Lenoir Cycle - Constant Pressure Heat Rejection (3-1)

Constant Pressure Heat Rejection (3-1)

The final stage (3-1) involves a constant pressure heat rejection back to the original state. From the first law of thermodynamics we find: .

From the definition of work:, we recover the following for the heat rejected during this process: .

As a result, we can determine the heat rejected as follows: ${}_3Q_1 = mc_p \left( {T_1 - T_3 } \right)$ from the definition of constant pressure specific heats for an ideal gas: $c_p = \frac{{\gamma R}} {{\gamma - 1}}$ .

The overall efficiency of the cycle is determined by the total work over the heat input, which for a Lenoir cycle equals $\eta _{th} = \frac{{{}_2W_3 + {}_3W_1 }} {{{}_1Q_2 }}$ . Note that we gain work during the expansion process but lose some during the heat rejection process.

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