Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S4 / 2

An ideal Rankine cycle has four main steps: liquid water is pumped reversibly to high pressure; heat is added at high pressure to produce hot vapour or supercritical working fluid; the fluid expands approximately isentropically through a turbine to produce work; and low-pressure heat rejection in a condenser returns it to liquid. For steady flow, the main work and heat transfers are differences of specific enthalpy.

A steam implementation of the Carnot cycle would require compression of a wet liquid-vapour mixture and reversible isothermal heat transfer, both difficult to achieve with practical machinery. Complete condensation before the Rankine pump makes compression work small because liquid specific volume is small. High-pressure heating and superheating can then occur in a boiler without demanding isothermal addition of all the heat. Superheating also reduces turbine-exit moisture, although the calculated exit may still be wet. The pump, turbine, condenser and boiler can operate continuously as separate flow devices. The Rankine cycle sacrifices some ideal thermal efficiency in return for a practicable steam-plant configuration; no real engine can exceed the Carnot efficiency for the same extreme reservoir temperatures.

Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet specific entropy is $s_3=6.282\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. At the condenser pressure the outlet mixture has $s_4=s_f+\chi (s_g-s_f)$, where $\chi$ is the vapour mass fraction. HenceThe outlet is approximately $71.5\%$ vapour and $28.5\%$ liquid by mass. Its specific enthalpy isThe source's common unit label omits the kelvin denominator for entropy: $s$ must be interpreted in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, while $h$ is in $\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.

For the liquid pump take the explicit incompressible-water estimate $v_f\simeq 10^{-3}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$. Its work input isso $h_2=h_1+w_p=111.998\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. With $h_3=3463$ and the preceding $h_4=1838.74$ in the same units,ThusIf pump work is neglected because no liquid density was tabulated, the corresponding approximation is $(3463-1838.74)/(3463-84)=48.1\%$. Stating the approximation explains the small difference. The temperature-extreme Carnot efficiency is about $1-293.15/873.15=66.4\%$, above both results; finite-temperature-range heat addition in the Rankine cycle does not attain that limit.

Post-combustion carbon capture and storage separates carbon dioxide from an already formed flue gas, often by absorption in a regenerable solvent. The recovered gas must be purified, dehydrated, compressed, transported and injected into a suitable geological store. Caprock integrity, trapping mechanisms, monitoring and long-term liability matter as much as capture at the chimney.

Dilute carbon dioxide mixed with nitrogen requires a large gas-processing plant. Solvent regeneration consumes heat; compression consumes power; impurities can cause solvent degradation and corrosion. Retrofitting adds equipment, cooling demand and cost, and lowers net electrical output per unit fuel. Storage capacity, injection performance, leakage risk and public acceptance must be demonstrated, while upstream fuel-cycle emissions remain.

A capture fraction is not the same as the reduction per delivered kilowatt-hour. For illustrative capture $\chi =0.9$ and a plant-efficiency decrease from $0.4$ to $0.3$, residual stack emissions per delivered unit scale as $(1-\chi )(0.4/0.3)=0.133$, corresponding to an $86.7\%$ reduction rather than $90\%$. A full life-cycle assessment includes the additional fuel and upstream emissions. These figures illustrate the accounting and are not claimed as universal plant performance.