Gas and Vapour Power and Refrigeration Cycles

Air-standard Diesel cycle

The air-standard Diesel cycle is the ideal model of a compression-ignition engine. It consists of isentropic compression (1-2), constant-pressure heat addition (2-3), isentropic expansion (3-4), and constant-volume heat rejection (4-1). The compression ratio is r = V_1/V_2, and the cut-off ratio is \rho = V_3/V_2 > 1, which is the ratio of volumes at the end and the start of the constant-pressure heating.

Under cold-air-standard assumptions with constant specific heats, the thermal efficiency is \eta = 1 - \dfrac{1}{r^{\gamma-1}}\cdot\dfrac{\rho^{\gamma}-1}{\gamma(\rho-1)}, where \gamma = c_p/c_v. The first factor comes from isentropic compression, and the second factor is greater than 1 for any \rho > 1. For the same compression ratio and \gamma, the Diesel cycle is therefore less efficient than the Otto cycle.

To evaluate the efficiency, first find T_2 = T_1 r^{\gamma-1} and T_3 = \rho T_2 from the isobaric heating, then substitute into the formula. When only the cylinder volumes are given, r and \rho follow directly from the volume ratios.

Since the expansion 3-4 ends at volume V_1, the exit temperature is T_4 = T_3(\rho/r)^{\gamma-1}. This relation gives the exit temperature from the peak temperature and the volume ratios, and the pressures are not needed.

The heat added per unit mass is q_{in} = c_p(T_3 - T_2), the heat rejected is q_{out} = c_v(T_4 - T_1), and the net work is w_{net} = q_{in} - q_{out}. A common error is to report the heat supplied as the net work.

If the constant-volume rejection is replaced by a constant-pressure rejection that returns the gas to the initial pressure, the isentropic expansion must continue to a lower pressure. A lower pressure on an isentrope means a larger volume and a lower temperature, so the new exit temperature is below the original T_4.

Rankine cycle net work

In the ideal Rankine cycle, the net work per unit mass is the turbine work minus the pump work, w_{net} = (h_3 - h_4) - (h_2 - h_1). The pump work for compressed liquid is approximately v\,\Delta P, because liquid water is nearly incompressible. When the pump work is neglected, the feedwater enthalpy equals the saturated liquid enthalpy at the condenser pressure, and the net work equals the turbine work.

The net power output is the mass flow rate times the net work per unit mass, \dot{W}_{net} = \dot{m}\,w_{net}.

The specific steam consumption (SSC) is the reciprocal of the net work per unit mass, SSC = 1/w_{net} in kg/kJ. Because the relation is a reciprocal, a percentage reduction in net work does not give the same percentage change in SSC. A 20% fall in w_{net} multiplies the SSC by 1/0.8 = 1.25, so the SSC rises by 25%.

Adding superheat to an ideal Rankine cycle leaves the pump work unchanged, because the pump work depends only on the pressure rise. The turbine work output and the cycle efficiency both increase, since the mean temperature of heat addition rises. The moisture content at the turbine exit decreases.

Lowering the condenser pressure at fixed turbine inlet conditions increases the turbine work, since the isentropic expansion goes farther. It also increases the pump work, because the pressure rise \Delta P across the pump grows. The exit quality falls, so the moisture content at the turbine exit rises, and the heat rejected in the condenser decreases.

Air-standard Otto cycle efficiency

The air-standard Otto cycle consists of two isentropic processes and two constant-volume heat-transfer processes. Its thermal efficiency depends only on the compression ratio r and the specific heat ratio \gamma, and is given by \eta = 1 - \dfrac{1}{r^{\gamma-1}}. The exponent is \gamma - 1 applied to the compression ratio, not to a pressure ratio.

Since both heat addition and rejection occur at constant volume, the efficiency can also be written as \eta = 1 - T_1/T_2, where T_1 is the temperature at the start of compression and T_2 is the temperature at the end of compression, because T_2/T_1 = r^{\gamma-1}. Under cold-air-standard assumptions, a given pair of these temperatures therefore fixes the efficiency directly.