Efficiency of a heat engine

<i>PV</i> diagram of an example of a heat engineThe efficiency of a heat engine is the ratio of work output divided by heat input. The process from a to b occurs at constant volume; the heat added, Qab, therefore depends on the moles of gas, the molar specific heat at constant volume, CV, and the change in temperature. The process from b to c occurs at constant pressure, so the heat added, Qbc, instead depends on the molar specific heat at constant pressure, CP: Q ab =n C V ( T b T a ), Q bc =n C P ( T c T b ) Read the text aloud
The equations above and below use the molar specific heat—which you learned about on page 1339. Typical values for air are CV = 20.8 J/(mol K) and CP = 29.1 J/(mol K). Read the text aloud
Assume an ideal gas for which PV = nRT. The temperatures at points a, b, c, and d are therefore determined by the pressures and volumes: T a = P 1 V 1 nR ,   T b = P 2 V 1 nR ,   T c = P 2 V 2 nR ,   T d = P 1 V 2 nR . Read the text aloud
After substituting for the temperatures we find the heat inputs Qab and Qbc: Q ab = C v R V 1 ( P 2 P 1 )                          Q bc = C p R P 2 ( V 2 V 1 ) Read the text aloud
Calculating the efficiency of a heat engineConsider a large car engine, in which each piston has a 10 cm diameter, stroke of 10 cm, and clearance of 2 cm and operates between 1 and 5 atmospheres (100,000 to 500,000 Pa). Apply the above equations to calculate the input heat as Q ab + Q bc =157 J+1,374 J=1,531 J. The work done in the thermodynamic process is the area inside the PV diagram, or ( P 2 P 1 )( V 2 V 1 )=314 J. The input heat is almost five times the work output of the engine! Furthermore, to keep the pressure constant from b to c requires that the temperature be Tc = 15,363ºF! A real engine could not sustain this high a temperature. Read the text aloud
This thermodynamic cycle has a terrible efficiency of 21%. Heat not converted to work is rejected to the surroundings during steps c → d and d → a. According to the first law, the heat rejected is the heat added minus the work done, or 1,271 J (79%). Soon we will see how to get better efficiency, but, according to basic thermodynamic principles, no heat engine operating above absolute zero can convert 100% of heat into output work. Read the text aloud

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