Thermodynamics and Marine Engines

Internal combustion engine cycles

The air-standard Otto cycle models a spark-ignition engine. It consists of isentropic compression, constant-volume heat addition, isentropic expansion and constant-volume heat rejection. Its thermal efficiency depends only on the compression ratio r and the ratio of specific heats \gamma, as \eta_{Otto} = 1 - \dfrac{1}{r^{\gamma-1}}, so a higher compression ratio gives a higher efficiency. The same efficiency can be written as \eta = 1 - (T_4 - T_1)/(T_3 - T_2), because q_{in} = c_v(T_3 - T_2) and q_{out} = c_v(T_4 - T_1). The two forms are equivalent once Pv = RT and R = c_v(\gamma - 1) are used.

The net specific work of a cycle equals the mean effective pressure multiplied by the swept specific volume, w_{net} = \text{mep}\,(v_1 - v_2). Here v_1 = RT_1/p_1 and v_2 = v_1/r. When the mean effective pressure is given directly, the specific heat ratio is not needed for this calculation.

• compression ignition — ignition in a diesel engine occurs because the air is compressed to a temperature above the self-ignition temperature of the fuel. No spark or external heat is needed. Example: a fuel jet injected into air at about 800 K ignites without a spark.
• supercharging — supplying intake air above ambient pressure with a compressor, so that more air mass is trapped per cycle and the power output rises.

The air-standard Diesel cycle adds heat at constant pressure up to the cut-off point. Isentropic compression gives T_2 = T_1 r^{\gamma-1}. Heat addition at constant pressure then gives T_3 = T_2 + q_{23}/c_p, and T_3 is the peak temperature of the cycle.

The Diesel efficiency is \eta_{Diesel} = 1 - \dfrac{1}{r^{\gamma-1}}\cdot\dfrac{r_c^{\gamma}-1}{\gamma(r_c-1)}, where r_c is the cut-off ratio. The power output is the efficiency multiplied by the heat input rate, \dot{Q}_{in} = \dot{m}\,c_p\,\Delta T.

The air-standard dual cycle adds heat in two stages, first at constant volume and then at constant pressure. Isentropic compression and expansion, and constant-volume heat rejection, complete the cycle.

For equal compression ratios the Otto cycle has a higher efficiency than the Diesel cycle, because the cut-off factor in the Diesel expression is greater than unity. In the usual comparison with equal heat input, the Otto peak pressure, reached at constant volume, exceeds the Diesel peak pressure, which equals the compression-end pressure.

Vapour compression refrigeration cycle

The simple vapour compression cycle consists of isentropic compression 1-2, isobaric condensation 2-3, isenthalpic throttling 3-4 and isobaric evaporation 4-1. On a T-s diagram the throttling line is not vertical, because entropy increases during throttling. On a p-h diagram the throttling line is vertical, since the enthalpy stays constant, while the isobaric processes are horizontal lines.

The coefficient of performance is \mathrm{COP} = q_L/w = (h_1 - h_4)/(h_2 - h_1), where h_4 = h_3 because throttling is isenthalpic. Example: if h_1 - h_4 = 120 kJ/kg and h_2 - h_1 = 30 kJ/kg, then \mathrm{COP} = 4.

In an ideal cycle with saturated states, the refrigeration effect is q_L = h_1 - h_4, where h_1 = h_g at the evaporator temperature and h_4 = h_f at the condenser temperature. The compressor work is w = q_L/\mathrm{COP}. Adding this work to h_1 gives the enthalpy at the compressor discharge, which is the condenser inlet enthalpy.

• intercooling in a two-stage cycle — the refrigerant is compressed to an intermediate pressure, cooled at constant pressure, and then compressed to the condenser pressure. The p-h diagram therefore shows two compression segments joined by an isobaric cooling step at the intermediate pressure.
• flash intercooling — the condensate is throttled to an intermediate pressure in a flash chamber. The saturated liquid is throttled again to evaporator pressure, and the flash vapour is mixed with the first-stage discharge. The p-h diagram shows two pressure levels with this flash separation.

Carnot cycle and Clausius inequality

The Clausius inequality states that \oint \delta Q/T \le 0 for any cycle, where the equality holds only for reversible cycles. For a reversible cycle the integral equals \oint dS, and entropy is a property, so the integral over any closed reversible cycle is zero.

A Carnot engine operating between reservoirs at T_H and T_L has the efficiency \eta = 1 - T_L/T_H, with temperatures in kelvin. The heat ratio equals the temperature ratio, Q_L/Q_H = T_L/T_H. Example: if Q_H = 100 kJ and T_L/T_H = 0.4, then Q_L = 40 kJ.

A new temperature scale defined through the Carnot relation T_L/T_H = Q_L/Q_H gives \eta = 1 - T_L/T_H on that scale. If the steam point and the ice point are separated by a known number of units, the steam point is found from \Delta T = T_H - T_L = \eta\,T_H. Hence the hot-reservoir value is T_H = \Delta T/\eta.

For a closed system taken through a reversible process 1-P-2 followed by an irreversible return 2-Q-1, the system's net entropy change over the cycle is zero. The surroundings gain entropy in the irreversible return, so their net change is positive. The total entropy generated is therefore positive.

A reversible heat engine reversed into a heat pump operates between the same reservoirs with the same heat magnitudes. For a reversible engine, Q_H = Q_L/(1-\eta), and the work input for the reversed device is W = Q_H - Q_L. The heat rejected to the high-temperature reservoir has the same magnitude as before.

Rankine cycle reheat and regeneration

Reheating sends partly expanded steam back to the boiler to be superheated before it expands further. This raises the enthalpy at the turbine exit and therefore the dryness fraction of the exhaust steam. Reheating also reduces the moisture that can erode the last turbine stages.

In a regenerative steam cycle, the condensate is raised to boiler pressure by the feed pump before it is heated. Feed water heaters that use waste heat are placed between the pump discharge and the boiler, where the water is already at boiler pressure and can be heated without vaporisation. Example: a heater placed after the boiler could not preheat feed water, because the water has already left the boiler.

The thermal efficiency of a Rankine cycle, \eta = 1 - q_{out}/q_{in}, improves when the mean temperature of heat addition increases or the mean temperature of heat rejection decreases. Raising the boiler pressure and superheating the steam both raise the mean temperature of heat addition. Lowering the condenser pressure reduces the temperature of heat rejection. An intercooler is a device for multistage gas compression and does not apply to a steam Rankine cycle.

Psychrometric processes

On a psychrometric chart with dry-bulb temperature on the horizontal axis and specific humidity on the vertical axis, a process moving to the right indicates heating, and a process moving upward indicates humidification. A process moving downward indicates dehumidification, and a process moving to the left indicates cooling. A line that moves both right and up therefore represents heating with humidification.

When air is cooled at constant dry-bulb temperature, its sensible heat does not change, and the heat removed is the latent heat of the moisture condensed. The heat transfer rate is \dot{Q} = \dot{m}_a\,(\omega_1 - \omega_2)\,h_{fg}, where \omega is the specific humidity and \dot{m}_a is the dry-air mass flow rate. The wet-bulb temperatures are not needed in this calculation.

The dew point is the temperature at which vapour in the air begins to condense at constant humidity ratio. When the dew point equals the dry-bulb temperature, the vapour pressure equals the saturation pressure at that temperature. The air is then saturated, and the relative humidity is \phi = p_v/p_g = 1, or 100 %.

Thermodynamic properties of a system

A property such as pressure, temperature or density is a point function, whose value depends only on the equilibrium state of the system and not on the path taken to reach it. The differential of a point function is an exact differential, so its change between two states is the same for every path. Heat and work are path functions, and their differentials are inexact.

For a reversible process, the entropy change of a closed system is dS = \delta Q_{rev}/T. In a reversible adiabatic process \delta Q = 0, so the entropy change is zero. For a reversible isothermal process, \Delta S = Q/T, which is positive when heat is added and negative when heat is rejected.

Marine nuclear propulsion boilers

A nuclear propulsion plant generates steam by reactor heat, not by fuel firing. The boiling water reactor boiler is the reactor-based steam supply arrangement used in nuclear ship propulsion. Water tube, Cochran and double evaporation boilers are conventional steam generators and are not the boiler types of the nuclear plant.

In a steam-to-steam generator, steam from the main boiler heats the secondary feed water through a closed heat-transfer surface. The primary feed never mixes with the contaminated secondary water, so the primary system is protected from contamination. A double evaporation boiler uses the same closed-surface principle, in which the steam of the first evaporator heats the second evaporator without direct mixing.

Degree of reaction of turbine stage

The degree of reaction of a turbine stage is the fraction of the total stage enthalpy drop that occurs in the moving blades, R = \dfrac{\Delta h_{m}}{\Delta h_{m} + \Delta h_{f}}, where \Delta h_m and \Delta h_f are the drops across the moving and fixed blades. The fraction in the fixed blades is 1 - R = \dfrac{\Delta h_f}{\Delta h_m + \Delta h_f}, which is not the degree of reaction. A ratio of the two blade drops is also not the degree of reaction.

Stoichiometric air-fuel ratio of combustion

The stoichiometric air-fuel ratio is found from the balanced combustion equation. Air is taken as one mole of oxygen with 3.76 moles of nitrogen per mole of oxygen. For methane, the reaction is \mathrm{CH_4} + 2(\mathrm{O_2} + 3.76\,\mathrm{N_2}) \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O} + 7.52\,\mathrm{N_2}. The mass ratio is found by multiplying the moles of air by its molar mass, which is about 28.97 kg/kmol, and dividing by the mass of fuel.

First law for closed systems

For a closed system, the first law is \Delta U = Q - W, where W is the work done by the system. In a piston-cylinder arrangement with a free piston, the gas pressure equals the atmospheric pressure plus the piston weight per unit area, P = P_{atm} + mg/A. Work done by the gas during a constant-pressure lift is W = P\,\Delta V = P\,A\,\Delta L.

Steady flow energy equation for nozzle

For a steady adiabatic nozzle with no work interaction and negligible potential energy change, the energy equation reduces to h_1 + V_1^2/2 = h_2 + V_2^2/2. For a perfect gas with constant c_p, this gives T_2 = T_1 - (V_2^2 - V_1^2)/(2c_p). The velocity at each section follows from continuity, V = \dot{V}/A, where \dot{V} is the volume flow rate.

Constant pressure heating of steam

When saturated liquid is heated at constant pressure, it first evaporates at the saturation temperature T_{sat}. The entropy of vaporisation is s_{fg} = h_{fg}/T_{sat}. The vapour is then superheated at constant pressure, and its entropy change is \Delta s = c_p \ln(T_2/T_1) with temperatures in kelvin. The total entropy change for mass m is the sum of these two parts.

Marine diesel engine cooling systems

Large marine diesel engines reject a very high heat load and therefore use a closed fresh-water circuit in which a pump forces the coolant through the engine. The heat is removed in a sea-water cooler. Thermosyphon circulation depends on density differences and gives too little flow for such a load. Evaporative cooling loses coolant, and air circulation has far too little heat transfer capacity for large engines.

Steady flow energy equation for combustor

For a steady-flow combustor with no work interaction and negligible kinetic and potential energy changes, the energy balance gives \dot{Q} = \dot{m}\,c_p\,(T_2 - T_1). The specific heat at constant pressure follows from c_p = c_v + R. Example: if c_v = 0.718 kJ/kg K and R = 0.287 kJ/kg K, then c_p = 1.005 kJ/kg K.

Marine fresh water generation

Ships carry only a limited stock of fresh water, so seawater must be converted into distilled water on board. The flash evaporator is a common shipboard distiller for this purpose. Seawater is preheated by a low-grade heat source such as engine jacket water or exhaust gas, and it is then admitted into a chamber held at a pressure below the saturation pressure corresponding to its temperature. Part of the water flashes into vapour almost instantly, and this vapour is condensed on cooling tubes to give distilled water with very low salt content. The unevaporated brine is discharged overboard.