Fluid Mechanics and Machines

Answer: C. A fluid deforms continuously under the action of a shear stress, however small

A fluid, by definition, keeps deforming for as long as any shear stress is applied, whereas a solid reaches a static deformation. Liquids can sustain compressive normal stress, and liquids do not fill their container.

Answer: D. the smallest length of interest is much larger than the molecular mean free path

Treating the fluid as continuous with smoothly varying properties requires a small Knudsen number, i.e. the characteristic length is much greater than the mean free path of the molecules.

Answer: B. the fluid layer in contact with the wall has the same velocity as the wall

Viscous fluid adheres to the surface, so its velocity relative to the boundary is zero at the wall. This gives a velocity gradient and hence wall shear stress.

Answer: A. individual fluid particles are followed as they move through space

The Lagrangian approach tracks identified fluid particles and their histories, as in particle mechanics. Observation at fixed points is the Eulerian approach.

Answer: B. Eulerian approach

The Eulerian approach describes the flow as fields (velocity, pressure) at fixed locations in space as time passes.

Answer: C. a fixed region of space chosen for analysis through whose boundary fluid may flow

A control volume is a defined region in space used in the Eulerian approach; mass, momentum and energy cross its control surface. A fixed mass of fluid is a system.

Answer: A. Pa·s

μ = τ/(du/dy) has units (N/m²)/(1/s) = N·s/m² = Pa·s.

Answer: D. L²T⁻¹

ν = μ/ρ = (ML⁻¹T⁻¹)/(ML⁻³) = L²T⁻¹, with SI unit m²/s.

Answer: D. liquids decreases and of gases increases

In liquids cohesive forces dominate and weaken as temperature rises; in gases molecular momentum exchange dominates and grows with temperature.

Answer: C. 40 Pa

Newton's law of viscosity: τ = μ du/dy = 0.8 × 50 = 40 Pa.

Answer: A. 100 N

F = μ A U/h = 0.1 × 0.5 × 2 / 0.001 = 100 N.

Answer: B. 2.5 × 10⁻⁶ m²/s

ν = μ/ρ = 0.002/800 = 2.5 × 10⁻⁶ m²/s.

Answer: D. shear stress is directly proportional to the rate of shear strain

For a Newtonian fluid τ = μ du/dy with constant μ at given temperature and pressure; water, air and light oils are examples.

Answer: C. Paint or blood

Paint and blood show decreasing apparent viscosity with increasing shear rate. Water, air and kerosene are Newtonian.

Answer: A. behaves like a rigid body until a yield stress is exceeded and then flows with a linear stress–strain-rate relation

A Bingham plastic (e.g. toothpaste, drilling mud) needs τ above the yield stress τ_y, after which τ = τ_y + μ_p du/dy.

Answer: B. the cohesive forces between liquid molecules at the free surface

Molecules at a free surface experience a net inward cohesive pull, so the surface behaves like a stretched membrane with tension σ (N/m).

Answer: B. 4σ/d

Force balance on half the droplet: Δp·πd²/4 = σ·πd, so Δp = 4σ/d. A soap bubble has two surfaces and gives 8σ/d.

Answer: C. 144 Pa

Δp = 4σ/d = 4 × 0.072 / 0.002 = 144 Pa.

Answer: D. 14.7 mm

h = 4σ cosθ/(ρ g d) = 4 × 0.072 / (1000 × 9.81 × 0.002) = 0.0147 m ≈ 14.7 mm.

Answer: A. cohesion between mercury molecules exceeds adhesion to glass, giving a contact angle greater than 90°

When cohesion dominates, the liquid does not wet the wall (θ > 90°), so the meniscus is convex and the level falls in the tube.

Answer: A. is the same in all directions

At a point in a static fluid the pressure is isotropic: it acts equally in all directions, independent of orientation.

Answer: B. p = p₀ + ρ g h

The hydrostatic equation dp/dz = −ρg gives p = p₀ + ρ g h for depth h below the free surface.

Answer: C. 98.1 kPa

p = ρ g h = 1000 × 9.81 × 10 = 98 100 Pa = 98.1 kPa (gauge). Adding 101.3 kPa would give the absolute value, 199.4 kPa.

Answer: D. 61.3 kPa

Vacuum pressure = p_atm − p_abs = 101.3 − 40 = 61.3 kPa.

Answer: D. 6.37 m

h = p/(ρ g) = 50 000 / (800 × 9.81) = 6.37 m.

Answer: A. Bourdon gauge

A Bourdon gauge senses pressure through the tendency of a curved, flattened tube to straighten, moving a pointer.

Answer: C. 26.7 kPa

p = ρ_Hg g h = 13 600 × 9.81 × 0.2 = 26 683 Pa ≈ 26.7 kPa.

Answer: B. 12.4 kPa

Δp = (ρ_Hg − ρ_w) g h = (13 600 − 1000) × 9.81 × 0.1 = 12 361 Pa ≈ 12.4 kPa. Ignoring the water gives 13.3 kPa.

Answer: C. the pressure at the centroid multiplied by the area

F = ρ g h̄ A, where h̄ is the depth of the centroid; this is the average pressure times area.

Answer: B. 88.3 kN

F = ρ g h̄ A = 9810 × 1.5 × (2×3) = 88 290 N ≈ 88.3 kN.

Answer: D. below the centroid, except for a horizontal surface where they coincide

h_cp = h̄ + I_G sin²θ/(A h̄) ≥ h̄, since the pressure increases with depth. For a horizontal surface the pressure is uniform, so they coincide.

Answer: A. 3.71 m

h̄ = 2 + 1.5 = 3.5 m; I_G/(A h̄) = (3³/12)/(3×3.5) = 0.214 m, so h_cp = 3.714 m.

Answer: B. fluid properties at a point do not change with time

Steady flow means ∂(any property)/∂t = 0 at every point. Uniformity is about change with position, not time.

Answer: C. the velocity vector is the same at every point at a given instant

Uniform flow has ∂V/∂s = 0 at a given instant, as in flow in a long straight pipe of constant diameter.

Answer: D. steady non-uniform

Discharge constant gives steady flow, while the area change makes the velocity vary with position, so it is non-uniform.

Answer: A. coincide

In steady flow every particle passing a point follows the same track, so the three kinds of lines are identical.

Answer: B. drawn so that the velocity vector is tangent to it at every point

A streamline is tangent to the instantaneous velocity at each of its points, so there is no flow across it.

Answer: A. density change along the flow is negligible, which for gases holds for Mach number below about 0.3

Incompressibility is a flow property. For gases, density variation is under 5% when Ma < 0.3, so air flows at low speeds are treated as incompressible.

Answer: D. 0.35

c = √(γRT) = √(1.4 × 287 × 288) = 340 m/s; Ma = 120/340 = 0.35, so compressibility effects start to matter.

Answer: C. two space coordinates, with no velocity in the third direction

2-D flow has velocity components in a plane, varying in that plane, and none perpendicular to it (e.g. flow over a very long cylinder).

Answer: B. fluid particles rotate about their own axes while moving

A flow is rotational when the curl of velocity is non-zero, i.e. particles spin about their own axes. Circular streamlines alone do not imply rotational flow (free vortex is irrotational).

Answer: C. Free vortex (outside the core)

In a free vortex v = C/r, so the curl is zero everywhere except at the centre. Forced vortex, rigid-body rotation and boundary-layer flow have non-zero vorticity.

Answer: A. inertia force to viscous force

Re = ρVD/μ = inertia force/viscous force; low Re means viscous-dominated, high Re means inertia-dominated flow.

Answer: D. 1800, laminar

Re = ρVD/μ = 900 × 1.2 × 0.05 / 0.03 = 1800 < 2000, so the flow is laminar.

Answer: D. 25 000

Re = VD/ν = 0.5 × 0.05 / 10⁻⁶ = 25 000, well above 4000, hence turbulent.

Answer: C. 2000

Pipe flow is laminar for Re < 2000, transitional from about 2000 to 4000, and turbulent above that.

Answer: B. flatter over the central region with a steeper gradient at the wall

Turbulent mixing equalises velocity in the core, giving a flatter profile and larger wall shear than the laminar parabola.

Answer: A. 0.133 m

D_h = 4A/P = 4 × (0.1×0.2) / (2×(0.1+0.2)) = 0.0800/0.6 = 0.133 m.

Answer: D. satisfies continuity automatically

Defining u = ∂ψ/∂y and v = −∂ψ/∂x makes ∂u/∂x + ∂v/∂y = 0 identically; ψ exists for any 2-D incompressible flow, rotational or not.

Answer: B. irrotational

Since V = ∇φ and curl(grad φ) = 0, a potential function exists only for irrotational flow.

Answer: C. intersect at right angles

Streamlines (ψ = const) and equipotential lines (φ = const) are orthogonal, which forms the basis of the flow net.

Answer: A. 3 m²/s

Flow between streamlines equals the difference of their stream function values: 5 − 2 = 3 m²/s per metre depth.

Answer: B. 6.71 m/s

u = ∂ψ/∂y = 3x = 6 m/s, v = −∂ψ/∂x = −3y = −3 m/s; speed = √(36+9) = 6.71 m/s.

Answer: A. incompressible and irrotational

∂u/∂x + ∂v/∂y = 2 − 2 = 0 (incompressible); ∂v/∂x − ∂u/∂y = 0 (irrotational).

Answer: D. ∂u/∂x + ∂v/∂y = 5 ≠ 0

Continuity for 2-D incompressible flow requires ∂u/∂x + ∂v/∂y = 0, but here the sum is 3 + 2 = 5.

Answer: C. 4 s⁻¹

ω_z = 2 − (−2) = 4 s⁻¹, twice the angular velocity 2 rad/s of the rigid-body rotation.

Answer: D. twice its angular velocity

Vorticity is the curl of velocity, which equals twice the local angular velocity of the fluid element.

Answer: A. the line integral of the velocity component tangent to the curve

Γ = ∮ V·dl, the integral of the tangential velocity around the closed loop.

Answer: C. 1.26 m²/s

Γ = ∮V·dl = ωr × 2πr = 5 × 0.2 × 2π × 0.2 = 1.26 m²/s (equivalently vorticity 2ω times area πr²).

Answer: B. the integral of vorticity over the enclosed area

∮V·dl = ∬(∇×V)·dA, so zero vorticity means zero circulation around any loop that does not enclose a singularity.

Answer: B. 12 m/s²

a_x = u ∂u/∂x = (2x)(2) = 4x = 12 m/s². Local acceleration is zero since the flow is steady.

Answer: A. rate of change of velocity with time at a fixed point

Local acceleration is ∂V/∂t at a fixed location and is zero for steady flow; the part due to movement to a new position is convective.

Answer: C. 0.5

A streamline is a curve of constant ψ. ψ = 2 × 1 = 2, so y = 2/4 = 0.5.

Answer: D. Fluid particles move in smooth parallel layers with no macroscopic mixing

Laminar flow is orderly, with layers sliding over one another and momentum transferred only by molecular viscosity.

Answer: D. ρ₁A₁V₁ = ρ₂A₂V₂

Conservation of mass for steady flow requires the mass flow rate ρAV to be the same at all sections. A₁V₁ = A₂V₂ holds only when ρ is constant.

Answer: A. 8 m/s

A₁V₁ = A₂V₂ ⇒ V₂ = 2 × (200/100)² = 8 m/s, since area scales with the square of diameter.

Answer: C. 0.0628 m³/s

Q = AV = (π/4)(0.2)² × 2 = 0.0314 × 2 = 0.0628 m³/s.

Answer: B. ∂u/∂x + ∂v/∂y + ∂w/∂z = 0

For incompressible flow the divergence of velocity is zero: ∇·V = 0.

Answer: D. dp/ρ + g dz + V dV = 0

Applying Newton's second law to a fluid element along a streamline with no friction gives dp/ρ + g dz + V dV = 0.

Answer: A. steady, frictionless and along a streamline

Bernoulli's equation requires steady, incompressible, inviscid flow along a streamline with no energy added or removed.

Answer: B. length

Each term is energy per unit weight, i.e. head, with dimension L (metres).

Answer: C. kinetic energy per unit weight of fluid

V²/2g is the velocity (kinetic) head, kinetic energy per unit weight.

Answer: A. 145 kPa

p₂ = p₁ + ½ρ(V₁² − V₂²) − ρg(z₂ − z₁) = 200 000 − 6000 − 49 050 ≈ 144 950 Pa ≈ 145 kPa.

Answer: B. V²/2g

EGL = p/ρg + V²/2g + z, HGL = p/ρg + z, so the vertical gap equals the velocity head V²/2g.

Answer: D. slopes downward in the flow direction because of friction losses

Head losses reduce total energy along the flow, so the EGL falls continuously, except where a pump adds head.

Answer: C. 9.90 m/s

Torricelli: V = √(2gh) = √(2 × 9.81 × 5) = 9.90 m/s.

Answer: C. 1.67 L/s

Q = Cd A √(2gh) = 0.6 × 3.142×10⁻⁴ × √(2 × 9.81 × 4) = 1.67×10⁻³ m³/s = 1.67 L/s.

Answer: A. stagnation pressure, from which velocity is found when static pressure is known

At the tip of a Pitot tube the fluid is brought to rest, so it senses p + ½ρV²; velocity follows from the difference with static pressure.

Answer: B. 1.98 m/s

V = √(2gΔh) = √(2 × 9.81 × 0.2) = 1.98 m/s.

Answer: D. the velocity increases at the throat

By continuity the velocity is higher in the throat and, by Bernoulli, the static pressure falls, giving a measurable pressure difference.

Answer: A. 12.8 L/s

V₂ = √[2Δp / (ρ(1 − (A₂/A₁)²))] = √[40 000 / (1000 × 0.9375)] = 6.53 m/s; Q = A₂V₂ = 1.963×10⁻³ × 6.53 = 12.8 L/s.

Answer: D. ρQ (V_out − V_in)

ΣF = ṁ (V_out − V_in) = ρQ(V_out − V_in): the rate of change of momentum flux through the control volume.

Answer: B. 785 N

F = ρ A V² = 1000 × (π/4)(0.05)² × 20² = 785 N, since all of the jet's momentum is destroyed by the plate.

Answer: C. 2000 N

Velocity reverses, so the change is 2V: F = ρQ(V − (−V)) = 2 × 1000 × 0.05 × 20 = 2000 N.

Answer: C. n − m

The π theorem gives (n − m) independent dimensionless π terms.

Answer: B. 4

Number of π terms = n − m = 7 − 3 = 4.

Answer: A. inertia force to gravity force

Fr = V/√(gL) = inertia/gravity; it governs free-surface flows such as flow over spillways and around ships.

Answer: D. Reynolds number

Pressure-driven flow in closed conduits is dominated by viscous forces, so Reynolds number similarity is required.

Answer: D. surface tension force

We = ρV²L/σ = inertia force/surface tension force; it matters for droplets, bubbles and capillary waves.

Answer: A. 2 m/s

Froude similarity: V ∝ √L, so V_m = V_p × √(1/25) = 10/5 = 2 m/s.

Answer: B. L_r^2.5

Q = AV, A ∝ L², V ∝ L^0.5, so Q ∝ L^2.5.

Answer: C. 20 m/s

With the same ν, Re = VL/ν constant ⇒ V_m = V_p (L_p/L_m) = 2 × 10 = 20 m/s.

Answer: C. Dynamic viscosity

Dynamic viscosity has dimensions ML⁻¹T⁻¹; the others are dimensionless ratios of forces.

Answer: B. flow speed to speed of sound

Ma = V/c, where c = √(γRT) for an ideal gas; it governs compressibility effects.

Answer: D. pressure force to inertia force

Eu = Δp/(ρV²), the ratio of pressure force to inertia force.

Answer: B. parabolic with maximum velocity at the centre

Solving the Navier–Stokes equation for steady laminar pipe flow gives u = u_max[1 − (r/R)²], a paraboloid.

Answer: A. 2

u_max = 2 V_avg for the parabolic profile.

Answer: D. linear, zero at the centre and maximum at the wall

From a force balance τ = (Δp/L)(r/2), varying linearly from zero on the axis to τ₀ at the wall.

Answer: C. Q = π D⁴ Δp /(128 μ L)

Integrating the parabolic profile gives Q = πR⁴Δp/(8μL) = πD⁴Δp/(128μL).

Answer: C. 16 times

Q ∝ D⁴, so doubling D multiplies Q by 2⁴ = 16.

Answer: D. 3.93 × 10⁻⁵ m³/s

Q = πD⁴Δp/(128μL) = π(0.02)⁴(2000)/(128 × 0.1 × 2) = 3.93×10⁻⁵ m³/s.

Answer: A. 64/Re

For laminar flow f = 64/Re (Darcy–Weisbach), independent of pipe roughness. 16/Re is the Fanning value.

Answer: B. 0.04

f = 64/Re = 64/1600 = 0.04.

Answer: D. 4.08 m

h_f = f L V²/(2gD) = 0.02 × 100 × 4 /(2 × 9.81 × 0.1) = 4.08 m.

Answer: B. 4.53 m

h_f = 32 × 0.1 × 10 × 0.5 /(900 × 9.81 × 0.0004) = 16/3.532 = 4.53 m (Re = 90, laminar).

Answer: A. 2 Pa

τ₀ = Δp·D/(4L) = 1000 × 0.04/(4 × 5) = 2 Pa.

Answer: C. velocity to the first power

With f = 64/Re ∝ 1/V, h_f ∝ fV² ∝ V, whereas in turbulent flow h_f ∝ V^1.75 to V².

Answer: D. 2/3

Velocity profile is parabolic; the mean is two-thirds of the centre-line maximum, whereas pipe flow gives one-half.

Answer: A. 15 000 Pa/m

Δp/L = 12 μ V̄ / h² = 12 × 0.05 × 0.1 /(0.002)² = 15 000 Pa/m.

Answer: B. linear

With dp/dx = 0 the shear stress is constant, so du/dy is constant and u varies linearly from 0 to U.

Answer: C. friction along the pipe length

Major losses are frictional losses along straight pipe lengths; losses at fittings, entries and exits are minor losses.

Answer: C. K V²/2g

Minor loss h_m = K V²/(2g), where K is the loss coefficient of the fitting.

Answer: D. 0.459 m

h_L = (V₁ − V₂)²/(2g) = 9/19.62 = 0.459 m (Borda–Carnot).

Answer: A. 0.459 m

All the kinetic energy is lost: h = V²/2g = 9/19.62 = 0.459 m (K = 1).

Answer: B. 0.102 m

h = 0.5 V²/2g = 0.5 × 4/19.62 = 0.102 m.

Answer: B. 25 m

K V²/2g = f Le V²/(2gD) ⇒ Le = K D/f = 5 × 0.1/0.02 = 25 m.

Answer: A. 99% of the free-stream velocity

Since the profile merges asymptotically with the free stream, δ is taken where u = 0.99 U.

Answer: D. 3.16 mm

Re_x = 0.5 × 0.2 /10⁻⁶ = 10⁵; δ = 5 × 0.2 /√10⁵ = 1/316.2 = 3.16 mm.

Answer: C. x^(1/2)

Blasius solution gives δ = 5x/√Re_x ∝ x^0.5, while the turbulent layer grows as ≈ x^0.8.

Answer: D. 0.75 m

x_cr = Re_cr ν/U = 5×10⁵ × 1.5×10⁻⁵/10 = 0.75 m.

Answer: A. a fuller velocity profile and higher wall shear stress

Turbulent mixing brings high-momentum fluid toward the wall, increasing du/dy at the wall and skin friction.

Answer: C. an adverse pressure gradient causes the velocity gradient at the wall to become zero

Under an adverse pressure gradient (dp/dx > 0) the near-wall fluid loses momentum; at separation (∂u/∂y)_wall = 0 and the flow reverses.

Answer: B. promote a turbulent boundary layer that delays separation and reduces pressure drag

A turbulent boundary layer resists separation, shrinking the wake and reducing form drag.

Answer: D. the distance by which the outer streamlines are displaced due to the boundary layer

δ* = ∫(1 − u/U)dy measures the outward shift of the external flow caused by the velocity deficit near the wall.

Answer: D. in a nozzle before the water strikes the runner

In an impulse turbine such as the Pelton wheel, the nozzle turns the head into a high-speed jet at atmospheric pressure, and the runner only changes the jet's momentum.

Answer: B. Pelton wheel

Pelton (and Turgo) turbines are impulse machines; Francis, Kaplan and propeller turbines are reaction machines.

Answer: A. axial

A Kaplan turbine is an axial-flow reaction turbine with adjustable runner blades.

Answer: C. Pelton wheel

Pelton wheels have the lowest specific speed and are used for high heads with small flows; Kaplan turbines suit low heads and large flows.

Answer: A. above atmospheric, falling through the runner

In reaction turbines, part of the head remains as pressure energy at the runner inlet, and the pressure drops across the runner while the water is fully enclosed.

Answer: C. reaction turbines

Reaction turbines discharge through a draft tube so that the runner can sit above the tail water and the exit kinetic energy is partly recovered.

Answer: D. convert exit kinetic energy to pressure and give suction below the runner

A gradually diverging draft tube lowers the exit velocity, which creates sub-atmospheric pressure at the runner outlet, increasing the effective head and permitting an elevated setting.

Answer: B. Draft tube

A Pelton wheel operates in air at atmospheric pressure so no draft tube is needed; it uses a nozzle with spear, buckets and a jet deflector.

Answer: B. regulate the jet area and thus the discharge

Moving the spear needle in or out of the nozzle changes the flow area and discharge while keeping the jet velocity almost constant.

Answer: A. divert the jet from the buckets on sudden load rejection

Slowly closing the spear avoids water hammer in the penstock, so the deflector quickly diverts the jet away from the buckets to limit overspeed.

Answer: D. regulate the flow and direct water onto the runner at the right angle

The adjustable guide vanes control discharge with load changes and set the inlet flow angle onto the runner.

Answer: C. radially inward at inlet, nearly axial at outlet

Water enters the runner from the spiral casing radially and leaves axially into the draft tube (mixed flow).

Answer: A. distribute water uniformly around the guide vanes

The casing area decreases along the flow path so that equal discharge and velocity are delivered to the stay and guide vanes around the whole circumference.

Answer: B. keep the speed constant by regulating the flow as load changes

Generators must run at synchronous speed, so the governor adjusts the spear (Pelton), guide vanes (Francis) or guide vanes and blades (Kaplan) when the load changes.

Answer: D. both guide vanes and runner blades are adjustable

Adjustable runner blades together with adjustable guide vanes keep efficiency high over a wide range of load and head.

Answer: C. produces unit power under unit head

N_s = N√P/H^(5/4) is the speed of a similar turbine producing 1 kW (or 1 hp) under 1 m head.

Answer: C. 70.7

N_s = 500 × √2000 / 100^1.25 = 500 × 44.72 / 316.2 = 70.7.

Answer: D. 100 rpm

N_u = N/√H = 600/6 = 100 rpm.

Answer: B. 86.8 m/s

V₁ = Cv √(2gH) = 0.98 × √(2 × 9.81 × 400) = 0.98 × 88.58 = 86.8 m/s.

Answer: A. 31.4 m/s

u = πDN/60 = π × 1.2 × 500/60 = 31.4 m/s.

Answer: C. one-half

Efficiency η = 2u(V − u)(1 + k cosφ)/V² is maximum at u = V/2.

Answer: B. 981 kW

P = ρ g Q H = 1000 × 9.81 × 2 × 50 = 981 000 W = 981 kW.

Answer: D. 87.7%

η_o = shaft power/water power = 860/981 = 0.877.

Answer: A. 85%

η_h = V_w1 u₁/(gH) = 20 × 25/(9.81 × 60) = 0.85.

Answer: B. at the runner outlet or draft tube inlet, where pressure is lowest

Minimum pressure occurs at the runner exit and draft tube inlet; when it drops to the vapour pressure, vapour bubbles form and collapse.

Answer: A. pitting of blades, noise, vibration and fall in efficiency

Collapsing vapour bubbles produce high local pressure impacts, which erode the surface and cause noise and vibration.

Answer: D. 0.14

σ = (10.3 − 0.3 − 3)/50 = 7/50 = 0.14.

Answer: C. lowering the turbine setting relative to the tail water

A lower setting raises the pressure at the runner exit above the vapour pressure, increasing σ.

Answer: A. 29.7 m³/s

Q = (π/4)(D₀² − D_b²)V_f = (π/4)(9 − 1.44) × 5 = 29.7 m³/s.

Answer: B. speed and head

Operating (constant-speed, constant-head) curves are measured by varying the gate opening (load) while holding the speed and head fixed.

Answer: D. higher, because blade adjustment keeps efficiency flat

Adjustable blades follow the changing flow, so Kaplan efficiency remains high over a wide range of loads.

Answer: C. the speed reached at full gate when the load is suddenly lost

If the load is rejected and the governor fails, the runner accelerates to the runaway speed, which can be 1.8 to 3 times the normal speed depending on type.

Answer: B. rotodynamic pumps and positive displacement pumps

Rotodynamic pumps (centrifugal, axial, mixed flow) add energy by momentum change; positive displacement pumps (reciprocating, gear, screw) trap and push fixed volumes.

Answer: D. Gear pump

A gear pump traps fluid between teeth and casing and carries it to the delivery side; the others are rotodynamic.

Answer: A. forced vortex action of the rotating impeller

The impeller imparts rotational motion to the liquid, which is thrown outward and gains pressure and kinetic energy in the casing.

Answer: C. convert part of the exit kinetic energy into pressure energy

The increasing cross-section of the volute reduces fluid velocity, raising pressure toward the delivery pipe.

Answer: B. keep the suction pipe full by acting as a non-return valve

The foot valve (with strainer) prevents the water from draining back into the sump when the pump is stopped, helping priming.

Answer: D. remove air from the suction pipe and casing before starting

A centrifugal pump cannot create enough suction to expel air; if the casing holds air, it runs without pumping (air binding), so it must be filled with liquid before start.

Answer: A. the power input is minimum at zero flow

For radial-flow centrifugal pumps, power demand rises with flow, so starting against a closed valve minimises motor starting load.

Answer: C. backward-curved vanes

Backward-curved vanes give higher efficiency and a stable head–discharge curve, with more of the energy appearing as pressure.

Answer: C. 30.6 m

H = 20 × 15 / 9.81 = 30.6 m.

Answer: D. 22.8 m/s

u₂ = πD₂N/60 = π × 0.3 × 1450/60 = 22.8 m/s.

Answer: A. the total head the pump works against, including all losses

H_m = H_s + H_d + h_fs + h_fd + V_d²/2g: static lift plus all losses and delivery velocity head.

Answer: B. 21.0 kW

P = ρ g Q H/η = 9810 × 0.05 × 30/0.7 = 21 021 W ≈ 21.0 kW.

Answer: A. 4.91 kW

P = ρ g Q H = 9810 × 0.02 × 25 = 4905 W ≈ 4.91 kW.

Answer: C. 68.4%

η_o = η_man × η_mech × η_vol = 0.8 × 0.9 × 0.95 = 0.684.

Answer: B. unit discharge under unit head

N_s = N√Q/H^(3/4) is the speed of a similar pump delivering 1 m³/s against 1 m head.

Answer: D. 34.3

N_s = 1450 × √0.05 / 20^0.75 = 1450 × 0.2236 / 9.457 = 34.3.

Answer: D. 2, 4, 8

Affinity laws: Q ∝ N, H ∝ N², P ∝ N³.

Answer: B. 81% of the original

H ∝ D², so (0.9)² = 0.81.

Answer: C. the total suction head above the vapour pressure head

NPSH_a = p_s/ρg + V_s²/2g − p_v/ρg, the energy margin above vaporisation at the suction flange.

Answer: A. 6.0 m

NPSH_a = H_atm − H_v − H_s − h_fs = 10.3 − 0.5 − 3 − 0.8 = 6.0 m.

Answer: A. NPSH available greater than NPSH required

The available suction margin must exceed what the pump needs (NPSHr) to keep local pressure above vapour pressure.

Answer: C. at the impeller eye (vane inlet)

Pressure is lowest where fluid enters the impeller vanes, so vapour bubbles form there when NPSH is inadequate.

Answer: B. 10.3 m

Atmospheric pressure supports a water column of about 10.3 m; practical lifts are 6 to 7 m once vapour pressure and losses are included.

Answer: D. falls steadily as discharge increases

For backward-curved vanes the head decreases as flow increases, with maximum head at zero discharge (shut-off).