Diploma Civil and Mechanical Engineering for Engineering Assistant · Chapter 3

Fluid Mechanics

What to remember

Pressure at depth h is p = ρgh, and it acts equally in all directions (Pascal's law). Absolute pressure = atmospheric + gauge pressure.
Continuity: A₁V₁ = A₂V₂ (Q = AV). Bernoulli: p/ρg + V²/2g + z = constant along a streamline for steady, ideal, incompressible flow.
Reynolds number Re = ρVd/μ: laminar flow below about 2000, turbulent above about 4000. Head loss in a pipe by Darcy-Weisbach: hf = λLV²/2gd.

Properties of fluids

Mass density ρ = mass ÷ volume. Water: 1000 kg/m³. Specific weight w = ρg; for water 9.81 kN/m³. Specific gravity = density of fluid ÷ density of water. Mercury has specific gravity 13.6.
Viscosity is the internal resistance to flow. Newton's law of viscosity: τ = μ(du/dy). μ is the dynamic viscosity: unit Pa·s or N·s/m²; 1 poise = 0.1 Pa·s. Kinematic viscosity ν = μ/ρ: unit m²/s; 1 stokes = 10⁻⁴ m²/s.
Viscosity of liquids decreases with temperature. Viscosity of gases increases with temperature.
An ideal fluid has zero viscosity and is incompressible. A Newtonian fluid follows Newton's law of viscosity (water, air, oil).
Surface tension (σ, N/m) is the force per unit length at the free surface. It makes drops spherical.
Capillarity: rise or fall of liquid in a thin tube. Rise h = 4σ cos θ ÷ (ρgd). Water rises in a glass tube (θ is small); mercury is depressed.
Worked example 1: water (σ = 0.0735 N/m, θ = 0°) in a 1 mm tube: h = 4 × 0.0735 ÷ (1000 × 9.81 × 0.001) = 0.030 m = 30 mm.
Compressibility: bulk modulus K = −dp ÷ (dV/V). Liquids are nearly incompressible.
Vapour pressure is the pressure at which a liquid boils at a given temperature. If local pressure drops to vapour pressure, vapour bubbles form (cavitation).

Fluid pressure and measurement

Pascal's law: pressure applied to an enclosed fluid is transmitted equally in all directions. A hydraulic press works on this law.
Hydrostatic pressure: p = ρgh. Pressure increases linearly with depth.
Worked example 2: at 10 m depth in water, p = 1000 × 9.81 × 10 = 98,100 Pa ≈ 98.1 kPa (gauge).
Atmospheric pressure (standard) = 101.325 kPa = 760 mm of mercury ≈ 10.33 m of water.
Absolute pressure = atmospheric + gauge. Vacuum pressure = atmospheric − absolute.
Pressure head: h = p/ρg. A pressure expressed as a height of liquid column.
Measuring devices:

Device	Use
Barometer	Atmospheric pressure
Piezometer	Small positive gauge pressure of a liquid
Simple U-tube manometer	Moderate gauge pressure, with a heavy liquid (mercury)
Differential manometer	Pressure difference between two points
Inclined manometer	Small pressures with greater accuracy
Bourdon gauge	Mechanical gauge for high pressures

Total pressure on a submerged plane surface: F = ρg·A·x̄, where x̄ is the depth of the centroid. The centre of pressure is always at or below the centroid: h* = x̄ + I_G ÷ (A·x̄).
Worked example 3: a vertical rectangular gate 2 m wide and 3 m deep with its top edge at the water surface. F = 1000 × 9.81 × 6 × 1.5 = 88,290 N ≈ 88.3 kN. The centre of pressure is at 2/3 of the depth = 2 m below the surface.
Buoyancy (Archimedes' principle): the upward force equals the weight of fluid displaced. A floating body displaces fluid equal to its own weight. The centre of buoyancy is the centroid of the displaced volume.
Metacentre (M) is the point about which a floating body starts to oscillate. Metacentric height GM = BM − BG, where BM = I ÷ V (I = moment of inertia of the waterline area, V = displaced volume). The body is stable if M is above G, unstable if M is below G.
Worked example 4: a fully submerged 2 m³ block in water has buoyant force 1000 × 9.81 × 2 = 19,620 N.

Fluid flow and Bernoulli's equation

Types of flow: steady (conditions at a point do not change with time) and unsteady; uniform (velocity same at all points at one instant) and non-uniform; laminar and turbulent; compressible and incompressible; rotational and irrotational.
Streamline is a line whose tangent gives the velocity direction. A streamtube is a bundle of streamlines.
Discharge: Q = A × V (m³/s). Continuity equation: A₁V₁ = A₂V₂ for incompressible flow.
Worked example 5: a 200 mm pipe carrying 2 m/s narrows to 100 mm. Area ratio is 4, so V₂ = 8 m/s. Q = (π/4)(0.2)² × 2 = 0.0628 m³/s.
Bernoulli's equation: p/ρg + V²/2g + z = constant. The three terms are the pressure head, velocity (kinetic) head and datum (potential) head. Assumptions: steady, incompressible, non-viscous flow along a streamline, with no energy added or lost.
Hydraulic gradient line (HGL) = p/ρg + z. Total energy line (TEL) = p/ρg + V²/2g + z. TEL lies above HGL by V²/2g.

Flow measuring devices

Venturimeter: a converging section, throat and diverging section placed in a pipe. Pressure difference between the inlet and the throat gives the discharge: Q = Cd × a₁a₂√(2gh) ÷ √(a₁² − a₂²). Cd is about 0.97 to 0.99. The gradual diverging cone recovers pressure, so the loss is small.
Orifice meter: a thin plate with a hole in a pipe; cheaper but with a larger loss than a venturimeter.
Pitot tube: measures velocity at a point from stagnation pressure: V = √(2gh).
Worked example 6: a rise of 0.2 m in the pitot tube gives V = √(2 × 9.81 × 0.2) ≈ 1.98 m/s.
Orifice (a small opening in a tank): theoretical velocity V = √(2gH) (Torricelli's theorem). Actual discharge Q = Cd × a × √(2gH).
Hydraulic coefficients: Cc (contraction) = area of jet ÷ area of orifice, about 0.62. Cv (velocity) = actual ÷ theoretical velocity, about 0.97 to 0.99. Cd = Cc × Cv, about 0.6 for a sharp-edged orifice. The vena contracta is the section of minimum jet area, at about half a diameter from the orifice.
Worked example 7: H = 19.62 m gives V = √(2 × 9.81 × 19.62) = 19.62 m/s.
Notches and weirs measure discharge in open channels:

Device	Discharge	Depends on head as
Rectangular notch/weir	Q = (2/3) Cd L √(2g) H^(3/2)	H^(3/2)
Triangular (V) notch	Q = (8/15) Cd √(2g) tan(θ/2) H^(5/2)	H^(5/2)

A V-notch is preferred for small discharges, because the head changes more for the same change of flow.

Flow through pipes

Reynolds number: Re = ρVd/μ = Vd/ν. It is the ratio of inertia force to viscous force. Laminar flow: Re < 2000. Transition: 2000 to 4000. Turbulent flow: Re > 4000.
Worked example 8: V = 1 m/s, d = 0.1 m, ν = 10⁻⁶ m²/s gives Re = 10⁵ (turbulent).
Darcy-Weisbach equation: hf = λLV² ÷ 2gd, with λ the Darcy friction factor. For laminar flow, λ = 64/Re. (Some texts write hf = 4fLV²/2gd with f = λ/4.) Head loss is proportional to V² in turbulent flow and to V in laminar flow.
Worked example 9: λ = 0.02, L = 100 m, d = 0.1 m, V = 2 m/s: hf = 0.02 × 100 × 4 ÷ (2 × 9.81 × 0.1) = 4.08 m.
Hagen-Poiseuille law (laminar flow): the pressure drop is proportional to μLV/d². In laminar flow the maximum velocity at the centre is twice the average velocity.
Minor losses: sudden enlargement (V₁ − V₂)²/2g; sudden contraction about 0.5V²/2g; entrance about 0.5V²/2g; exit V²/2g; bends and fittings.
Pipes in series: same discharge and total head loss = sum of the losses. Pipes in parallel: same head loss and total discharge = sum of the discharges. An equivalent pipe replaces a compound pipe with equal loss for equal discharge.
Water hammer: a sudden pressure rise when flow in a pipe is stopped suddenly (valve closure). It is reduced by slow closing, surge tanks and air vessels.
Power transmitted through a pipe is maximum when the head lost in friction is one-third of the supply head; maximum efficiency is then 66.7%.
Open channel flow (Chezy and Manning): Chezy V = C√(mi) (m = hydraulic mean depth = A/P, i = bed slope). Manning V = (1/n) R^(2/3) S^(1/2). The best rectangular section has b = 2d; the best trapezoidal section is half of a regular hexagon.

Exam traps

Dynamic viscosity (Pa·s) and kinematic viscosity (m²/s) are different. ν = μ/ρ.
Viscosity of a liquid falls with heating, but that of a gas rises.
Absolute pressure = gauge + atmospheric. Vacuum = atmospheric − absolute.
The centre of pressure is below the centroid (for a vertical surface), not at it.
A body is stable when the metacentre is above the centre of gravity.
Laminar flow occurs below Re = 2000. Turbulent flow is above about 4000. The range between is transitional.
Discharge through a rectangular notch varies as H^(3/2), but through a V-notch as H^(5/2).
Cd = Cc × Cv, and Cd is about 0.6, while Cv is about 0.97.

One-liners

1. 1 atm = 101.325 kPa = 760 mm Hg ≈ 10.33 m of water.
2. Specific gravity of mercury is 13.6.
3. 1 poise = 0.1 Pa·s.
4. Hydrostatic pressure p = ρgh.
5. Continuity equation: A₁V₁ = A₂V₂.
6. Velocity of efflux V = √(2gH).
7. Venturimeter works on Bernoulli's equation and measures pipe discharge.
8. Pitot tube measures point velocity.
9. Laminar friction factor λ = 64/Re.
10. Water hammer is caused by sudden valve closure.
11. A floating body is stable when GM is positive.
12. Maximum transmitted power occurs when friction head loss equals one-third of the supply head.

Practice questions

The unit of dynamic viscosity in SI is
1. kg/m³
2. N/m
3. Pa·s (N·s/m²)
4. m²/s
Answer
C. Pa·s (N·s/m²)
Dynamic viscosity μ is measured in N·s/m².
Kinematic viscosity is equal to
1. shear stress divided by density
2. dynamic viscosity divided by density
3. density divided by dynamic viscosity
4. dynamic viscosity multiplied by density
Answer
B. dynamic viscosity divided by density
ν = μ/ρ.
With rise in temperature, the viscosity of a liquid
1. first increases then falls to zero
2. increases
3. decreases
4. stays the same
Answer
C. decreases
Molecular cohesion falls as temperature rises, so liquids flow more easily.
Pascal's law states that pressure in an enclosed fluid at rest is
1. greatest at the top
2. zero at the centre
3. transmitted only downward
4. transmitted equally in all directions
Answer
D. transmitted equally in all directions
Pressure at a point acts equally in all directions.
Absolute pressure is equal to
1. atmospheric pressure − gauge pressure
2. gauge pressure − vacuum pressure
3. gauge pressure only
4. atmospheric pressure + gauge pressure
Answer
D. atmospheric pressure + gauge pressure
p(abs) = p(atm) + p(gauge).
Standard atmospheric pressure is equal to
1. 760 mm of mercury
2. 1000 mm of mercury
3. 76 mm of mercury
4. 7.6 m of mercury
Answer
A. 760 mm of mercury
One standard atmosphere is 760 mm Hg, or 101.325 kPa.
The specific gravity of mercury is about
1. 0.136
2. 13.6
3. 1.36
4. 136
Answer
B. 13.6
Mercury is 13.6 times as dense as water.
A body floats in stable equilibrium when its metacentre is
1. below the centre of gravity
2. at the centre of buoyancy only
3. above the centre of gravity
4. at the free surface always
Answer
C. above the centre of gravity
A positive metacentric height gives a restoring moment.
The continuity equation for incompressible flow is
1. A₁V₁ = A₂V₂
2. A₁/V₁ = A₂/V₂
3. p₁V₁ = p₂V₂
4. A₁ + V₁ = A₂ + V₂
Answer
A. A₁V₁ = A₂V₂
Mass flow is conserved, so discharge is constant.
Bernoulli's equation represents conservation of
1. energy
2. momentum
3. mass only
4. volume only
Answer
A. energy
The sum of pressure, kinetic and potential heads is constant.
Which device is used to measure discharge in a pipe?
1. Piezometer
2. Venturimeter
3. Hydrometer
4. Barometer
Answer
B. Venturimeter
The venturimeter uses the pressure drop at the throat to find discharge.
A pitot tube is used to measure
1. discharge in an open channel
2. viscosity
3. velocity at a point
4. surface tension
Answer
C. velocity at a point
It converts stagnation pressure into velocity.
Flow in a pipe is laminar when the Reynolds number is below about
1. 10,000
2. 4000
3. 100,000
4. 2000
Answer
D. 2000
Re below 2000 means laminar flow.
Water hammer in a pipe is caused by
1. a large pipe diameter
2. sudden closure of a valve
3. slow opening of a valve
4. leakage in the pipe
Answer
B. sudden closure of a valve
A sudden stop of flow creates a pressure wave.
The coefficient of discharge of an orifice equals
1. Cc ÷ Cv
2. Cc + Cv
3. Cc × Cv
4. Cv − Cc
Answer
C. Cc × Cv
Cd = Cc × Cv.
The gauge pressure at a depth of 5 m in water (g = 9.81 m/s²) is
1. 49.05 kPa
2. 98.1 kPa
3. 4.905 kPa
4. 490.5 kPa
Answer
A. 49.05 kPa
p = ρgh = 1000 × 9.81 × 5 = 49,050 Pa.
A gauge reads 50 kPa and the atmospheric pressure is 100 kPa. The absolute pressure is
1. 100 kPa
2. 150 kPa
3. 5000 kPa
4. 50 kPa
Answer
B. 150 kPa
Absolute = gauge + atmospheric = 150 kPa.
A pressure of 98.1 kPa in water corresponds to a head of
1. 1 m
2. 100 m
3. 9.81 m
4. 10 m
Answer
D. 10 m
h = p/ρg = 98,100 ÷ (1000 × 9.81) = 10 m.
An oil has specific gravity 0.8. Its density is
1. 80 kg/m³
2. 800 kg/m³
3. 8000 kg/m³
4. 1250 kg/m³
Answer
B. 800 kg/m³
Density = 0.8 × 1000 = 800 kg/m³.
A fluid has μ = 0.002 Pa·s and ρ = 1000 kg/m³. Its kinematic viscosity is
1. 2 × 10⁻⁶ m²/s
2. 2 m²/s
3. 5 × 10⁻⁷ m²/s
4. 2 × 10⁻³ m²/s
Answer
A. 2 × 10⁻⁶ m²/s
ν = 0.002 ÷ 1000 = 2 × 10⁻⁶ m²/s.
A column of mercury 0.5 m high (S = 13.6) exerts a pressure of about
1. 6.67 kPa
2. 4.9 kPa
3. 667 kPa
4. 66.7 kPa
Answer
D. 66.7 kPa
p = 13,600 × 9.81 × 0.5 = 66,708 Pa.
A pipe of 300 mm carries water at 1 m/s and narrows to 100 mm. The velocity in the narrow section is
1. 1 m/s
2. 27 m/s
3. 9 m/s
4. 3 m/s
Answer
C. 9 m/s
Area ratio = (300/100)² = 9, so V₂ = 9 m/s.
The discharge in a 0.1 m diameter pipe flowing at 2 m/s is about
1. 0.157 m³/s
2. 0.0157 m³/s
3. 0.00785 m³/s
4. 0.0628 m³/s
Answer
B. 0.0157 m³/s
Q = (π/4)(0.1)² × 2 = 0.0157 m³/s.
The velocity of efflux from an orifice under a head of 5 m (g = 9.81 m/s²) is about
1. 4.9 m/s
2. 49 m/s
3. 98 m/s
4. 9.9 m/s
Answer
D. 9.9 m/s
V = √(2gH) = √98.1 ≈ 9.9 m/s.
Oil with ρ = 800 kg/m³ and μ = 0.04 Pa·s flows at 1 m/s in a 50 mm pipe. The Reynolds number is
1. 1000
2. 2000
3. 40,000
4. 100
Answer
A. 1000
Re = ρVd/μ = 800 × 1 × 0.05 ÷ 0.04 = 1000, which is laminar.
The friction factor λ = 64/Re for the above laminar flow (Re = 1000) is
1. 0.64
2. 0.0064
3. 0.064
4. 0.032
Answer
C. 0.064
λ = 64 ÷ 1000 = 0.064.
A fully submerged body of volume 0.5 m³ in water (g = 9.81 m/s²) experiences a buoyant force of
1. 9810 N
2. 490.5 N
3. 2452 N
4. 4905 N
Answer
D. 4905 N
F = ρgV = 1000 × 9.81 × 0.5 = 4905 N.
A floating body weighs 800 N. The buoyant force on it is
1. 8000 N
2. 400 N
3. 800 N
4. 1600 N
Answer
C. 800 N
For floating equilibrium, the buoyant force equals the weight.
The total pressure on a vertical rectangular gate 2 m wide and 3 m deep, top at the water surface, acts at a depth of
1. 2 m
2. 1 m
3. 1.5 m
4. 3 m
Answer
A. 2 m
The centre of pressure of a rectangle with its top at the surface is at 2/3 of the depth.
The coefficient of contraction is 0.62 and the coefficient of velocity is 0.97. The coefficient of discharge is about
1. 0.65
2. 0.60
3. 1.59
4. 0.35
Answer
B. 0.60
Cd = 0.62 × 0.97 = 0.601.
Water rises in a tube of 2 mm diameter (σ = 0.0735 N/m, θ = 0°, g = 9.81 m/s²) by about
1. 15 mm
2. 30 mm
3. 7.5 mm
4. 60 mm
Answer
A. 15 mm
h = 4σ/ρgd = 0.294 ÷ (1000 × 9.81 × 0.002) = 0.015 m.
The hydraulic mean depth of a rectangular channel 4 m wide with 1 m depth of flow is
1. 0.25 m
2. 4 m
3. 0.667 m
4. 1 m
Answer
C. 0.667 m
m = A/P = 4 ÷ (4 + 2) = 0.667 m.
In laminar flow in a circular pipe with average velocity 1.5 m/s, the maximum velocity is
1. 0.75 m/s
2. 1.5 m/s
3. 2.25 m/s
4. 3 m/s
Answer
D. 3 m/s
Maximum velocity is twice the average in laminar pipe flow.
If the head over a rectangular notch is doubled, the discharge becomes about
1. 2 times
2. 2.83 times
3. 5.66 times
4. 4 times
Answer
B. 2.83 times
Q ∝ H^(3/2), so 2^1.5 = 2.83.
If the head over a triangular notch is doubled, the discharge becomes about
1. 4 times
2. 2.83 times
3. 5.66 times
4. 2 times
Answer
C. 5.66 times
Q ∝ H^(5/2), so 2^2.5 = 5.66.
Maximum power transmission through a pipe occurs when friction head loss equals
1. one-half of the supply head
2. one-third of the supply head
3. two-thirds of the supply head
4. the supply head
Answer
B. one-third of the supply head
The efficiency at that condition is 66.7%.
Which statements are correct? 1. Viscosity of gases increases with temperature. 2. Viscosity of liquids increases with temperature.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Liquid viscosity decreases with temperature.
Which statements are correct? 1. The centre of pressure lies below the centroid of a submerged vertical plane. 2. Total pressure on a plane is ρgAx̄.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are correct.
Which statements are correct? 1. Cavitation occurs when pressure falls to the vapour pressure. 2. Cavitation occurs at very high pressure.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Vapour bubbles form at low pressure, not high pressure.
Which statements are correct? 1. The total energy line lies above the hydraulic gradient line by V²/2g. 2. Bernoulli's equation applies to viscous flow without any correction.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Bernoulli's equation as stated applies to ideal flow. Losses must be added for real fluids.
Which statements are correct? 1. For laminar flow the head loss varies as V². 2. For turbulent flow the Darcy-Weisbach loss varies as V².
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Laminar head loss varies as V and turbulent loss roughly as V².
Match: (a) Venturimeter (b) Pitot tube (c) V-notch (d) Manometer. Which pairing is correct?
1. (c) pipe discharge, (d) velocity
2. (a) point velocity, (c) pressure
3. (b) discharge in open channel, (d) viscosity
4. (a) pipe discharge, (b) point velocity
Answer
D. (a) pipe discharge, (b) point velocity
The venturimeter gives pipe discharge, and the pitot tube gives velocity at a point.
A pipe has λ = 0.02, L = 100 m, d = 0.1 m and V = 2 m/s (g = 9.81 m/s²). The head loss is about
1. 8.16 m
2. 40.8 m
3. 2.04 m
4. 4.08 m
Answer
D. 4.08 m
hf = λLV²/2gd = 0.02 × 100 × 4 ÷ (2 × 9.81 × 0.1) = 4.08 m.
Two pipes of equal diameter and friction carry water in parallel. The head loss in each pipe is
1. added together, and the discharge is the same in both
2. equal, and the total discharge is the sum
3. equal, and the discharge is the same in both
4. different, and the discharge is the sum
Answer
B. equal, and the total discharge is the sum
In parallel pipes the head loss is the same, and discharges add.
The ratio of the specific weight of water (9.81 kN/m³) and an oil of specific gravity 0.9 is
1. 1.9
2. 0.9
3. 1.11
4. 9
Answer
C. 1.11
Oil specific weight = 8.829 kN/m³, so 9.81 ÷ 8.829 = 1.11.

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1.The atmospheric pressure at sea level is _________ atm.
2.The force exerted by an object perpendicular to the surface is called:
3.The force employed on a unit area of the surface is called _____.
4.The effect of thrust on a surface depends on the area on which it acts. Force acting on a smaller area exerts a ___________ compared to a larger area.
5.The density of an ideal gas can be doubled by halving the ______________.
6.Solid CO 2 gets converted directly into gaseous state on decrease of pressure to:
7.The speed of sound depends primarily on the nature and ____________ of the medium.
8.The speed of light is ______.
9.The magnitude of the buoyant force depends on the _______ of the fluid.
10.The mass per unit volume is called

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