How unit 3 is examined
This unit covers fluid properties, flow types, Newton's law of viscosity, Pascal's law, Bernoulli's equation and the working of turbines and pumps; viscosity numericals and the Pelton wheel, reciprocating pump and turbine questions carry most marks.
Fluid properties pressure, density and viscosity etc.
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Definition. A fluid is a substance that deforms continuously under any shear force, however small; <mark>viscosity is the internal resistance of a fluid to flow, arising from friction between its adjacent layers.</mark>
Key points.
- Density is mass per unit volume, $\rho = m/V$; for water it is $1000\ \text{kg/m}^3$.
- Pressure is normal force per unit area, $p = F/A$, measured in $\text{N/m}^2$ (Pa); $1\ \text{bar} = 10^5\ \text{N/m}^2$.
- Dynamic viscosity $\mu$ has unit $\text{N·s/m}^2$ (Pa·s); $1\ \text{poise} = 0.1\ \text{N·s/m}^2$. Kinematic viscosity is $\nu = \mu/\rho$, in $\text{m}^2/\text{s}$.
- Importance: in lubrication a viscous oil film prevents metal-to-metal contact between shaft and bearing, and its thickness decides friction loss.
- Viscosity also fixes the pressure drop in pipelines, the power a pump needs, the behaviour of hydraulic oils, and the atomisation of fuel.
Asked: [7 marks] (Nov 2022) Define viscosity. What is the importance of viscosity while selecting fluids in various engineering applications?
Types of fluids
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Definition. Fluids are classified by their stress behaviour (Newtonian or not) and by the nature of their flow (uniform, laminar and so on).
Key points.
- A Newtonian fluid obeys $\tau = \mu\,du/dy$ with constant $\mu$; water, air and oil are examples.
- A non-Newtonian fluid has a non-linear or variable relation between $\tau$ and $du/dy$; paint, blood and toothpaste are examples.
- An ideal fluid is frictionless and incompressible; a real fluid has viscosity.
- Reynolds number decides laminar or turbulent flow in a pipe: below about 2000 it is laminar, above about 4000 turbulent.
| Basis | First type | Second type |
|---|---|---|
| Uniform / non-uniform | Velocity same at every point at a given instant, $\partial V/\partial s = 0$ (constant-diameter pipe) | Velocity changes from point to point, $\partial V/\partial s \neq 0$ (tapering pipe) |
| Laminar / turbulent | Particles move in smooth parallel layers, no mixing; $Re < 2000$ | Particles move chaotically with eddies and mixing; $Re > 4000$ |
| Newtonian / non-Newtonian | $\tau \propto du/dy$, straight line through origin | $\tau$ is not proportional to $du/dy$ |
Answer frame. Open by defining each pair in one line; then give the table row by row with one example each; close with the Reynolds-number limits.
Asked: [7 marks] (Dec 2024) Define and distinguish between: i) uniform and non-uniform flow, ii) laminar and turbulent flow, iii) Newtonian and non-Newtonian flow.
Newton's law of viscosity
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Definition. <mark>Newton's law of viscosity states that the shear stress between adjacent fluid layers is directly proportional to the velocity gradient (rate of shear strain) perpendicular to the flow.</mark>
Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u3-01" viewBox="0 0 338 252" width="338" height="252" role="img" aria-label="Fluid film of thickness dy between a moving plate (velocity u) and a fixed plate; velocity varies linearly from 0 to u."><style>#dsfig-u3-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u3-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u3-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u3-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u3-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u3-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u3-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u3-01 .t{fill:#16181D;font-weight:500}#dsfig-u3-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u3-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u3-01 .dot{fill:#16181D}#dsfig-u3-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u3-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u3-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u3-01 .ah{fill:#454C5A}#dsfig-u3-01 .ah.hi{fill:#2340B8}#dsfig-u3-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u3-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u3-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u3-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u3-01 .e{stroke:#B1B7C3}html.dark #dsfig-u3-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u3-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u3-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u3-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u3-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u3-01 .t{fill:#E6E8ED}html.dark #dsfig-u3-01 .t.inv{fill:#0F1115}html.dark #dsfig-u3-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u3-01 .dot{fill:#E6E8ED}html.dark #dsfig-u3-01 .ann{fill:#8FA3FF}html.dark #dsfig-u3-01 .lbl{fill:#858D9C}html.dark #dsfig-u3-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u3-01 .ah{fill:#B1B7C3}html.dark #dsfig-u3-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u3-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u3-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u3-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah11" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh11" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,40 L279,40"/><path class="e" d="M59,212 L279,212"/><g class="wl"><rect x="113.5" y="31" width="111" height="18" rx="9"/><text class="t" x="169" y="40" dy=".35em" text-anchor="middle">Moving_plate_u</text></g><g class="wl"><rect x="124.3" y="203" width="89.4" height="18" rx="9"/><text class="t" x="169" y="212" dy=".35em" text-anchor="middle">Fixed_plate</text></g><circle class="n" cx="40" cy="40" r="18"/><text class="t" x="40" y="40" dy=".35em" text-anchor="middle">M</text><circle class="n" cx="298" cy="40" r="18"/><text class="t" x="298" y="40" dy=".35em" text-anchor="middle">N</text><circle class="n" cx="40" cy="212" r="18"/><text class="t" x="40" y="212" dy=".35em" text-anchor="middle">P</text><circle class="n" cx="298" cy="212" r="18"/><text class="t" x="298" y="212" dy=".35em" text-anchor="middle">Q</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Fluid film of thickness dy between a moving plate (velocity u) and a fixed plate; velocity varies linearly from 0 to u.</figcaption></figure>
Formula. $$\tau = \mu\frac{du}{dy}$$
Key points.
- Here $\tau = F/A$ is the shear stress, $du/dy$ is the velocity gradient (shear rate), and the constant $\mu$ is the dynamic viscosity.
- The fluid layer touching the moving plate sticks to it and moves at $u$, while the layer at the fixed plate has zero velocity (no-slip condition).
- For a thin film the profile is linear, so $du/dy = u/y$ with $y$ the film thickness.
- Significance: $\mu$ measures how strongly a fluid resists flow, so it fixes friction force, power loss and lubrication behaviour.
- Fluids obeying the law with constant $\mu$ are Newtonian; a plot of $\tau$ against $du/dy$ is a straight line through the origin with slope $\mu$.
- Viscosity of liquids falls as temperature rises, while that of gases rises.
Example. Plate 0.025 mm from a fixed plate moves at 60 cm/s and needs $\tau = 2\ \text{N/m}^2$; find $\mu$.
| Step | Working |
|---|---|
| Given | $dy = 0.025\times10^{-3}$ m, $du = 0.6$ m/s, $\tau = 2$ |
| Formula | $\mu = \tau\,dy/du$ |
| Substitute | $\mu = 2\times0.025\times10^{-3}/0.6$ |
$\mu = 8.33\times10^{-5}\ \text{N·s/m}^2$
Example (power). Plate area $1.5\times10^6\ \text{mm}^2 = 1.5\ \text{m}^2$, speed 0.4 m/s, gap 0.15 mm, $\mu = 1\ \text{poise} = 0.1$. $\tau = 0.1\times0.4/(0.15\times10^{-3}) = 266.67$; $F = \tau A = 266.67\times1.5 = 400$ N; $P = Fu = 400\times0.4 = 160$ W.
Example (shaft in sleeve). $\mu = 10\ \text{poise} = 1\ \text{N·s/m}^2$, $D = 0.5$ m, $N = 200$ rpm, $L = 0.1$ m, $t = 0.002$ m.
| Step | Working |
|---|---|
| Surface speed | $u = \pi D N/60 = \pi\times0.5\times200/60 = 5.236$ m/s |
| Shear stress | $\tau = \mu u/t = 1\times5.236/0.002 = 2618\ \text{N/m}^2$ |
| Shear force | $F = \tau\,\pi D L = 2618\times\pi\times0.5\times0.1 = 411.2$ N |
| Power lost | $P = Fu = 411.2\times5.236$ |
Power lost $\approx 2153$ W (2.15 kW)
Answer frame. For the law: state it, write $\tau=\mu\,du/dy$, draw the two-plate figure, list points 1-6, close with the significance of $\mu$. For a numerical: convert to SI first (poise, mm, cm/s, rpm), write the formula, substitute, box the answer with unit.
Pitfall: Forgetting to convert poise, mm and cm/s to SI before substituting; also use area $\pi D L$ (not $\pi D$) for the shaft. Some answer keys print 41.12 N and 215.3 W, which drops a factor of 10; the correct working gives 411.2 N and 2153 W.
Asked: [7 marks] (Nov 2022, Jun 2022) A plate 0.025 mm from a fixed plate moves at 60 cm/s and requires 2 N/m^2 to maintain this speed; determine the fluid viscosity. Also: flat plate of area 1.5x10^6 mm^2 pulled at 0.4 m/s relative to another plate 0.15 mm away, fluid viscosity 1 poise; find force and power. Asked: [8 marks] (Jun 2023) Lubricating oil of viscosity 10 poise between shaft (D = 0.5 m, 200 rpm) and sleeve (length 100 mm); calculate the power lost if the oil film is 2 mm thick. Asked: [7 marks] (Dec 2024) State Newton's law of viscosity and explain its significance; describe how viscosity relates to shear stress and shear rate in Newtonian fluids. Asked: [14 marks] (Jun 2025) Write short note on any three: i) Newton's law of viscosity, ii) Zeroth law of thermodynamics, iii) Micrometer, iv) Centrifugal compressor.
Pascal's law
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Definition. <mark>Pascal's law states that the pressure at a point in a fluid at rest is the same in all directions; equivalently, pressure applied to an enclosed fluid is transmitted equally and undiminished to every part of it.</mark>
Key points.
- Formula: in a hydraulic press, $F_1/A_1 = F_2/A_2$, so a small force on a small piston gives a large force on a large piston.
- It holds only for a static fluid, where no shear stress exists.
- Applications are the hydraulic press, hydraulic jack, brakes and lifts.
Bernoulli's equation for incompressible fluids
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Definition. <mark>For steady, frictionless, incompressible flow along a streamline, the total energy per unit weight (pressure head + velocity head + datum head) is constant.</mark>
Formula. $$\frac{p}{\rho g}+\frac{v^2}{2g}+z=\text{constant}$$
Assumptions.
- The flow is steady and the fluid incompressible.
- The fluid is inviscid (frictionless) and the flow irrotational.
- The equation is applied along a single streamline, and only gravity and pressure forces act.
Derivation.
- Take a small element of length $ds$ and area $dA$ along a streamline, inclined at $\theta$ to the horizontal.
- Newton's second law along $s$: $p\,dA-(p+dp)\,dA-\rho g\,dA\,ds\cos\theta=\rho\,dA\,ds\cdot a_s$.
- With $a_s = v\,dv/ds$ and $\cos\theta\,ds=dz$, dividing by $\rho\,dA$ gives Euler's equation: $$\frac{dp}{\rho}+v\,dv+g\,dz=0$$
- Integrate for constant $\rho$: $p/\rho+v^2/2+gz=\text{const}$; divide by $g$ to get the head form above.
Example. Water flows through 30 cm and 20 cm sections at 80 L/s. Section 1 is 8 m and section 2 is 5 m above datum; $p_1=4$ bar. Find $p_2$.
| Step | Working |
|---|---|
| Areas | $A_1=0.0707$, $A_2=0.0314\ \text{m}^2$ |
| Velocities | $V_1=0.08/0.0707=1.132$, $V_2=0.08/0.0314=2.546$ m/s |
| Bernoulli | $\frac{4\times10^5}{1000\times9.81}+\frac{1.132^2}{19.62}+8=\frac{p_2}{\rho g}+\frac{2.546^2}{19.62}+5$ |
| Heads | $40.77+0.065+8=48.84$; $0.330+5=5.33$ |
| Result | $p_2/\rho g=43.51$ m |
$p_2 = 43.51\times9810 \approx 4.27\times10^5\ \text{N/m}^2 \approx 4.27$ bar
Repeat question (Dec 2023): $d_1=25$ cm, $d_2=10$ cm, $Q=30$ L/s, $z_1=10$ m, $z_2=2$ m, $p_1=40\ \text{N/cm}^2=4\times10^5$. Then $V_1=0.611$, $V_2=3.820$ m/s, $p_2/\rho g=48.05$ m, so $p_2\approx4.71\times10^5\ \text{N/m}^2\approx4.71$ bar.
Answer frame. For the derivation: state Pascal's law in one line, list assumptions, draw the streamline element, derive Euler's equation, integrate, and close with the boxed Bernoulli equation. For the numerical: convert to SI, find $A$ and $V$ from $Q=AV$, apply Bernoulli, solve $p_2$.
Pitfall: Using litres or bar without conversion, and forgetting the velocity head or datum terms. The key printed 4.29 bar; exact arithmetic gives 4.27 bar.
Asked: [8 marks] (Dec 2023, Jun 2025) Water flows through a pipe of diameters 30 cm and 20 cm at sections 1 and 2, flow 80 L/s, section 1 at 8 m and section 2 at 5 m above datum, pressure at 1 is 4 bar; find pressure at 2. Also: diameters 25 cm and 10 cm, 30 L/s, 10 m and 2 m above ground, 40 N/cm^2 at section 1; find pressure at 2. Asked: [7 marks] (Jun 2022) State Pascal's law. Derive Bernoulli's equation. What are the assumptions made in Bernoulli's equation?
Working principle of hydraulic machines, pumps, turbines, reciprocating pumps
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Definition. Hydraulic machines convert hydraulic energy into mechanical energy (turbines) or mechanical energy into hydraulic energy (pumps).
Hydraulic turbines and classification
Key points.
- Principle: a turbine converts the potential and kinetic energy of flowing water into mechanical rotation of a runner, which drives a generator.
- By action of water: in an impulse turbine (Pelton) all the head is first converted to kinetic energy in a nozzle and the runner works at atmospheric pressure; in a reaction turbine (Francis, Kaplan) only part is converted, so pressure drops across the runner and the casing is full of water.
- By direction of flow: tangential (Pelton), radial or mixed (Francis), axial (Kaplan).
- By head: high head above 250 m (Pelton), medium 60-250 m (Francis), low below 60 m (Kaplan).
- By specific speed: low (Pelton), medium (Francis), high (Kaplan).
- Applications: Pelton in hilly hydro plants with small discharge, Francis in medium-head plants, Kaplan in river barrages with large discharge.
Answer frame. Open with the energy-conversion principle; give the classification as a table or tree; then point-wise head, flow and application for each type; close with the choice of turbine by head.
Pelton wheel
Key points.
- It is a tangential-flow impulse turbine used for high head and low discharge.
- Water from the penstock reaches the nozzle, where pressure energy becomes a high-velocity jet.
- A spear (needle) inside the nozzle moves to vary the jet area and so regulate the flow.
- The runner is a disc carrying double-hemispherical (double-ellipsoidal) buckets with a central splitter ridge.
- The jet strikes the splitter, divides into two halves, and is deflected through about $165^\circ$; the change of momentum gives an impulse force and torque on the runner.
- The casing prevents splashing and guides water to the tailrace; a braking jet stops the runner quickly.
<figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u3-02" viewBox="0 0 474 252" width="474" height="252" role="img" aria-label="Pelton wheel. Pen = penstock, Noz = nozzle with spear, Jet = high-velocity jet, Run = runner with double hemispherical buckets inside casing, Tail = tailrace."><style>#dsfig-u3-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u3-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u3-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u3-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u3-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u3-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u3-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u3-02 .t{fill:#16181D;font-weight:500}#dsfig-u3-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u3-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u3-02 .dot{fill:#16181D}#dsfig-u3-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u3-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u3-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u3-02 .ah{fill:#454C5A}#dsfig-u3-02 .ah.hi{fill:#2340B8}#dsfig-u3-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u3-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u3-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u3-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u3-02 .e{stroke:#B1B7C3}html.dark #dsfig-u3-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u3-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u3-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u3-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u3-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u3-02 .t{fill:#E6E8ED}html.dark #dsfig-u3-02 .t.inv{fill:#0F1115}html.dark #dsfig-u3-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u3-02 .dot{fill:#E6E8ED}html.dark #dsfig-u3-02 .ann{fill:#8FA3FF}html.dark #dsfig-u3-02 .lbl{fill:#858D9C}html.dark #dsfig-u3-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u3-02 .ah{fill:#B1B7C3}html.dark #dsfig-u3-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u3-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u3-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u3-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah12" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh12" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,40 L148,40" marker-end="url(#ah12)"/><path class="e" d="M188,40 L277,40" marker-end="url(#ah12)"/><path class="e" d="M317,40 L406,40" marker-end="url(#ah12)"/><path class="e" d="M427,59 L427,184" marker-end="url(#ah12)"/><circle class="n" cx="40" cy="40" r="18"/><text class="t" x="40" y="40" dy=".35em" text-anchor="middle">Pen</text><circle class="n" cx="169" cy="40" r="18"/><text class="t" x="169" y="40" dy=".35em" text-anchor="middle">Noz</text><circle class="n" cx="298" cy="40" r="18"/><text class="t" x="298" y="40" dy=".35em" text-anchor="middle">Jet</text><circle class="n" cx="427" cy="40" r="18"/><text class="t" x="427" y="40" dy=".35em" text-anchor="middle">Run</text><rect class="n" x="402" y="197" width="50" height="30" rx="15"/><text class="t" x="427" y="212" dy=".35em" text-anchor="middle">Tail</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Pelton wheel. Pen = penstock, Noz = nozzle with spear, Jet = high-velocity jet, Run = runner with double hemispherical buckets inside casing, Tail = tailrace.</figcaption></figure>
Answer frame. Open with "Pelton wheel is a tangential-flow impulse turbine for high head"; draw the labelled sketch; develop construction (nozzle, spear, runner, buckets, casing, brake jet) then working; close with the jet momentum change producing torque.
Draft tube
Key points.
- A draft tube is an expanding (diverging) pipe joining the runner exit of a reaction turbine to the tailrace, ending below tailrace water level.
- It lets the turbine be set above tailrace level without losing head, for easy maintenance.
- Its gradually increasing area reduces the exit velocity, so kinetic energy leaving the runner is regained as pressure, creating suction at the runner outlet and raising the working head.
- Types: conical, simple elbow, moody spreading, and elbow with varying cross-section.
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Answer frame. Define it, draw runner-tube-tailrace, give the two functions, then the types.
Reciprocating pump
Definition. <mark>A reciprocating pump is a positive displacement pump in which a piston or plunger moves to and fro inside a cylinder, drawing liquid in and forcing it out through non-return valves.</mark>
Key points.
- Components: cylinder, piston with rod, connecting rod and crank driven by a motor, suction pipe with suction valve and strainer, delivery pipe with delivery valve, and sump.
- Suction stroke: the crank turns and the piston moves away from the cylinder head, creating a partial vacuum; atmospheric pressure on the sump pushes liquid up the suction pipe, the suction valve opens and liquid fills the cylinder.
- Delivery stroke: the piston moves back towards the head, pressurising the liquid; the suction valve shuts, the delivery valve opens and liquid is forced up the delivery pipe.
- In a single-acting pump one delivery occurs per crank revolution, so the discharge is intermittent.
- Flow is unsteady, so air vessels are fitted to smooth it; discharge is $Q = ALN/60$ for single-acting.
<figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u3-04" viewBox="0 0 560 338" width="560" height="338" role="img" aria-label="Single-acting reciprocating pump. Sump = water source with strainer, Sv = suction valve, Cyl = cylinder with piston, rod, connecting rod and crank, Dv = delivery valve, Del = delivery pipe."><style>#dsfig-u3-04 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u3-04 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u3-04 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u3-04 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u3-04 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u3-04 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u3-04 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u3-04 .t{fill:#16181D;font-weight:500}#dsfig-u3-04 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u3-04 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u3-04 .dot{fill:#16181D}#dsfig-u3-04 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u3-04 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u3-04 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u3-04 .ah{fill:#454C5A}#dsfig-u3-04 .ah.hi{fill:#2340B8}#dsfig-u3-04 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u3-04 .wl .t{font-size:12px;font-weight:700}#dsfig-u3-04 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u3-04 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u3-04 .e{stroke:#B1B7C3}html.dark #dsfig-u3-04 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u3-04 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u3-04 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u3-04 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u3-04 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u3-04 .t{fill:#E6E8ED}html.dark #dsfig-u3-04 .t.inv{fill:#0F1115}html.dark #dsfig-u3-04 .kd{stroke:#E6E8ED}html.dark #dsfig-u3-04 .dot{fill:#E6E8ED}html.dark #dsfig-u3-04 .ann{fill:#8FA3FF}html.dark #dsfig-u3-04 .lbl{fill:#858D9C}html.dark #dsfig-u3-04 .ptr{fill:#8FA3FF}html.dark #dsfig-u3-04 .ah{fill:#B1B7C3}html.dark #dsfig-u3-04 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u3-04 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u3-04 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u3-04 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah14" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh14" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M58.4,279.6 L111.2,226.8" marker-end="url(#ah14)"/><path class="e" d="M141.8,201.5 L237.5,137.6" marker-end="url(#ah14)"/><path class="e" d="M273,120 L364.1,89.6" marker-end="url(#ah14)"/><path class="e" d="M402,77 L493.1,46.6" marker-end="url(#ah14)"/><rect class="n" x="15" y="283" width="50" height="30" rx="15"/><text class="t" x="40" y="298" dy=".35em" text-anchor="middle">Sump</text><circle class="n" cx="126" cy="212" r="18"/><text class="t" x="126" y="212" dy=".35em" text-anchor="middle">Sv</text><circle class="n" cx="255" cy="126" r="18"/><text class="t" x="255" y="126" dy=".35em" text-anchor="middle">Cyl</text><circle class="n" cx="384" cy="83" r="18"/><text class="t" x="384" y="83" dy=".35em" text-anchor="middle">Dv</text><circle class="n" cx="513" cy="40" r="18"/><text class="t" x="513" y="40" dy=".35em" text-anchor="middle">Del</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Single-acting reciprocating pump. Sump = water source with strainer, Sv = suction valve, Cyl = cylinder with piston, rod, connecting rod and crank, Dv = delivery valve, Del = delivery pipe.</figcaption></figure>
Answer frame. Define the pump; draw cylinder, piston, crank, connecting rod, both pipes and valves; explain suction then delivery stroke; close with the intermittent, positive displacement nature.
Priming of a centrifugal pump
Key points.
- Priming is filling the suction pipe, casing and delivery pipe up to the delivery valve with liquid before starting the pump.
- It is needed because air is far lighter than water, so the head produced is $h=p/\rho g$ with a tiny $\rho$, giving negligible pressure rise.
- That pressure is too low to lift water from the sump, so the pump cannot draw liquid.
- Without priming the pump runs dry, overheats and delivers nothing; a foot valve keeps the water from draining back.
Asked: [8 marks] (Nov 2022, Jun 2023, Jun 2025) Explain the working principle, construction and working of the Pelton wheel with a neat sketch. Asked: [8 marks] (Jun 2022, Dec 2023) What is the reciprocating pump? Explain the working and construction of a single-acting reciprocating pump with a neat sketch. Asked: [7 marks] (Dec 2024, Jun 2025) Explain how hydraulic turbines convert fluid energy into mechanical energy; types and applications. Also: basic principle of turbine; classify turbines by type of energy, direction of flow, head and specific speed. Asked: [6 marks] (Dec 2023) What is priming? Why is it required in a centrifugal pump? Asked: [6 marks] (Jun 2023) What is a draft tube? Why is it used in a reaction turbine? Describe with a neat sketch.
Last-minute revision
- Viscosity is the internal resistance of a fluid to flow; $\mu$ is in N·s/m^2 and 1 poise = 0.1 N·s/m^2.
- Newton's law: $\tau=\mu\,du/dy$, with $\tau=F/A$.
- Shaft in sleeve: $u=\pi DN/60$, $F=\tau\pi DL$, $P=Fu$.
- Viscosity of liquids falls with temperature, of gases rises.
- Pascal's law: pressure in a static fluid is equal in all directions; $F_1/A_1=F_2/A_2$.
- Bernoulli: $p/\rho g+v^2/2g+z=$ const, for steady, inviscid, incompressible flow along a streamline.
- $Q=A_1V_1=A_2V_2$; $1$ bar $=10^5$ N/m^2; $\rho_{water}=1000$ kg/m^3.
- Pelton = impulse, tangential, high head; Francis = reaction, medium head; Kaplan = reaction, axial, low head.
- Reciprocating pump is positive displacement with suction and delivery valves and intermittent flow.
- Priming removes air, since air cannot build enough head to lift water.
- A draft tube regains kinetic energy as pressure and lets the runner sit above the tailrace.
Memory hooks
- Turbine head order: Pelton is the Peak (high), Francis in the Field (medium), Kaplan in the Kanal (low).
- Newton's law: "Tau equals Mu times Slope of velocity".
- Bernoulli terms: Pressure, Velocity, Position, three heads that add to a constant.
- Reciprocating pump: Suck then Push, and the valves open one at a time.
- Poise to SI: divide by 10.
Coverage checklist
- Fluid properties pressure, density and viscosity etc.: viscosity definition and importance (Nov 2022).
- Types of fluids: uniform, laminar and Newtonian pairs (Dec 2024).
- Newton's law of viscosity: plate numericals (Nov 2022, Jun 2022), shaft power loss (Jun 2023), statement and significance (Dec 2024), short note (Jun 2025).
- Pascal's law: statement (asked with Bernoulli, Jun 2022).
- Bernoulli's equation for incompressible fluids: pipe pressure numerical (Dec 2023, Jun 2025), derivation (Jun 2022).
- Only working principle of Hydraulic machines, pumps, turbines, Reciprocating pumps: Pelton wheel, reciprocating pump, turbine types, priming, draft tube.