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 2 m wide sits in the wall of a tank of oil, 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 = 2.5 mb = 2 mρ = 880 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 100 mm to 50 mm. In the wide section the pressure is 160 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 100 mmd₂ = 50 mmp₁ = 160 kPav₁ = 3 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A closed water tank has compressed air above the water at a gauge pressure of 120 kPa. Water leaves through an opening with a diameter of 20 mm, 2 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².

p_air = 120 kPa (gauge)h = 2 md = 20 mmρ = 1000 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 0.8 m/s through a horizontal pipe, 1000 m long, with an inside diameter of 50 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 = 0.8 m/sL = 1000 mD = 50 mmf = 0.03ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 300 m long with an inside diameter of 300 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), 4 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 100 L/s?

Q = 100 L/sL = 300 mD = 300 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump with an efficiency of 78% delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 60 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 80 L/sH = 60 mη = 78%g = 9.81 m/s²

P_shaft = kW

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 23 m higher. The pipe is 100 m long with an inside diameter of 100 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 5 (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 250 kPa. Find the head H the pump must add.

Q = 20 L/sΔz = 23 mL = 100 mD = 100 mmf = 0.018ΣK = 5p_tank = 250 kPa

H = m

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 = 5.349 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 + 2.5/2 = 5.25 m

A = H·b = 2.5·2 = 5 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·5.25·5 = 226.6 kN

Centre of pressure

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

I_c = b·H³/12 = 2·2.5³/12 = 2.604 m⁴

y_cp = y_c + I_c/(y_c·A) = 5.25 + 2.604/(5.25·5) = 5.349 m

Answer

y_cp = 5.349 m

3.Continuity & Bernoulli

Answer: p₂ = 92.5 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 = 3·(100/50)² = 12 m/s

½ρ(v₁² − v₂²) = ½·1000·(3² − 12²)/1000 = −67.5 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 160 + (−67.5) = 92.5 kPa

Answer

p₂ = 92.5 kPa

4.Continuity & Bernoulli

Answer: Q = 5.25 L/s

Bernoulli from the surface to the jet

v = √(2·(120000/1000 + 9.81·2)) = 16.71 m/s

Flow rate

Q = A·v = 0.0003142·16.71 = 0.00525 m³/s = 5.25 L/s

Answer

Q = 5.25 L/s

5.Pipe friction & head loss

Answer: Δp = 192 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·(1000/0.05)·0.8²/(2·9.81) = 19.57 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 = 1000·9.81·19.57 = 192000 Pa = 192 kPa

Answer

Δp = 192 kPa

6.Pipe friction & head loss

Answer: H = 2.622 m

Velocity

V = Q/A = 0.1/0.07069 = 1.415 m/s

V²/(2g) = 0.102 m

Loss coefficients

f·L/D = 0.018·300/0.3 = 18

ΣK = 0.5 + 4·0.3 + 5 = 6.7

Energy equation

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

H = (1 + 18 + 6.7)·0.102 = 2.622 m

Answer

H = 2.622 m

7.Pumps & pump power

Answer: P_shaft = 60.37 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 = 1000·9.81·0.08·60 = 47.09 kW

Shaft power

P_shaft = P_h/η = 47.09/0.78 = 60.37 kW

Answer

P_shaft = 60.37 kW

8.Pumps & pump power

Answer: H = 56.09 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) = (18 + 5)·0.3305 = 7.602 m

Pressure head

p/(ρ·g) = 250000/(1000·9.81) = 25.48 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 23 + 25.48 + 7.602 = 56.09 m

Answer

H = 56.09 m

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