David Lignell
Class 6
Net work is area under curve 3-4 minus area under curve 1-2
Question: how to best compute the net work?
Write the net work in terms of the internal energies at points 1, 2, 3, 4.
\[η = \frac{W_{net}}{Q_{in}}\] \[W_{net} = (u_3-u_4) - (u_2-u_1)\] \[W_{net} = c_v(T_3-T_4) - c_v(T_2-T_1)\] \[Q_{in} = u_3 - u_2 = c_v(T_3-T_2)\] \[η = \frac{T_3 - T_4 - T_2 + T_1}{T_3-T_2} = 1 - \frac{T_4 - T_1}{T_3 - T_2}\] Factor out \(T_1\) from the numerator and \(T_2\) from the denominator: \[η = 1 - \left(\frac{T_1}{T_2}\right)\left(\frac{T_4/T_1 - 1}{T_3/T_2 - 1}\right)\]
\[η = 1 - \left(\frac{T_1}{T_2}\right)\left(\frac{T_4/T_1 - 1}{T_3/T_2 - 1}\right)\] Now, the compression and expansion steps are adiabatic, so \(TV^{γ-1}=\mbox{const}\). Hence,
\[η = 1 - \frac{1}{r^{γ-1}}\]
How does this efficiency compare to Carnot?
Key difference with the Otto cycle is the constant pressure heat addition rather than constant volume.
Question: do you expect the efficiency of the Diesel cycle to be higher or lower than for the Otto cycle?
\[η = 1 - \frac{1}{r^{γ-1}}\left(\frac{α^γ-1}{γ(α-1)}\right),\] where \(r=V_1/V_2\) is the compression ratio and \(α=V_3/V_2=T_3/T_2\) is the cut-off ratio. \(\alpha\) can be written in terms of the high and low temperatures and the compression ratio: \(\alpha=T_3/(T_1r^{\gamma-1})\).