Thermal engines efficiency and energy losses

A heat engine receives heat $Q_h$ from a hot reservoir, performs useful work $W$, and rejects heat $Q_c$ to a cold reservoir. By the first law of thermodynamics, $W = Q_h - Q_c$, so the efficiency is $\eta = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}.$ Because rejected heat is always positive, $\eta < 1$: it is impossible to convert heat completely into work in a cyclic process. The ideal limit is set by the Carnot cycle: $\eta_C = 1 - \frac{T_c}{T_h},$ where temperatures are expressed in kelvins. In real installations, additional losses appear: friction, heat conduction, incomplete combustion of fuel, and heating of exhaust gases, so the actual efficiency is noticeably lower than $\eta_C$. To account for losses, it is useful to compare the real efficiency with the Carnot efficiency and separately track where the rejected heat goes.

In one cycle, an engine receives $Q_h = 800\ \text{J}$ from a hot reservoir and performs useful work $W = 240\ \text{J}$. Find the engine efficiency and the heat $Q_c$ rejected to the cold reservoir.

An ideal Carnot engine operates between $T_h = 500\ \text{K}$ and $T_c = 300\ \text{K}$. A real engine on the same reservoirs receives $Q_h = 800\ \text{J}$ and performs $W = 200\ \text{J}$ per cycle. Compare its efficiency with the Carnot limit.

An automobile engine burns fuel, releasing $2.0\ \text{MJ}$ of heat per minute; the useful mechanical work during the same time is $500\ \text{kJ}$. Find the efficiency, the useful power, and the heat rejected to the environment per minute. Estimate how far it is from the Carnot limit at $T_h = 900\ \text{K}$ and $T_c = 350\ \text{K}$, and list the main loss channels.

Carry out a safe investigation using a computer simulation or a spreadsheet. Set temperatures $T_h$ and $T_c$ in kelvins, and calculate the Carnot limit $\eta_C = 1 - T_c/T_h$. Then set $Q_h$ and choose $W$ so that the real efficiency $\eta = W/Q_h$ is below $\eta_C$. Plot $\eta$ versus $T_h/T_c$ or fill a table for three temperature pairs. Note which additional losses would have to be included in a real installation. Do not use open flames, pressurized steam, or homemade boilers: observation and calculation are performed only virtually.

Show that the supplied heat is divided into useful work, heat rejected to the cold reservoir, and additional losses in a real installation.

A Sankey-style energy flow diagram: the flow $Q_h$ enters from the left and splits into the useful work flow $W$ and the rejected heat flow $Q_c$. Loss arrows branch downward from the $W$ branch (friction, aerodynamic drag, pumping losses), and exhaust and cooling arrows branch from the $Q_c$ branch. Nearby labels: $\eta = W/Q_h$ and $\eta_C = 1 - T_c/T_h$.

A car drives $100$ km and uses $8$ L of petrol. Burning $1$ L of petrol releases about $33$ MJ, so the fuel supplies $Q_{\text{in}} = 8 \times 33 = 264$ MJ. At an efficiency of about $25\,\%$ the useful work is $W = 0{,}25 \times 264 = 66$ MJ, and the remaining $198$ MJ leave with the exhaust gases at roughly $700$ °C and through the radiator coolant at $90$ °C. At $90$ km/h the trip lasts about $1{,}1$ h, so the engine dumps heat into the surroundings at a rate of roughly $50$ kW.

The definition of efficiency $\eta = \dfrac{W}{Q_{\text{in}}}$ and the loss balance $Q_{\text{out}} = Q_{\text{in}} - W$ are at work here: part of the fuel's internal energy must be given to a colder environment, and that given-away energy is exactly the loss.

In a power plant steam enters the turbine at $540$ °C ($813$ K) and condenses in the condenser at $30$ °C ($303$ K). The limiting Carnot efficiency is $1 - \dfrac{303}{813} = 0{,}63$, that is $63\,\%$, yet a real plant delivers only about $40\,\%$ to the grid. To supply $1$ kWh $= 3{,}6$ MJ of electrical energy at an efficiency of $0{,}40$, the plant must burn $3{,}6 / 0{,}40 = 9{,}0$ MJ of heat, which is about $375$ g of coal with a heat of combustion of $24$ MJ/kg.

The scenario tests the Carnot formula $\eta_{\text{Carnot}} = 1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}}$ (temperatures in kelvin) and shows the gap between the ideal cycle and a real plant, where turbine friction, incomplete combustion and heat leaks through pipe insulation are added.

A cyclist produces $300$ W of useful power at the pedals while consuming chemical energy from food at about $1200$ W; the efficiency of this "engine" is $\eta = 300 / 1200 = 0{,}25$, that is $25\,\%$. During a $30$ min ride ($1800$ s) the useful work is $300 \times 1800 = 540$ kJ, while the total energy used is $1200 \times 1800 = 2160$ kJ, so $1620$ kJ are released as heat from a body at $37$ °C into air at $20$ °C — part of it leaves with sweat and breath.

The example uses energy conservation for a thermal engine: $Q_{\text{in}} = W + Q_{\text{out}}$ and $\eta = W/Q_{\text{in}}$, together with the second-law requirement that part of the heat be dumped into a cold environment, otherwise no work is possible.