How much wood… ? U -W Q

Evaluating heat engines with PV diagrams

Process Meaning Implications Example Isobaric P=0 +Q  V 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙=𝑄−𝑃𝑉 Isovolumetric V=0 W Q  P  T 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =𝑄 Isothermal T=0 PV = constant 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =0 𝑄=−𝑊 Adiabatic Q=0 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =−𝑃𝑉

Process Implications Example In English Isobaric +Q  V 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙=𝑄−𝑃𝑉 To expand (contract) a gas at constant pressure, heat must be added (taken away). Isovolumetric W Q  P  T 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =𝑄 Adding (subtracting) heat to a gas at constant volume increases (decreases) pressure. Isothermal PV = constant 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =0 𝑄=−𝑃𝑉 When energy is added to a gas at constant temperature, it will expand and its pressure will drop. Adiabatic 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 =𝑃𝑉 Doing work on a gas decreases its volume, and increases its pressure and temperature.

How much work is done to gas. How much does internal energy change How much work is done to gas? How much does internal energy change? How much heat is added? Conventions: W+ = work done TO gas by environment i.e., gas is compressed, final volume is smaller W- = work done BY gas on environment i.e., gas expands, final volume is larger Q+ = heat flows INTO gas Q- = heat flows FROM gas

Isobaric process: work 𝑊=−𝑃𝑉 Final Initial W= - (2 Pa) (10 m3 – 1 m3) = -18 J

Isobaric process: change in internal energy 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑛𝑅𝑇 𝑇= 𝑃 𝑉𝑓−𝑉𝑖 𝑛𝑅 Final 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑃(𝑉𝑓−𝑉𝑖) Initial Where T = P(Vf-Vi)/nR So, U = 3/2 nR P (Vf-Vi) / nR Or, U = 3/2 P (Vf-Vi) Here, U = 3/2 (2 Pa) (9 m3) = 27 J

Isobaric process: heat 𝑈=𝑄+𝑊 𝑄=𝑈−𝑊 Final Initial Take-away: Isobaric processes do work on the environment, but require that heat be added. Q= 27 J – 18 J = 9 J

Isovolumetric process: work 𝑊=−𝑃𝑉 Final V = 0 m3, so W = 0 J Initial

Isovolumetric process: change in internal energy 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑛𝑅𝑇 Final 𝑇= 𝑉 𝑃𝑓−𝑃𝑖 𝑛𝑅 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑉(𝑃𝑓−𝑃𝑖) Where T = V(Pf-Pi)/nR So, U = 3/2 nR V (Pf-Pi) / nR Or, U = 3/2 V (Pf-Pi) Here, U = 3/2 (1 m3) (20 Pa – 1 Pa) = 28.5 J Initial

Isovolumetric process: heat 𝑈=𝑄+𝑊 Final 𝑄=𝑈−𝑊 Take-away: Isovolumetric processes do no work on the environment; additional pressure requires heat be added to the system. Q= 28.5 J – 0 J = 28.5 J Initial

Isothermal process: change in internal energy 𝑈𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙= 3 2 𝑛𝑅𝑇 Final Initial T = 0 K, so U = 0 J

Isothermal process: work and heat 𝑈=𝑄+𝑊 Initial @𝑈=0, 𝑄=−𝑊 Take-away: Heat added to isothermal process does work on environment with no change in temperature. Final

The process makes a difference! Final Initial Initial Final Work in isovolumetric / isobaric process < work in isothermal process

Rank processes from those that do most work to those that do least work. Initial Final B C D Area under the curve does most work

Efficiency Ratio of what you get out by what you put in: 𝑒= 𝑊 𝑜𝑢𝑡 𝑊 𝑖𝑛 In the case of heat engines, 𝑊 𝑖𝑛 = 𝑄 𝐻 (energy input from, e.g., gasoline) So, 𝑒= 𝑊 𝑜𝑢𝑡 𝑄 𝐻 where, since energy is conserved, 𝑊 𝑜𝑢𝑡 = 𝑄 𝐻 − 𝑄 𝐿 So, 𝑒= 𝑄 𝐻 − 𝑄 𝐿 𝑄 𝐻 =1− QL QH

Example An automobile engine has an efficiency of 20% and produces an average of 23,000 J of mechanical work per second. How much heat input is required? How much is discharged as waste? a) QH = W / e = 23000 J / 0.20 = 115 kJ b) e= 1 – QL / QH, so QL = QH(1-e) = 115 kJ (1-0.2) = 92 kJ

Ideal efficiency In an ideal engine (i.e., perfectly reversible process), QH  TH and QL  TL So, 𝑒𝑖𝑑𝑒𝑎𝑙=1− TL TH Note: this equation is sometimes called Carnot efficiency in honor of its discoverer, a French physicist Sadi Carnot.

Coefficient of performance For cooling devices, 𝐶𝑂𝑃≡ 𝑄 𝐿 𝑊 = 𝑄 𝐿 (𝑄 𝐻 − 𝑄 𝐿 ) For heating devices, 𝐶𝑂𝑃≡ 𝑄 𝐻 𝑊 Since we care about TL for cooling devices, COP defined as a function of TL for cooling devices.

Why heat pumps? How much thermal energy is delivered by 1500-W electric heater? How much thermal energy is delivered by 1500-W heat pump with a COP of 3.0? 1500 W COP = QH/W, so QH = (COP) (W) = 4500 W… NOT a violation of conservation of energy; extra heat coming from outside