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

d = 1 mH = 1 mb = 1.5 mρ = 1000 kg/m³g = 9.81 m/s²

F = kN

2.Hydrostatic forceHard

A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.

d = 2.5 mH = 2 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²

F = kN

3.Continuity & BernoulliMedium

Water flows at 10 L/s through a horizontal pipe that narrows from 80 mm to 40 mm. The pressure in the narrow section is 290 kPa. Ignoring losses, find the pressure p₁ in the wide section.

Q = 10 L/sd₁ = 80 mmd₂ = 40 mmp₂ = 290 kPaρ = 1000 kg/m³

p₁ = kPa

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 2.5 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 = −2.5 mg = 9.81 m/s²p_atm = 101.3 kPa

p_C = kPa

5.Pipe friction & head lossMedium

Water flows at 60 L/s through a pipe 100 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 = 60 L/sL = 100 mD = 150 mmf = 0.022g = 9.81 m/s²

h_f = m

6.Pipe friction & head lossHard

Two open reservoirs are joined by a pipe 500 m long with an inside diameter of 300 mm and a Darcy friction factor f = 0.015. The water surface of the upper reservoir is 39 m above that of the lower one. 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. Find the flow rate Q in L/s.

Δz = 39 mL = 500 mD = 300 mmf = 0.015K_entrance = 0.5K_bend = 0.3 (×2)K_valve = 5K_exit = 1.0

Q = L/s

7.Pumps & pump powerMedium

A pump delivers 25 L/s of water (ρ = 1000 kg/m³) against a head of 14 m. The pump efficiency is 80% 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 = 25 L/sH = 14 mη_pump = 80%η_motor = 95%g = 9.81 m/s²

P_el = 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 9 m higher. The pipe is 100 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022; the minor-loss coefficients add up to ΣK = 5 (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 250 kPa. Find the head H the pump must add.

Q = 75 L/sΔz = 9 mL = 100 mD = 200 mmf = 0.022ΣK = 5p_tank = 250 kPa

H = m

3Solutions

1.Hydrostatic force

Answer: F = 22.07 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 + 1/2 = 1.5 m

A = H·b = 1·1.5 = 1.5 m²

Resultant force

F = ρ·g·y_c·A = 1000·9.81·1.5·1.5 = 22.07 kN

Answer

F = 22.07 kN

2.Hydrostatic force

Answer: F = 181.3 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 + 2/2 = 3.5 m

A = H·b = 2·3 = 6 m²

Resultant force

F = ρ·g·y_c·A = 880·9.81·3.5·6 = 181.3 kN

Answer

F = 181.3 kN

3.Continuity & Bernoulli

Answer: p₁ = 319.7 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 = 0.01/0.005027 = 1.989 m/s

v₂ = Q/A₂ = 0.01/0.001257 = 7.958 m/s

Bernoulli

p₁ = p₂ + ½ρ(v₂² − v₁²) = 290 + ½·1000·(7.958² − 1.989²)/1000 = 290 + 29.68 = 319.7 kPa

Answer

p₁ = 319.7 kPa

4.Continuity & Bernoulli

Answer: p_C = 66.97 kPa

Velocity in the hose

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

v = √(2g·2.5) = √(2·9.81·2.5) = 7.004 m/s

Bernoulli from the surface to C

p_atm = p_C + ½ρv² + ρgz_C

p_C = 101.3 − 1000·(7.004²/2 + 9.81·1)/1000 = 101.3 − 34.34 = 66.97 kPa

Answer

p_C = 66.97 kPa

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

5.Pipe friction & head loss

Answer: h_f = 8.618 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 = 0.06/0.01767 = 3.395 m/s

Velocity head

V²/(2g) = 3.395²/(2·9.81) = 0.5876 m

Darcy–Weisbach

h_f = f·(L/D)·V²/(2g) = 0.022·(100/0.15)·3.395²/(2·9.81) = 8.618 m

Answer

h_f = 8.618 m

6.Pipe friction & head loss

Answer: Q = 345.1 L/s

Loss coefficients

f·L/D = 0.015·500/0.3 = 25

ΣK = 0.5 + 2·0.3 + 5 + 1.0 = 7.1

Energy equation

Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.

39 = (25 + 7.1)·V²/(2·9.81)

V = √(2·9.81·39/32.1) = 4.882 m/s

Flow rate

Q = V·π·D²/4 = 4.882·0.07069 = 0.3451 m³/s = 345.1 L/s

Answer

Q = 345.1 L/s

7.Pumps & pump power

Answer: P_el = 4.518 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 = 1000·9.81·0.025·14 = 3.433 kW

Shaft power

P_shaft = 3.433/0.8 = 4.292 kW

Electrical power

P_el = 4.292/0.95 = 4.518 kW

Answer

P_el = 4.518 kW

8.Pumps & pump power

Answer: H = 39.13 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) = (11 + 5)·0.2905 = 4.648 m

Pressure head

p/(ρ·g) = 250000/(1000·9.81) = 25.48 m

Pump head

Both surfaces are practically at rest, so there is no velocity head term.

H = Δz + p/(ρ·g) + h_L = 9 + 25.48 + 4.648 = 39.13 m

Answer

H = 39.13 m

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