GTU BE03000201 Engineering Thermodynamics Summer 2026 Solved Paper & Solutions

A system is in thermodynamic equilibrium when its properties do not change with time and it has no tendency to change, as long as the surroundings stay the same. Three kinds of equilibrium must exist together:

1. Thermal equilibrium: the temperature is the same everywhere in the system, so no heat flows inside it.
2. Mechanical equilibrium: there is no unbalanced force inside the system or between it and the surroundings, so the pressure is uniform.
3. Chemical equilibrium: there is no chemical reaction or transfer of matter from one part to another.

Only then does the system have definite properties.

Matter can be studied in two ways.

For example, the pressure of a gas is the average force of molecules hitting the wall (microscopic), but it is simply read on a gauge (macroscopic). Engineering thermodynamics uses the macroscopic point of view.

Thermodynamic property

A property is any measurable characteristic of a system that describes its state, such as pressure, temperature, volume, internal energy, enthalpy and entropy. Its value depends only on the state of the system, not on how the system reached that state.

Property is a point function

• A property has a definite value at each state. Its change between two states depends only on the end states, not on the path:

• So its differential is exact ($dP$, $dV$, $dT$), and over a complete cycle the net change of any property is zero: $\oint dX = 0$.
• Heat and work are not properties. They depend on the path (path functions), so we write $\delta Q$ and $\delta W$, not $dQ$ and $dW$.

Types of properties

Test: divide a system into two halves. An intensive property stays the same in each half, and an extensive property becomes half.

State, process and cycle

• State: the condition of a system, fixed by its properties. For a simple compressible system (such as a gas), two independent properties fix the state (state postulate). Example: $P$ and $T$ fix the state of a gas.
• Process: a change of the system from one state to another. A process in which the system stays nearly in equilibrium at every moment is a quasi-static process, and only such a process can be drawn as a line on a P-V diagram.
• Cycle: a series of processes that brings the system back to its first state, so all properties return to their first values.

Other properties

• Primary properties can be measured directly: $P$, $V$, $T$.
• Derived properties are found from them: $U$, $H = U + PV$, $S$.

A steady flow process is a flow process in which the conditions at every point in the control volume do not change with time. The conditions may differ from point to point, but at any fixed point they stay constant.

Conditions for steady flow:

1. The mass flow rate at the inlet equals the mass flow rate at the outlet, so no mass is stored.
2. The properties (pressure, temperature, velocity) at every point do not change with time.
3. The rates of heat and work transfer are constant.

Examples: a turbine, compressor, nozzle or boiler running at a fixed load.

Steady flow energy equation (SFEE) for 1 kg of fluid:

Here $h$ is enthalpy, $V$ is velocity, $z$ is height, $q$ is heat supplied and $w$ is work done by the fluid.

i) Boiler

• Water enters and steam leaves. No work is done: $w = 0$.
• Changes in kinetic and potential energy are negligible.

The heat supplied is equal to the rise in enthalpy of the water as it turns into steam.

ii) Compressor

• Work is done on the fluid, so $w$ is negative.
• Changes in kinetic and potential energy are negligible. Some heat $q$ may be lost to the surroundings through the cooling fins.

If the compressor is well insulated ($q = 0$), the work input is $w_{in} = h_2 - h_1$, the rise in enthalpy of the air.

Kelvin-Planck statement

It is impossible to build a device that works in a cycle, exchanges heat with a single reservoir, and produces an equal amount of work.

So every heat engine must reject some heat to a cold reservoir, and no engine can be 100 percent efficient. A machine that breaks this statement is a perpetual motion machine of the second kind (PMM2).

Clausius statement

It is impossible to build a device that works in a cycle and whose only effect is to transfer heat from a colder body to a hotter body.

So a refrigerator or a heat pump always needs work input. Heat never flows from cold to hot by itself.

Equivalence of the two statements

The two statements are equivalent: if one is broken, the other is broken too.

Case 1: Breaking Clausius breaks Kelvin-Planck

• A pump moves $Q_2$ from cold to hot with no work (Clausius broken).
• An ordinary engine takes $Q_1$ from the hot reservoir, gives $W = Q_1 - Q_2$ and rejects $Q_2$ to the cold reservoir.
• The cold reservoir gives $Q_2$ and gets $Q_2$ back, so its net change is zero.
• Together, the two devices take $Q_1 - Q_2$ from the hot reservoir alone and turn it fully into work. This breaks the Kelvin-Planck statement.

Case 2: Breaking Kelvin-Planck breaks Clausius

• An engine takes $Q_1$ from the hot reservoir and turns it fully into work $W = Q_1$ (Kelvin-Planck broken).
• This work drives an ordinary refrigerator, which takes $Q_2$ from the cold reservoir and gives $Q_1 + Q_2$ to the hot reservoir.
• The net effect is that $Q_2$ moves from the cold reservoir to the hot one with no outside work. This breaks the Clausius statement.

So the two statements say the same thing about nature.

Carnot theorem has two parts:

1. No heat engine working between two given reservoirs can be more efficient than a reversible engine working between the same two reservoirs.
2. All reversible engines working between the same two reservoirs have the same efficiency, whatever their working fluid.

So the efficiency of a reversible engine depends only on the temperatures of the source and sink:

It sets the upper limit of efficiency for every real engine and is the basis of the thermodynamic (absolute) temperature scale.

Given: $m = 1$ kg, $T_1 = 30 + 273 = 303$ K, $T_2 = 80 + 273 = 353$ K, $P = 1$ bar (constant), $C_p = 4.2$ kJ/kg-K.

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

The entropy rises because heat is added to the water.

Statement

The entropy of an isolated system can never decrease. It increases in every real (irreversible) process, and stays constant only in an ideal reversible process.

Derivation

By the Clausius inequality, for any process between two states,

The equal sign holds for a reversible process, and the greater than sign for an irreversible one. An isolated system exchanges no heat with anything, so $\delta Q = 0$ and

Universe as an isolated system

Any system together with its surroundings forms an isolated system, called the universe. So

The entropy of a system alone can decrease (for example when a hot body cools), but the entropy of its surroundings must then rise by at least as much.

Example: heat transfer between two bodies

Heat $Q$ flows from a hot body at $T_1$ to a cold body at $T_2$ ($T_1 > T_2$):

It is positive, so the process is possible. Heat flowing from cold to hot would make it negative, so that can never happen by itself.

Meaning and applications

1. Direction of processes: a process can happen only in the direction that makes $\Delta S_{universe} \geq 0$. If $\Delta S_{universe} < 0$, the process is impossible.
2. Test of reversibility: $\Delta S_{universe} = 0$ means the process is reversible, and $\Delta S_{universe} > 0$ means it is irreversible.
3. Measure of irreversibility: the larger the entropy generated, the more work is lost ($I = T_0\,\Delta S_{universe}$).
4. Equilibrium: an isolated system reaches equilibrium when its entropy reaches the maximum possible value.

So the entropy of the universe keeps increasing as real processes happen.

Air standard efficiency is the thermal efficiency of an engine cycle worked out by assuming that the working fluid is air behaving as an ideal gas, instead of the real mixture of fuel, air and burnt gases.

Assumptions:

1. Air is an ideal gas with constant specific heats.
2. All processes are reversible. Compression and expansion are isentropic.
3. Combustion is replaced by heat addition from an outside source, and exhaust by heat rejection.
4. The mass of air stays the same (closed cycle).

It is the highest efficiency a real engine on that cycle could reach. For the Otto cycle, $\eta = 1 - \dfrac{1}{r^{\gamma - 1}}$.

So the Rankine cycle is less efficient in theory but is the practical cycle for steam power plants.

A feed water heater (FWH) heats the feed water, on its way from the condenser to the boiler, using steam bled (extracted) from the turbine at one or more middle stages. A cycle with feed water heaters is a regenerative cycle.

Why it improves the cycle

• In a simple Rankine cycle, the cold feed water is heated in the boiler from a low temperature, so the mean temperature of heat addition is low.
• With an FWH, the water enters the boiler hotter, so heat is added at a higher mean temperature, and the efficiency rises.
• The bled steam gives its latent heat to the feed water instead of rejecting it in the condenser.

Types

1. Open (direct contact) FWH

The bled steam and the feed water mix directly. The water leaves as saturated liquid at the bleed pressure.

• Simple, cheap and gives good heat transfer.
• Removes dissolved air and gases (works as a deaerator).
• Needs a pump after every heater, because the heater pressure is lower than the boiler pressure.

2. Closed (surface type) FWH

The feed water flows through tubes, and the bled steam condenses outside the tubes. They do not mix.

• The feed water and bled steam can be at different pressures, so only one pump is needed.
• The condensed steam (drip) is either pumped forward into the feed line or cascaded backward to a lower pressure heater or the condenser through a trap.
• Costlier and more complex. Heat transfer is less effective, so the feed water leaves slightly cooler than the steam.

Advantages

1. Higher thermal efficiency.
2. Less thermal stress in the boiler, since hot water enters it.
3. Smaller condenser, since less steam reaches it.
4. Less moisture in the last stages of the turbine.

Disadvantages

The plant becomes more complex and costly, and the work output per kg of steam falls, because bled steam does not expand fully.

A regenerative cycle is a Rankine cycle in which the feed water is heated before it enters the boiler, using heat from the cycle itself.

• In practice, some steam is bled from the turbine at middle stages and used in feed water heaters to heat the feed water.
• The feed water enters the boiler hotter, so the mean temperature of heat addition rises.
• This raises the thermal efficiency of the cycle, although the work per kg of steam falls a little because the bled steam does not expand fully.

Brake thermal efficiency is the ratio of the brake power (the useful power available at the crankshaft) to the power supplied by the fuel.

Here $BP$ is the brake power in kW, $\dot{m}_f$ is the fuel used in kg/s, and $CV$ is the calorific value of the fuel in kJ/kg.

Meaning: it shows how much of the fuel energy becomes useful work at the shaft. The rest is lost to the exhaust, cooling water, friction and radiation.

Relation with other efficiencies:

where $\eta_{ith}$ is the indicated thermal efficiency and $\eta_{mech}$ is the mechanical efficiency.

Typical values: about 25 to 30 percent for petrol engines and 30 to 40 percent for diesel engines.

Example: an engine gives 10 kW and uses 2.5 kg/h of fuel with $CV = 42000$ kJ/kg. Fuel power $= \dfrac{2.5}{3600} \times 42000 = 29.17$ kW, so $\eta_{bth} = \dfrac{10}{29.17} = 34.3\%$.

Simple Brayton cycle

The Brayton cycle is the ideal cycle of a gas turbine. Air is compressed isentropically, heated at constant pressure in a combustion chamber, expanded isentropically in a turbine and cooled at constant pressure (exhaust). Its efficiency is

where $r_p$ is the pressure ratio. Three changes improve it.

1. Regeneration

• A regenerator (heat exchanger) uses the hot exhaust gas leaving the turbine to heat the compressed air before the combustion chamber.
• Less fuel is needed for the same turbine inlet temperature, so the efficiency rises. The net work does not change.
• It works only when the turbine exhaust is hotter than the compressor exit, which is true at low pressure ratios.
• Effectiveness: $\varepsilon = \dfrac{\text{actual temperature rise of air}}{\text{maximum possible rise}}$.

2. Intercooling

• Air is compressed in two stages (LP and HP compressors), and cooled in an intercooler between them, back to about the inlet temperature.
• Compressing cooler air needs less work, so the net work output rises.
• Alone, it lowers the efficiency, because the air leaves the HP compressor cooler, so more fuel is needed. With a regenerator, the exhaust heat makes up for this.

3. Reheating

• The gas is expanded in two turbines (HP and LP). Between them, it is reheated in a second combustion chamber, back to about the maximum temperature.
• The turbine work rises, so the net work output rises.
• Alone, it also lowers the efficiency, because the exhaust leaves hotter. With a regenerator, this hot exhaust heats the compressed air, so the efficiency rises.

Flow through the complete plant

Combined effect

With intercooling, reheating and regeneration together, the net work and the efficiency both rise a lot. With many stages of intercooling and reheating, the cycle approaches the Ericsson cycle, whose efficiency equals the Carnot efficiency.