Thermodinamika + Neraca Energi 8/24/2011

Transkripsi

1 THERMODYNAMICS & ENERGY BALANCE Lecture Note Principles of Food Engineering (ITP 330) Dosen : Prof Dr Purwiyatno Hariyadi, MSc Dept of Food Science & Technology Faculty of Agricultural Technology Bogor Agricultural University BOGOR 2002 THERMODYNAMICS AND ENERGY BALANCE Learning Objectives Understand the conceptual basis of the Law of Thermodynamics Understand the fundamental energy balance concepts Be able to list and discuss important terms related to energy transfer Be able to list and discuss energy balance applications in food processing and handling operations Be able to conceptually describe how energy balance determinations or calculations are obtained Purwiyatno Hariyadi/ITP/Fateta/IPB 1

2 WHAT IS THERMODYNAMICS? Thermodynamics is the branch of science which studies the transformation of energy from one form to another Thermodynamics - Science which is concerned with changes in the forms or location of energy and may be thought in terms of energy dynamics Thermodynamics of process : > looks at the energy transformations which occur as a result of process How much heat is evolved during a process? What determines the spontaneous process? What determines the extent of process? DESCRIPTION OF THE SYSTEM1 Composed of a finite portion of matter and is defined in terms of the boundaries which enclose it Boundaries may be real or imaginary Region surrounding boundaries may be referred to as its environment May consider a plant or any portion thereof as a boundary mass System energy Surrounding=environment Purwiyatno Hariyadi/ITP/Fateta/IPB 2

3 DESCRIPTION OF THE SYSTEM2 Two (common) types of systems are: open system closed system mass System energy Open system - boundaries permit the crossing of matter - energy may cross the boundaries of the open system with the flow of mass or separately Closed System - boundaries do not permit the crossing of matter - energy may cross the boundaries of closed systems DESCRIPTION OF THE SYSTEM3 Steady state conditions: > mass of the system remains unchanged > rate of flow leaving system is constant and equal to that entering the system Transient (unsteady) state conditions: > mass of the system may remain unchanged > heat of the system changes with time Purwiyatno Hariyadi/ITP/Fateta/IPB 3

4 DESCRIPTION OF THE SYSTEM 4 Energy which crosses the boundary is classified as either heat or work heat mass System work Heat is the form of energy that is transferred from the environment external to the system by way of diffusion due to a temperature gradient Positive sign - refers to heat entering system Negative sign - heat leaving system PROPERTIES OF THE SYSTEM 1 Property - Observable, measurable, or calculable characteristic of a substance which depends only upon the state of the substance State of a given system is its condition or its position with respect to other systems Equation of state - relationship between > pressure, > specific volume, and > temperature Purwiyatno Hariyadi/ITP/Fateta/IPB 4

5 PROPERTIES OF THE SYSTEM 2 Equation of state of a perfect/ideal gas (Boyle, Charles, Guy-Lussac) : PV = nrt; where: P = absolute pressure, kpa/m 2 V = volume, m 3 n = number of molecules, kgmole R = universal gas constant [=]???? T = absolute temperature, o K Standard Condition? At 273 o K, 760 mm Hg ( kpa), 1 gmole occupy 22,4 L 1 kgmole occupy 224 m 3 PROPERTIES OF THE SYSTEM 3 R = lit(atm)/(gmole o K) = 8315 Nm/kgmole o K = 1545 ft(lbf)/(lbmole o R Typical properties of a system for a given state are : > pressure, > volume, > temperature, > velocity, and > the elevation of the system Purwiyatno Hariyadi/ITP/Fateta/IPB 5

6 PROPERTIES OF THE SYSTEM 4 Van der Waal s Equation of state : 2 n a P + ( V nb) 2 = nrt V where: P = absolute pressure V = volume, m 3 n = number of molecule R = gas constant T = absolute temp a, b = constant Gas a Pa(m 3 /kgmole) 2 b m 3 /kgmole Air Ammonia CO Water vapor PURE SUBSTANCES 1 Pure substance is a single substance which retains an unvarying molecular structure Examples include: > pure oxygen > ammonia > dry air (in the gaseous state) - largely composed of oxygen and nitrogen with fixed percentages of each Purwiyatno Hariyadi/ITP/Fateta/IPB 6

7 PURE SUBSTANCES 2 A pure substance may exist in any of three phases including solid, liquid, or gas =f(p (P, V, T) Melting - change of phase from solid to liquid Vaporization - change of phase from liquid to gas Condensation - change of phase from vapor to liquid Sublimation - substance passing from the solid directly to a gaseous phase (dry ice) PURE SUBSTANCES 3 kpa) Pressure ( solid liquid gas Triple point (T) H 2 O T (4,6 Torr, 001 o C) CO 2 T(54 Torr, - 57 o C) Temperature (K) Purwiyatno Hariyadi/ITP/Fateta/IPB 7

8 PURE SUBSTANCES 4 Pressure (kpa) solid Condensation liquid gas Melting Vaporization Sublimation Temperature (K) PURE SUBSTANCES 5 Pressure (kpa) solid liquid Critical Point gas The higher the pressure the higher the saturation temperature Critical point : gas and liquid become indistinguishable density and other properties become identica Temperature (K) Purwiyatno Hariyadi/ITP/Fateta/IPB 8

9 PURE SUBSTANCES 6 Gas or Vapor? > = identical!!! Vapor : - gas which exists below its critical temperature - condensable by compresion at constant T Gas : - non condensable gas - gas above the critical point PURE SUBSTANCES Vapor Pressure Vapor-liquid Equlibrium Vaporization and condensation at constant T and P are equilibrium process - equilibrium pressure = vapor pressure - at a given T : > there is only one P at which liquid and vapor coexist (in equilibrium) Pressure e (kpa) Vapor and liquid in equilibrium Temperature (K) Purwiyatno Hariyadi/ITP/Fateta/IPB 9

10 PURE SUBSTANCES Vapor Pressure P=900 mm Hg P=500 mm Hg P=250 mm Hg All vapor Vapor All H 2 O liquid liquid 190 o H F 2 O 190 o F 190 o F a) Pressure (kp Transformation of liquid water into water vapor at constant T Temperature (K) PURE SUBSTANCES Vapor Pressure P=147 psia P=147 psia P=147 psia All vapor 213 o F Vapor liquid 212 o F H 2 O 211 o F All liquid H 2 O a) Pressure (kp Transformation of liquid water vapor into water at constant P Temperature (K) Purwiyatno Hariyadi/ITP/Fateta/IPB 10

11 Internal Energy, E System may be losing and gaining energy Total energy of the system? > internal energy, E Internal energy : total energy of system (the sum of all the system's energy) Chemical, nuclear, heat, gravitational, etc It is impossible to measure the total internal energy of our system > intrinsic property So why define a quantity which we cannot measure? We can measure changes in the internal energy Thermodynamics is all about changes in energy : The change in internal energy of a system a very useful experimental quantity Change of Internal Energy, E E may change in 3 different ways : heat passes into or out of the system; work is done on or by the system; mass enters or leaves the system Again : Closed system : no transfer of mass is possible : E may only change due to heat and work Isolated system : heat, work and mass transfer are all impossible no change in E Open system : E may change due to transfer of heat, mass and work between system and surroundings Purwiyatno Hariyadi/ITP/Fateta/IPB 11

12 Closed system If δq and δw are the increments of heat and work energy crossing the system s boundaries : or de = δq - δw ΔE E = Q -W The First Law of Thermodynamics = law of conservation of energy ISOTHERMAL EXPANSION OF AN IDEAL GAS AGAINST A FIXED ESTERNAL PRESSURE P atm P atm Work?? = force x distance = pressure x area x distance = P atm x A x (h2-h1) h1) =P =P atm ΔV h1 h2 Purwiyatno Hariyadi/ITP/Fateta/IPB 12

13 ISOTHERMAL EXPANSION OF AN IDEAL GAS AGAINST A FIXED ESTERNAL PRESSURE Remember! Positive sign Negative sign If W = -P atm ΔV If then - heat entering system - work done on the system (compression) - heat leaving system - work done by the system (expansion) > P [=] Pa > V [=] m 3 > W [=] J Enthalpy (H) Another intrinsic thermodynamic variable H = E + PV or, in differential form : dh = de + PdV + VdP PdV = δw > dh = de + δw W + VdP δw W + de = δq > dh = δq Q + VdP for constant pressure process (dp=0) dh = δq Q or ΔH H = Q Specific heat at constant P (C p ) > C p = dq dt Enthalpy = Heat content < > ΔH H = Q = C p dt p Purwiyatno Hariyadi/ITP/Fateta/IPB 13

14 Enthalpy (H) Enthalpy = Heat content > ΔH H = Q= C p dt > ΔH = mc pav (T 2 -T 1 ) ΔH H : positive > heat is absorbed (endothermic) ΔH H : negative > heat is envolved (exothermic) Back to Ineternal energy : de = δq - δw Constant Volume process : δw W =0 > de = δq ΔE E = Q Specific heat at constant V (C v ) > C V = dq dt > ΔE E = C V dt V Relationship between C p and C v de = dq - PdV teking the derivative with resoect to T : de dt = dq dt P dv P dt C p C V 1 mole of Ideal gas PV = RT at constant pressure : (dv/dt) = R/P R C V = C P -R > C P /C V = γ > C P /R = γ/( /(γ-1) Purwiyatno Hariyadi/ITP/Fateta/IPB 14

15 STEAM TABLE Gas ready to start to condense : saturated gas > dew point Liquid ready to start to vaporize : saturated liquid > bubble/boiling point Mixture of liquid and vapor at equilibrium (called a wet gas) > both liquid and vapor are saturated Pressure (kp a) Temperature (K) STEAM TABLE Degree of superheat and Steam quality Pressure (k kpa) 100 psia 3278 o F Temperature (K) Steam 500 o F, 100 psia Degree of superheat = = 1722 o F Wet vapor : consists of saturated vapor + saturated liquid Steam quality = weight fraction of vapor Purwiyatno Hariyadi/ITP/Fateta/IPB 15

16 SAT- STEAM TABLE Appendix A3 (Toledo, p 572-3) Temp ( O F) Absolute presure lb/in 2 Spec Vol (ft 3 /lb) Ethalpy (BTU/lb) Sat Evap Sat Sat Evap Sat liquid v fg vapor liquid h fg vapor v f v g h f h g SAT-STEAM STEAM TABLE Appendix A4 (Toledo, p 574-5) 5) Temp ( O C) Absolute presure kpa Sat liquid h h f Ethalpy (MJ/kg) Evap h fg Sat vapor h h g Purwiyatno Hariyadi/ITP/Fateta/IPB 16

17 SAT-STEAM STEAM TABLE Example (1) At 290 o F and psia the specific volume of a wet steam mixture is 405 ft 3 /lb What is the quality of the steam? Look at the Table (A3) v f = ft 3 /lb v g = ft 3 /lb basis : 1 lb of wet steam mixture let x = vapor weight fraction > (1-x) = liquid weight fraction ft 74641ft 3 [ (1 x)lb liquid] + [ x lb vapor] = 405 ft 1lb liquid 1lb vapor x x = X =? 405 Gas Mixture P t = P a + P b + P c P n > Dalton s Law of Partial Pressures P t = total presure P t P a, P b, P c and P n = partial pressure n i = f(p i ) > P i V = n i RT V t = V a + V b + V c V n > Amagat s Law of Partial Volumes P t = total volume P a, P b, P c and P n = partial volume n i = f(v i ) > P V i = n i RT Purwiyatno Hariyadi/ITP/Fateta/IPB 17

18 Gas Mixture/Sat-steam table example (Toledo, p 119) Head space of can at 20 o C Pressure : 10 in Hg vacuum Atmospheric pressure = 30 in Hg Volume head space = 164 cm 3 Calculate the quantity of air in head space! Head space consists of air and water vapor P t = P air + P water P t = 10 in Hg vacuum = P bar -P gage = (30-10)= 20 in Hg ( Pa/in Hg) = 67,728 Pa P water =? From Steam Table (appendix A4) : at 20 o C, vapor pressure of water = P water = Pa P air P air air = P t -P water air = 67, = 65,3924 Pa Gas Mixture/Sat-team team table example (Toledo, p 119) Head space of can at 20 o C Pressure : 10 in Hg vacuum Atmospheric pressure = 30 in Hg Volume head space = 164 cm 3 Calculate the quantity of air in head space! n air air = (P air V)/RT use SI unit n air n air P V air = RT = 7 air 4 40 x 10 = kgmoles T = = 293 K P air = 65,3924 Pa V = 164 cm 3 = 164 cm 3 (10-6 )m 3 /cm 3 = 2 x 10-5 m 3 R = 8315 Nm/kgmoleK N 63,392 4 )(1 64 x 10 2 m Nm (8315 ( kgmoles K 5 )( 293 m K ) 3 ) Purwiyatno Hariyadi/ITP/Fateta/IPB 18

19 Gas Mixture/Sat-steam table example (Toledo, p 128) n Sealing condition for canning process : Temperature : 80 o C; P atmospheric = 758 mmhg Calculate the vacuum (mm Hg) inside the can when the content cool down to 20 o C Answer : Assume the headspace consists of air and H 2 O vapor Appendix A4 Vapor pressure of H 2 O at 80 o C = kpa = Pa Vapor pressure of H 2 O at 20 o C = kpa = 2,3366 Pa PV = RT P t = P air + P H2O P air = P t -P H2O Condition 1 : T = 80 o C and P t = 758 mm Hg= 101,064 Pa P air = (101,064-46,3601) Pa (101,064 47,3601)Pa x V m = Nm 8315 ( ) K kgmolek air = V kgmole Gas Mixture/Sat-steam table example (Toledo, p 128) Sealing condition for canning process : Temperature : 80 o C; P atmospheric = 758 mmhg Calculate the vacuum (mm Hg) inside the can when the content cool down to 20 o C Answer : Condition 2 : T = 20 o C and P t =? n air = V kgmole n air PV = RT 1 = 8315 Px V Nm kgmolek ( ) K = V kgmole PV = V P = P = 44,575 Pa absolute P = 332 mm Hg absolute Vacuum = = 426 mm Hg Purwiyatno Hariyadi/ITP/Fateta/IPB 19

20 SUPERHEATED STEAM TABLE Appendix A2 (Toledo, p 571) Superheated steam : steam (water vapor) at T higher than boiling point Temp ( o F) 1 psi Ts=10174 o F v h Abs Pressure (psi) 5 psi Ts=16224 o F v h Ts : saturation Temp at deignated pressure v : spec volume (ft 3 /lb) h : enthalpy (BTU/lb) Sat-steam table example (Toledo, p 148) How much heat is required to convert 1 lb H2O (70 o F) to steam at psia (250 o F) - steam at psia > boiling point=212 o F (Sat steam Table) > at 250 o F > 212 o F : superheated! - heat required = h g (250 o F, psia) -h f (70 o F) = 1168 BTU/lb BTU/lb = BTU/lb How much heat would be given off by cooling superheated steam at psia (500 o F) to 250 o F at the same pressure? - basis 1 lb of steam - heat given off = h g (14696 psia, 500 o F) -h g (14696 psia, 250 o F) = = 1186 BTU/lb - superheated steam is not very efficeient heating medium! Purwiyatno Hariyadi/ITP/Fateta/IPB 20

21 HUKUM THERMODINAMIKA I : > Konservasi Energi > Kesetimbangan Energi Masukan sistem Keluaran Energi masuk = Energi keluar + Akumulasi Kondisi Steady State = tidak terjadi akumulasi : > Energi masuk = Energi keluar ENERGI > PANAS= uap, air, padatan, dll > MEKANIK > ELEKTRIK > ELEKTROMAGNETIK > HIDROLIK > DLL Steps in Energy Balance Preparation = Steps in Mass Balance Preparation Draw a sketch or diagram describing process Identify information available Identify boundaries of system with dotted lines Identify all input (inflows) and output (outflows) Use symbols or letters to identify unknown items/quantities Write energy e balance a equation : choose appropriate basis of calculation do total and/or component energy balance Solve resulting algebraic equation(s) Purwiyatno Hariyadi/ITP/Fateta/IPB 21

22 KESETIMBANGAN PANAS contoh 1 Hitung air yang diperlukan untuk mensuplai alat pindah panas yang digunakan untuk mendinginkan pasta tomat (100 kg/jam) dari 90 o C ke 20 o C Pasta tomat: 40% padatan Naiknya suhu air pendingin = = 10 o C Pasta Tomat q kg/jam 90 o C 40% padatan q 4 Air hangat T 2 (T 2 > T 1 ; T 2 -T 1 = = 10 o C) T 2 = T o C air dingin (T 1 ), W Kg q 3 q 2 Pasta tomat 20 o C KESETIMBANGAN PANAS contoh 1 Misal: T 1 = 20 o C T ref : 20 o C T 2 = 30 o C J Cp air = 4187 KgK K Cp Pasta tomat = 3349 M Formula Siebel = 3349(06) = J/KgK Kandungan panas masuk: J q1 = 100 Kg = KgK o ( ) K MJ Kandungan panas keluar: J o q = 100 Kg ( 20 20) K = KgK 2 0 Purwiyatno Hariyadi/ITP/Fateta/IPB 22

23 KESETIMBANGAN PANAS contoh 1 Air masuk, W kg q J = K KgK o ( ) 0 = 3 Wkg 4187 J q = Wkg 4187 = KgK o ( ) K, ( w ) J Kesetimbangan Panas air dingin (T 1 ), WKg Pasta Tomat q kg/jam 90 o C 40% padatan q 4 Air hangat T 2 (T 2 > T 1 ; T 2 -T 1 = = 10 o C) T 2 = T o C q 3 q 2 Pasta tomat 20 o C q 1 + q 3 = q 2 + q 4 KESETIMBANGAN PANAS contoh 1 Atau: q 1 + q 3 = q 2 + q 4 q 2 = q MJ = q J = 41,870 (w) J w = 4759 Kg Σ Panas yang hilang dari pasta tomat = Σ Panas yang diserap oleh air pendingin J J kg 1 1 KgK KgK + o 100 kg ( ) K = W 4187 ( T T ) K 100 (284676) (70) = 41,870 W W = 4759 Kg Purwiyatno Hariyadi/ITP/Fateta/IPB 23

24 KESETIMBANGAN PANAS contoh 2 Pemblansiran hancuran tomat dengan uap 1 Hancuran tomat: 949% H 2 O 51% padatan 70 o F 2 Uap yang digunakan: uap jenuh pada 1 atm (212 o F) 3 Kondensat uap akan mengencerkan hancuran tomat dan suhu hancuran tomat keluar = 190 o F BTU 4 C padatan tomat = 05 o lb F Hitung: Konsentrasi total padatan hancuran tomat yang dihasilkan KESETIMBANGAN PANAS contoh 2 Basis: 100 lb 212 o F uap Hancuran tomat 70 o F 949% H 2 O 51% padatan Hancuran tomat masuk H 2 O Hancuran tomat panas 190 o F 949 lb air, 70 o F h 1 = BTU (daftar uap) lb 51 lb padatan, 70 o BTU F h 2 = C p (T -T o ) = 05 (70-0) = 35 lb T 0 =T =T ref =0 =0 o F Purwiyatno Hariyadi/ITP/Fateta/IPB 24

25 KESETIMBANGAN PANAS contoh 2 Uap masuk 212 o F uap Hancuran tomat 70 o F 949% H 2 O 51% padatan H 2 O X lb, h 3 = BTU lb (Tabel Uap) Produk Hancuran tomat panas 190 o F (949 + x) lb air, 190 o F h 4 = 158 BTU (Tabel Uap) lb 51 lb padatan, 190 o F h 5 = C p (190-0) = 85 Total keseimbangan entalpi: h 1 + h 2 + h 3 = h 4 + h 5 BTU lb KESETIMBANGAN PANAS contoh 3 Udara, 211 o C, 0002 H 2 O/udara kering (w/w) Uap jenuh 1211 o C Apel 211 o C 80% H 2 O 454 Kg/jam PEMANAS 767 o C H 2 O daur ulang Udara 433 o C 004 H 2 O/ud (w/w) Apel kering 10% H 2 O 377 o C Notasi: q 1 : entalpi air dalam udara masuk (uap pada 1211 o C) q 1 2 : entalpi udara kering pada 211 o C q 3 : entalpi air dalam apel masuk (air pada 211 o C) q 4 : entalpi padatan dalam buah apel masuk pada 211 o C q : masukan panas q 5 : entalpi air dalam udara keluar (uap pada 433 o C) q 6 : entalpi udara kering keluar (433 o C) q 7 : entalpi air pada apel keluar (377 o C) q 8 : entalpi padatan dalam apel keluar (377 o C) Purwiyatno Hariyadi/ITP/Fateta/IPB 25

26 KESETIMBANGAN PANAS contoh 3 Kesetimbangan Entalpi : q + q 1 + q 2 + q 3 + q 4 = q 5 + q 6 + q 7 + q 8 Kesetimbangan massa untuk padatan apel : (02) (454) = x (09) x= berat apel kering x = 1009 Kg/hr Kesetimbangan air: Air hilang dari apel = air diterima oleh udara pengering = 3551 Kg/jam Per kilogram udara kering ( ) = 0038 Mis W = massa udara yang kering (Kg) Total air yang diterima = 0038 (w) kg 3531 = 0038 w w = Kg udara kering/jam Kg air Kg udara kering KESETIMBANGAN PANAS contoh 3 q 1 = entalpi air dalam udara masuk (uap pada 211 o C) Tabel uap h q = MJ/kg (interpolasi) q 1 = Kg air Kg ud kering q 1 = mj Kg ( 92921kg udkering ) mj Kg q 2 = entalpi udara kering pada 211 o C q 2 = mc p dt - mc p (T 2 -T ref ) q 2 Dari tabel 25 o C: C pm = 1008 J/KgK 50 o C: C pm = 1007 J/KgK Asumsi: C pm pada 211 o C = 1008 J/KgK = J KgK ( 92921kg udkering ) 1008 ( 211-0) K q 2 = Purwiyatno Hariyadi/ITP/Fateta/IPB 26

27 KESETIMBANGAN PANAS contoh 3 q 3 = entalpi air dalam apel masuk (air pada 211 o C) Tabel uap h f = MJ/kg (interpolasi) q 3 = 454 (08) (008999) = mj q 4 = entalpi padat dalam apel (211 o C) q 4 = (454) (02) (83736) (211-0) = mj J C p padatan = KgK q = entalpi air dalam udara kering (433 o 5 C) Kg air q 5 = (92921 kg ud Kering) (004 ) (h 9 pada 433 o C) Kg udkering Tabel uap h 9 = mj/kg Uap jenuh 140 o C KESETIMBANGAN PANAS contoh 4 Puree buah, 20 o C 12% padatan evaporator Puree buah, 100 Kg/jam 40 o C 40% padatan Kondensat 110 o C uap, 40 o C Air dingin, 20 o C KONDENSOR Kondensat, 37 o C Air hangat, 30 o C a hitung laju aliran masing-masing produk (kondensat) b hitung konsumsi uap (uap jenuh yangdipakai, 140 o C, akan berkondensasi pada 110 o C) C total padatan = 210 kj/kgk C air = 419 kj/kgk C pada kondensor: hitung laju aliran air dingin (gunakan Tabel Uap Purwiyatno Hariyadi/ITP/Fateta/IPB 27

28 TERIMAKASIH SELAMAT BELAJAR Purwiyatno Hariyadi/ITP/Fateta/IPB 28