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

d = 0.5 mH = 2 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 1 m wide sits in the wall of a tank of glycerin, with its top edge 1.5 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 1.5 mH = 1.5 mb = 1 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

Water flows at 39 L/s through a horizontal pipe that narrows from 100 mm to 75 mm. The pressure in the narrow section is 270 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 39 L/sd₁ = 100 mmd₂ = 75 mmp₂ = 270 kPaρ = 1000 kg/m³

p₁ = kPa

4.Continuity & BernoulliHard

A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 50 mm and a gauge pressure of 130 kPa. The nozzle exit diameter is 30 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².

d₁ = 50 mmp₁ = 130 kPa (gauge)d₂ = 30 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 1.2 m/s through a horizontal pipe, 150 m long, with an inside diameter of 250 mm. Use the Darcy friction factor f = 0.018 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 1.2 m/sL = 150 mD = 250 mmf = 0.018ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.022. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 5) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 30 L/s?

Q = 30 L/sL = 100 mD = 150 mmf = 0.022K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 5K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

In a test, a pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 48 m while its shaft takes 57.4 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.

Q = 100 L/sH = 48 mP_shaft = 57.4 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 34 m higher. The pipe is 200 m long with an inside diameter of 200 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 = 80 L/sΔz = 34 mL = 200 mD = 200 mmf = 0.018ΣK = 8p_tank = 150 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 29.43 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 = 0.5 + 2/2 = 1.5 m

A = H·b = 2·1 = 2 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.5·2 = 29.43 kN

Answer

F = 29.43 kN

2.Hydrostatic force

Answer: y_cp = 2.333 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 = 1.5 + 1.5/2 = 2.25 m

A = H·b = 1.5·1 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·2.25·1.5 = 41.72 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1.5³/12 = 0.2813 m⁴

y_cp = y_c + I_c/(y_c·A) = 2.25 + 0.2813/(2.25·1.5) = 2.333 m

Answer

y_cp = 2.333 m

3.Continuity & Bernoulli

Answer: p₁ = 296.6 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.

Velocities

v₁ = Q/A₁ = 0.039/0.007854 = 4.966 m/s

v₂ = Q/A₂ = 0.039/0.004418 = 8.828 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 270 + ½·1000·(8.828² − 4.966²)/1000 = 270 + 26.64 = 296.6 kPa

Answer

p₁ = 296.6 kPa

4.Continuity & Bernoulli

Answer: H = 15.22 m

Exit velocity

β = 30/50 = 0.6

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·130000/(1000·0.8704)) = 17.28 m/s

Height of the jet

Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.

H = v₂²/(2g) = 17.28²/(2·9.81) = 15.22 m

Answer

H = 15.22 m

5.Pipe friction & head loss

Answer: Δp = 7.776 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) = 0.018·(150/0.25)·1.2²/(2·9.81) = 0.7927 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 = 1000·9.81·0.7927 = 7776 Pa = 7.776 kPa

Answer

Δp = 7.776 kPa

6.Pipe friction & head loss

Answer: Δz = 3.197 m

Velocity

V = Q/A = 0.03/0.01767 = 1.698 m/s

V²/(2g) = 0.1469 m

Loss coefficients

f·L/D = 0.022·100/0.15 = 14.67

ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1

Total head loss

h_f = 14.67·0.1469 = 2.154 m

h_m = 7.1·0.1469 = 1.043 m

Δz = h_f + h_m = 3.197 m

Answer

Δz = 3.197 m

7.Pumps & pump power

Answer: η = 82.03 %

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 = 1000·9.81·0.1·48 = 47.09 kW

Efficiency

η = P_h/P_shaft = 47.09/57.4 = 82.03%

Answer

η = 82.03%

8.Pumps & pump power

Answer: H = 57.88 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) = (18 + 8)·0.3305 = 8.593 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 = 34 + 15.29 + 8.593 = 57.88 m

Answer

H = 57.88 m

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