Thermodynamics and Power Cycles

Psychrometrics and relative humidity

Psychrometrics studies mixtures of dry air and water vapour, and its central quantity is the relative humidity \phi, defined as the ratio of the partial pressure of vapour to the saturation pressure at the same temperature: \phi = p_v/p_{sat}(T). Because vapour cannot exceed saturation at a given temperature, \phi lies between 0 and 1, and \phi = 1 means saturated air. Relative humidity must not be confused with the humidity ratio (specific humidity) \omega = 0.622\,p_v/(p - p_v), which is a mass-based quantity in kg of vapour per kg of dry air. The factor 0.622 is the ratio of the molecular masses of water and dry air. On the standard psychrometric chart, dry-bulb temperature is drawn on the horizontal axis and specific humidity on the vertical axis, which is perpendicular to it. Wet-bulb temperature and enthalpy appear as oblique lines, and relative humidity appears as a family of curves that rise toward the saturation line. When two moist-air streams mix adiabatically at steady state with constant c_p, the energy balance gives a mass-weighted mean temperature, T_m = (\dot m_1 T_1 + \dot m_2 T_2)/(\dot m_1 + \dot m_2). Mass flow rates are obtained from volume flow rates by \dot m = \rho \dot V with \rho = P/(RT).

Thermodynamic property relations and tank filling

Thermodynamic property relations follow from the fact that internal energy U is a state function, so dU is an exact differential. For dU = T\,dS - P\,dV, the partial derivatives identify T = (\partial U/\partial S)_V and P = -(\partial U/\partial V)_S. Equality of mixed second derivatives gives Maxwell relations such as (\partial T/\partial V)_S = -(\partial P/\partial S)_V. The same method applies to other work pairs. For a rubber band with dU = T\,dS + \tau\,dL, the tension is \tau = (\partial U/\partial L)_S, and the Maxwell relation is (\partial T/\partial L)_S = (\partial \tau/\partial S)_L. Each derivative must hold the correct variable fixed, and pairing the wrong variables yields an invalid relation. A rigid tank filled from a supply line at constant P_i and T_i obeys an energy balance with no work term. When the tank starts empty, this gives m_2 u_2 = m_i h_i - Q_{loss}. For an insulated, initially evacuated tank with constant properties, T_2 = k\,T_i with k = c_p/c_v, independent of the final pressure, and heat loss lowers T_2 further. The final mass then follows from m_2 = P_2 V/(RT_2). For a saturated liquid–vapour mixture of quality x, v = x\,v_g + (1-x)\,v_f, so in a rigid tank of known volume the total mass is m = V/v.

Carnot heat engine efficiency

A Carnot cycle, made of two isothermal and two isentropic processes, is the most efficient cycle that can operate between two fixed reservoirs. Its efficiency depends only on the absolute reservoir temperatures, \eta = 1 - T_L/T_H, with temperatures in kelvin, and no real engine can exceed it. The maximum work obtainable from a heat input \dot Q_H is therefore \dot W_{max} = (1 - T_L/T_H)\,\dot Q_H. A proposed set of heat and work values for an engine is admissible only if it satisfies the first law, Q_H - Q_L = W, and the second law in the form Q_L/T_L - Q_H/T_H \ge 0, which requires the net entropy change of the reservoirs to be non-negative. A reversed Carnot cycle used as a heat pump has \text{COP}_{HP} = T_H/(T_H - T_L), so the power input equals the heat delivered to the warm space divided by this COP. When a heat engine drives such a heat pump, the total heat delivered to a common warm space is the sum of the heat contributed by the engine and by the heat pump.

Reheat Rankine cycle work and efficiency

In an ideal reheat Rankine cycle, the steam is partly expanded in a high-pressure turbine, returned to the boiler to be reheated at roughly constant pressure, and then expanded in a low-pressure turbine. The net work is the sum of both turbine works minus the pump work, w_{net} = w_{HPT} + w_{LPT} - w_{pump}. The heat input is the sum of the boiler heat and the reheat heat, and the thermal efficiency is \eta = w_{net}/q_{in}. For the whole cycle, the first law gives the heat rejected in the condenser as q_{out} = q_{in} - w_{net}, so knowing the efficiency and the net work fixes both the heat input and the heat rejected. The pump work is small compared with the turbine work, but it must be subtracted in exact calculations.

Vapour compression refrigeration cycle

A vapour compression refrigeration cycle consists of evaporation, compression, condensation, and throttling. The throttling valve is isenthalpic, so the enthalpy leaving the condenser equals the enthalpy entering the evaporator. The refrigeration effect is q_L = h_1 - h_4 and the compressor work is w = h_2 - h_1, giving \text{COP}_R = q_L/w. Energy conservation over the cycle gives the heat rejected in the condenser as q_H = q_L + w. The quality entering the evaporator is x = (h_4 - h_f)/h_{fg} at the evaporator pressure, so the refrigerant is a wet mixture at that point.

Clausius inequality

The Clausius inequality states that for any thermodynamic cycle, \oint \delta Q/T \le 0, where \delta Q is the heat crossing the boundary at the boundary temperature T. Equality holds for a reversible cycle, and the strict inequality holds for an irreversible cycle. The inequality therefore holds for any cycle, reversible or irreversible, and it is not restricted to reversible processes. It is a statement about closed cycles, since the integral requires a closed path. The equality for reversible cycles is the basis for defining entropy as a property through dS = \delta Q_{rev}/T.

Polytropic process work

A polytropic process of an ideal gas follows PV^n = constant. For n \ne 1, the boundary work is W = (P_2 V_2 - P_1 V_1)/(1 - n), and the final pressure follows from P_2 = P_1 (V_1/V_2)^n. The special cases are n = 0 (isobaric), n = 1 (isothermal), n = \gamma (isentropic), and n \to \infty (isochoric). For a reversible polytropic process of an ideal gas, the heat transfer follows from the first law as Q = W(\gamma - n)/(\gamma - 1), where W is the work done by the gas. The sign of the result indicates heat added or removed, so the absolute value is taken when only the magnitude is asked.

Air-standard Otto cycle efficiency

The air-standard Otto cycle consists of two isentropic and two constant-volume processes. With compression ratio r = V_1/V_2 and specific heat ratio \gamma, its thermal efficiency is \eta = 1 - r^{1-\gamma}, which depends only on r and \gamma and not on the heat input. The compression ratio of an engine is r = (V_d + V_c)/V_c, where V_d is the displacement volume and V_c is the clearance volume. Efficiency rises as r increases, so a given efficiency fixes r directly through r^{\gamma-1} = 1/(1-\eta). The isentropic relations T_2 = T_1 r^{\gamma-1} and P_2 = P_1 r^{\gamma} give the state after compression. Because heat addition occurs at constant volume, T_3 = T_2 + q_{in}/c_v and the maximum pressure is P_3 = P_2\,T_3/T_2.

Constant-volume energy balance with stirrer work

For a rigid, insulated container, the volume is fixed so there is no boundary work, and with no heat transfer the energy balance reduces to \Delta U = W_{shaft}. A stirrer delivers shaft work into the gas, which raises its internal energy at constant volume, so m\,c_v\,\Delta T = W_{shaft}. For an ideal gas in a rigid vessel, pressure is proportional to absolute temperature, so P_2/P_1 = T_2/T_1. If the container does not take part in the interaction, its own heat capacity is excluded from the balance. Stirring work in an adiabatic rigid tank is dissipated within the fluid, so the temperature and the enthalpy h = c_p T of an ideal gas increase. Because the dissipation is irreversible, the entropy of the gas also increases even though Q = 0.

Boundary work against constant external pressure

Boundary work is W = \int P\,dV evaluated at the moving boundary. When a piston moves against a constant external pressure P_0, the work crossing the boundary is W = P_0\,\Delta V, regardless of the internal state of the gas. In a sudden, irreversible expansion, the work done by the piston on the surroundings is therefore P_0\,A\,\Delta L, and it differs from the reversible \int P\,dV of the gas. A massless piston transmits the external pressure unchanged to the surroundings. For an isobaric process of an ideal gas, the work per unit mass is w = P(v_2 - v_1) = R(T_2 - T_1). If a stop halts the piston, cooling at constant volume afterwards does no boundary work.

Rankine cycle principles

The ideal Rankine cycle consists of isentropic pumping, isobaric heating in the boiler, isentropic expansion in the turbine, and isobaric heat rejection in the condenser. The turbine exit pressure is the saturation pressure at the condenser temperature, P_{exit} = P_{sat}(T_{cond}). Efficiency rises with a higher mean temperature of heat addition, which is achieved by superheating the steam or by raising the boiler pressure, and it also rises with a lower condenser pressure, which lowers the heat rejection temperature. A statement that efficiency increases as boiler pressure decreases is therefore false. A turbine with isentropic efficiency \eta_T produces an actual enthalpy drop of \eta_T (h_{in} - h_{2s}), and its power is \dot m times this drop. The isentropic exit quality is x_s = (s_{in} - s_f)/(s_g - s_f) at the exit pressure, which fixes h_{2s}.

Isothermal processes in thermodynamic cycles

An isothermal process keeps the temperature constant during the process. For an ideal gas, internal energy and enthalpy depend only on temperature, so \Delta U = 0 and the first law gives Q = W for an isothermal process. A single isothermal expansion of this kind is an ordinary process and does not violate the first or the second law, because the Kelvin–Planck restriction applies to cyclic devices that convert heat completely into work. Standard cycles containing isothermal processes include the Carnot cycle, with two isothermal and two isentropic processes, and the ideal Stirling cycle, with two isothermal and two constant-volume processes. The Otto, Diesel, and Brayton cycles contain only adiabatic (isentropic) and constant-volume or constant-pressure processes, so no pair containing one of them qualifies.

Compressibility factor of real gas

A real gas deviates from ideal-gas behaviour, and the compressibility factor Z = Pv/(RT) measures this deviation, with Z = 1 for an ideal gas. In molar form, \bar v = Z\bar R T/P with \bar R = 8.314 kJ/(kmol·K), so the temperature is T = P\bar v/(Z\bar R) when the given specific volume is per kmol. Converting kelvin to degrees Celsius means subtracting 273.15 from the Kelvin value. A Z below 1 means the actual volume is smaller than the ideal-gas value at the same P and T, which is typical when intermolecular attraction dominates.

Triple point and phase boundary equations

The triple point is the unique state at which solid, liquid, and vapour coexist in equilibrium. On the P–T diagram, the solid–vapour (sublimation) curve and the liquid–vapour (vaporisation) curve meet at this point. Each boundary is often represented by an empirical relation of the form \ln P = A - B/T, derived from the integrated Clausius–Clapeyron equation with an ideal-gas vapour and approximately constant latent heat. At the triple point, both curves share the same P and T, so the triple point temperature is found by equating the two expressions for \ln P and solving for T.