AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri. Chapter 6 Energy Balances on Chemical Processes

Transcription

1 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Chapter 6 Energy Balances on Chemical Processes

2 Thermodynamics system surroundings system boundary

3 Forms of Energy 1. Energy Possessed within the system 1.1 Internal Energy 1.2 Kinetic Energy 1.3 Potential Energy 2. Energy transferred across the boundary 2.1 Heat 2.2 Work

4 Forms of energy that possessed within the system Define E = total energy (kj), e = E/m (kj/kg) In an absence of exerted energy, E = U + KE + PE when, or e = u + ke + pe U = internal energy (kj), u = U/m (kj/kg) KE = kinetics energy (kj), ke = KE/m (kj/kg) PE = potential energy (kj), pe = PE/m (kj/kg)

5 Internal energy (U) Kinetics energy (KE) Potential energy (PE) He v He He He He He He He He He He He He He KE mv 2 2 PE mgz z Related to molecular structure and activities in a system When m = mass of a car v = velocity of a car When m = mass of an airplane g = gravitational acceleration z = elevation of an airplane

6 Point Function (State Function) Properties such as T, P, v are point function (or state function), they have exact differentials designated by d For example; Volume change by process A is dv A ; dv A = V 2 V 1 = ΔV A Volume change by process B is dv B ; dv B = V 2 V 1 = ΔV B It means the volume change by process A and B are the same although the system takes different path.

7 Energy Transfer by Work Work (W) is an energy transfer associated with a force (F) acting through a displacement (s), when both are measured in the same direction. It is a path function. δw = F ds (1) Integration of Eq. (1) from state 1 to state 2 gives amount of work done 2 2 W = F ds (2) 1 1

8 Work has magnitude and direction! Formal sign convention of work: Work done by the system is positive (W > 0) Work done on the system is negative (W < 0) surroundings Work done by system system Work done on system

9 The time rate at which work is done by or on the system is called power P = δw /dt (Watt, hp) (3) Efficiency is defined as E = actual work maximum work

10 Energy Transfer by Heat (Q) A form of energy transferred solely by a temperature difference between system and surroundings A path function A positive sign of heat refers to heat transfer into a system (from surroundings) A negative sign of heat refers to heat transfer from a system (to surroundings)

11 Define: Q = heat amount (kj or Btu) q = heat transfer per unit mass q Q m (kj/kg or Btu/lb m ) Q = rate of heat transfer (Watt or Hp) Q Q dt (4)

12 A Process with heat transfer surroundings system Q out (-) Q in (+) The system is under an adiabatic process when Q = 0 Adiabatic process Isothermal process

13 Modes of heat transfer Heat transfer by conduction Heat transfer by convection Heat transfer by radiation

14 Enthalpy (H,h) Enthalpy is a combined property of internal energy (U) and pressure-volume product (PV, or Flow Work). H = U + PV (kj) or h = u + Pv (kj/kg)

15 Flow work Control volume involves mass flow across their boundary and some work is required to push the mass into or out of the control volume. This work is called flow work or flow energy. If Then, P = fluid pressure A = cross-sectional area of the fluid L = distance of fluid movement F = PA W flow = PAL = PV (5) W flow is a combination property for flowing system only!

16 Specific heat (C v and C p ) Specific heat is the energy required to raise the temperature of unit mass of substance by 1 degree Specific heat at constant volume, C v, occurred when the system s volume is constant C v u T v (kj/kg K) or (kj/kmol K) Specific heat at constant pressure, C p, occurred when the system s pressure is constant (6) C p h T p (kj/kg K) or (kj/kmol K) (7)

17 Internal energy, enthalpy and specific heats of ideal gas From ideal gas EOS: Pv = RT (8) Enthalpy: h = u +Pv (9) Combine (1) and (2) h = u +RT (10) Since u = u(t), then h = h(t)

18 Internal energy, enthalpy and specific heats of ideal gas (cont.) Then specific heats are function of T only: du = C v (T) dt (11) And dh = C p (T) dt (12) Regarding Eq.(23) h = u +RT (13) dh = du +RdT (14) Replacing Eq.(12) and (13) into (14) C p = C v +R (15)

19 Internal energy, enthalpy and specific heats of solid and liquid Solid and liquid are incompressible substance; dv = 0 and C p = C v = C (16) Internal energy change (du): du = C av (T 2 T 1 ) (17) Enthalpy change (dh): dh = du + vdp = C av (dt) + vdp (18)

20 The 1 st law of Thermodynamics (the law of energy balance) E E in E out (19) when and E = Energy change of the system Contributed by the change of internal energy, kinetic energy and potential energy E in E out = Energy transfer across the system boundary Contributed by work, heat and mass flow across the boundary

21 Application of Energy Balances for Non-reactive System Simplified Energy Balance Equation: (U + PE + KE) system = Q - W + (H + PE + KE) flows (20) When U = internal energy KE = kinetic energy PE = potential energy Q = heat transfered to the system W = work done on the system H = enthalpy = U + PV

22 The three common systems found in engineering applications are 1. Closed system 2. Open system with steady-state flow 3. Open system with unsteady-state operation

23 Energy changes on Non-reactive systems Changes in pressure at constant temperature Changes in temperature at constant pressure Phase change at constant pressure and temperature Mixing or dissolving at constant pressure and temperature

24 Application of Energy Balances for Non-reactive System Closed (and stationary) Systems: There will be no PE nor KE for the system. There is no material flow across the boundary. U system = Q - W (21)

25 Example 1: reaction Energy Balance on a closed system without 10-lbm of CO 2 at 80F are stored in a fire extinguisher that has a volume of 4.0 ft 3. How much heat must be removed from the extinguisher so that 40% of CO 2 becomes liquid? CO 2 CO 2 Initial state: CO 2 gas T 1 = 80 F m 1 = 10 lbm V 1 = 4.0 ft 3 Final state: 40% liquid m 2 = 10 lbm V 2 = 4.0 ft 3

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27 Application of Energy Balances for Non-reactive System Open system with steady-state flow: There is no change of system s energy Total energy flows in is equal to total energy flows out E in = E out (22) (H+PE+KE) incoming flows + Q in + W done on system = (H+PE+KE) outgoing flows + Q out + W done by system

28 Example 2: Mixing 2 streams of sucrose solutions with different concentrations; Feed 1 with mass fraction of sucrose of 0.1 and Feed 2 with mass fraction of sucrose of 0.5, will give the 100 kg sucrose solution product with 0.3 mass fraction of sucrose, as specified by customer s demand. Given that the specific enthalpy for Feed 1, Feed2 and Product are 200 kj/kg, 300 kj/kg, and 275 kj/kg, respectively. Find (1) ratio of Feed 1 and Feed 2 to obtain the product specified, and (2) the amount of heat transfer between the system and surroundings.

29 Solution: Basis: 100 kg Product Assuming that this is a steady-state flow system, Total material balance is F 1 + F 2 = P Sucrose balance is 0.1 F F 2 = 0.3(100 kg) Therefore, F 1 = 50 kg and F 2 = 50 kg Ratio of Feed1 to Feed 2 is 1 ANS Furthermore, energy balance shall be used to find hear transfer amount. Total energy balance is h 1 F 1 + h 2 F 2 + Q = -h p P (200 kj/kg)(50 kg) + (300 kj/kg)(50 kg) + Q = (275 kj/kg)(100 kg) Q = 2,500 kj ANS

30 Example 3: Energy Balance on an open, steady-state system without reaction Hot gas 100 kmol/min at 500 C is being cooled to 300 C by transferring heat to the liquid water that enters at 20 C and exits at 213 C, in an insulated heat exchanger. The cooling water does not mix with the gas. Calculate the mass flow rate of water stream. m W,1 T W,1 m G,3 1 m 3 4 G,4 T G,3 2 T G,4 gas %mol CO 2 20 N 2 10 CH 4 30 H 2 O 40 m W,2 T W,2

31 Given information for Example 3 Data retrieved from reference CD Component in hot gas h from 3 to 4 (kj/kmol) CO 2-9,333 N 2-6,215 CH 4-11,307 H 2 O -7,441 Data retrieved from steam table T ( C) H f (kj/kg)

32 Application of Energy Balances for Non-reactive System Open system with unsteady-state operation: Mass (or mole) of the system change with time. Energy possessed by system changes with time. (U + PE + KE) system = Q - W + (H + PE + KE) flows Neglect the changes of KE and PE, U final U initial = Q W + H incoming flows - H outgoing flows (23)

33 Sensible Heat Sensible heat is a specific enthalpy change associated with temperature change of pure substance due to the process in a single phase system. Where

34 Heat Capacity of a Mixture For a mixture of gases or liquids, calculate the total enthalpy changes as the sum of the enthalpy changes for the pure mixture components. (24)

35 Latent Heat Latent heat is a specific enthalpy change associated to phase transition at constant pressure and temperature. The specific enthalpy change when substance changes from liquid to vapor phase, or oppositely, is a Latent heat of vaporization; h vap. The specific enthalpy change when substance changes from solid to liquid phase, or oppositely, is a Latent heat of fusion; h fus or Heat of melting; h melt

36 Example 4: Energy Balance on a condenser

37 Data required for Example 4 Substance Cp (kj/mol C) Acetone (liq) x10-5 T Acetone (vap) x10-5 T 12.78x10-8 T 2 Nitrogen (gas) x10-5 T x10-8 T 2 h vap,ac = kj/mol

38 Heat of Solution (or Mixing) The heat of solution; h sol, is defined as enthalpy change for a process in which 1 mole of solute is dissolved in r moles of a liquid solvent at constant temperature. As r becomes large (solution is dilute), h sol approaches a limiting value. The heat of mixing has the same meaning as the heat of solution when the process involves mixing two fluids rather than dissolving solid or gas in a liquid.

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40 Example 5: Concentration of Sulfuric acid A 5.0 wt% H 2 SO 4 solution at 60 F is to be concentrated to 40.0 wt% by evaporation of water. The concentrated solution and water vapor emerge from the evaporator at 180 F and 1 atm. Calculate the rate at which heat must be transferred to the evaporator to process 1,000 lb m /h of the feed solution.

41 Application of Energy Balances for Systems with Chemical Reactions Concerning the chemical reactions, more terms of energy involved to the reactions are applied; The Standard Heat (Enthalpy) of Formation The Heat (Enthalpy) of Reaction The Sensible Heat (Enthalpy Change) The Heat (Enthalpy) of Combustion

42 The Standard Heat of Formation ( H F) The standard heat of formation is the enthalpy change associated with the formation of 1 mol of a compound from its constituent elements and products in their standard state of 25 C and 1 atm. Example: The standard heat of formation for CO 2 is obtained from the reaction of C-element (C) and O-element (O 2 ), C(s) + O 2 (g) CO 2 (g) [1] H F,CO2 is heat transferred from a steady-state reaction [1]. At standard state, Q = kj/kmol CO 2 = H F,CO2

43 Quiz: Calculate the standard heat of reaction for the following reaction. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Given that; Compound Standard specific heat of formation (kj/gmol) NH O 2 0 NO H 2 O

44 The Standard Heat of Reaction ( H RXN ) The standard heat of reaction is the enthalpy change when stoichiometric quantities of reactants in the standard state (25 C and 1 atm) react completely to produce the products. For example: Consider the steady-state flow process for Benzene (C 6 H 6 ) production from Cyclohexane (C 6 H 12 ), at 25 C and 1 atm. C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) [2] The energy balance is reduced to be Q = H RXN = (1) H F,C6H12 + (-1) H F,C6H6 + (-3) H F,H2

45 At 25 C and 1 atm C 6 H 6 C 6 H 12 Reactor H 2 C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) [2] Compound Standard specific heat of formation (kj/gmol) C 6 H H 2 0 C 6 H

46 When the reaction is not completed. Regarding the process to produce cyclohexane from benzene, if the extent of reaction is given as, n i,out = n i,in + i H RXN = [(1) H F,C6H12 + (-1) H F,C6H6 + (-3) H F,H2 ] Suppose the fractional conversion is given as equal to 0.80, and benzene is a limiting reactant, fed to the process for 1 mole. Find the heat of reaction actually transferred.

47 Example 6: What can we do with H RXN? For the Bhopal process to produce carbaryl (C 12 H 11 O 2 N), methyl isocyanate (C 2 H 3 NO) which is a very toxic gas is generated during the process. C 3 H 2 NH 2 + COCl 2 C 2 H 3 NO + 2HCl (a) C 2 H 3 NO + C 10 H 8 O C 12 H 11 O 2 N (b) To eliminate the generation of methyl isocyanate, an alternative process for green chemistry is proposed. C 10 H 8 O + COCl 2 C 11 H 7 O 2 Cl C 11 H 7 O 2 Cl + CH 3 NH 2 C 12 H 11 O 2 N + HCl (d) (c) Show the alternative process is safer than the Bhopal process.

48 Example 7: What if the reaction does not take place at standard conditions? An inventor believes that he has developed a new catalyst that can make the gas phase reaction as shown here; CO 2 + 4H 2 2H 2 O + CH 4 The conversion of CO 2 is 40%. Assume that 1.5 mole of CO 2 enters the reactor at 700 C together with 4 moles of H 2 at 100 C. Determine the heat of reaction if the exit gas leaves at 1 atm and 500 C.

49 Heat of Reaction and Sensible Heat CO 2 at 700 C H 2 at 100 C Sensible Heats Sensible Heats Actual Reaction Product(s) at 500 C (H T H 25 ) Reactants Reaction at 25 C and 1 atm (H 25 H T ) Products H RXN

50 Example 8: Batch and reactive process For the commercial production of citric acid (C 6 H 8 O 7 ) in a batch process, three different phases occur for which the stoichiometry are slightly different. Initial phase, between hr; 1 gmol glucose gmol O 2 (g) 3.81 gmole biomass gmole citric acid gmol CO 2 (g) gmole polyols Medium phase, between hr; 1 gmol glucose gmol O 2 (g) 1.54 gmole biomass gmole citric acid gmol CO 2 (g) gmole polyols Late phase, between hr; 1 gmol glucose gmol O 2 (g) gmole polyols 0.86 gmole citric acid gmol CO 2 (g)

51 In anaerobic batch process, 30% glucose solution at 25 C is introduced into a fermentation vessel. Citric acid is to be produced by using the fungus Aspergillus niger. Stoichiometric sterile air is mixed with a culture solution by a 100 hp aerator. Only 60% overall of the glucose supplied is eventually converted to citric acid. The initial phase is run at 32 C, the medium phase is run at 35 C, and the late phase is run at 25 C. Base on the given data, how much net heat has to be added or removed from the fermenter during the production of a batch of 10,000 kg of citric acid. Aerator 100 hp 30% glucose 25 C 32 C 35 C 25 C 10,000 kg Citric acid At the beginning Initial phase Medium phase Late phase At the End

52 Example 9: M& E Balances for Multiple Units Process Air 60 F 450 F Heat Exchanger Reactor 1 CO(g) 60 F CO(g) CO 2 (g) N 2 (g) 1400 F Reactor F 800 F Steam Gen Products 1000 F SO 2 2,200 lbmol/hr 77 F Sat. steam At 450 psig H 2 O 60 F

53 Example 9: (Cont.) CO is burned with 80% of the theoretical air in reactor 1. The combustion gases are used to generate steam, and also to transfer heat to the reactants in reactor 2. A portion of the combustion gases that are used to heat the reactants in reactor 2 are recycled. SO 2 is oxidized in reactor 2. Calculate the moles of CO burned per hour in reactor 1. The gas involved in the SO 2 oxidation do not come in direct contact with the combustion gas used to heat the SO 2 reactants and products. Data for reactor 2 are: Reactants Mole fraction T ( F) Products Mole fraction T ( F) SO O Total SO O SO Total 1.000

54 The Heat of Combustion ( H C) Standard heat of combustion was commonly used to calculate the enthalpy changes when chemical reactions occur. The conventional used are as follows: 1. The compound is oxidized with oxygen or some other substances to the products CO 2 (g), H 2 O(l), HCl(aq), and so on. 2. The reference conditions are still 25 C and 1 atm. 3. Zero values of H C are assigned to certain substances such as CO 2 (g), H 2 O(l), HCl(aq) and O 2 (g). 4. Stoichiometric quantities react completely. H C (25 C) i Products n H C, i i Reactan ts n H C, i

55 Example 10: Conversion of H C to the corresponding H F The heat of combustion of liquid lactic acid (C 3 H 6 O 3 ) is -1,361 kj/gmol [Electronic J. of Biotech., 7 No. 2, 2004]. What is the corresponding standard heat of formation? The formation of lactic acid can be considered as; 3C(s) + 3H 2 (g) + 1.5O 2 (g) C 3 H 6 O 3

56 Psychrometric Charts and Applications Psychrometric charts, or Humidity charts give information on humidity of mixture of airwater system The applications of psychrometric charts are humidification, dehumidification, drying and Cooling tower.

57 Humidity (H) Humidity means the ratio of mass of water vapor(moisture) per unit mass of dry air H = m water m dry air = 18P water 29(P total P water )

58 Relative Humidity (RH) For any gas mixture, When, RH = P vapor P P vapor = partial pressure of vapor in gas mixture P* = vapor pressure of the vapor component Note: Vapor pressure is a pressure of vapor component when it is saturated in gas mixture at a given temperature and mixture.

59 %Relative Humidity (%RH) Consider the moisture (water) in air, When, %RH = P water P water P water = partial pressure of moisture in air mixture P water * = vapor pressure of the moisture at the given temperature

60 Wet-bulb Temperature (T wb ) Wet-bulb temperature is the equilibrium temperature of water vapor in the air. Dry-bulb Temperature (T db ) Dry-bulb temperature is system temperature.

61 Dew Point Temperature (T dp ) Dew point temperature is the temperature at which the water vapor in the air will first condense at constant pressure and composition.

62 Psychrometric Charts

63 Psychrometric Charts Reading

64 Psychrometric Charts Reading

65 Psychrometric Charts Reading

66 Psychrometric Charts Reading

67 Example 11: Determine Properties of Moist Air from the Psychrometric Chart List all the properties you can find on the psychrometric chart for moist air at a dry-bulb temperature of 90 F and a wet-bulb temperature of 70 F. A) Dew point: When the air at the said condition is cooled at constant pressure, it eventually reaches a temperature at which the moisture begins to condense at about 60 F. B) Relative humidity: By interpolation between 30% RH and40%rh, the relative humidity of the air at the said condition is about 37%RH.

68 Example 11 Determine Properties of Moist Air from the Psychrometric Chart (Cont.) List all the properties you can find on the psychrometric chart for moist air at a dry-bulb temperature of 90 F and a wet-bulb temperature of 70 F. C) Humidity: The humidity can be read from the right-hand ordinate as lbm H 2 O/lbm dry air. D) Humid volume (specific volume): By interop;ation between ft 3 /lbm, the humid volume is about 14.1 ft 3 /lbm.

69 Example 11 Determine Properties of Moist Air from the Psychrometric Chart (Cont.) List all the properties you can find on the psychrometric chart for moist air at a dry-bulb temperature of 90 F and a wet-bulb temperature of 70 F. E) Enthalpy: The enthalpy value for saturated air with a wet-bulb temperature of 70 F is 34.1 Btu/lbm.

70 Example 12: Combined Material Balance and Energy Balance for a Cooling Tower You have been requested to redesign a water cooling tower that has a blower with a capacity of 8.30 x 10 6 ft 3 /hr of moist air. The moist air enters at 80 F and a wet-bulb temperature of 65 F. The exit air is to leave at 95 F and 90 F wet-bulb. How much water can be cooled in lbm/hr if the water to be cooled enters the tower at 120 F, leaves the tower at 90 F and is not recycled?