547 Pa = 14.55 kPa

Answer

Δp = 14.55 kPa

6.Pipe friction & head loss

Answer: H = 9.993 m

Velocity

V = Q/A = 0.075/0.03142 = 2.387 m/s

V²/(2g) = 0.2905 m

Loss coefficients

f·L/D = 0.018·300/0.2 = 27

ΣK = 0.5 + 3·0.3 + 5 = 6.4

Energy equation

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

H = (1 + 27 + 6.4)·0.2905 = 9.993 m

Answer

H = 9.993 m

7.Pumps & pump power

Answer: η = 80.11 %

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 = 1000·9.81·0.06·46 = 27.08 kW

Efficiency

η = P_h/P_shaft = 27.08/33.8 = 80.11%

Answer

η = 80.11%

8.Pumps & pump power

Answer: H = 37.23 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) = (15 + 6)·0.3305 = 6.941 m

Pressure head

p/(ρ·g) = 150000/(1000·9.81) = 15.29 m

Pump head

Both surfaces are practically at rest, so there is no velocity head term.

H = Δz + p/(ρ·g) + h_L = 15 + 15.29 + 6.941 = 37.23 m

Answer

H = 37.23 m

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