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 oil, with its top edge 2.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2.5 mH = 3 mb = 3 mρ = 880 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 25 mm. The water level is kept 4.5 m above the orifice and the discharge coefficient is 0.62. Find the flow rate Q. Use g = 9.81 m/s².

d = 25 mmh = 4.5 mC_d = 0.62g = 9.81 m/s²

Q = L/s

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 150 mm to 100 mm while rising 7 m. At section 1 the pressure is 330 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 150 mmd₂ = 100 mmp₁ = 330 kPav₁ = 1.5 m/sΔz = 7 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 0.8 m/s through a pipe 200 m long with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss h_f.

V = 0.8 m/sL = 200 mD = 250 mmf = 0.028g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018. The water surface of the upper reservoir is 21 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 0.2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.

Δz = 21 mL = 100 mD = 200 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 0.2K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump delivers 15 L/s of water (ρ = 1000 kg/m³) against a head of 62 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 94%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 15 L/sH = 62 mη_pump = 60%η_motor = 94%g = 9.81 m/s²

P_el = 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 15 m higher. The pipe is 300 m long with an inside diameter of 100 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 upper tank is closed and the air above its water is at a gauge pressure of 200 kPa. Find the head H the pump must add.

Q = 20 L/sΔz = 15 mL = 300 mD = 100 mmf = 0.02ΣK = 8p_tank = 200 kPa

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 = 4.188 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 = 2.5 + 3/2 = 4 m

A = H·b = 3·3 = 9 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·4·9 = 310.8 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) = 4 + 6.75/(4·9) = 4.188 m

Answer

y_cp = 4.188 m

3.Continuity & Bernoulli

Answer: Q = 2.86 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) = √(2·9.81·4.5) = 9.396 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) = 0.62·0.0004909·9.396 = 0.00286 m³/s = 2.86 L/s

Answer

Q = 2.86 L/s

4.Continuity & Bernoulli

Answer: p₂ = 256.8 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 1.5·(150/100)² = 3.375 m/s

½ρ(v₁² − v₂²) = ½·1000·(1.5² − 3.375²)/1000 = −4.57 kPa

ρgΔz = 1000·9.81·7/1000 = 68.67 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 330 + (−4.57) − 68.67 = 256.8 kPa

Answer

p₂ = 256.8 kPa

5.Pipe friction & head loss

Answer: h_f = 0.7307 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.

Velocity head

V²/(2g) = 0.8²/(2·9.81) = 0.03262 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.028·(200/0.25)·0.8²/(2·9.81) = 0.7307 m

Answer

h_f = 0.7307 m

6.Pipe friction & head loss

Answer: Q = 184.9 L/s

Loss coefficients

f·L/D = 0.018·100/0.2 = 9

ΣK = 0.5 + 4·0.3 + 0.2 + 1.0 = 2.9

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

21 = (9 + 2.9)·V²/(2·9.81)

V = √(2·9.81·21/11.9) = 5.884 m/s

Flow rate

Q = V·π·D²/4 = 5.884·0.03142 = 0.1849 m³/s = 184.9 L/s

Answer

Q = 184.9 L/s

7.Pumps & pump power

Answer: P_el = 16.18 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 = 1000·9.81·0.015·62 = 9.123 kW

Shaft power

P_shaft = 9.123/0.6 = 15.21 kW

Electrical power

P_el = 15.21/0.94 = 16.18 kW

Answer

P_el = 16.18 kW

8.Pumps & pump power

Answer: H = 57.86 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) = (60 + 8)·0.3305 = 22.47 m

Pressure head

p/(ρ·g) = 200000/(1000·9.81) = 20.39 m

Pump head

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

H = Δz + p/(ρ·g) + h_L = 15 + 20.39 + 22.47 = 57.86 m

Answer

H = 57.86 m

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