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

d = 1.5 mH = 2 mb = 1.5 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 oil, with its top edge 3 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 3 mH = 1 mb = 1 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 25 mm. In the wide section the pressure is 370 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p₂ in the narrow section.

d₁ = 100 mmd₂ = 25 mmp₁ = 370 kPav₁ = 0.5 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 100 mm and a gauge pressure of 70 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 = 100 mmp₁ = 70 kPa (gauge)d₂ = 25 mmρ = 1000 kg/m³g = 9.81 m/s²

H = m

5.Pipe friction & head lossMedium

Water flows at a mean velocity of 2.6 m/s through a pipe 200 m long with an inside diameter of 150 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss h_f.

V = 2.6 m/sL = 200 mD = 150 mmf = 0.025g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 300 m long with an inside diameter of 250 mm and a Darcy friction factor f = 0.015. The minor losses are a square-edged entrance (K = 0.5), 4 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 60 L/s?

Q = 60 L/sL = 300 mD = 250 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×4)K_valve = 5K_exit = 1.0

Δz = m

7.Pumps & pump powerMedium

A pump delivers 30 L/s of water (ρ = 1000 kg/m³) against a head of 12 m. The pump efficiency is 72% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 30 L/sH = 12 mη_pump = 72%η_motor = 95%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 50 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 200 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 3 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 50 L/sΔz = 7 mL = 200 mD = 150 mmf = 0.02ΣK = 3

H = m

3Solutions

1.Hydrostatic force

Answer: F = 73.58 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 = 1.5 + 2/2 = 2.5 m

A = H·b = 2·1.5 = 3 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·2.5·3 = 73.58 kN

Answer

F = 73.58 kN

2.Hydrostatic force

Answer: y_cp = 3.524 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 = 3 + 1/2 = 3.5 m

A = H·b = 1·1 = 1 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·3.5·1 = 30.21 kN

Centre of pressure

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

I_c = b·H³/12 = 1·1³/12 = 0.08333 m⁴

y_cp = y_c + I_c/(y_c·A) = 3.5 + 0.08333/(3.5·1) = 3.524 m

Answer

y_cp = 3.524 m

3.Continuity & Bernoulli

Answer: p₂ = 338.1 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 = 0.5·(100/25)² = 8 m/s

½ρ(v₁² − v₂²) = ½·1000·(0.5² − 8²)/1000 = −31.88 kPa

p₂ = p₁ + ½ρ(v₁² − v₂²) = 370 + (−31.88) = 338.1 kPa

Answer

p₂ = 338.1 kPa

4.Continuity & Bernoulli

Answer: H = 7.164 m

Exit velocity

β = 25/100 = 0.25

v₂ = √(2p₁/(ρ(1 − β⁴))) = √(2·70000/(1000·0.9961)) = 11.86 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) = 11.86²/(2·9.81) = 7.164 m

Answer

H = 7.164 m

5.Pipe friction & head loss

Answer: h_f = 11.48 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) = 2.6²/(2·9.81) = 0.3445 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.025·(200/0.15)·2.6²/(2·9.81) = 11.48 m

Answer

h_f = 11.48 m

6.Pipe friction & head loss

Answer: Δz = 1.957 m

Velocity

V = Q/A = 0.06/0.04909 = 1.222 m/s

V²/(2g) = 0.07615 m

Loss coefficients

f·L/D = 0.015·300/0.25 = 18

ΣK = 0.5 + 4·0.3 + 5 + 1.0 = 7.7

Total head loss

h_f = 18·0.07615 = 1.371 m

h_m = 7.7·0.07615 = 0.5863 m

Δz = h_f + h_m = 1.957 m

Answer

Δz = 1.957 m

7.Pumps & pump power

Answer: P_el = 5.163 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 = 1000·9.81·0.03·12 = 3.532 kW

Shaft power

P_shaft = 3.532/0.72 = 4.905 kW

Electrical power

P_el = 4.905/0.95 = 5.163 kW

Answer

P_el = 5.163 kW

8.Pumps & pump power

Answer: H = 19.11 m

Velocity

V = Q/A = 0.05/0.01767 = 2.829 m/s

V²/(2g) = 0.408 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (26.67 + 3)·0.408 = 12.11 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + h_L = 7 + 12.11 = 19.11 m

Answer

H = 19.11 m

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