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

d = 3 mH = 1.5 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 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 = 1 mb = 1.5 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 75 mm. In the wide section the pressure is 340 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 100 mmd₂ = 75 mmp₁ = 340 kPav₁ = 2.25 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A horizontal venturi meter (inlet 100 mm, throat 50 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 340 mm. The discharge coefficient is 0.96. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².

d₁ = 100 mmd₂ = 50 mmΔh = 340 mmC_d = 0.96ρ_Hg = 13 600 kg/m³g = 9.81 m/s²

Q = L/s

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2 m/s through a pipe 800 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2 m/sL = 800 mD = 150 mmf = 0.018g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 100 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.02) 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), 2 bends (K = 0.3 each) and a valve (K = 10). 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 = 100 mD = 300 mmf = 0.02K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 10

H = m

7.Pumps & pump powerMedium

In a test, a pump delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 72 m while its shaft takes 75.3 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 80 L/sH = 72 mP_shaft = 75.3 kWg = 9.81 m/s²

η = %

8.Pumps & pump powerHard

A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 7 m higher. The pipe is 500 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.022; 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 150 kPa. Find the head H the pump must add.

Q = 75 L/sΔz = 7 mL = 500 mD = 250 mmf = 0.022ΣK = 8p_tank = 150 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 55.18 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 + 1.5/2 = 3.75 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.75·1.5 = 55.18 kN

Answer

F = 55.18 kN

2.Hydrostatic force

Answer: y_cp = 3.028 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 + 1/2 = 3 m

A = H·b = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·3·1.5 = 38.85 kN

Centre of pressure

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

I_c = b·H³/12 = 1.5·1³/12 = 0.125 m⁴

y_cp = y_c + I_c/(y_c·A) = 3 + 0.125/(3·1.5) = 3.028 m

Answer

y_cp = 3.028 m

3.Continuity & Bernoulli

Answer: p₂ = 334.5 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 = 2.25·(100/75)² = 4 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.25² − 4²)/1000 = −5.469 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 340 + (−5.469) = 334.5 kPa

Answer

p₂ = 334.5 kPa

4.Continuity & Bernoulli

Answer: Q = 17.85 L/s

Pressure difference from the manometer

Δp = (13600 − 1000)·9.81·0.34 = 42026 Pa

Throat velocity

Continuity and Bernoulli together give the ideal throat velocity.

β = 50/100 = 0.5

v₂ = √(2·42026/(1000·(1 − 0.0625))) = 9.469 m/s

Actual flow rate

Q = C_d·A₂·v₂ = 0.96·0.001963·9.469 = 0.01785 m³/s = 17.85 L/s

Answer

Q = 17.85 L/s

5.Pipe friction & head loss

Answer: h_f = 19.57 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) = 2²/(2·9.81) = 0.2039 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.018·(800/0.15)·2²/(2·9.81) = 19.57 m

Answer

h_f = 19.57 m

6.Pipe friction & head loss

Answer: H = 1.077 m

Velocity

V = Q/A = 0.075/0.07069 = 1.061 m/s

V²/(2g) = 0.05738 m

Loss coefficients

f·L/D = 0.02·100/0.3 = 6.667

ΣK = 0.5 + 2·0.3 + 10 = 11.1

Energy equation

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

H = (1 + 6.667 + 11.1)·0.05738 = 1.077 m

Answer

H = 1.077 m

7.Pumps & pump power

Answer: η = 75.04 %

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 = 1000·9.81·0.08·72 = 56.51 kW

Efficiency

η = P_h/P_shaft = 56.51/75.3 = 75.04%

Answer

η = 75.04%

8.Pumps & pump power

Answer: H = 28.48 m

Velocity

V = Q/A = 0.075/0.04909 = 1.528 m/s

V²/(2g) = 0.119 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (44 + 8)·0.119 = 6.187 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 = 7 + 15.29 + 6.187 = 28.48 m

Answer

H = 28.48 m

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