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

d = 2.5 mH = 2.5 mb = 1 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 0.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 0.5 mH = 2 mb = 2 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

Water flows through a horizontal pipe that narrows from 150 mm to 75 mm. In the wide section the pressure is 370 kPa and the velocity is 2.75 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 150 mmd₂ = 75 mmp₁ = 370 kPav₁ = 2.75 m/sρ = 1000 kg/m³

p₂ = kPa

4.Continuity & BernoulliHard

A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 75 mm and a gauge pressure of 50 kPa. The nozzle exit diameter is 25 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 = 75 mmp₁ = 50 kPa (gauge)d₂ = 25 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 2.1 m/s through a horizontal pipe, 50 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 = 2.1 m/sL = 50 mD = 250 mmf = 0.018ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 2 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. What difference in water level Δz between the reservoirs gives a flow rate of 50 L/s?

Q = 50 L/sL = 150 mD = 200 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 0.2K_exit = 1.0

Δz = 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 26 m. Use g = 9.81 m/s². Find the shaft power P_shaft needed, in kW.

Q = 80 L/sH = 26 mη = 78%g = 9.81 m/s²

P_shaft = kW

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 15 m higher. The pipe is 100 m long with an inside diameter of 200 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 = 75 L/sΔz = 15 mL = 100 mD = 200 mmf = 0.02ΣK = 8p_tank = 200 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 91.97 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 = 2.5 + 2.5/2 = 3.75 m

A = H·b = 2.5·1 = 2.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·3.75·2.5 = 91.97 kN

Answer

F = 91.97 kN

2.Hydrostatic force

Answer: F = 51.8 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 = 2·2 = 4 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·1.5·4 = 51.8 kN

Answer

F = 51.8 kN

3.Continuity & Bernoulli

Answer: p₂ = 313.3 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 = 2.75·(150/75)² = 11 m/s

½ρ(v₁² − v₂²) = ½·1000·(2.75² − 11²)/1000 = −56.72 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 370 + (−56.72) = 313.3 kPa

Answer

p₂ = 313.3 kPa

4.Continuity & Bernoulli

Answer: H = 5.161 m

Exit velocity

β = 25/75 = 0.3333

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·50000/(1000·0.9877)) = 10.06 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) = 10.06²/(2·9.81) = 5.161 m

Answer

H = 5.161 m

5.Pipe friction & head loss

Answer: Δp = 7.938 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·(50/0.25)·2.1²/(2·9.81) = 0.8092 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 = 1000·9.81·0.8092 = 7938 Pa = 7.938 kPa

Answer

Δp = 7.938 kPa

6.Pipe friction & head loss

Answer: Δz = 1.749 m

Velocity

V = Q/A = 0.05/0.03142 = 1.592 m/s

V²/(2g) = 0.1291 m

Loss coefficients

f·L/D = 0.015·150/0.2 = 11.25

ΣK = 0.5 + 2·0.3 + 0.2 + 1.0 = 2.3

Total head loss

h_f = 11.25·0.1291 = 1.452 m

h_m = 2.3·0.1291 = 0.2969 m

Δz = h_f + h_m = 1.749 m

Answer

Δz = 1.749 m

7.Pumps & pump power

Answer: P_shaft = 26.16 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 = 1000·9.81·0.08·26 = 20.4 kW

Shaft power

P_shaft = P_h/η = 20.4/0.78 = 26.16 kW

Answer

P_shaft = 26.16 kW

8.Pumps & pump power

Answer: H = 40.62 m

Velocity

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

V²/(2g) = 0.2905 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (10 + 8)·0.2905 = 5.229 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 + 5.229 = 40.62 m

Answer

H = 40.62 m

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