GTU BE03000201 Engineering Thermodynamics Winter 2025 Solved Paper & Solutions

PMM1 (perpetual motion machine of the first kind) is an imaginary machine that would produce work continuously without taking any energy from outside, or give out more energy than it takes in.

Why it is impossible: it violates the first law of thermodynamics. For any cycle,

So the net work of a cycle can only come from the net heat supplied to it. If no heat (or other energy) is supplied, $\oint \delta Q = 0$, and so $\oint \delta W = 0$. Energy cannot be created from nothing, so a machine that gives work with no energy input can never be built.

Sign convention: heat supplied to the system is positive, and work done by the system is positive.

Given: the system rejects heat, so $Q = -120$ kJ. Work is done on the system, so $W = -80$ kJ.

The minus sign means the internal energy falls. The system loses 120 kJ as heat but gets only 80 kJ back as work, so its stored energy drops by the difference.

Diesel cycle is the ideal air standard cycle of a compression ignition (CI) engine. Heat is added at constant pressure and rejected at constant volume.

P-V and T-s diagrams

Processes

• 1 to 2: isentropic (reversible adiabatic) compression of air.
• 2 to 3: heat $Q_1$ added at constant pressure (fuel burns as it is injected).
• 3 to 4: isentropic expansion (power stroke).
• 4 to 1: heat $Q_2$ rejected at constant volume.

Derivation of efficiency

Let the compression ratio be $r = \dfrac{V_1}{V_2}$ and the cut-off ratio be $\rho = \dfrac{V_3}{V_2}$. For 1 kg of air:

Write every temperature in terms of $T_1$:

• 1 to 2 (isentropic): $T_2 = T_1\,r^{\gamma - 1}$
• 2 to 3 (constant pressure, so $V/T$ is constant): $T_3 = T_2\,\rho = T_1\,\rho\,r^{\gamma - 1}$
• 3 to 4 (isentropic, $V_4 = V_1$): $T_4 = T_3\left(\dfrac{V_3}{V_1}\right)^{\gamma - 1} = T_3\left(\dfrac{\rho}{r}\right)^{\gamma - 1} = T_1\,\rho^{\gamma}$

Put these in the efficiency:

Points to note

• The efficiency rises with the compression ratio $r$ and falls as the cut-off ratio $\rho$ grows (a larger load means a larger $\rho$).
• The factor $\dfrac{\rho^{\gamma} - 1}{\gamma(\rho - 1)}$ is always more than 1, so for the same $r$ the Diesel cycle is less efficient than the Otto cycle. In practice a diesel engine runs at a much higher $r$ (15 to 22), so it is more efficient.
• When $\rho \to 1$ (almost no fuel), the factor becomes 1 and the efficiency becomes equal to the Otto cycle value, $1 - \dfrac{1}{r^{\gamma - 1}}$.
• Only air is compressed in a diesel engine, so a high $r$ does not cause knocking. The heat of the compressed air ignites the injected fuel.

Test: divide a system in equilibrium into two equal halves. A property that stays the same in each half is intensive. A property that becomes half is extensive.

An extensive property per unit mass is intensive, for example the specific volume $v = V/m$.

Principle of increase of entropy: the entropy of an isolated system can never decrease. It increases in every real (irreversible) process and stays constant only in a reversible process.

Where it comes from: by the Clausius inequality, for any process $dS \geq \dfrac{\delta Q}{T}$. An isolated system exchanges no heat, so $\delta Q = 0$ and $dS \geq 0$.

Applied to any system: a system and its surroundings together form an isolated system (the universe), so

Interpretation:

1. It shows the direction of natural processes. A process can happen only if it makes $\Delta S_{universe} \geq 0$. Heat flows from hot to cold, never the other way by itself.
2. $\Delta S_{universe}$ measures irreversibility: the larger it is, the more work is lost.
3. An isolated system reaches equilibrium when its entropy is the maximum possible.

Given: $m = 2$ kg of water, $T_1 = 30 + 273 = 303$ K, $T_2 = 80 + 273 = 353$ K, $P = 1$ bar (constant), $C_p = 4.18$ kJ/kg·K.

Assumption: water boils at about $99.6^\circ$C at 1 bar, so it stays liquid from $30^\circ$C to $80^\circ$C, and $C_p$ is taken as constant.

The entropy rises because heat is added to the water. The value is positive, as expected for heating.

The Carnot cycle is not used in steam power plants for two main reasons:

1. Compression of wet steam is not practical. The Carnot cycle needs a water and steam mixture to be compressed isentropically to saturated liquid. A pump cannot handle such a two-phase mixture. It would need a large, costly compressor and much more work.
2. Constant temperature heating is limited to the wet region. It is possible only inside the saturation dome, so the maximum temperature stays below the critical temperature (about $374^\circ$C), which keeps the efficiency low. The turbine exhaust is also very wet, and the droplets erode the blades.

A thermodynamic system is a fixed quantity of matter or a region in space chosen for study. Everything outside it is the surroundings, and the real or imaginary surface between them is the boundary.

• In an open system, matter flows in and out, so the analysis uses the steady flow energy equation.
• In a closed system, the mass is fixed but its boundary can move, as when a piston moves.
• An isolated system has no interaction at all with its surroundings.

Gouy-Stodola theorem

Statement: the irreversibility (the work lost, or the exergy destroyed) in a process is equal to the temperature of the surroundings multiplied by the entropy generated.

Here $T_0$ is the temperature of the surroundings (atmosphere) in kelvin.

Proof

A closed system changes from state 1 to state 2. It exchanges heat $Q_0$ only with the surroundings at $T_0$ and does work $W$.

First law:

Second law (entropy balance): the surroundings lose $Q_0$ at $T_0$, so

Actual work:

Reversible work: in a reversible process $S_{gen} = 0$, so

Subtract:

How it quantifies irreversibility

1. Find the entropy generated in each process or device by an entropy balance: $S_{gen} = \Delta S_{system} + \Delta S_{surroundings}$.
2. Multiply by $T_0$ to get the work lost, in kJ. This turns an entropy number into an energy number that an engineer can compare with the work output.
3. Do this for every part of a plant (boiler, turbine, condenser). The part with the largest $T_0 S_{gen}$ wastes the most work and should be improved first.

Example: 1000 kJ of heat flows from a body at 500 K to one at 300 K, with $T_0 = 300$ K.

So 400 kJ of possible work is lost because heat flowed through a finite temperature difference. If a reversible engine had been placed between the two bodies, it could have produced this 400 kJ as work.

First law for a closed system undergoing a change of state: when a closed system goes from state 1 to state 2, the heat supplied is equal to the increase in its stored energy plus the work done by it.

When the changes in kinetic and potential energy are negligible, $E = U$, so

Points to note:

• $Q$ and $W$ depend on the path, but their difference $Q - W$ is the same for every path between the two states. So $\Delta U$ depends only on the end states, and internal energy is a property.
• For a cycle, $\Delta U = 0$, which gives $\oint \delta Q = \oint \delta W$.

Throttling process: the flow of a fluid through a restriction, such as a partly opened valve, a small orifice or a porous plug, which causes a large drop in pressure without any work being done. It is an irreversible process.

Analysis by the first law (steady flow energy equation):

For throttling:

• No work: $w = 0$ (no shaft or moving parts).
• No heat: $q = 0$ (the process is quick and the valve is small, so it is nearly adiabatic).
• No change in kinetic or potential energy: the velocities and heights before and after are nearly the same.

So the equation becomes

The enthalpy stays constant, so throttling is an isenthalpic process. For an ideal gas, $h$ depends only on temperature, so the temperature also stays the same. For a real gas or a vapour, the temperature usually falls (the Joule-Thomson effect).

Uses: the expansion valve of a refrigerator, and the throttling calorimeter for finding the dryness fraction of steam.

Given: inlet $P_1 = 1$ MPa, $T_1 = 300^\circ$C, $h_1 = 3051.2$ kJ/kg, $V_1 = 50$ m/s. Exit $P_2 = 0.2$ MPa, $V_2 = 300$ m/s. No heat loss.

For a nozzle: $q = 0$ (no heat loss), $w = 0$ (no work), and $z_1 = z_2$ (horizontal). So

The drop in enthalpy, $3051.2 - 3007.45 = 43.75$ kJ/kg, is turned into the increase in kinetic energy of the steam. The exit pressure is not needed for this result. It would be used only to find the exit temperature from the steam tables.

Mean effective pressure (MEP) is the constant average pressure which, if it acted on the piston during the whole power stroke, would produce the same net work as the actual cycle.

Significance:

• It lets us compare engines of different sizes. A higher MEP means more work from the same swept volume, so a smaller and lighter engine can give the same power.
• The power of an engine is found from it: $P = \dfrac{p_m L A N}{60}$ (with $N$ in working strokes per minute).
• The indicated MEP comes from the indicator diagram, and the brake MEP from the brake power. Their ratio is the mechanical efficiency.

Given: compression ratio $r = 8$, $\gamma = 1.4$.

The efficiency rises with the compression ratio, but in a petrol engine $r$ cannot go much above about 10 because the mixture starts to knock.

In a simple Rankine cycle, raising the boiler pressure raises the efficiency, but the steam leaving the turbine becomes too wet. In the reheat cycle, the steam is expanded in two turbines, and between them it is sent back to the boiler and heated again at a lower pressure.

T-s diagram

(The saturation dome is not drawn. a to b is boiling and b to 1 is superheating.)

• 6 to 1: heating, boiling and superheating in the boiler at high pressure.
• 1 to 2: isentropic expansion in the high pressure (HP) turbine.
• 2 to 3: reheating at constant (intermediate) pressure in the reheater, back to about the first temperature.
• 3 to 4: isentropic expansion in the low pressure (LP) turbine to the condenser pressure.
• 4 to 5: condensation. 5 to 6: pumping into the boiler.

Efficiency

How reheating helps

1. Drier steam at the turbine exit: without reheat, expansion from 1 would end deep inside the dome with a dryness fraction below about 0.85. Reheating moves point 4 to the right, so the steam is much drier. This protects the last turbine blades from erosion by water droplets. This is the main purpose of reheating.
2. Higher mean temperature of heat addition: the extra heat 2 to 3 is added at a high temperature. If the reheat pressure is chosen well (about 20 to 25 percent of the boiler pressure), the mean temperature at which heat is added goes up, so the efficiency rises a little (about 4 to 5 percent).
3. More work per kg of steam: the second expansion adds $(h_3 - h_4)$, so less steam is needed for the same power.
4. It allows a higher boiler pressure, which itself raises the efficiency, without making the exhaust too wet.

If the reheat pressure is too low, heat is added at a low temperature and the efficiency can fall.