# Why can't the Carnot cycle be used to operate automobiles, if its the most efficient?

Since the cycle proposed by Carnot (consisting of two isothermal and two adiabatic processes) is the most efficient one on which to run a heat engine, why do we use other, less efficient ones (like the Otto or the Diesel cycle) for automobiles?

On the P-V diagram for the Carnot cycle the upper and lower curves are very close together. As a result, this cycle has very little area enclosed within the cycle diagram and very little work is done per cycle. Therefore, to produce a reasonably useful quantity of work the engine must be very massive, heavy, and expensive relative to other engines based on other cycles like the Otto cycle and the Diesel cycle which enclose more area on the P-V diagram and do much more work per cycle.

A larger engine with a relatively larger amount of surface area will sustain greater losses to the environment through that surface area reducing its efficiency and partially or completely offsetting any other advantages it may have. This larger engine will also have to move more mass of working fluid through it to produce the same work output as some other engine cycle with greater work output per cycle increasing the internal flow losses and again reducing the real world efficiency of the Carnot enginen relative to others. A heavier engine will also penalize a vehicle with greater per mile energy consumption once again offsetting any other advantages the Carnot cycle may present.

Bottom line, real Carnot engines can not be designed to compete with real Otto or Diesel engines on an overall economic basis.

• Can't you just make it turn over faster? – Ryan Thorngren May 8 at 13:29

The Carnot cycle is so efficient because it minimizes the increase in entropy. This comes at the expense of needing to follow a specific path on the PV diagram that theoretically should take infinite time (otherwise you have a difference in temperature between source and sink which increases entropy).

In a real engine, where you want to extract a finite amount of work per unit time, you need to set up a thermal difference between source and sink; this decreases the efficiency below what you can get from a Carnot cycle.

See for example this earlier question

• I understand that the perfectly reversible processes cannot be replicated in reality, but could you elaborate on the point about the source and sink being at the same temperature? I thought that the ideal cycle needed a thermal difference in any case, to produce work – Benjamin Hornigold Sep 29 '15 at 12:13
• All power cycles need a thermal difference between the high source temperature and the low sink temperature to generate any power. Any temperature difference between the high source and the working fluid accepting the heat, or the low temperature sink and the working fluid rejecting the heat causes the production of entropy and an efficiency loss. Heat transfer is slow compared to expansion and compression. The Stirling cycle and the Carnot cycle (external combustion) require heat to be transferred in from an external source and must run slowly to give enough time for heat transfer to occur. – L. White May 9 at 18:04