AEE Civil Engineering Core · Chapter 9

Fluid Mechanics and Hydraulics (Pipes, Notches, Weirs)

What to remember

Fluid statics gives p = ρgh. Force on a plane surface is F = ρg·A·h̄ and acts below the centroid, at the centre of pressure.
Flow measurement is based on Bernoulli's equation plus continuity. Orifices, venturimeters, notches and weirs all give Q = Cd × (ideal discharge).
Pipe flow is governed by losses: major loss by Darcy-Weisbach (hf = fLV²/2gD) and minor losses as a multiple of V²/2g. Laminar flow has f = 64/Re.

1. Fluid properties

Mass density ρ = mass/volume (water about 1000 kg/m³). Specific weight w = ρg (water about 9810 N/m³). Specific gravity is the ratio to water and has no unit.
Dynamic viscosity μ: Newton's law is τ = μ(du/dy). Unit: N·s/m² or Pa·s. Kinematic viscosity ν = μ/ρ, unit m²/s.
Viscosity of liquids falls as temperature rises. Viscosity of gases rises with temperature.
Surface tension σ (N/m). Pressure inside a droplet is p = 4σ/d. Inside a soap bubble it is p = 8σ/d because there are two surfaces.
Capillary rise in a tube: h = 4σcosθ/(ρgd). Water wets glass and rises. Mercury does not wet glass and is depressed.
Bulk modulus K = −dp/(dV/V). A large K means the fluid is nearly incompressible.
An ideal fluid has no viscosity. A Newtonian fluid has shear stress directly proportional to velocity gradient.

2. Fluid statics

Pressure at a point is the same in all directions (Pascal's law). Pressure increases with depth: p = ρgh.
Absolute pressure = gauge pressure + atmospheric pressure. Standard atmosphere is about 101.3 kPa, equal to about 10.3 m of water or 760 mm of mercury.
Manometers measure pressure using liquid columns. A piezometer is the simplest. A U-tube differential manometer gives pressure difference: Δp = (ρm − ρ)g·h when the manometric liquid has density ρm.
Total pressure on a plane surface: F = ρg·A·h̄, where h̄ is the depth of the centroid.
Centre of pressure: h* = h̄ + I_G sin²θ/(A·h̄). For a vertical surface, h* = h̄ + I_G/(A·h̄). The centre of pressure always lies below the centroid.
For a rectangle b × d: I_G = bd³/12. For a triangle: I_G = bh³/36. For a circle: I_G = πd⁴/64.
Buoyancy: the upthrust equals the weight of fluid displaced (Archimedes). It acts at the centre of buoyancy, the centroid of the displaced volume.
Metacentre (M): the point where the line of buoyancy meets the original vertical axis for a small tilt. GM = BM − BG, with BM = I/V.
A floating body is stable if M is above G (GM positive), neutral if M coincides with G, and unstable if M is below G.

3. Fluid kinematics and dynamics

Continuity: A₁V₁ = A₂V₂ for incompressible flow.
Steady flow: properties at a point do not change with time. Uniform flow: velocity does not change with distance.
Stream function ψ exists for any 2-D incompressible flow. Velocity potential φ exists only for irrotational flow. Lines of constant ψ and constant φ cross at right angles.
Bernoulli's equation (steady, incompressible, frictionless, along a streamline): p/ρg + V²/2g + z = constant. The three terms are pressure head, velocity head and datum head.
With real fluids, a loss term hL is added between the two sections.
Venturimeter: Q = Cd·a₁a₂√(2gh)/√(a₁² − a₂²). The head h is the difference of piezometric head between inlet and throat. Cd is about 0.95 to 0.99.
Orifice meter works on the same principle but with greater head loss. A Pitot tube measures velocity: V = √(2gh).
Dimensional analysis: Buckingham's π-theorem says a problem with n variables and m fundamental dimensions gives (n − m) dimensionless groups.
Reynolds number Re = ρVD/μ (inertia/viscous). Froude number Fr = V/√(gL) (inertia/gravity). Models of open-channel structures use Froude similarity.

4. Flow through orifices and mouthpieces

An orifice is a small opening in the side or bottom of a tank. It is small if the head is large compared with its size.
Theoretical velocity (Torricelli): V = √(2gH). Actual velocity V = Cv√(2gH).
Vena contracta is the section of minimum jet area, at about half the orifice diameter downstream.
Coefficients: Cc = area of jet/area of orifice (about 0.62). Cv (about 0.97 to 0.99). Cd = Cc × Cv (about 0.6 to 0.62).
Discharge: Q = Cd·a·√(2gH).
Large rectangular orifice: Q = (2/3)·Cd·b·√(2g)·(H₂^1.5 − H₁^1.5).
Time to empty a tank of constant area A through an orifice of area a from H₁ to H₂: t = 2A(√H₁ − √H₂)/(Cd·a·√(2g)).
A mouthpiece is a short tube fixed to an orifice. A convergent-divergent mouthpiece gives the highest discharge coefficient. A Borda (re-entrant) mouthpiece running free has Cc = 0.5.

5. Notches and weirs

A notch is an opening in the side of a tank or channel. A weir is a similar structure built across a river or canal. H is the head over the crest.

Type	Discharge formula	Remark
Rectangular notch/weir	Q = (2/3)·Cd·L·√(2g)·H^1.5	Most common
Triangular (V) notch	Q = (8/15)·Cd·tan(θ/2)·√(2g)·H^2.5	Best for small flows
Trapezoidal	Rectangular part + two half V-notches	Compromise
Cipolletti weir	Q = 1.86·L·H^1.5	Side slope 1 horizontal : 4 vertical
Broad-crested weir	Q = 1.705·Cd·L·H^1.5	Crest width large, 0.5H or more

Francis formula for a rectangular weir with end contractions: Q = 1.84·(L − 0.1nH)·H^1.5, with n = 2 for two contractions and n = 0 for a suppressed weir.
Velocity of approach is accounted for by replacing H with H + Va²/2g.
Error in H: for a rectangular notch, dQ/Q = 1.5·dH/H. For a triangular notch, dQ/Q = 2.5·dH/H. A V-notch is therefore more sensitive for small flows, which is why it is used for low discharge.
Narrow-crested (sharp-crested) weir has crest thickness less than 0.5H. A broad-crested weir has maximum discharge at critical depth, H being 3/2 times the critical depth.
Ogee weir has an S-shaped profile following the underside of a falling nappe. Spillways of dams in Andhra Pradesh and elsewhere use this profile. Barrages on the Godavari and Krishna act as large weir structures that regulate irrigation flows.
A submerged (drowned) weir has the downstream water above the crest.

6. Flow in pipes

Reynolds number classifies flow: laminar if Re is below about 2000, transitional between about 2000 and 4000, turbulent above about 4000.
Darcy-Weisbach equation: hf = f·L·V²/(2g·D), with Darcy f = 4 × Fanning friction factor. Another form: hf = 4f′LV²/(2gD), where f′ is the Fanning coefficient.
Laminar flow: f = 64/Re. Hagen-Poiseuille gives Δp = 32μVL/D² and Q = πΔp·D⁴/(128μL). Maximum velocity is twice the mean velocity. Shear stress at the wall is largest and is zero at the centre.
Turbulent flow in smooth pipes: f depends on Re (Blasius f = 0.316/Re^0.25 for Re up to about 10⁵). In rough pipes at high Re, f depends only on relative roughness. The Moody chart shows these zones.
Minor losses are written as K·V²/2g:
Sudden enlargement: (V₁ − V₂)²/2g
Sudden contraction: about 0.5·V₂²/2g (using Cc, the loss is (1/Cc − 1)²V₂²/2g)
Pipe entrance (sharp): 0.5V²/2g
Pipe exit: V²/2g
Pipes in series: discharge is the same, total head loss is the sum. Pipes in parallel: head loss is the same, discharges add.
Equivalent pipe (Dupuit's equation): L/D⁵ = L₁/D₁⁵ + L₂/D₂⁵ + … (same friction factor).
Hydraulic gradient line (HGL) is the line of p/ρg + z. Total energy line (TEL) is higher than HGL by V²/2g. Both slope downward in the flow direction, and TEL never rises unless a pump adds energy.
Siphon: the highest point should not have pressure below vapour pressure, so its practical height is limited to about 7.6 m of water for safety.
Power transmission through a pipe: maximum power is delivered when head lost in friction is one-third of the supply head (hf = H/3). Maximum efficiency at that condition is about 66.7%.
Water hammer: sudden closure of a valve gives a pressure rise Δp = ρ·c·V (Joukowsky). The wave speed c in water in a rigid pipe is about 1400 m/s. Surge tanks and relief valves protect penstocks.

7. Worked examples

Example 1 (orifice). An orifice of area 0.0001 m² under a head of 5 m with Cd = 0.6, g = 10 m/s² (clean numbers for hand work). Q = 0.6 × 0.0001 × √(2 × 10 × 5) = 0.6 × 0.0001 × 10 = 0.0006 m³/s.

Example 2 (V-notch). A right-angled V-notch (θ = 90°, tan 45° = 1), Cd = 0.6, H = 0.2 m, g = 9.81. Q = (8/15) × 0.6 × 1 × 4.43 × 0.2^2.5 = (8/15)(0.6)(4.43)(0.017889) ≈ 0.0254 m³/s.

Example 3 (laminar pipe). Oil flows in a pipe at Re = 1600. f = 64/1600 = 0.04.

Example 4 (power transmission). For a supply head of 90 m, maximum power transmission needs hf = 90/3 = 30 m.

Exam traps

Centre of pressure is below the centroid, never above, for a vertical submerged plane.
Darcy f is four times the Fanning coefficient. Check which friction factor a formula uses.
Laminar f = 64/Re does not depend on roughness. Turbulent f in rough pipes does not depend on Re.
Cd = Cc × Cv, not Cc/Cv. Cv is close to 1, while Cc is near 0.62.
Rectangular notch: Q ∝ H^1.5. V-notch: Q ∝ H^2.5. Do not interchange.
Cipolletti weir has side slope 1 horizontal to 4 vertical. It compensates for end contraction, so no end-contraction correction is applied.
Maximum power through a pipe is at hf = H/3, while maximum efficiency is at hf tending to zero.
Stream function exists for any 2-D incompressible flow. Velocity potential needs irrotational flow.
Gauge pressure plus atmospheric pressure gives absolute pressure. Vacuum is below atmospheric.

One-liners

1. Pascal's law: pressure at a point is equal in all directions in a fluid at rest.
2. Newton's law of viscosity: τ = μ·du/dy.
3. Metacentric height GM = BM − BG; stable if positive.
4. Venturimeter throat has lowest pressure and highest velocity.
5. Typical Cd of a sharp-edged orifice is about 0.6.
6. Triangular notch is best for measuring small discharges.
7. Laminar flow in a pipe has Re below about 2000.
8. Hagen-Poiseuille: Q ∝ D⁴ for fixed pressure drop in laminar flow.
9. Sudden enlargement loss is (V₁ − V₂)²/2g.
10. Pipe exit loss is V²/2g.
11. Water hammer pressure rise is ρcV.
12. The broad-crested weir discharge is maximum when the head is 3/2 times the depth at the crest.

Practice questions

Which is the SI unit of dynamic viscosity?
1. m²/s
2. Pa·s
3. kg/m³
4. N/m
Answer
B. Pa·s
Dynamic viscosity has unit N·s/m², which is Pa·s; m²/s is kinematic viscosity.
The law that relates shear stress in a fluid to the velocity gradient is known as
1. Newton's law of viscosity
2. Archimedes' principle
3. Bernoulli's theorem
4. Pascal's law
Answer
A. Newton's law of viscosity
Newton's law of viscosity states τ = μ(du/dy).
The excess pressure inside a soap bubble of diameter d and surface tension σ is
1. σ/d
2. 4σ/d
3. 8σ/d
4. 2σ/d
Answer
C. 8σ/d
A bubble has two surfaces, so p = 8σ/d; a liquid droplet has p = 4σ/d.
For a submerged vertical plane surface, the centre of pressure lies
1. above the centroid of the surface
2. at the free surface
3. exactly at the centroid
4. below the centroid of the surface
Answer
D. below the centroid of the surface
h* = h̄ + I_G/(A·h̄), which is always greater than h̄.
A floating body is in stable equilibrium when its metacentre is
1. below its centre of gravity
2. below the centre of buoyancy
3. at the centre of buoyancy
4. above its centre of gravity
Answer
D. above its centre of gravity
Positive metacentric height (M above G) gives a restoring couple.
In a horizontal venturimeter, the pressure is lowest at the
1. throat
2. inlet section
3. outlet of the diffuser
4. start of the convergent cone
Answer
A. throat
The velocity is highest at the throat, so the pressure is lowest there.
The coefficient of discharge of an orifice is related to the coefficients of contraction and velocity by
1. Cd = Cv / Cc
2. Cd = Cc × Cv
3. Cd = Cc / Cv
4. Cd = Cc + Cv
Answer
B. Cd = Cc × Cv
Actual discharge = Cc·a × Cv·√(2gH), so Cd = Cc × Cv.
Flow in a circular pipe is laminar when the Reynolds number is below about
1. 10,000
2. 100,000
3. 2000
4. 400
Answer
C. 2000
Laminar flow in pipes is found for Re below about 2000.
For laminar flow in a circular pipe, the Darcy friction factor is
1. 64/Re
2. 32/Re
3. 0.316/Re^0.25
4. 16/Re
Answer
A. 64/Re
From the Hagen-Poiseuille law, f = 64/Re.
Which notch is most suitable for measuring very small discharges accurately?
1. Broad-crested weir
2. Triangular (V) notch
3. Rectangular notch
4. Ogee spillway
Answer
B. Triangular (V) notch
Q ∝ H^2.5, so a small change in discharge gives a large change in head.
The head loss at the exit of a pipe discharging into a large tank is
1. (V₁ − V₂)²/2g
2. 0.1 V²/2g
3. V²/2g
4. 0.5 V²/2g
Answer
C. V²/2g
All the kinetic energy of the pipe flow is lost in the tank.
The side slope of a Cipolletti trapezoidal weir is
1. 4 horizontal to 1 vertical
2. 2 horizontal to 1 vertical
3. 1 horizontal to 1 vertical
4. 1 horizontal to 4 vertical
Answer
D. 1 horizontal to 4 vertical
The 1H:4V slope compensates for end contractions.
Water hammer in a pipeline is caused mainly by
1. slow opening of a valve
2. laminar flow
3. sudden closure of a valve
4. low temperature of water
Answer
C. sudden closure of a valve
Rapid change of velocity creates pressure waves.
Maximum power transmission through a pipe of given supply head H occurs when the friction head loss is
1. H/2
2. H/3
3. H/4
4. 2H/3
Answer
B. H/3
Power ∝ (H − hf)·Q and hf ∝ Q², which gives a maximum at hf = H/3.
The stream function in fluid flow exists for
1. irrotational flow only
2. any two-dimensional incompressible flow
3. three-dimensional compressible flow
4. only for turbulent flow
Answer
B. any two-dimensional incompressible flow
ψ is defined through continuity; φ needs irrotationality.
In fully developed laminar flow in a circular pipe, the maximum velocity is how many times the mean velocity?
1. 2
2. 1.5
3. 1.25
4. 3
Answer
A. 2
The profile is parabolic with u_max = 2 × V_mean.
To account for velocity of approach in weir discharge, the head H is replaced by
1. H − Va²/2g
2. H + Va/2g
3. H × Va
4. H + Va²/2g
Answer
D. H + Va²/2g
Kinetic head of the approaching flow adds to the head over the crest.
The head loss due to sudden enlargement from velocity V₁ to V₂ is
1. (V₁ − V₂)²/2g
2. (V₁² − V₂²)/2g
3. (V₁ + V₂)²/2g
4. 0.5 V₁²/2g
Answer
A. (V₁ − V₂)²/2g
Carnot-Borda formula for loss at sudden expansion.
What is the gauge pressure at a depth of 10 m in fresh water? (ρ = 1000 kg/m³, g = 9.81 m/s²)
1. 981 kPa
2. 10 kPa
3. 9.81 kPa
4. 98.1 kPa
Answer
D. 98.1 kPa
p = ρgh = 1000 × 9.81 × 10 = 98,100 Pa.
A vertical rectangular gate 2 m wide and 3 m deep has its top edge at the water surface. The total hydrostatic force is (ρ = 1000, g = 9.81)
1. 29.4 kN
2. 176.6 kN
3. 88.3 kN
4. 58.9 kN
Answer
C. 88.3 kN
F = ρg·A·h̄ = 1000 × 9.81 × 6 × 1.5 = 88,290 N.
For the same gate (2 m wide, 3 m deep, top at the free surface), the centre of pressure is at what depth below the surface?
1. 1.0 m
2. 2.5 m
3. 2.0 m
4. 1.5 m
Answer
C. 2.0 m
h* = 1.5 + (2 × 27/12)/(6 × 1.5) = 1.5 + 0.5 = 2.0 m.
Water (σ = 0.073 N/m, zero contact angle) rises in a glass tube of 2 mm diameter by about (ρ = 1000, g = 9.81)
1. 14.9 mm
2. 29.8 mm
3. 7.4 mm
4. 3.7 mm
Answer
A. 14.9 mm
h = 4σ/(ρgd) = 0.292/(1000 × 9.81 × 0.002) = 0.0149 m.
An orifice of area 0.001 m² has Cd = 0.6 under a constant head of 5 m. Taking g = 10 m/s², the discharge is
1. 0.0006 m³/s
2. 0.06 m³/s
3. 0.01 m³/s
4. 0.006 m³/s
Answer
D. 0.006 m³/s
Q = 0.6 × 0.001 × √(2 × 10 × 5) = 0.6 × 0.001 × 10.
If the head over a rectangular sharp-crested weir is doubled, the discharge becomes about
1. 4 times
2. 2.83 times
3. 5.66 times
4. 2 times
Answer
B. 2.83 times
Q ∝ H^1.5, so the ratio is 2^1.5 = 2.83.
If the head over a triangular notch becomes four times, the discharge becomes
1. 8 times
2. 32 times
3. 10 times
4. 16 times
Answer
B. 32 times
Q ∝ H^2.5, so the ratio is 4^2.5 = 32.
Water (ν = 10⁻⁶ m²/s) flows at 0.5 m/s in a 0.1 m diameter pipe. The Reynolds number is
1. 500
2. 5,000
3. 500,000
4. 50,000
Answer
D. 50,000
Re = VD/ν = 0.5 × 0.1/10⁻⁶ = 5 × 10⁴ (turbulent).
Using Darcy's equation with f = 0.02, L = 100 m, D = 0.1 m, V = 2 m/s and g = 10 m/s², the head loss is
1. 8 m
2. 2 m
3. 4 m
4. 0.4 m
Answer
C. 4 m
hf = fLV²/(2gD) = 0.02 × 100 × 4/(20 × 0.1) = 4 m.
Water flows from a 0.05 m pipe into a larger pipe so that velocity drops from 4 m/s to 1 m/s. With g = 10 m/s², the head loss at the sudden enlargement is
1. 0.45 m
2. 0.15 m
3. 0.8 m
4. 0.9 m
Answer
A. 0.45 m
hL = (4 − 1)²/(2 × 10) = 9/20 = 0.45 m.
A pipe of length 960 m and diameter 0.2 m is to be replaced by an equivalent pipe of diameter 0.1 m (same friction factor). The equivalent length is
1. 30 m
2. 480 m
3. 15 m
4. 60 m
Answer
A. 30 m
L/D⁵ is constant, so L′ = 960 × (0.1/0.2)⁵ = 960/32 = 30 m.
A venturimeter has inlet area 0.005 m², throat area 0.003 m², Cd = 1 and differential head 0.2 m of water. With g = 10 m/s², the discharge is
1. 0.075 m³/s
2. 0.0075 m³/s
3. 0.015 m³/s
4. 0.00375 m³/s
Answer
B. 0.0075 m³/s
Q = a₁a₂√(2gh)/√(a₁² − a₂²) = (0.005 × 0.003 × 2)/0.004 = 0.0075.
A sudden, instantaneous valve closure stops water (ρ = 1000 kg/m³, wave speed 1400 m/s) flowing at 2 m/s. The pressure rise is
1. 1.4 MPa
2. 28 MPa
3. 2.8 MPa
4. 0.28 MPa
Answer
C. 2.8 MPa
Δp = ρcV = 1000 × 1400 × 2 = 2.8 × 10⁶ Pa.
A rectangular weir with two end contractions is 3 m long and the head is 0.5 m. In the Francis formula the effective length is
1. 3.0 m
2. 2.8 m
3. 2.5 m
4. 2.9 m
Answer
D. 2.9 m
L − 0.1nH = 3 − 0.1 × 2 × 0.5 = 2.9 m.
Consider the statements about viscosity. 1. The viscosity of liquids decreases as temperature rises. 2. The viscosity of gases increases as temperature rises. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
In liquids cohesion falls with heating; in gases molecular momentum exchange rises.
Consider the statements. 1. Absolute pressure equals gauge pressure plus atmospheric pressure. 2. Gauge pressure can never be negative. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Vacuum (suction) pressure is a negative gauge pressure, so statement 2 is false.
Consider the statements about an orifice. 1. The vena contracta lies a short distance downstream of the orifice. 2. The coefficient of velocity is always greater than 1. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Cv is less than 1 (about 0.97 to 0.99) because of friction.
Consider the statements about pipe flow. 1. In laminar flow the friction factor depends only on the Reynolds number. 2. In laminar flow the friction factor depends strongly on pipe roughness. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
f = 64/Re in laminar flow, independent of roughness.
Consider the statements. 1. Discharge over a rectangular notch varies as H^2.5. 2. Discharge over a triangular notch varies as H^2.5. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Rectangular notch Q ∝ H^1.5; triangular Q ∝ H^2.5.
Consider the statements. 1. For pipes in series, the discharge is the same in each pipe. 2. For pipes in parallel, the head loss is the same in each pipe. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
These are the two basic rules for compound pipes.
Consider the statements. 1. The total energy line lies above the hydraulic gradient line by V²/2g. 2. The total energy line rises along the flow direction in a uniform pipe without a pump. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Energy is lost to friction, so TEL falls; a pump is needed to raise it.
Consider the statements. 1. The velocity potential exists for every two-dimensional flow. 2. The stream function exists for every two-dimensional incompressible flow. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Velocity potential needs irrotational flow.
Consider the statements. 1. A Cipolletti weir needs a correction for end contractions. 2. The Francis formula is used for rectangular weirs with end contractions. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
The 1:4 side slopes already compensate for end contractions in a Cipolletti weir.
Consider the statements about capillarity. 1. Mercury is depressed in a clean glass tube. 2. Capillary rise is inversely proportional to the tube diameter. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Mercury does not wet glass; h = 4σcosθ/(ρgd).
Which pair of device and use is correctly matched?
1. Piezometer – discharge in a pipe
2. V-notch – large flood flows
3. U-tube manometer – fluid viscosity
4. Pitot tube – local velocity of flow
Answer
D. Pitot tube – local velocity of flow
A pitot tube measures V = √(2gh) at a point.
Which pair of dimensionless number and ratio of forces is correct?
1. Reynolds number – inertia to gravity force
2. Froude number – inertia to viscous force
3. Froude number – inertia to elastic force
4. Reynolds number – inertia to viscous force
Answer
D. Reynolds number – inertia to viscous force
Re = inertia/viscous; Fr = inertia/gravity.
Which pair of name and topic is correctly matched?
1. Dupuit – hydrostatic force
2. Hagen-Poiseuille – turbulent flow
3. Torricelli – capillarity
4. Joukowsky – water hammer pressure
Answer
D. Joukowsky – water hammer pressure
Δp = ρcV is Joukowsky's relation; Hagen-Poiseuille is for laminar flow; Torricelli gives V = √(2gH); Dupuit gives equivalent pipe.

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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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