Diesel engines

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 4.20.

The efficiency of an internal combustion engine is

\displaystyle e=\frac{W}{Q_{h}}=1-\left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1} \ \ \ \ \ (1)

 

The ratio {V_{1}/V_{2}} is known as the compression ratio (remember {V_{1}>V_{2}} so this is always greater than 1), and is the ratio of the minimum to the maximum volume within a cylinder in the engine. It might seem that we could increase the efficiency as much as we like (well, up to the limit imposed by the second law of thermodynamics, anyway) simply by increasing the compression ratio. However, if the air/petrol mix is compressed beyond a certain point, it spontaneously ignites resulting in the ignition occurring before the volume has reached its minimum value of {V_{2}}, so there is a practical limit due to the chemistry.

A Diesel engine avoids this problem by compressing only the air and then injecting the fuel vapour when the desired pressure is reached, at which point the mixture spontaneously ignites. The Diesel cycle differs from the Otto cycle in that the ignition occurs at constant pressure rather than constant volume, so the diagram looks like this:

As in the Otto cycle, the red and blue curves are adiabats, so no heat is exchanged during these steps. The efficiency of the Diesel cycle is most easily worked out by calculating the heat input {Q_{h}} along the yellow step 2 to 3 and the heat expelled during the green step 4 to 1. The efficiency is then

\displaystyle e=\frac{W}{Q_{h}}=\frac{Q_{h}-Q_{c}}{Q_{h}}=1-\frac{Q_{c}}{Q_{h}} \ \ \ \ \ (2)

 

[We could also calculate the net work by integrating along the red and blue curves like we did for the Otto cycle.]

The energy required for the constant pressure process 2 to 3 is the enthalpy difference between the two points. Remember that when a gas expands at constant pressure, both its volume and temperature increase, so the total energy required is the sum of the extra internal energy {\Delta U} and the work {P\Delta V} the gas does to expand. We have

\displaystyle Q_{h} \displaystyle = \displaystyle \Delta U+P\Delta V\ \ \ \ \ (3)
\displaystyle \displaystyle = \displaystyle \frac{f}{2}Nk\left(T_{3}-T_{2}\right)+P_{2}\left(V_{3}-V_{2}\right)\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle \frac{1}{\gamma-1}P_{2}\left(V_{3}-V_{2}\right)+P_{2}\left(V_{3}-V_{2}\right)\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle \frac{\gamma}{\gamma-1}P_{2}\left(V_{3}-V_{2}\right)\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \frac{\gamma}{\gamma-1}P_{2}V_{2}\left(\frac{V_{3}}{V_{2}}-1\right) \ \ \ \ \ (7)

where {f} is the number of degrees of freedom per molecule and {\gamma=\left(f+2\right)/f} is the adiabatic exponent. Subscripts refer to the numbered points on the graph.

The heat expelled occurs at constant volume, so we have

\displaystyle Q_{c} \displaystyle = \displaystyle \frac{f}{2}Nk\left(T_{4}-T_{1}\right)\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle \frac{V_{1}}{\gamma-1}\left(P_{4}-P_{1}\right) \ \ \ \ \ (9)

Since the red and blue curves are adiabats, we have

\displaystyle P_{4}V_{1}^{\gamma} \displaystyle = \displaystyle P_{2}V_{3}^{\gamma}\ \ \ \ \ (10)
\displaystyle P_{1}V_{1}^{\gamma} \displaystyle = \displaystyle P_{2}V_{2}^{\gamma} \ \ \ \ \ (11)

Therefore

\displaystyle Q_{c}=\frac{V_{1}P_{2}}{\gamma-1}\left[\left(\frac{V_{3}}{V_{1}}\right)^{\gamma}-\left(\frac{V_{2}}{V_{1}}\right)^{\gamma}\right] \ \ \ \ \ (12)

The efficiency is found by substituting into 2:

\displaystyle e \displaystyle = \displaystyle 1-\frac{V_{1}\left[\left(\frac{V_{3}}{V_{1}}\right)^{\gamma}-\left(\frac{V_{2}}{V_{1}}\right)^{\gamma}\right]}{\gamma V_{2}\left(\frac{V_{3}}{V_{2}}-1\right)}\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle 1-\frac{V_{1}\left[\left(\frac{V_{2}}{V_{1}}\right)^{\gamma}\left(\frac{V_{3}}{V_{2}}\right)^{\gamma}-\left(\frac{V_{2}}{V_{1}}\right)^{\gamma}\right]}{\gamma V_{2}\left(\frac{V_{3}}{V_{2}}-1\right)}\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle 1-\left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}\frac{\left(\frac{V_{3}}{V_{2}}\right)^{\gamma}-1}{\gamma\left(\frac{V_{3}}{V_{2}}-1\right)} \ \ \ \ \ (15)

The ratio {V_{1}/V_{2}} is the compression ratio as before, and the other ratio {V_{3}/V_{2}} is the cutoff ratio. The Diesel efficiency is always lower than the Otto efficiency 1 which we can see as follows.

Since {V_{3}\ge V_{2}}, the smallest possible value for {V_{3}/V_{2}} is 1. Using l’Hôpital’s rule we have

\displaystyle \lim_{x\rightarrow1}\frac{x^{\gamma}-1}{\gamma\left(x-1\right)}=\lim_{x\rightarrow1}\frac{\gamma x^{\gamma-1}}{\gamma}=1 \ \ \ \ \ (16)

Since {\gamma=1+\frac{2}{f}>1}, the function {\left(x^{\gamma}-1\right)/\left(x-1\right)} is monotonically increasing for {x>1} so

\displaystyle \left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}\frac{\left(\frac{V_{3}}{V_{2}}\right)^{\gamma}-1}{\gamma\left(\frac{V_{3}}{V_{2}}-1\right)} \displaystyle > \displaystyle \left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}\ \ \ \ \ (17)
\displaystyle e_{Diesel} \displaystyle < \displaystyle e_{Otto} \ \ \ \ \ (18)

For {V_{1}/V_{2}=18} and {V_{3}/V_{2}=2} the efficiency is (using {\gamma=\frac{7}{5}} for air)

\displaystyle e_{Diesel}=0.632 \ \ \ \ \ (19)

For the same compression ratio the Otto efficiency is

\displaystyle e_{Otto}=0.685 \ \ \ \ \ (20)

One thought on “Diesel engines

  1. Pingback: Steam engines; the Rankine cycle | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *