CHEE 221: Chemical Processes and Systems

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CH 221: Chemical Processes and Systems Module 4. nergy Balances without Reaction Part a: Introduction to nergy Balances (Felder & Rousseau Ch 7.0 7.4

nergy and nergy Balances: very chemical rocess volves the transfer of energy: Distillation (hase changes energy added for volatilization at reboiler and energy removed at condenser combustion ower generation reactors breag and formg chemical bonds fluid transort umg nergy balances are used to: determe the amount of energy that flows to or out of each rocess unit that must be added or removed calculate the energy requirement (and $ for the rocess and along with caital costs assess equiment alternatives. assess ways of reducg energy requirements (e.g. exchanges with the rocess order to imrove rocess rofitability CH 221 2

Units of nergy nergy has units of force times distance (mass length 2 /time 2 N m (=Joule dyne cm (=erg ft lb f SI units CGS system British engeerg system nergy origally defed as the amount of heat required to raise the temerature of a secified mass of water by one degree at 1 atm Unit Symbol Mass of H 2 O Temerature Interval ilocalorie cal 1 g C calorie cal 1 g C British thermal unit Btu 1 lb m F Unit conversions found side the front cover of F&R CH 221 3

Forms of nergy The total energy of a system has three comonents: 1. Ketic nergy ( energy due to the translational motion of the system as a whole 1 mu 2 1 m u 2 2 2 J [g][m/s] W J/s 2 2 [g/s][m/s] 2. Potential nergy ( energy due to the osition of the system a otential field (e.g. earth s gravitational field (g = 9.8 m/s 2 m u mass (g velocity (m/s mgz J [g][m/s 2 ][m] 2 mgz W J/s [g/s][m/s ][m] m h height mass (g of obect (m 3. Internal nergy (comg u F&R Ch 7.1 CH 221 4

xamle: K and P Benzene (SG = 0.879 flows to a rocess unit through a 2 Schedule 40 ie at a rate of 1000 L/h. Calculate for this stream oules/second. The same benzene is now umed to a holdg vessel that is 20 m above the level of the ig. Determe the rate of crease otential energy. F&R Ch 7.2 CH 221 5

Forms of nergy (cont d 3. Internal nergy (U all energy ossessed by system other than etic and otential energy cludg the energy arisg from the: rotational and vibrational motion of molecules with the system teractions between molecules with the system motion and teractions of electrons and nuclei with molecules Internal energy (U is related to enthaly (H U and H are a function of temerature chemical comosition hysical state (solid liquid or gas and only wealy a function of ressure U and H are relative quantities (relative to a reference state absolute values are not secified or nown values must be defed with resect to their reference state this is OK sce we are always terested changes U and H H U PV F&R Ch 7.1 CH 221 6

Intensive vs xtensive Variables; Secific Proerty Intensive Variables deendent of the size of the system e.g. temerature ressure density comosition (mass or mole fraction xtensive Variables deend on the size of the system e.g. mass number of moles volume (mass or molar flow rate and volumetric flow rate etic energy ( 1 2 mu otential energy and 2 ternal energy Secific Proerty a quantity that is obtaed by dividg an extensive roerty by the total amount of the material. denoted by ^ secific volume ( V units of m 3 /g enthaly and ternal energy commonly reorted as secific values U (J/g Ĥ (J/g H U PV F&R Ch 7.4b CH 221 7

xamle: Secific Internal nergy and nthaly The secific ternal energy of steam at 165 ºC and 7 bars ressure is 2571 J/g and the secific volume at the same temerature and ressure is 273 L/g. Calculate the secific enthaly of steam at this temerature and ressure and the rate at which enthaly is transorted by a stream at 165 ºC and 7 bars with a molar flow rate of 20 g mol/h. H UPV CH 221 8

Transfer of nergy In a closed system (no mass transferred across the system boundaries (i.e. batch system energy can still be transferred between the system and the surroundgs two ways: 1. Heat (Q energy that flows due to a temerature difference between the system and its surroundgs always flows from high to low temerature defed to be ositive if it flows to a system (i.e. ut 2. Wor (W energy that flows resonse to any drivg force (e.g. alied force torque other than temerature defed as ositive if it flows from the system (i.e. outut chemical rocesses wor may come for examle from a movg iston or movg turbe A system does not ossess heat or wor. Heat or wor only refer to energy that is beg transferred to the system. F&R Ch 7.1 CH 221 9

First Law of Thermodynamics The First Law of Thermodynamics states that energy can neither be created or destroyed (ust lie total mass Accumulation = In Out + Generation Consumtion But generation=0 and consumtion=0 sce energy cannot be created or destroyed so the general balance becomes: Accumulation = In Out m ( U m out ( out U out out Q W F&R Ch 7.1 CH 221 10

nergy Balances on Closed Systems Closed System no material crosses the system boundary over a eriod of time (e.g. batch rocess. General balance equation is: Accumulation = Inut Outut Although no mass crosses the boundaries energy ut0 and energy outut0 sce energy can be transferred across the boundary. Therefore the balance becomes: fal system energy itial system = energy net energy transferred itial system energy U fal system energy U energy transferred Q W f i i f i f U Q W 1 st Law of Thermodynamics for a Closed System ( = fal itial F&R Ch 7.3 CH 221 11

Notes on nergy Balances for a Closed System U Q W Possible Simlifications: if T system = T surroundgs then Q = 0 sce no heat is beg transferred due to temerature difference if the system is erfectly sulated then Q = 0 (system is adiabatic sce no heat is beg transferred between the system and the surroundgs if system is not acceleratg then = 0 if system is not risg or fallg then = 0 if energy is not transferred across the system boundary by a movg art (e.g. iston imeller rotor then W = 0 if system is at constant temerature (system is isothermal no hase changes or chemical reactions are tag lace and only mimal ressure changes then U = 0 F&R Ch 7.3 CH 221 12

xamles of Closed Systems xamle 1: Heatg water a sealed contaer xamle 2: Comressg a gas a cylder. U Q W U Q W U Q U W CH 221 13

Problem 7.9 F&R Write and simlify the closed system energy balance for each of the followg rocesses and state whether nonzero heat and wor terms are ositive or negative. Beg by defg the system. (a The contents of a closed flas are heated from 25Cto 80C. (b A tray filled with water at 20C is ut to a freezer. The water turns to ice at 5C. (Note: When a substance exands it does wor on its surroundgs and when it contracts the surroundgs do wor on it. (c A chemical reaction taes lace a closed adiabatic (erfectly sulated rigid contaer. (d Reeat art (c only suose the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically the temerature the reactor creased. CH 221 14

nergy Balances on Oen Systems at Steady State Oen System material crosses the system boundary as the rocess occurs (e.g. contuous rocess at steady state. In an oen system wor must be done to ush ut fluid at a ressure P and flow rate V to the system ( PV wor and wor is done by the outut fluid at ressure P out and flow rate V out on the surroundgs as it leaves the system. V (m 3 /s V (m 3 out /s PV Process Unit W P (N/m 2 (N/m 2 out PoutV out W P out Net rate of flow wor done by the system: W fl W out W P out V out P V For several ut and outut W fl outut P V ut P V F&R Ch 7.4a CH 221 15

Flow Wor and Shaft Wor The total rate of wor ( W done by a system on its surroundgs is divided to to arts where: W W s W fl where W s W fl = shaft wor rate of wor done by the rocess fluid on a movg art with the system (e.g. iston turbe rotor = flow wor rate of wor done by the fluid at the system outlet mus the rate of wor done on the fluid at the system let F&R Ch 7.4a CH 221 16

Steady State Oen System nergy Balance The general balance equation for an oen system (i.e. contuous rocess at steady state is: Inut = Outut m ( U m out ( out U out out Q energy ut U W energy outut U out out out U energy transferred Q ( W Q U CH 221 17 s W fl ( W W out out out s fl F&R Ch 7.4c

CH 221 18 Steady State Oen System nergy Balance If is the total rate of energy transort for ut and outut energy outut outut outut outut P V H m U m U ( ( ( ut ut ut ut P V H m U m U ( ( ( ( fl s outut ut W W Q ( fl s ut outut W W Q ( ( ut outut s fl s V P P V W Q W W Q energy transferred

Steady State Oen System nergy Balance P V terms cancel H Q W 1 st Law of Thermodynamics for an Oen System at Steady State ( = outut ut s CH 221 19 H F&R Ch 7.4c outut outut m outut m m H u 2 gz / 2 ut m ut ut m H m u gz 2 / 2

Notes on nergy Balances for an Oen System H Q W s Possible Simlifications: if T system = T surroundgs then Q = 0 sce no heat is beg transferred due to temerature difference if the system is erfectly sulated then Q = 0 (system is adiabatic sce no heat is beg transferred between the system and the surroundgs if energy is not transferred across the boundary by a movg art (e.g. iston imeller rotor then W s 0 if flow and outflow are of the same velocity then 0 if there is no large vertical distance between the lets and outlets of a system then 0 if system is at constant temerature (system is isothermal no hase changes or chemical reactions are tag lace and only mimal ressure changes then H 0 F&R Ch 7.4c CH 221 20

xamle F&R 7.4 2 Five hundred ilograms er hour of steam drives a turbe. The steam enters the turbe at 44 atm and 450C at a lear velocity of 60 m/s and leaves at a ot 5 m below the turbe let at atmosheric ressure and a velocity of 360 m/s. The turbe delivers shaft wor at a rate of 70 W and the heat loss from the turbe is estimated to be 10 4 cal/h. Calculate the secific enthaly change associated with the rocess. CH 221 21

Summary on nergy Balances The First Law of Thermodynamics for a closed (i.e. batch system is: U and the First Law of Thermodynamics for an oen system at steady state (i.e. contuous system is: H CH 221 22 Q W Q W Changes etic and otential energy can be calculated but are usually small for chemical systems. Heat and wor uts are given the roblem or are what you must solve for. The maor tas is calculatg changes U or H: Ch7: Usg tabulated values (steam tables Ch8: Phase changes with no reaction Ch9: nergy balances with reaction s