3.020 Lecture 4 Prof. Rafael Jaramillo 1 Heat engines, abstracted after slides • Cyclic machine • Returns to same state after each cycle • Thermal voirs reser- • Total work and heat • Maintained at hot (TH ) and cold (Tc ) throughout • Work in and out • Heat in and out • Eﬃciency η≡ WN ET QIN • Typical representation: “ Heat engine with eﬃciency η operating between TH and Tc ” • Each cycle: QIN absorbed from TH QOU T “rejected” to Tc WOU T performed Question: What is QIN − WOut − QOU T ? 1 slides: identifying TH , TC for real engines 2 Calculating a cyclic process • Work and heat are process variables • Thermodynamics doesn’t describe real-world process What to do ?? ⇒ Describe hypothetical process for which system remains in equilibrium at all times. - can use state variables and equations of state (if available) - in practice, such a cycle would take inﬁnite time. Power = 3 Work −→ 0 Cyclic tiem period Carnot cycle with an ideal gas (I) Isothermal expansion at TH (II) Adiabatic expansion to Tc (III) Isothermal compression at Tc (IV) Adiabatic compresion to TH Note: Wout is the area enclosed by the cycle (true of any cycle, not just Carnot) 2 3.1 Isotherms P V = nRT P dV + V dP = 0 V nRT dV = − dP = − 2 dP P P nRT P dV = − dP P Z Z nRT Pf inal Vinitial ⇒ δW = + dP = +nRT ln = +nRT ln Vf inal P Pinitial R Sanity Check : expansion does work on surroundings, so δW < 0 Vi < 1, Vf 3.2 ln Vi Vf <0 X Isotherms, continued ⇒ For ideal gas, internal energy U is a function of T only, dU = nCv dT dU = 0for isothermal process δQ = −δW Q = −W 3.3 recall that δW = −P dV single variable calculus! Adiabats δQ = 0, δW = −P dW - For ideal gas: – how to calculate ? dU = δW + δQ = −P dV = nCv dT |{z} 0 W = nCv (Tf inal − Tinitial ) - Adiabatic curves are described by: γ T V = const, Pf inal = Pinitial Vinitial Vf inal γ , γ= Cp Cv Deltoﬀ ch. 4, Lectures 6-7 3 • Adiabats are steeper than isotherms on (P,V) plane becasue γ > 1 Q: Why is γ > 1 ? 3.4 Adding all contributions to W and Q W � I nRTH ln V1 /V2 II nCv (TC − TH ) � III nRTC ln V3 /V4 IV nCv (TH − TC ) 3.5 Q � −nRTH ln V1 /V2 � −nRTC ln V3 /V4 Calculating the Carnot eﬃciency � � � WT OT = − nRTH ln V1 /V2 + nRTC ln V3 /V4 total work done by the engine � QIN = −nRTH ln V1 /V2 heat absorbed at TH WT OT Tc ln(V3 /V4 ) η= =1+ TH ln(V1 /V2 ) QIN Using property of adiabat T V γ−1 = const can show that V3 /V4 = (V1 /V2 )−1 ηcarnot = 1 − TC /TH 3.6 Considering heat transfers heat absorbed Carnot engine QIN = nRTH ln V less eﬃcient en- QIN = nRTH ln V gine that burns the same quantity of fuel v= heat released QOU T,C = nRTC ln V QOU T > QOU T,C required by conservation of energy if WT OT < WT OT,C 4 V2 V1 = V3 V4 3.7 Considering quantity δQ/T - Quantity: H - Carnot: H δQ T δQ T - Less eﬃcient: −→ integral around the cycle = nRTH ln V TH H δQ T = − nRTC ln V TC nRTH ln V TH =0 T − QTOU < 0. C because QOU T > QOU T,C ⇒ We will soon see that this is related to entropy generation by the less eﬃcient cycle 5

Lecture 4 Notes: Heat Engines and Energy Conversion Efficiency

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