Supplemental Material for Elementary Principles of Chemical Processes. Chapter 7 Energy and Energy Balances

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1 Chapter 7 Energy and Energy Balance Name: Date: In Chapter 7, it will be hown how energy balance can be ued to olve problem involving chemical procee. Different type of problem including the type of energy and how it i tranferred in and out of a ytem, which can be either open or cloed. The following problem build upon the fundamental covered in your text a applied to hydrogen proceing, hydrogen a an energy carrier, and the ue of hydrogen in fuel cell Energy Balance on a Turbine Ue of Tabulated Enthalpy Data Energy Balance on a One-Component Proce Energy Balance on a Two-Component Proce 7.6- Simultaneou Material and Energy Balance

2 Example Energy Balance on a Turbine A ytem combining a olid-oxide fuel cell with a ga turbine ha been proved to achieve higher operating efficiencie at high preure. The exhaut gae from the fuel cell are entering a team turbine a hown in the following figure. SOLID-OXIDE FUEL CELL P 1 =1 MPa T 1 =800 K e = 105 H O,1 2 m u = 50 1 P 2 = 0.1 MPa T 2 = 400 K e = 105 H O,2 2 m u = J Q = g Determine the haft wor done by the turbine. Ẇ Strategy Thi problem can be olved by performing energy balance on the turbine, uing the information given in the proce diagram to obtain the different quantitie involved in the energy balance. Solution The amount of energy tranferred to or from a ytem a heat or haft wor i equal to the rate of change of the enthalpy, and the potential and inetic energie. Thi definition can be written a follow: H + E + E = Q W p The rate of change of enthalpy i given by: ( ˆ ˆ 2 1) H = H H The enthalpie of the team entering and exiting the turbine can be found in Table 2-05 of Perry Chemical Engineer Handboo, 8 th Edition. P(MPa) T(K) (/)

3 Subtituting the value from the table into the equation for the rate of change of enthalpy give: e H = H = W Since there i a difference in the velocity of the team between the inlet and outlet tream, the difference in the inetic energy mut be obtained a follow: m E = u u ( ) 2 1 Entering the ma flow rate and the velocitie into thi equation yield: e 1 18 g e m m 1 E = 80 2 J E = W The potential energy i due to a difference in the height. Since both inlet and outlet tream are at the ame level, the term for the potential energy in the energy balance can be neglected. The only term that remain to be calculated i the heat lo term Q, which can be obtained by multiplying the ma flow rate by the heat lo in J g : Q = mq By ubtituting the information from the proce diagram we get: e 1 18 g J 1 Q = 1000 e 1 g 1000 J Q = W Now that all the term in the energy balance are nown, we can ubtitute thee quantitie and olve for the haft wor Ẇ : ( ) W = Q E E p H = W.69 W W W = W

4 7.5-1 Ue of Tabulated Enthalpy Data Methanol i the fuel ued for producing electricity in direct-methanol fuel cell. The following table i howing thermodynamic propertie of thi fuel. Thi data wa obtained from Table 2-24 of Perry Chemical Engineer Handboo, 8 th Edition. Phae T, K P, Pa L ˆV,, Liquid x Liquid x Vapor Vapor Calculate the change in enthalpy and internal energy from 285 K to 15 K, and from 0 K to 45 K. Strategy We need to ue the value from the table and the definition of internal energy to find the olution to thi problem. Solution The change in the enthalpy will be given by the following equation: Hˆ = Hˆ Hˆ f i where: = Enthalpy of methanol at the final temperature f = Enthalpy of methanol at the initial temperature i Firt we will calculate the change in enthalpy from 285 K to 15 K (liquid phae). Subtituting the value from the table into the equation for the change in enthalpy, we have: = K - 15 K = 285 K - 15 K

5 For the temperature change from 0 to 45 (vapor phae), the change in enthalpy can be calculated a follow: = 0 K - 45 K = K - 45 K The value we jut obtained for the change in enthalpy will be ued for calculating the change in internal energy for methanol in both the liquid and vapor phae. The internal energy definition i given by the following equation: The term Uˆ = H ˆ (PV) ˆ (PV) ˆ can be rewritten a follow (PV) ˆ = (P Vˆ P V ˆ ) f f i i Where the ub index f refer to the final condition, and the ub index i to the initial condition of methanol. Inerting thi equation into the equation for Û yield: Uˆ = H ˆ (P Vˆ P V ˆ ) f f i i We can inert the quantitie from the table to determine the change in internal energy for the liquid phae when increaing it temperature from 285 K to 15 K. 2 L 1 m Û = Pa x K 15K 1000 L L 1 m Pa 1000 L Û = 285K 15K The ame procedure i followed to difference in the internal energy of methanol in the vapor phae from 0 to 45 K. L L 1 m Û = Pa Pa L Û = K 45K

6 7.6-1 Energy Balance on a One-Component Proce In a team-methane reforming proce for hydrogen production, water i fed to a boiler which will produce uperheated team. Thi team i ued to heat the reactant before entering the reactor tube. A diagram of the proce occurring in the boiler i hown below: g H2O 425 T = 0 C P = 4.8 MPa Superheated Steam T = 60 C P = 4.8 MPa Strategy The tarting point for olving thi problem will be the general energy balance equation. We will alo need the enthalpie of both the liquid water entering the boiler and the team produced. Solution We can tart by applying the energy balance equation to thi proce: H + E + E = Q W p For the proce taing place in thi problem, the following aumption can be made: E = 0 Since there i no information given about the velocitie of the inlet and p outlet tream the velocity difference will be neglected E = 0 There i no coniderable difference in height between the inlet and outlet tream W = 0 There are no mechanical part moving Thu, the energy balance equation will be decribed by: Q = H The change in the enthalpy H = n(h ˆ H ˆ ) out in H can be calculated a follow:

7 The enthalpy of the liquid water entering the ytem can be obtained by interpolation uing the data from Table 2-05 from Perry' Chemical Engineer Handboo, 8 th Edition, ummarized in the following table. 1 MPa 4.8 MPa 5 MPa 00 K = 0.15 K,P = 1 = 0.15 K,P = 4.8 = 0.15 K,P = 5 MPa 400 K We can interpolate down at contant preure, then acro at contant temperature. From the table hown above, we need to calculate firt the value of the ar enthalpy at a contant temperature of 0.15 K at a preure of 1 MPa. We etup the linear interpolation a: ˆ ˆ Tmid T H T= 0.15 K,P= 1 MPa T = low Thigh T ˆ ˆ low T H Tlow Entering the value from the table into thi equation and olving = 0.15 K,P = 1 MPa yield: 0.15 K 00 K = 0.15 K,P= 1 MPa K 00 K 0.15 K,P= 1 MPa The ame procedure i repeated at the temperature of 0.15 K, only that thi time will be done for the preure of 5 MPa. ˆ T H mid Tlow = T T Hˆ T= 0.15 K,P= 5 MPa Hˆ high T Tlow Entering the value from the table into thi equation and olving = 0.15 K,P = 5 MPa yield: 0.15 K 00 K = K,P= 5 MPa K 00 K

8 @T= 0.15 K,P= 5 MPa = Now, the only remaining tep to be done i to obtain the value of the ar enthalpy of liquid water at the temperature and preure of the feed tream: ˆ P H mid Plow = P P Hˆ T= 0.15 K,P= 4.8 MPa high P Plow Plow Subtituting the value from the table into thi equation, we can olve for = = to 0.15 K,P 4.8 MPa MPa 1 MPa 0.15 K,P= 4.8 MPa 5 MPa 1 MPa 0.15 K,P= 4.8 MPa = If we ubtitute the calculated enthalpie into the initial table, we get: 1 MPa 4.8 MPa 5 MPa 00 K 0.15 K 400 K

9 To determine the enthalpy of the uperheated team exiting thi proce, a imilar procedure will be followed, only that the value for vapor water at T = 60 C and P = 4.8 MPa will be determined tough interpolation. The following table i howing the enthalpy value for the team: 600 K MPa 4.8 MPa 5 MPa = 6.15 K,P = 1 = 6.15 K,P = 4.8 = 6.15 K,P = 5 MPa 700 K After doing the interpolation procedure and entering the value into the table, we get: 600 K K 700 K Now we can determine the heat required to boil the feed water, by calculating the enthalpy change: Q = H = n(h ˆ H ˆ ) out 1 MPa 4.8 MPa 5 MPa in Entering the correponding quantitie into thi equation give: g H O 1 H O 1000 H O 1 18 g H O 1 H O Q = H = Q = W

10 7.6-2 Energy Balance on a Two-Component Proce Hydrogen can be produced by team-methane reforming. The feed to thi proce i heated to a temperature of 450 C before entering the reaction chamber. What i the required flow rate of each ga if the heater conume 2.5 W of power to bring the reaction gae to the operation condition? The feed to the reaction chamber contain a team/methane ar ratio of. e H2O n, Strategy We need to determine the unnown parameter by performing an energy balance on the heater and ue the information about the team/methane ratio given in the problem tatement. Solution H2O,in T = 210 C P = 1.6 MPa H= e CH n, CH 4,in T = 0 C P = 1.6 MPa H= The energy equation i given by: H + E + E = Q W p 4 Q Steam/Methane mixture e n out, T = 450 C P = 1.6 MPa H H2O = H CH4 =26.9 The change in the inetic and potential energie will be negligible ince there are no ignificant difference in the velocity and the height between the inlet and outlet tream. Since there are no mechanical part moving in the heater, the haft wor will be alo neglected. Thu, the equation i reduced to: Q = H The change in enthalpy for thi proce will be given by: H = H ˆ ( Hˆ + H ˆ ) out mix H O, in H O,in CH, in CH,in

11 Subtituting thi equation into the energy balance equation yield: Q = H ˆ ( Hˆ + H ˆ ) out mix H O, in H O,in CH, in CH,in For the team-methane reforming proce, the ratio of the number of e of team to the number of e of methane i equal to. Thi can be decribed by the following equation: = H O, in CH, in 2 4 Thi equation can be ued to determine the ar flow rate of the mixed gae exiting the heater: = + = + out CH, in H O, in CH, in CH, in Subtituting the lat two equation in the general energy equation applied for thi problem give: Q = 4 H ˆ ( Hˆ + H ˆ ) CH, in mix H O,in CH, in CH,in Another quantity that remain unnown in thi equation i the enthalpy of the mixture exiting the heater, which can be calculated a follow: mix H ˆ = y H ˆ + y H ˆ mix CH CH H O H O Since the only component preent in the outlet tream are water and methane, and the ratio of e of team to e of methane i nown, the ar fraction can be obtained a follow: y + y = 1 CH H O 4 2 y + y = 1 CH CH 4 4 Solving for the ar fraction of methane y CH4, we get: y CH 4 e CH = 4 and the ar fraction of water can be obtained by ubtituting the ar fraction into the equation for the um of the ar fraction: y = 1 y = 1 H O CH 2 4

12 y e H O 2 = H O 2 Now the enthalpy of the mixture can be calculated by ubtituting the ar fraction and enthalpie of methane and water to yield: ˆ ˆ ˆ H = y H + y H = 26.9 mix CH CH H O H O = mix Subtituting the nown enthalpie and the conumed power into the energy balance equation allow u to olve for the ar flow rate of methane being fed to the heater: 2.5 W = 4n n n CH, in CH, in CH, in CH, in W = = n CH, in 4 1 n = CH, in 4 Recalling the ar ratio of the component in the proce: = H O, in CH, in 2 4 Entering the ar flow rate of methane into thi equation give: = n H O, in 2 n = H O, in 2

13 7.6- Simultaneou Material and Energy Balance Synthei ga i produced in a central-cale coal gaification proce at a preure of 27 bar and a temperature of 260 F. Thi ynga tream will be mixed with team before entering a hift reactor to further produce hydrogen. A diagram of thi proce i hown in the figure: 6 H2O 1 = T = 260 C P = 2.4 bar H 1 =6.501 MIXING POINT Synga n 2, T = 260 C P = 27 bar = Steam/Synga mixture e n, T = 260 C P = 27 bar H = = Determine the unnown flow rate for the mixer operating at adiabatic condition. Strategy In order to olve thi problem, material balance mut be done around the mixing point. Since the compoition of each of the tream are unnown, we alo need to perform an energy balance on the ytem. Solution We can tart by writing an overall material balance on the ytem: Input = Output + = 1 2 A it can be een in thi equation, there are two unnown variable. Hence, we will need another equation in order to be able to olve thi problem. There i no information given in the problem tatement about the compoition of the tream but the enthalpie given in the proce diagram can be ued to perform an energy balance on the ytem: Q W = H + E + E p

14 The following aumption can be done for the proce decribed in the problem tatement: Q = 0 (adiabatic ytem) W = 0 (the tream are jut being added but there are no mechanical part moving) E = 0 p E = 0 (there are no ignificant height difference) (the velocity gradient between the input and output tream i negligible) Thu, the energy equation will be given by: H = 0 The change in the enthalpy for thi ytem will be given by: H = Hˆ Hˆ = 0 outlet i i i i inlet Writing thi equation in term of the variable in thi problem yield: ( ˆ ) H = H ˆ + H = Now we have both the energy and material balance equation, which can be olved imultaneouly to determine the flow rate ṅ and ṅ. Solving for ṅ from the overall 2 2 material balance, we get: = 2 1 Subtituting thi equation into the energy balance equation give: H ˆ + Hˆ = 0 2 We can enter the quantitie for all nown variable in thi equation to get: 6 n n 0 + =

15 Thi equation can be olved for the flow rate ṅ a follow: 6 n n = = n = Now the value of the flow rate ṅ can be ubtituted into the material balance equation to yield: n = n = 2 6