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

d = 4 mH = 3 mb = 1.5 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 glycerin, with its top edge 2 m below the surface. Find the depth y_cp of the centre of pressure below the surface.

d = 2 mH = 1 mb = 1.5 mρ = 1260 kg/m³g = 9.81 m/s²

y_cp = m

3.Continuity & BernoulliMedium

A pitot-static tube in a wind tunnel reads a difference of 1300 Pa between the stagnation pressure and the static pressure. The air density is 1.2 kg/m³. Find the air speed v.

p₀ − p = 1300 Paρ = 1.2 kg/m³

v = m/s

4.Continuity & BernoulliHard

A hose with an inner diameter of 32 mm siphons water from an open tank. The highest point C of the hose is 1 m above the water surface, and the outlet (open to the air) is 3.5 m below the water surface. Ignore losses and take g = 9.81 m/s². Find the absolute pressure p_C at point C. The atmospheric pressure is 101.3 kPa.

d = 32 mmz_C = +1 mz₂ = −3.5 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water (ρ = 1000 kg/m³) flows at 0.8 m/s through a horizontal pipe, 400 m long, with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.

V = 0.8 m/sL = 400 mD = 100 mmf = 0.025ρ = 1000 kg/m³

Δp = kPa

6.Pipe friction & head lossHard

A pipe 300 m long with an inside diameter of 300 mm (Darcy friction factor f = 0.018) 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), 3 bends (K = 0.3 each) and a valve (K = 5). Use g = 9.81 m/s². How far H must the reservoir surface be above the pipe outlet for a flow rate of 150 L/s?

Q = 150 L/sL = 300 mD = 300 mmf = 0.018K_entrance = 0.5K_bend = 0.3 (×3)K_valve = 5

H = m

7.Pumps & pump powerMedium

A pump delivers 10 L/s of water (ρ = 1000 kg/m³) against a head of 32 m. The pump efficiency is 65% and the electric motor driving it has an efficiency of 92%. Use g = 9.81 m/s². Find the electrical power P_el the motor draws, in kW.

Q = 10 L/sH = 32 mη_pump = 65%η_motor = 92%g = 9.81 m/s²

P_el = kW

8.Pumps & pump powerHard

A pump moves 60 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 9 m higher. The pipe is 300 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 = 6 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 60 L/sΔz = 9 mL = 300 mD = 200 mmf = 0.02ΣK = 6

H = m

3Solutions

1.Hydrostatic force

Answer: F = 242.8 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 = 4 + 3/2 = 5.5 m

A = H·b = 3·1.5 = 4.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·5.5·4.5 = 242.8 kN

Answer

F = 242.8 kN

2.Hydrostatic force

Answer: y_cp = 2.533 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 + 1/2 = 2.5 m

A = H·b = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1260·9.81·2.5·1.5 = 46.35 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) = 2.5 + 0.125/(2.5·1.5) = 2.533 m

Answer

y_cp = 2.533 m

3.Continuity & Bernoulli

Answer: v = 46.55 m/s

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.

Bernoulli to the stagnation point

The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.

p₀ − p = ½ρv²

v = √(2·1300/1.2) = 46.55 m/s

Answer

v = 46.55 m/s

4.Continuity & Bernoulli

Answer: p_C = 57.15 kPa

Velocity in the hose

Surface and outlet are both open to the air, so the pressure terms cancel.

v = √(2g·3.5) = √(2·9.81·3.5) = 8.287 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(8.287²/2 + 9.81·1)/1000 = 101.3 − 44.15 = 57.15 kPa

Answer

p_C = 57.15 kPa

If this pressure falls to the vapour pressure of the water, the siphon stops working.

5.Pipe friction & head loss

Answer: Δp = 32 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) = 0.025·(400/0.1)·0.8²/(2·9.81) = 3.262 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 = 1000·9.81·3.262 = 32000 Pa = 32 kPa

Answer

Δp = 32 kPa

6.Pipe friction & head loss

Answer: H = 5.83 m

Velocity

V = Q/A = 0.15/0.07069 = 2.122 m/s

V²/(2g) = 0.2295 m

Loss coefficients

f·L/D = 0.018·300/0.3 = 18

ΣK = 0.5 + 3·0.3 + 5 = 6.4

Energy equation

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

H = (1 + 18 + 6.4)·0.2295 = 5.83 m

Answer

H = 5.83 m

7.Pumps & pump power

Answer: P_el = 5.249 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 = 1000·9.81·0.01·32 = 3.139 kW

Shaft power

P_shaft = 3.139/0.65 = 4.83 kW

Electrical power

P_el = 4.83/0.92 = 5.249 kW

Answer

P_el = 5.249 kW

8.Pumps & pump power

Answer: H = 15.69 m

Velocity

V = Q/A = 0.06/0.03142 = 1.91 m/s

V²/(2g) = 0.1859 m

Head losses

h_L = (f·L/D + ΣK)·V²/(2g) = (30 + 6)·0.1859 = 6.693 m

Pump head

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

H = Δz + h_L = 9 + 6.693 = 15.69 m

Answer

H = 15.69 m

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