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

d = 3 mH = 2.5 mb = 3 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 75 mm to 50 mm. In the wide section the pressure is 240 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 75 mmd₂ = 50 mmp₁ = 240 kPav₁ = 1.5 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

Water flows up through a pipe that narrows from 100 mm to 40 mm while rising 3 m. At section 1 the pressure is 240 kPa and the velocity is 2.25 m/s. Ignoring losses, find the pressure p₂ at section 2.

d₁ = 100 mmd₂ = 40 mmp₁ = 240 kPav₁ = 2.25 m/sΔz = 3 mρ = 1000 kg/m³

p₂ = kPa

5.Pipe friction & head lossMedium

Water flows at 10 L/s through a pipe 1000 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.022 and g = 9.81 m/s². Find the friction head loss h_f.

Q = 10 L/sL = 1000 mD = 150 mmf = 0.022g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

A pipe 100 m long with an inside diameter of 200 mm (Darcy friction factor f = 0.015) 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 = 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 = 200 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 10

H = m

7.Pumps & pump powerMedium

A pump delivers 60 L/s of water (ρ = 1000 kg/m³) against a head of 28 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 88%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 60 L/sH = 28 mη_pump = 60%η_motor = 88%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 38 m higher. The pipe is 150 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 = 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 = 20 L/sΔz = 38 mL = 150 mD = 100 mmf = 0.018ΣK = 8p_tank = 150 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 312.7 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 + 2.5/2 = 4.25 m

A = H·b = 2.5·3 = 7.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·4.25·7.5 = 312.7 kN

Answer

F = 312.7 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 = 235.4 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 = 1.5·(75/50)² = 3.375 m/s

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

p₂ = p₁ + ½ρ(v₁² − v₂²) = 240 + (−4.57) = 235.4 kPa

Answer

p₂ = 235.4 kPa

4.Continuity & Bernoulli

Answer: p₂ = 114.2 kPa

Continuity and Bernoulli

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

v₂ = v₁·(d₁/d₂)² = 2.25·(100/40)² = 14.06 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.25² − 14.06²)/1000 = −96.35 kPa

ρgΔz = 1000·9.81·3/1000 = 29.43 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) − ρgΔz = 240 + (−96.35) − 29.43 = 114.2 kPa

Answer

p₂ = 114.2 kPa

5.Pipe friction & head loss

Answer: h_f = 2.394 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·0.15²/4 = 0.01767 m²

V = Q/A = 0.01/0.01767 = 0.5659 m/s

Velocity head

V²/(2g) = 0.5659²/(2·9.81) = 0.01632 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.022·(1000/0.15)·0.5659²/(2·9.81) = 2.394 m

Answer

h_f = 2.394 m

6.Pipe friction & head loss

Answer: H = 5.868 m

Velocity

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

V²/(2g) = 0.2905 m

Loss coefficients

f·L/D = 0.015·100/0.2 = 7.5

ΣK = 0.5 + 4·0.3 + 10 = 11.7

Energy equation

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

H = (1 + 7.5 + 11.7)·0.2905 = 5.868 m

Answer

H = 5.868 m

7.Pumps & pump power

Answer: P_el = 31.21 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 = 1000·9.81·0.06·28 = 16.48 kW

Shaft power

P_shaft = 16.48/0.6 = 27.47 kW

Electrical power

P_el = 27.47/0.88 = 31.21 kW

Answer

P_el = 31.21 kW

8.Pumps & pump power

Answer: H = 64.86 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) = (27 + 8)·0.3305 = 11.57 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 = 38 + 15.29 + 11.57 = 64.86 m

Answer

H = 64.86 m

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