Review Engine efficiency e: fraction of the heat Lecture 10 input that is converted to work W Q Q e = =1+ C =1 C Q Q Q H H H ALL heat engines have e < 1 The Carnot cycle What is the greatest efficiency an engine can have? Pre-reading: §20.6 YF §19.1
The Carnot cycle The Carnot cycle The Carnot cycle is a hypothetical, idealised heat The Carnot cycle has two isothermal and two engine that has the maximum possible efficiency. adiabatic processes, both reversible.
In order to maximise efficiency, we have to avoid irreversible processes such as heat transfer with a temperature change.
Thus every heat transfer must be isothermal at either TH or TC.
Efficiency of a Carnot engine Efficiency of a Carnot engine Q T ln(V /V ) In order to calculate the efficiency, we need to So C = C c d QH TH ln(Vb/Va) find the ratio QC/QH. ✓ ◆ 1. Isothermal expansion ab: Now, for the two adiabatic processes Q = 0 and heat Q supplied from hot reservoir at constant H 1. Adiabatic expansion bc: temperature T H 1 1 Vb TH V = TC V QH = Wab = nRTH ln b c Va 2. Adiabatic compression da: 2. Isothermal compression cd: 1 1 TH Va = TC V heat QC rejected to cold reservoir at constant d 1 1 temperature TC So V Vb Vc Vb Vc Q = W = nRT ln c = or = C cd C Va Vd V V Vd ✓ ◆ ✓ ◆ a d
Efficiency of a Carnot engine Efficiency of a Carnot engine Hence the two logarithms cancel out, and we get Q T T T e =1+ C =1 C = H C QC TC QC TC QH TH TH = | | = Q H T H or Q T | H | H The bigger the temperature difference, the greater the efficiency. So the efficiency is
QC TC TH TC e.g. jet engines are e =1+ =1 = QH TH TH made of ceramic, which can withstand The efficiency of a Carnot engine depends only on the temperature difference of the two heat temperatures in reservoirs. excess of 1000 °C.
Entropy in a Carnot engine Example For a Carnot cycle,
Water at the surface of the ocean near the equator QC TC QH QC = so + =0 has a temperature of 298 K, whereas 700 m QH TH TH TC below the surface the temperature is 280 K. Two adiabatic processes: ΔS = 0 Two isothermal processes: ΔS = Q/T If you build a heat engine using these two layers of water as the heat reservoirs, what is the So total entropy change is maximum possible efficiency? QH QC Stotal = SH + SC = + =0 TH TC so e is maximum possible efficiency for a Carnot heat engine.
Example Which of the following designs are feasible?
T = 600 K TH = 600 K H TH = 500 K Q Q 1000 J 1000 J H 1000 J H QH W 600 J 400 J W
W 500 J QC QC 600 J 600 J 500 J QC T = 200 K T = 300 K T = 300 K C C C REVIEW