8.21 The Physics of Energy Fall 2009

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1 MIT OpenCourseWare The Physics of Energy Fall 2009 For information about citing these materials or our Terms of Use, visit:

2 8.21 Lecture 10 Phase Change Energy Conversion I September 30,

3 Why this now? Thermodynamics of heat of heat extraction Phase Phase change in pure in pure substances The The vapor vapor compression cycle: cycle: heat heat pumps, refrigeration, air conditioners The The Rankine steam steam cycle cycle and and steam steam turbines Some Some implementations of the of the Rankine cycle cycle } Part I } Part II 3

Several aims Explain how simple thermodynamic cycles can move heat from low to high temperatures Explain why cycles that make a fluid change from vapor to liquid and back dominate practical

4 Several aims Explain how simple thermodynamic cycles can move heat from low to high temperatures Explain why cycles that make a fluid change from vapor to liquid and back dominate practical applications, and how they work Construct and evaluate the dominant cooling ( vapor compression ) and power ( Rankine ) cycles in use today Heat extraction devices are everywhere Household energy use Image removed due to copyright restrictions. Similar graph can be found on web page EnergyMgt/EnergyPieChart_500.jpg Kitchen energy use Household air conditioning 5.1% of all U. S. energy use (2001) Household refrigeration 4.4% Commercial AC and refrigeration 7% Plus industrial cooling and actively cooled transport! Image courtesy of EPA. 4

5 Heat pumps are gaining popularity for home heating Use work to move heat from cold environment to warm environment Heat pump image removed due to copyright restrictions. Same principle as refrigerator, except aim is to warm rather than to cool! Could describe heat extraction devices absent phase change, but Heat extraction devices almost universally employ phase change thermodynamic cycle Which are chosen for thermodynamic properties (eg. liquid vapor near ambient temperatures) eg. Freon Turbine generator image removed due to copyright restrictions. Please see: Geopresentation/images/img038.jpg And phase change power generation is ubiquitous 5

Image and text removed due to copyright restrictions. Many projects involve building solar thermal plants, which use cheaper technology than the solar panels often seen on roofs.

6 Image and text removed due to copyright restrictions. Many projects involve building solar thermal plants, which use cheaper technology than the solar panels often seen on roofs. In such plants, mirrors heat a liquid to create steam that drives an electricity-generating turbine. As in a fossil fuel power plant, that steam must be condensed back to water and cooled for reuse. Please see: energy-environment/30water.html 6

7 Why this now? The vapor compression cycle: heat pumps, refrigeration, air conditioners The Rankine steam cycle and steam turbines Some implementations of the Rankine cycle 7

8 Heat engines and heat extraction devices with flowing fluids Fluid flows in, bringing heat, flows out removing heat, work gets done, but In a cycle, the machine returns to its original state and does not store energy or entropy Conservation of energy around a cycle: machine does not store energy Q H = Q L + W Engine Coefficient of Performance (= Efficiency) Best possible: Entropy is conserved around a cycle if it s performed reversibly: machine does not store entropy CoP = W Q H T H T L T H Q L T L Q H T H =0 CoP refrigeration = Q L W T L T H T L CoP heat pump = Q H W T H T H T L 8

9 Reminder: Thermodynamics of an ideal engine Based on fluid executing a cycle Fluid begins and ends in the same state Fluid must have same energy and entropy at beginning and end of cycle Absorbs heat Q H at T H. Does work W Expels heat Q L at T L First Law Change in internal energy of fluid around cycle must vanish Q H = Q L + W Second Law The entropy of the universe can never decrease When a system absorbs heat Q at temperature T, then it gains entropy S Q. When it loses heat Q T at temperature T it loses entropy S Q T With entropy: You always get more than you want and get rid of less than you hope. And the equality holds only when heat transfer is reversible Compute S universe S universe Q L T L Q H T H 0 9

10 Combine 1st and 2nd laws Q H = Q L + W Q L T L Q H T H 0 Substitute and rearrange W Q H T H T L T H Efficiency = Coefficient of Performance CoP T H T L T H And maximum CoP is only reached when heat transfer is reversible: Minimize temperature and pressured gradients 10

Thermodynamics of an ideal heat extraction device Same as engine: fluid executes cycle; begins and ends in same state. Direction of heat and work flows are reversed: Absorbs heat Q L at T L.

11 Thermodynamics of an ideal heat extraction device Same as engine: fluid executes cycle; begins and ends in same state. Direction of heat and work flows are reversed: Absorbs heat Q L at T L. Work done on it, W Expels heat Q H at T H First Law Q H = Q L + W ) ) Second Law S universe = Q H T H Q L T L 0 Note signs: Here Q H /T H is entropy delivered to universe and Q L /T L is entropy removed from universe 11

12 Coefficient of Performance? It depends what you re trying to accomplish Air conditioning/refrigeration: remove heat from low temperature reservoir: CoP refrigeration = Q L W T L T H T L Heat pump: provide heat to high temperature reservoir: CoP heat pump = Q H W T H T H T L Relations among ideal CoP s CoP ideal heat pump = 1 CoP ideal engine CoP ideal heat pump is always greater than unity and can be very large CoP ideal heat pump = CoP ideal refrigerator +1 12

13 Carnot cooling cycle [1 2] Isentropic compression: Work is done on the gas. It heats up to T H. [2 3] Isothermal compression: Work is done on the gas. Heat equal to work is expelled as Q H [3 4] Isentropic expansion: Gas does work. It cools to T L [4 1] Isothermal expansion: Gas does work. Heat equal to work is absorbed as Q L. If steps are carried out reversibly, then it s guaranteed to reach thermodynamic limit for CoP Example: A refrigerator T L = 20 F T H = 70 F CoP = =

14 Cooling based on phase change Goes back to Michael Faraday in 1820: Liquid ammonia left to evaporate in air cools the air! Make cyclic by condensing ammonia elsewhere and expelling heat Commercialized by Willis Carrier Must review thermodynamics of phase change Liquid/vapor Phases separated by boiling or condensation curve T boiling (P ) or P boiling (T ) Other important points: Triple Point and Critical Point Snow flake image removed due to copyright restrictions. Water drop image removed due to copyright restrictions. Clouds image removed due to copyright restrictions. Please see: Please see: /11/water_drop_causing_a_ripple.jpg 14

15 Why this now? The vapor compression cycle: heat pumps, refrigeration, air conditioners The Rankine steam cycle and steam turbines Some implementations of the Rankine cycle 15

16 Phase Diagram Water but other substances are similar Phase changes Solid/liquid Enthalpy (latent heat) of melting/solidification Liquid/gas Enthalpy (latent heat) of vaporization/condensation Solid/gas Special points Enthalpy of sublimation Triple point: T = K, P = Pa Critical point: T = 647K, P = MPa Enthalpies of phase change Image from Enthalpy of melting (at 0 C) 334 kj/kg Enthalpy of vaporization (at 100 C) 2.26 MJ/kg 16

17 Why use phase change? 1. Large energy storage potential 2.26 MJ to vaporize 1 kg H 2 O at 100 C 2. Energy transfer at constant temperature and pressure! Bring liquid to boiling point, then T and P stay fixed until all liquid vapor! Copious heat transfer under reversible conditions! 3. Flexibility in inducing phase transition. Adjust pressure to select working T, for example. 4. Enhanced heat transfer in boiling 17

18 1. and 2. Large energy storage potential and constant T and P energy transfer. Take 1 kg water at 1 atm and 100 and add heat (example: resistive heating element) Temperature and pressure stay the same until 2.26 MJ has been added and all liquid vapor. Conditions are reversible: Phase change ceases as soon as heat is removed (turn off current) 3. Choice of operating set points (T and P ) Vapor pressure graph removed due to copyright restrictions. Please see: Most added heat goes into internal energy of vapor (small amount in p dv work). Compare adding heat to 1 kg of water vapor at 100. Heat capacity at constant pressure: 2kJ/kg K. Vapor pressure chart removed due to copyright restrictions. So to add same amount of heat to water vapor at constant pressure would raise temperature by 2.26 MJ/2 kj 1000 K! 18

4. Enhanced heat transfer in boiling Compare resistively heated wire in

advantages: (1) vapor bubble spontaneously migrate away from surface,

molecule carries full enthalpy of vaporization with it as it leaves the

0 Laminar flow over a cylinder 0 10 20 30 Temperature difference (T s -T f )

19 4. Enhanced heat transfer in boiling Compare resistively heated wire in asymptotically laminar liquid flow with same wire in boiling pool Two advantages: (1) vapor bubble spontaneously migrate away from surface, whereas fluid flow is minimal near surface due to viscosity (2) Each vapor molecule carries full enthalpy of vaporization with it as it leaves the heated surface. 3.0 Heat Flux (105 W/m 2 ) Heat Flux (W/m 2 ) Laminar flow over a cylinder Temperature difference (T s -T f ) in 0 C 10 6 Water Pool Boiling Heat Transfer Temperature difference (Ts -Tf) in 0 C 19

20 Following phase change in pv, ST, and quality So far we looked at phase change in the Need to look at it in the pt plane. pv and ST planes to get full description Why? Because melting) one point in the pt plane covers whole process of boiling (or Image from Lee Carkner, Department of Physics and Astronomy, Augustana College 20

21 Phase change in the pv -plane Walk along the isotherms 550 K 600 K 650 K Choose a T Mixed Phase Slowly lower P V increases a little CP Isotherm Until you reach P boiling (T ) C D Then all liquid turns to vapor A B With dramatic increase in volume at fixed P Then P again can decrease 21

22 Saturation dome for water Phase change curve Mixed phase below the dome Saturated vapor on the right part of curve Saturated liquid on the left part of curve Quality: Critical point Saturated vapor, quality =1 χ = m v m v + m l Mixed phase Saturated liquid, quality =0 22

23 Phase change in the TS-plane Walk along the isobars Critical point Saturated vapor, quality =1 Choose a P Slowly add heat (entropy) System does work against constant P And T increase (C p ) Until you reach T boiling (P ) Then all liquid turns to vapor Subcooled liquid Mixed phase Superheated vapor With dramatic increase in heat (enthalpy) at fixed T Then T again can increase Saturated liquid, quality =0 23

24 Properties of the mixed phase To a very good approximation, entensive properties of mixed phase are merely The proportional sum of the liquid and vapor properties. For example, entropy S = m vs v + m l S l m v + m l = χs v + (1 χ)s l Applies to energy, entropy, enthalpy, volume Why not exact? Interface properties: 50/50 mix of liquid and vapor water has slightly different properties than a fog. Where to get properties of saturated liquid and vapor? Not from perfect gas law! Thermodynamic tables! 24

25 Steam tables: For each pressure The saturation temperature boiling point Properties of saturated liquid and saturated vapor at the boiling point Table of properties at other temperatures Subcooled liquid Superheated vapor 25

Example A kilojoule of heat is added to a kilogram of liquid water at 1 atmosphere pressure and 100 C. What are the properties of the resultant mixture?

1 atmosphere is close enough to 1 10 5 N/m 2 to use tables on last slide h liq = 418 kj/kg h vap = 2675 kj/kg 2500 2000 Enthalpy per kilogram 2675 h mixture =

26 Example A kilojoule of heat is added to a kilogram of liquid water at 1 atmosphere pressure and 100 C. What are the properties of the resultant mixture? Since the system is at constant pressure, the heat is added as enthalpy. 1 atmosphere is close enough to N/m 2 to use tables on last slide h liq = 418 kj/kg h vap = 2675 kj/kg Enthalpy per kilogram 2675 h mixture = kj/kg, Which corresponds to a quality of χ = Once we know χ =0.44, (and 1 χ =0.56), Quality v = 0.44(1.69) ( ) = m 3 / kg s = 0.44(7.36) (1.30) = 3.97 kj/kg K 26

27 Digression: Thermodynamics with flowing fluids Need to deal with devices where materials flow in and out! Pipe heated by resistive coil: fluid flows in, heats, flows out. Throttle : fluid pushed through nozzle Fluid enters: ρ 1, p 1, T 1, u 1, h 1, s 1 Leaves: ρ 2, p 2, T 2, u 2, h 2, s 2 These are density (ρ), pressure (p), temperature (T ), specific energy (u), enthalpy (h), and entropy (s) Heating pipe P 1 P 2 Specific energy energy per unit mass,... Concept: control volume In a time t, apply first (and second) laws on a fixed domain Throttle In a time t, mass m 1 = ρ 1 A 1 (v 1 t) enters and m 2 = ρ 2 A 2 (v 2 t) leaves And m 1 = m 2 Entering mass brings energy U = u 1 ρ 1 A 1 v 1 t And similarly for enthalpy and entropy entering and leaving 8.21 Lecture 12: Phase change energy conversion II 27

Apply first law to control volume E in = Internal energy in + pdv work in = u 1 ρ 1 v 1 A 1 t + p 1 A 1 v 1 t ρ 1,p 1,v 1 u 1,h 1,s 1 Q A 2 E out = Internal energy out + pdv work out = u 1 ρ 2 v 2 A

28 Apply first law to control volume E in = Internal energy in + pdv work in = u 1 ρ 1 v 1 A 1 t + p 1 A 1 v 1 t ρ 1,p 1,v 1 u 1,h 1,s 1 Q A 2 E out = Internal energy out + pdv work out = u 1 ρ 2 v 2 A 2 t + p 2 A 2 v 2 t First law E out = E in + Q W A 1 u 2 ρ 2 v 2 A 2 t + p 2 A 2 v 2 t = u 1 ρ 1 v 1 A 1 t + p 1 A 1 v 1 t + Q W Divide out m = ρ 2 v 2 A 2 t = ρ 1 v 1 A 1 t W ρ 2,p 2,v 2 u 2,h 2,s 2 u + p ρ = U m + u 2 + p 2 ρ 2 = u 1 + p 1 ρ 1 + Q m W m p m/v = 1 (U + PV)=h m h 2 = h 1 + dq dm dw dm Result: Enthalpy out = Enthalpy in + Heat in Work out h 2 = h 1 + dq dm dw dm = h 1 + q w 8.21 Lecture 12: Phase change energy conversion II 28

29 Some examples! Heat exchanger (evaporator): Heat in, no work h 2 = h 1 + q Heat exchanger (condensor): Heat out, no work h 2 = h 1 q Throttle: No heat, no work h 2 = h 1 Pump (adiabatic): Work in, no heat h 2 = h 1 + w Turbine (adiabatic): Work out, no heat h 2 = h 1 w 8.21 Lecture 12: Phase change energy conversion II 29

30 Summary An engine cycle run backwards a refrigerator or a heat pump CoP heat pump = TH (TH TL) = 1 CoP engine CoP AC = TL (TH TL) Phase change takes place at constant temperature and pressure Phase change working fluid: (1) High heat capacity; (2) heat transfer at constant T; (3) wide range of (T, p) set points; (4) rapid energy transfer. Phase change in (T, p), (p,v) and (S,T) planes. Quality χ = m v m v + m l Saturated vapor, saturated liquid, superheated vapor, subcooled liquid Quality calculations: properties of the mixed phase are additive (enthalpy for example) h mixed (χ) =χh vapor + (1 χ)h liquid When fluid moves through a device follow the enthalpy! h out = h in + Q m W m WORK DONE HEAT ADDED 8.21 Lecture 12: Phase change energy conversion II 30

CHAPTER 8 ENTROPY. Blank

CHAPTER 8 ENTROPY. Blank CHAPER 8 ENROPY Blank SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L.