Skip to content
CampusPrep
BT-203 · Basic Mechanical Engineering/Quick Revision Short Notes

Basic Mechanical Engineering (BT-203) - Unit 4 Short Notes

How unit 4 is examined

Basic thermodynamics (systems, the laws, the constant-volume process) and steam boilers (types, mountings, accessories, efficiency numericals, draught, steam tables). The marks sit in the laws of thermodynamics, boiler classification and working, and the boiler-trial numericals.

Thermodynamic system, properties, state, process

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Medium weight</span>

Definition. A thermodynamic system is a specified quantity of matter or region of space chosen for study; everything outside it is the surroundings, and the real or imaginary surface between them is the boundary.

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-01" viewBox="0 0 259 166" width="259" height="166" role="img" aria-label="System, boundary and surroundings. Open: mass and energy cross. Closed: only energy crosses. Isolated: nothing crosses."><style>#dsfig-u4-01 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-01 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-01 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-01 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-01 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-01 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-01 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-01 .t{fill:#16181D;font-weight:500}#dsfig-u4-01 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-01 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-01 .dot{fill:#16181D}#dsfig-u4-01 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-01 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-01 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-01 .ah{fill:#454C5A}#dsfig-u4-01 .ah.hi{fill:#2340B8}#dsfig-u4-01 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-01 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-01 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-01 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-01 .e{stroke:#B1B7C3}html.dark #dsfig-u4-01 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-01 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-01 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-01 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-01 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-01 .t{fill:#E6E8ED}html.dark #dsfig-u4-01 .t.inv{fill:#0F1115}html.dark #dsfig-u4-01 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-01 .dot{fill:#E6E8ED}html.dark #dsfig-u4-01 .ann{fill:#8FA3FF}html.dark #dsfig-u4-01 .lbl{fill:#858D9C}html.dark #dsfig-u4-01 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-01 .ah{fill:#B1B7C3}html.dark #dsfig-u4-01 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-01 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-01 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-01 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah15" 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="ahh15" 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="M59,126 L186,126"/><g class="wl"><rect x="91.7" y="117" width="68.7" height="18" rx="9"/><text class="t" x="126" y="126" dy=".35em" text-anchor="middle">boundary</text></g><circle class="n" cx="40" cy="126" r="18"/><text class="t" x="40" y="126" dy=".35em" text-anchor="middle">Sys</text><rect class="n" x="187" y="111" width="50" height="30" rx="15"/><text class="t" x="212" y="126" dy=".35em" text-anchor="middle">Surr</text><circle class="n" cx="126" cy="40" r="18"/><text class="t" x="126" y="40" dy=".35em" text-anchor="middle">Bnd</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">System, boundary and surroundings. Open: mass and energy cross. Closed: only energy crosses. Isolated: nothing crosses.</figcaption></figure>

Key points.

  1. An open system exchanges both mass and energy with the surroundings, for example a turbine, pump, nozzle or boiler with steady flow.
  2. A closed system exchanges only energy (heat and work), not mass, for example gas in a piston-cylinder with the valves shut.
  3. An isolated system exchanges neither mass nor energy, for example the universe or a perfectly insulated thermos flask.
  4. A property is a measurable characteristic of the system (pressure, volume, temperature, internal energy); it depends only on the state, not on how the state was reached, so it is a point function.
  5. Properties are intensive (independent of mass: pressure, temperature) or extensive (proportional to mass: volume, energy).
  6. State is the condition of the system fixed by its properties; a process is the change from one state to another, and a cycle is a process series returning to the initial state.
  7. Work and heat are path functions: they are transient energy in transit, not properties, and their small amounts are written $\delta W$ and $\delta Q$ (inexact), unlike $dU$ (exact).
Feature Open Closed Isolated
Mass transfer Yes No No
Energy transfer Yes Yes No
Example Turbine Piston-cylinder Thermos flask

Proof: work and heat are path functions. On a p-V diagram take a system going from state 1 to 2 by path A and by path B. Work is $W=\int p\,dV$, the area under the curve, and the two areas differ, so $W_A\ne W_B$: work depends on path. By the first law $Q=\Delta U+W$; $\Delta U$ is the same for both paths (point function) but $W$ differs, so $Q_A\ne Q_B$ and heat is a path function too. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-02" viewBox="0 0 424 295" width="424" height="295" role="img" aria-label="p-V diagram idea: states 1 and 2 joined by paths A and B enclose different areas under the curves"><style>#dsfig-u4-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-02 .t{fill:#16181D;font-weight:500}#dsfig-u4-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-02 .dot{fill:#16181D}#dsfig-u4-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-02 .ah{fill:#454C5A}#dsfig-u4-02 .ah.hi{fill:#2340B8}#dsfig-u4-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-02 .e{stroke:#B1B7C3}html.dark #dsfig-u4-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-02 .t{fill:#E6E8ED}html.dark #dsfig-u4-02 .t.inv{fill:#0F1115}html.dark #dsfig-u4-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-02 .dot{fill:#E6E8ED}html.dark #dsfig-u4-02 .ann{fill:#8FA3FF}html.dark #dsfig-u4-02 .lbl{fill:#858D9C}html.dark #dsfig-u4-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-02 .ah{fill:#B1B7C3}html.dark #dsfig-u4-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah16" 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="ahh16" 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="M51.9,240.2 L198.9,56.4" marker-end="url(#ah16)"/><path class="e" d="M57,246.5 L193.2,178.4" marker-end="url(#ah16)"/><path class="e" d="M223.9,54.8 L370.9,238.6" marker-end="url(#ah16)"/><path class="e" d="M229,177.5 L365.2,245.6" marker-end="url(#ah16)"/><circle class="n" cx="40" cy="255" r="18"/><text class="t" x="40" y="255" dy=".35em" text-anchor="middle">S1</text><circle class="n" cx="384" cy="255" r="18"/><text class="t" x="384" y="255" dy=".35em" text-anchor="middle">S2</text><circle class="n" cx="212" cy="40" r="18"/><text class="t" x="212" y="40" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="212" cy="169" r="18"/><text class="t" x="212" y="169" dy=".35em" text-anchor="middle">B</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">p-V diagram idea: states 1 and 2 joined by paths A and B enclose different areas under the curves</figcaption></figure>

<mark>A property depends only on the state of the system, whereas work and heat depend on the path followed.</mark>

Answer frame. For the definition question, open with the system-boundary-surroundings definition, draw three small boxes (open, closed, isolated) with arrows for mass and energy, then close with the comparison table. For the proof, open by defining point and path function, draw the two-path p-V diagram, compare the areas, then use the first law for heat.

Asked: [7 marks] (Jun 2022, Dec 2024) Define thermodynamic system and explain open, closed and isolated systems. Asked: [6 marks] (Jun 2023) Prove that work and heat are path functions. Pitfall: Do not call heat or work a property of the system; they are only energy crossing the boundary.

Zeroth, First and second law of thermodynamics

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Zeroth law. If two bodies A and B are each in thermal equilibrium with a third body C, then A and B are in thermal equilibrium with each other. Thermal equilibrium means equal temperature and no net heat flow when in contact.

Thermometer use. Let C be the mercury of the thermometer, A the body whose temperature is wanted and B the calibrated scale or a reference body (ice, steam point). Mercury is put in contact with A until it stops expanding, so mercury and A are in equilibrium; the reading equals that seen when mercury is in equilibrium with the reference marks, so the law lets the same reading mean the same temperature for every body. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-03" viewBox="0 0 424 252" width="424" height="252" role="img" aria-label="Zeroth law. A = body, C = mercury (thermometer), B = scale or reference body. A and B are then in equilibrium."><style>#dsfig-u4-03 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-03 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-03 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-03 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-03 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-03 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-03 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-03 .t{fill:#16181D;font-weight:500}#dsfig-u4-03 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-03 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-03 .dot{fill:#16181D}#dsfig-u4-03 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-03 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-03 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-03 .ah{fill:#454C5A}#dsfig-u4-03 .ah.hi{fill:#2340B8}#dsfig-u4-03 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-03 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-03 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-03 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-03 .e{stroke:#B1B7C3}html.dark #dsfig-u4-03 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-03 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-03 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-03 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-03 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-03 .t{fill:#E6E8ED}html.dark #dsfig-u4-03 .t.inv{fill:#0F1115}html.dark #dsfig-u4-03 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-03 .dot{fill:#E6E8ED}html.dark #dsfig-u4-03 .ann{fill:#8FA3FF}html.dark #dsfig-u4-03 .lbl{fill:#858D9C}html.dark #dsfig-u4-03 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-03 .ah{fill:#B1B7C3}html.dark #dsfig-u4-03 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-03 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-03 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-03 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah17" 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="ahh17" 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="M53.4,198.6 L198.6,53.4"/><path class="e" d="M225.4,53.4 L370.6,198.6"/><path class="e hi" d="M59,212 L365,212"/><g class="wl"><rect x="102.5" y="117" width="47.1" height="18" rx="9"/><text class="t" x="126" y="126" dy=".35em" text-anchor="middle">equil</text></g><g class="wl"><rect x="274.5" y="117" width="47.1" height="18" rx="9"/><text class="t" x="298" y="126" dy=".35em" text-anchor="middle">equil</text></g><g class="wl hi"><rect x="188.5" y="203" width="47.1" height="18" rx="9"/><text class="t" x="212" y="212" dy=".35em" text-anchor="middle">equil</text></g><circle class="n" cx="40" cy="212" r="18"/><text class="t" x="40" y="212" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="212" cy="40" r="18"/><text class="t" x="212" y="40" dy=".35em" text-anchor="middle">C</text><circle class="n" cx="384" cy="212" r="18"/><text class="t" x="384" y="212" dy=".35em" text-anchor="middle">B</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Zeroth law. A = body, C = mercury (thermometer), B = scale or reference body. A and B are then in equilibrium.</figcaption></figure>

First law. Energy can neither be created nor destroyed, only converted from one form to another; it is the law of conservation of energy applied to heat and work.

  1. For a cycle, the net heat supplied equals the net work done: $\oint\delta Q=\oint\delta W$.
  2. For a non-cyclic process in a closed system, heat supplied equals the rise in internal energy plus the work done by the system: $Q=\Delta U+W$, or $\delta Q=dU+\delta W$.
  3. Sign convention: heat added to the system and work done by the system are positive.

Derivation of the first law for a closed system. Let the system go from state 1 to 2 by path A and return by path B, then again from 1 to 2 by path A and return by path C.

  1. For each cycle, $\oint\delta Q=\oint\delta W$, so $\oint(\delta Q-\delta W)=0$.
  2. Cycle 1-A-2-B-1 gives $(\delta Q-\delta W)_A+(\delta Q-\delta W)_B=0$; cycle 1-A-2-C-1 gives $(\delta Q-\delta W)_A+(\delta Q-\delta W)_C=0$.
  3. Subtracting, $(\delta Q-\delta W)_B=(\delta Q-\delta W)_C$: the quantity is the same for every path between the same states.
  4. So $\delta Q-\delta W$ depends only on the end states. It is the change of a property, called energy $E$ (a point function): $\delta Q-\delta W=dE$.
  5. Neglecting kinetic and potential energy changes, $dE=dU$, so $\delta Q=dU+\delta W$ and $Q=\Delta U+W$. This also proves that energy is a property of the system. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-04" viewBox="0 0 424 312.2" width="424" height="312.2" role="img" aria-label="Two cycles 1-A-2-B-1 and 1-A-2-C-1 between states 1 and 2"><style>#dsfig-u4-04 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-04 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-04 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-04 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-04 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-04 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-04 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-04 .t{fill:#16181D;font-weight:500}#dsfig-u4-04 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-04 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-04 .dot{fill:#16181D}#dsfig-u4-04 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-04 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-04 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-04 .ah{fill:#454C5A}#dsfig-u4-04 .ah.hi{fill:#2340B8}#dsfig-u4-04 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-04 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-04 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-04 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-04 .e{stroke:#B1B7C3}html.dark #dsfig-u4-04 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-04 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-04 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-04 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-04 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-04 .t{fill:#E6E8ED}html.dark #dsfig-u4-04 .t.inv{fill:#0F1115}html.dark #dsfig-u4-04 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-04 .dot{fill:#E6E8ED}html.dark #dsfig-u4-04 .ann{fill:#8FA3FF}html.dark #dsfig-u4-04 .lbl{fill:#858D9C}html.dark #dsfig-u4-04 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-04 .ah{fill:#B1B7C3}html.dark #dsfig-u4-04 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-04 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-04 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-04 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah18" 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="ahh18" 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="M51.3,256.9 L199.5,56.9" marker-end="url(#ah18)"/><path class="e" d="M223.3,55.3 L371.5,255.3" marker-end="url(#ah18)"/><path class="e" d="M368.4,261.3 L229.2,163.8" marker-end="url(#ah18)"/><path class="e" d="M196.4,162.7 L57.2,260.2" marker-end="url(#ah18)"/><path class="e" d="M365.1,270.3 L232.9,257.1" marker-end="url(#ah18)"/><path class="e" d="M193.1,256.9 L60.9,270.1" marker-end="url(#ah18)"/><circle class="n" cx="40" cy="272.2" r="18"/><text class="t" x="40" y="272.2" dy=".35em" text-anchor="middle">S1</text><circle class="n" cx="384" cy="272.2" r="18"/><text class="t" x="384" y="272.2" dy=".35em" text-anchor="middle">S2</text><circle class="n" cx="212" cy="40" r="18"/><text class="t" x="212" y="40" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="212" cy="151.8" r="18"/><text class="t" x="212" y="151.8" dy=".35em" text-anchor="middle">B</text><circle class="n" cx="212" cy="255" r="18"/><text class="t" x="212" y="255" dy=".35em" text-anchor="middle">C</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Two cycles 1-A-2-B-1 and 1-A-2-C-1 between states 1 and 2</figcaption></figure>

Internal energy. Internal energy $U$ is the microscopic energy stored in a substance: kinetic energy of molecular motion plus potential energy of molecular configuration and intermolecular bonds. It is a property (point function), extensive, and for an ideal gas depends on temperature only.

Second law.

  1. Kelvin-Planck: it is impossible to construct a device operating in a cycle that produces no effect other than extracting heat from a single reservoir and doing an equivalent amount of work. Thus no heat engine has 100 percent efficiency.
  2. Clausius: it is impossible to construct a device operating in a cycle that transfers heat from a lower-temperature body to a higher-temperature body without external work. Thus a refrigerator needs a work input.

Equivalence of the two statements (show that violating one violates the other).

  • Violating Clausius violates Kelvin-Planck: let a refrigerator that breaks Clausius move $Q_2$ from cold sink to hot source with no work. Run a normal engine between the same reservoirs, taking $Q_1$ from the source and rejecting $Q_2$ to the sink. Together, the sink is unchanged, and the net effect is $Q_1-Q_2$ taken from a single source and converted fully to work. This breaks Kelvin-Planck.
  • Violating Kelvin-Planck violates Clausius: let an engine that breaks Kelvin-Planck take $Q_1$ from the source and give $W=Q_1$ as work. Use $W$ to drive a normal refrigerator taking $Q_2$ from the sink and delivering $Q_1+Q_2$ to the source. Net effect: $Q_2$ moves from sink to source with no external work. This breaks Clausius.
  • Hence the two statements are equivalent. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-05" viewBox="0 0 259 338" width="259" height="338" role="img" aria-label="Violating Kelvin-Planck (engine E, W = Q1) drives a refrigerator R: net Q2 goes cold to hot with no work"><style>#dsfig-u4-05 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-05 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-05 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-05 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-05 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-05 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-05 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-05 .t{fill:#16181D;font-weight:500}#dsfig-u4-05 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-05 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-05 .dot{fill:#16181D}#dsfig-u4-05 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-05 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-05 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-05 .ah{fill:#454C5A}#dsfig-u4-05 .ah.hi{fill:#2340B8}#dsfig-u4-05 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-05 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-05 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-05 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-05 .e{stroke:#B1B7C3}html.dark #dsfig-u4-05 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-05 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-05 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-05 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-05 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-05 .t{fill:#E6E8ED}html.dark #dsfig-u4-05 .t.inv{fill:#0F1115}html.dark #dsfig-u4-05 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-05 .dot{fill:#E6E8ED}html.dark #dsfig-u4-05 .ann{fill:#8FA3FF}html.dark #dsfig-u4-05 .lbl{fill:#858D9C}html.dark #dsfig-u4-05 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-05 .ah{fill:#B1B7C3}html.dark #dsfig-u4-05 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-05 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-05 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-05 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah19" 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="ahh19" 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="M115.5,55.8 L51.6,151.5" marker-end="url(#ah19)"/><path class="e" d="M59,169 L191,169" marker-end="url(#ah19)"/><path class="e" d="M140.4,276.4 L200.4,186.5" marker-end="url(#ah19)"/><path class="e" d="M201.5,153.2 L137.6,57.5" marker-end="url(#ah19)"/><g class="wl"><rect x="69.8" y="95.5" width="26.4" height="18" rx="9"/><text class="t" x="83" y="104.5" dy=".35em" text-anchor="middle">Q1</text></g><g class="wl"><rect x="105.6" y="160" width="40.8" height="18" rx="9"/><text class="t" x="126" y="169" dy=".35em" text-anchor="middle">W=Q1</text></g><g class="wl"><rect x="155.8" y="224.5" width="26.4" height="18" rx="9"/><text class="t" x="169" y="233.5" dy=".35em" text-anchor="middle">Q2</text></g><g class="wl"><rect x="145.5" y="95.5" width="47.1" height="18" rx="9"/><text class="t" x="169" y="104.5" dy=".35em" text-anchor="middle">Q1+Q2</text></g><circle class="n" cx="126" cy="40" r="18"/><text class="t" x="126" y="40" dy=".35em" text-anchor="middle">Hot</text><circle class="n" cx="40" cy="169" r="18"/><text class="t" x="40" y="169" dy=".35em" text-anchor="middle">E</text><circle class="n" cx="212" cy="169" r="18"/><text class="t" x="212" y="169" dy=".35em" text-anchor="middle">R</text><rect class="n" x="101" y="283" width="50" height="30" rx="15"/><text class="t" x="126" y="298" dy=".35em" text-anchor="middle">Cold</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Violating Kelvin-Planck (engine E, W = Q1) drives a refrigerator R: net Q2 goes cold to hot with no work</figcaption></figure>

==The first law says energy is conserved, $Q=\Delta U+W$; the second law says heat cannot fully become work in a cycle and cannot flow cold to hot unaided.==

Answer frame. Zeroth law: open with the statement and definition of thermal equilibrium, draw the A-B-C figure, then map A, B, C to body, scale and mercury and close with temperature measurement. First law: open with the conservation statement, write the cycle and non-cycle forms, give the two-cycle derivation with the p-V figure, define $U$, close with the sign convention. Second law: state both statements, draw the two block diagrams for equivalence, close by saying both are the same law.

Asked: [7 marks] (Nov 2022, Dec 2023) State the Zeroth law of thermodynamics. How is mercury in a thermometer able to find the temperature of a body using the Zeroth law? Discuss thermal equilibrium and state the Zeroth law. Asked: [7 marks] (Dec 2024, Jun 2025) What does the First law state and how does it relate to conservation of energy? State the first law for a closed system undergoing a cyclic and a non-cyclic process. Asked: [6 marks] (Dec 2023) State the Kelvin-Planck and Clausius statements of the second law. Explain their equivalence. Asked: [6 marks] (Jun 2023) Derive an expression for the first law applied to a closed system. Define internal energy. Asked: [7 marks] (Jun 2025) Define internal energy and prove that energy is a property of a system. Pitfall: In the equivalence proof, always show both directions, each with its own diagram.

Thermodynamic processes at constant pressure, volume, enthalpy and entropy

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Low weight</span>

Definition. A constant volume (isochoric) process is one in which $V_1=V_2$, so the boundary does not move.

Key points.

  1. Work done is zero: $W=\int p\,dV=0$ since $dV=0$.
  2. Change in internal energy for a gas of mass $m$: $\Delta U=m c_v(T_2-T_1)$.
  3. From the first law, $Q=\Delta U+W=\Delta U=m c_v(T_2-T_1)$, so all heat supplied raises internal energy; pressure rises as $p_1/T_1=p_2/T_2$.
  4. Change in enthalpy: $\Delta H=m c_p(T_2-T_1)$, with $H=U+pV$.

Asked: [7 marks] (Jun 2025) What is a constant volume process? Derive expressions for work done, change in internal energy, heat transfer and change in enthalpy.

Classification and working of boilers

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Definition. A boiler (steam generator) is a closed vessel in which water is heated by burning fuel to produce steam at pressure above atmospheric.

Classification with examples. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-06" viewBox="0 0 538 130" width="538" height="130" role="img" aria-label="tree diagram"><style>#dsfig-u4-06 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-06 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-06 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-06 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-06 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-06 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-06 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-06 .t{fill:#16181D;font-weight:500}#dsfig-u4-06 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-06 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-06 .dot{fill:#16181D}#dsfig-u4-06 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-06 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-06 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-06 .ah{fill:#454C5A}#dsfig-u4-06 .ah.hi{fill:#2340B8}#dsfig-u4-06 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-06 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-06 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-06 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-06 .e{stroke:#B1B7C3}html.dark #dsfig-u4-06 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-06 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-06 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-06 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-06 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-06 .t{fill:#E6E8ED}html.dark #dsfig-u4-06 .t.inv{fill:#0F1115}html.dark #dsfig-u4-06 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-06 .dot{fill:#E6E8ED}html.dark #dsfig-u4-06 .ann{fill:#8FA3FF}html.dark #dsfig-u4-06 .lbl{fill:#858D9C}html.dark #dsfig-u4-06 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-06 .ah{fill:#B1B7C3}html.dark #dsfig-u4-06 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-06 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-06 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-06 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah20" 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="ahh20" 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><line class="e" x1="259" y1="37" x2="71" y2="101"/><line class="e" x1="259" y1="37" x2="170" y2="101"/><line class="e" x1="259" y1="37" x2="245.5" y2="101"/><line class="e" x1="259" y1="37" x2="336.5" y2="101"/><line class="e" x1="259" y1="37" x2="447" y2="101"/><rect class="n" x="225.5" y="22" width="67" height="30" rx="8"/><text class="t" x="259" y="37" dy=".35em" text-anchor="middle">Boiler</text><rect class="n" x="14" y="86" width="114" height="30" rx="8"/><text class="t" x="71" y="101" dy=".35em" text-anchor="middle">Tube content</text><rect class="n" x="144" y="86" width="52" height="30" rx="8"/><text class="t" x="170" y="101" dy=".35em" text-anchor="middle">Axis</text><rect class="n" x="212" y="86" width="67" height="30" rx="8"/><text class="t" x="245.5" y="101" dy=".35em" text-anchor="middle">Firing</text><rect class="n" x="295" y="86" width="83" height="30" rx="8"/><text class="t" x="336.5" y="101" dy=".35em" text-anchor="middle">Pressure</text><rect class="n" x="394" y="86" width="106" height="30" rx="8"/><text class="t" x="447" y="101" dy=".35em" text-anchor="middle">Circulation</text></svg></figure>

Basis Types Examples
Content in tubes Fire tube / Water tube Cochran, Lancashire / Babcock and Wilcox
Axis of shell Horizontal, vertical, inclined Lancashire / Cochran / some water tube
Furnace position Internally fired / Externally fired Cochran, Lancashire / Babcock and Wilcox
Pressure Low (below about 80 bar) / High Cochran, Lancashire / Benson, Velox
Circulation Natural / Forced Babcock and Wilcox / Velox, Benson

Internally versus externally fired.

Point Internally fired Externally fired
Furnace Inside the boiler shell Outside the shell, in brickwork
Examples Cochran, Lancashire Babcock and Wilcox
Pressure Low High
Capacity Small Large
Space and cost Compact, cheaper Bulky, costlier
Use Small process steam, workshops Power plants

Babcock and Wilcox boiler. It is a horizontal, externally fired, natural-circulation water-tube boiler.

  1. Parts: a horizontal steam and water drum, inclined water tubes (about $15^\circ$) between an uptake header and a downtake header, a mud box at the bottom of the downtake header, furnace with grate, baffles, and a superheater.
  2. Working: hot flue gases from the grate rise over the inclined tubes, guided by fire-brick baffles in a zig-zag path, and then pass over the superheater to the chimney.
  3. Water in the hot tubes becomes lighter and rises through the uptake header to the drum, while cooler heavier water descends through the downtake header, so natural circulation occurs.
  4. Steam collects in the upper part of the drum; it goes to the superheater and then the main steam stop valve.
  5. Sediment settles in the mud box and is blown off. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-07" viewBox="0 0 606.5 338" width="606.5" height="338" role="img" aria-label="Babcock and Wilcox. Grate = furnace, Tubes = inclined water tubes, Baff = baffles, Sup = superheater, Down = downtake header and mud box"><style>#dsfig-u4-07 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-07 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-07 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-07 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-07 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-07 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-07 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-07 .t{fill:#16181D;font-weight:500}#dsfig-u4-07 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-07 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-07 .dot{fill:#16181D}#dsfig-u4-07 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-07 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-07 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-07 .ah{fill:#454C5A}#dsfig-u4-07 .ah.hi{fill:#2340B8}#dsfig-u4-07 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-07 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-07 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-07 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-07 .e{stroke:#B1B7C3}html.dark #dsfig-u4-07 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-07 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-07 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-07 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-07 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-07 .t{fill:#E6E8ED}html.dark #dsfig-u4-07 .t.inv{fill:#0F1115}html.dark #dsfig-u4-07 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-07 .dot{fill:#E6E8ED}html.dark #dsfig-u4-07 .ann{fill:#8FA3FF}html.dark #dsfig-u4-07 .lbl{fill:#858D9C}html.dark #dsfig-u4-07 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-07 .ah{fill:#B1B7C3}html.dark #dsfig-u4-07 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-07 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-07 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-07 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah21" 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="ahh21" 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="M64.5,281.6 L142.8,229.5" marker-end="url(#ah21)"/><path class="e" d="M198.5,212 L270,212" marker-end="url(#ah21)"/><path class="e" d="M319.6,197.6 L409.5,137.6" marker-end="url(#ah21)"/><path class="e" d="M442.8,115.5 L532.7,55.5" marker-end="url(#ah21)"/><path class="e" d="M147.4,54.4 L63.3,110.5" marker-end="url(#ah21)"/><path class="e" d="M61.6,140.4 L142.8,194.5" marker-end="url(#ah21)"/><path class="e" d="M169,182.5 L169,68" marker-end="url(#ah21)"/><g class="wl"><rect x="87.7" y="246" width="33.6" height="18" rx="9"/><text class="t" x="104.5" y="255" dy=".35em" text-anchor="middle">gas</text></g><g class="wl"><rect x="81" y="74" width="47.1" height="18" rx="9"/><text class="t" x="104.5" y="83" dy=".35em" text-anchor="middle">water</text></g><g class="wl"><rect x="145.5" y="117" width="47.1" height="18" rx="9"/><text class="t" x="169" y="126" dy=".35em" text-anchor="middle">steam</text></g><rect class="n" x="11.5" y="283" width="57" height="30" rx="15"/><text class="t" x="40" y="298" dy=".35em" text-anchor="middle">Grate</text><rect class="n" x="140.5" y="197" width="57" height="30" rx="15"/><text class="t" x="169" y="212" dy=".35em" text-anchor="middle">Tubes</text><rect class="n" x="273" y="197" width="50" height="30" rx="15"/><text class="t" x="298" y="212" dy=".35em" text-anchor="middle">Baff</text><circle class="n" cx="427" cy="126" r="18"/><text class="t" x="427" y="126" dy=".35em" text-anchor="middle">Sup</text><rect class="n" x="531" y="25" width="50" height="30" rx="15"/><text class="t" x="556" y="40" dy=".35em" text-anchor="middle">Chim</text><rect class="n" x="144" y="25" width="50" height="30" rx="15"/><text class="t" x="169" y="40" dy=".35em" text-anchor="middle">Drum</text><rect class="n" x="15" y="111" width="50" height="30" rx="15"/><text class="t" x="40" y="126" dy=".35em" text-anchor="middle">Down</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Babcock and Wilcox. Grate = furnace, Tubes = inclined water tubes, Baff = baffles, Sup = superheater, Down = downtake header and mud box</figcaption></figure>

Cochran boiler. It is a vertical, multi-tubular, internally fired fire-tube boiler with a hemispherical crown.

  1. Parts: cylindrical shell with hemispherical crown, firebox with grate, combustion chamber, flue pipe, horizontal smoke tubes, chimney and mountings.
  2. Working: fuel burns on the grate in the firebox; hot gases go up the flue pipe into the combustion chamber, turn, pass through the horizontal smoke tubes giving heat to the surrounding water, and leave via the smoke box and chimney.
  3. Water surrounds the firebox and tubes; steam forms above the water level in the crown space. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u4-08" viewBox="0 0 477.5 338" width="477.5" height="338" role="img" aria-label="Cochran gas path: Fbox = firebox, Comb = combustion chamber, Tubes = smoke tubes, Smk = smoke box"><style>#dsfig-u4-08 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u4-08 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u4-08 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u4-08 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u4-08 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u4-08 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u4-08 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u4-08 .t{fill:#16181D;font-weight:500}#dsfig-u4-08 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u4-08 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u4-08 .dot{fill:#16181D}#dsfig-u4-08 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u4-08 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u4-08 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u4-08 .ah{fill:#454C5A}#dsfig-u4-08 .ah.hi{fill:#2340B8}#dsfig-u4-08 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u4-08 .wl .t{font-size:12px;font-weight:700}#dsfig-u4-08 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u4-08 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u4-08 .e{stroke:#B1B7C3}html.dark #dsfig-u4-08 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u4-08 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u4-08 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u4-08 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u4-08 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u4-08 .t{fill:#E6E8ED}html.dark #dsfig-u4-08 .t.inv{fill:#0F1115}html.dark #dsfig-u4-08 .kd{stroke:#E6E8ED}html.dark #dsfig-u4-08 .dot{fill:#E6E8ED}html.dark #dsfig-u4-08 .ann{fill:#8FA3FF}html.dark #dsfig-u4-08 .lbl{fill:#858D9C}html.dark #dsfig-u4-08 .ptr{fill:#8FA3FF}html.dark #dsfig-u4-08 .ah{fill:#B1B7C3}html.dark #dsfig-u4-08 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u4-08 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u4-08 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u4-08 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah22" 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="ahh22" 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="M40,268.5 L40,240" marker-end="url(#ah22)"/><path class="e" d="M61.6,197.6 L145.7,141.5" marker-end="url(#ah22)"/><path class="e" d="M195,126 L266.5,126" marker-end="url(#ah22)"/><path class="e" d="M327.5,126 L406,126" marker-end="url(#ah22)"/><path class="e" d="M427,107 L427,68" marker-end="url(#ah22)"/><rect class="n" x="11.5" y="283" width="57" height="30" rx="15"/><text class="t" x="40" y="298" dy=".35em" text-anchor="middle">Grate</text><rect class="n" x="15" y="197" width="50" height="30" rx="15"/><text class="t" x="40" y="212" dy=".35em" text-anchor="middle">Fbox</text><rect class="n" x="144" y="111" width="50" height="30" rx="15"/><text class="t" x="169" y="126" dy=".35em" text-anchor="middle">Comb</text><rect class="n" x="269.5" y="111" width="57" height="30" rx="15"/><text class="t" x="298" y="126" dy=".35em" text-anchor="middle">Tubes</text><circle class="n" cx="427" cy="126" r="18"/><text class="t" x="427" y="126" dy=".35em" text-anchor="middle">Smk</text><rect class="n" x="402" y="25" width="50" height="30" rx="15"/><text class="t" x="427" y="40" dy=".35em" text-anchor="middle">Chim</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">Cochran gas path: Fbox = firebox, Comb = combustion chamber, Tubes = smoke tubes, Smk = smoke box</figcaption></figure>

Fusible plug. It is a safety mounting fixed on the crown plate of the firebox, with a gunmetal or bronze body (hexagonal nut) holding a central plug of gunmetal filled with a low-melting fusible alloy (lead-tin). Normally the water covers it and keeps it cool. If the water level falls dangerously, the alloy melts, the plug drops out, and steam and water rush into the furnace, putting out the fire and warning the operator.

<mark>A boiler is classified by tube content, axis, furnace position, pressure and circulation, with the Babcock and Wilcox as the standard water-tube example and the Cochran as the standard fire-tube example.</mark>

Answer frame. Classification: open by defining a boiler, list the five bases in a table with one example each, close with fire tube versus water tube. Working questions: open with the type, draw the labelled sketch, describe parts, then gas and water paths, close with the uses; for Cochran add the fusible plug sketch and its melting action.

Asked: [7 marks] (Jun 2022) Discuss how steam generators are classified. Give an example of each classification. Asked: [7 marks] (Nov 2022) Differences between externally fired and internally fired steam generators, and the application of each. Asked: [8 marks] (Dec 2023) Explain the construction and working of the Babcock and Wilcox boiler with a neat sketch. Asked: [8 marks] (Jun 2023) Explain the working of the Cochran boiler and fusible plug with neat sketches. Pitfall: Do not label the Cochran as a water-tube boiler; it is a fire-tube boiler with hot gas inside the tubes.

Mountings and accessories of boilers

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Low weight</span>

Definition. Mountings are the fittings on the boiler that are essential for its safe working and control; accessories are the auxiliary units installed to increase efficiency.

Point Mountings Accessories
Purpose Safety and control Efficiency and economy
Nature Fitted on the shell, compulsory by law Separate units, optional
Examples Safety valve, water level indicator, pressure gauge, fusible plug, stop valve, feed check valve, blow-off cock, manhole Economiser, air preheater, superheater, feed pump, injector

Key points.

  1. The economiser heats feed water with flue gases, so less fuel is needed and efficiency rises.
  2. The air preheater warms combustion air with waste flue gas, improving combustion and efficiency.
  3. The superheater raises steam temperature above saturation, giving dry steam and better turbine performance.
  4. The feed pump delivers water into the boiler against its pressure, keeping the water level safe and reliable.

Asked: [7 marks] (Dec 2024) Differentiate between boiler mountings and accessories. Discuss the role of accessories in enhancing efficiency and safety.

Efficiency and performance analysis

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Medium weight</span>

Definition. Boiler performance is measured by evaporation, and efficiency is the ratio of heat absorbed by the water to form steam to the heat supplied by the fuel.

Formulas.

  • Actual evaporation per kg fuel: $m_a=\dfrac{\text{steam generated}}{\text{fuel burnt}}$.
  • Equivalent evaporation (from and at $100^\circ$C): $m_e=\dfrac{m_a(h-h_f)}{539}$ in kcal, or $\dfrac{m_a(h-h_f)}{2257}$ in kJ.
  • Boiler efficiency: $\eta=\dfrac{m_a(h-h_f)}{C}\times100$, where $C$ is the calorific value; also $\eta=\dfrac{m_e\times539}{C}$.
  • Wet steam: $h=h_f+x\,h_{fg}$; dry saturated: $h=h_g$; feed water enthalpy $\approx T$ kcal/kg.

Example 1 (Jun 2022). Feed $20^\circ$C, $p=10$ kgf/cm$^2$, $x=0.9$, coal 250 kg/h, water 2260 kg/h, water in the boiler fell by 240 kg; $h_f=181.3$, $h_{fg}=482$ kcal/kg.

Step Working
Steam per hour $2260+240=2500$ kg
$m_a$ $2500/250=10$
$h$ $181.3+0.9\times482=615.1$ kcal/kg
$h_{f0}$ $20$ kcal/kg
$m_e$ $10\times(615.1-20)/539$

Actual evaporation = 10 kg steam/kg coal; equivalent evaporation = 11.04 kg/kg coal.

Example 2 (Nov 2022). Coal 1630 kg, water 13000 kg in 24 h, $p=7$ kgf/cm$^2$ dry saturated, coal 3 percent moisture, 4 percent ash, $C=7200$, feed $35^\circ$C. From tables (approximate) $h_g\approx660$ kcal/kg at 7 kgf/cm$^2$, $h_f=35$.

Step Working
$m_a$ $13000/1630=7.975$
Heat absorbed $7.975\times(660-35)=4984$ kcal/kg coal
$\eta$ $4984/7200=0.692$
$m_e$ $4984/539=9.25$
Dry coal $9.25/0.97=9.53$
Combustibles $9.25/(1-0.03-0.04)=9.25/0.93=9.94$

Efficiency = 69.2 percent; $m_e$ per kg dry coal = 9.53 kg; per kg combustibles = 9.94 kg. (Values depend on the table used for $h_g$.)

<mark>Boiler efficiency is heat absorbed by the steam per kg of fuel, $m_a(h-h_f)$, divided by the calorific value of the fuel.</mark>

Answer frame. Open with the definitions of actual and equivalent evaporation, list Given, write the formulas, evaluate step by step with the table values, and close with the boxed answers and units.

Asked: [7 marks] (Jun 2022) Boiler test: feed $20^\circ$C, 10 kgf/cm$^2$, dryness 0.9, coal 250 kg/h, water 2260 kg/h, water level dropped 240 kg. Find actual and equivalent evaporation per kg coal. Asked: [7 marks] (Nov 2022) Trial: 1630 kg coal in 24 h, 13000 kg water evaporated at 7 kgf/cm$^2$ dry saturated; coal 3 percent moisture, 4 percent ash, CV 7200 kcal/kg; feed $35^\circ$C. Find boiler efficiency, equivalent evaporation per kg dry coal and per kg combustibles. Pitfall: Add the 240 kg fall in boiler water to the water supplied to get the steam made.

Natural and artificial draught

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>

Definition. Draught is the small pressure difference that causes the flue gases to flow through the furnace, tubes and chimney.

Key points.

  1. Natural draught is produced by the chimney: hot flue gases are lighter than outside air, so the density difference sets up flow; the draught is $h=H(\rho_a-\rho_g)g$ for chimney height $H$, and the cost is nil but it depends on the weather.
  2. Artificial draught uses a fan or steam jet; it is forced draught (fan before the furnace) or induced draught (fan after, near the chimney), and balanced draught uses both.
  3. Artificial draught gives a controllable, larger draught with a shorter chimney and burns poor fuel, but it needs power.

Steam properties, use of steam tables

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>

Definition. Steam properties (temperature, pressure, specific volume, enthalpy, entropy) are read from steam tables at given pressure or temperature; dryness fraction $x$ is the mass of dry vapour per kg of wet steam.

Key points.

  1. Wet steam has $x<1$: $h=h_f+x\,h_{fg}$ and $v=x\,v_g$ (approximately).
  2. Dry saturated steam has $x=1$: $h=h_g=h_f+h_{fg}$.
  3. Superheated steam is at temperature above saturation: $h=h_g+c_{ps}(T_{sup}-T_{sat})$.
  4. In the tables, look up the pressure to get $T_{sat}$, $h_f$, $h_{fg}$, $h_g$, then apply $x$.

Last-minute revision

  • System types: open (mass + energy cross), closed (energy only), isolated (neither).
  • Work and heat are path functions; properties are point functions.
  • Zeroth law: A and B each in equilibrium with C, so A and B are in equilibrium; it gives temperature.
  • First law: $Q=\Delta U+W$; for a cycle $\oint\delta Q=\oint\delta W$.
  • Kelvin-Planck: no engine with 100 percent efficiency; Clausius: refrigerator needs work.
  • Constant volume: $W=0$, $Q=\Delta U=mc_v\Delta T$, $\Delta H=mc_p\Delta T$.
  • Babcock and Wilcox: horizontal, externally fired, water-tube, natural circulation, tubes inclined at $15^\circ$.
  • Cochran: vertical, internally fired, fire-tube; fusible plug sits on the crown.
  • Mountings are for safety; accessories are for efficiency.
  • $m_e=m_a(h-h_f)/539$; $\eta=m_a(h-h_f)/C$.
  • Wet steam $h=h_f+xh_{fg}$.
  • Boiler trial: steam = water supplied + fall in boiler water.

Memory hooks

  • Mountings = must have (safety); accessories = add-ons (efficiency).
  • Zeroth law: A = C and B = C means A = B, like a thermometer.
  • Fire tube = fire in tubes (Cochran, Lancashire); water tube = water in tubes (Babcock, Benson).
  • Kelvin-Planck: Kelvin = Kills 100 percent engines; Clausius = Cold to hot needs Cash (work).
  • Isochoric: I sit still, no work.

Coverage checklist

  • Thermodynamic system, properties, state, process: system definition and types (Jun 2022, Dec 2024); work and heat are path functions (Jun 2023).
  • Zeroth, First and second law of thermodynamics: zeroth law and thermometer; first law statement, derivation, internal energy and energy as property; Kelvin-Planck, Clausius and equivalence.
  • thermodynamic processes at constant pressure, volume, enthalpy & entropy: constant volume process (Jun 2025).
  • Classification and working of boilers: classification, internal versus external firing, Babcock and Wilcox, Cochran and fusible plug.
  • mountings and accessories of boilers: mountings versus accessories (Dec 2024).
  • Efficiency and performance analysis: two boiler trial numericals (Jun 2022, Nov 2022).
  • natural and artificial draught: unasked, definition and types.
  • steam properties, use of steam tables: unasked, dryness fraction and enthalpy relations.
Go to where you left off?

Quick Add to Notes

Save questions, your own notes and screenshots into notes filed by unit. It takes a free account.

Create free account

Have an account? Log in