ENGR 224 - hermodynamics #1 - Diagram for a Cascade VCR Cycle (21 ts) Baratuci Final 13-Jun-11 On a full sheet of paper, construct a complete Diagram for the cascade cascade vapor-compression refrigeration cycle shown in the flow diagram, below. Include all the points, lines, curves and regions and label them clearly and properly. he working fluid in both cycles is ammonia. All the processes are adiabatic, except the evaporator and condenser, which exchange heat with the cold and hot reservoirs, respectively. he isentropic efficiency of each compressor is 82%. treams 4 & 8 are saturated liquids. treams 2 & 6 are saturated vapors.
ENGR 224 - hermodynamics #2 - Isentropic Efficiency of a ropane Compressor (25 ts) Baratuci Final 13-Jun-11 ropane gas at 40 o C and 150 ka is compressed adiabatically in a continuous flow compressor. emperature and pressure gauges at the compressor effluent read 165 o C and 1700 ka, respectively. Determine the power output per kg of flowing propane and the isentropic efficiency of the compressor. R = 8.314 J/mole-K. he surroundings are at 20 o C and 100 ka. ropane: C 1.6794 kj/kg-k MW 44.1 g/mole Read : Given : 1 40 o C 2 165 o C 313.15 K 438.15 K 1 150 ka 2 1700 ka gauge Q 0 kj/kg atm 100 ka R 8.314 kj/kmol-k 2 1800 ka abs surr 20 o C Find : a.) W??? kj/kg b.),c??? Diagram : 1 = 150 ka 1 = 40 o C surr = 20 o C atm = 100 ka Compressor 1 = 40 o C 2 = 1800 ka Assumptions : 1-2 - 3 - ropane behaves as an ideal gas throughout this process. he heat capacity of the propane is constant throughout this process. Changes in kinetic and potential energies are negligible/ Equations / Data / olve : art a.) Begin by applying the 1st Law to the compressor. Assuming that changes in kinetic and potential enegies are negligible, the 1st Law for open systems is : Qˆ Wˆ ˆ ˆ H2 H1 Because the compressor is adiabatic, Q = 0. Because the heat capacity of the propane is constant, H = C. As a result, Eqn 1 becomes : Cˆ 1 2 1 2 Eqn 1 Eqn 2 lugging values into Eqn 2 yields : W 209.93 kj/kg art b.) Let's begin with the definition of the isentropic efficiency of a compressor.,c Ŵ Ŵ,isen,act Eqn 3 Applying the 1st Law to both the asctual and isentropic compressor allows us to modify Eqn 1, as follows.,c,isen 2 1,act 2 1 Eqn 4
Because the heat capacity of the propane is constant, H = C. his helps us simplify Eqn 4 considerably.,c Cˆ Cˆ,isen 2 1 2 1 2 1,act 2 1 2 1 2 1 Eqn 5 Before we can use Eqn 5 to complete this problem, we need to determine the value of 2, the temperature of the propane that would come out of the compressor if it isentropically compressed the propane from 1 to 2. We can determine 2 by applying Gibbs 2nd Equation to the hypothetical isentropic compressor. If we assume the propane behaves as an ideal gas throughout this process, then Gibbs 2nd Eqnuation is : ˆ ˆ ˆ ˆ 2 2 2 1 C Ln Ln 0 1 MW 1 We can solve Eqn 6 for 2, as follows. 2 R 2 Ĉ Ln Ln MW 1 1 R Eqn 6 2 2 2 Ln Ln Ln MW 1 1 1 MW Eqn 8 2 2 1 1 MW MW Eqn 9 2 Eqn 10 2 1 1 lugging values into Eqn 10 and then Eqn 5 completes the solution of this problem. 2 413.90 K 140.75 o C,C 0.8060 Verify : None of the assumptions can be verified based on the given information or the results. Answers : a.) W 210 kj/kg a.),c 0.806 Note : You could also determine 2 from the following relationship that applies for isentropic processes on ideal gases with constant heat capacities. 2 2 1 1 1 Eqn 11 C C 1 1 o Where : o o o CV C R 1 R / C ˆ 1 R / MW C Eqn 12 lugging values into Eqns 12 & 11 yields : 1.1265 2 413.90 K 140.75 o C o o R C CV 1 o Also note : MW C C 2 2 2 2 2 Eqn 13 1 1 1 1 1
ENGR 224 - hermodynamics #3 - Reversibility of Individual rocesses in a Regeneration Rankine Cycle (54 ts) Baratuci Final 13-Jun-11 Consider the process flow diagram for a Regeneration Rankine Cycle with a closed feed-water heater (FWH), shown below. he flow rate and properties of every stream in the process are listed in the table below the flow diagram. All of the processes are adiabatic except the boiler and the condenser, which of course exchange heat with the hot and cold thermal reservoirs, respectively. he hot reservoir has a temperature of 500 o C and the cold reservoir has a temperature of 25 o C. Changes in kinetic and potential energies are negligible in every process in this system. All of the mass flow rates are correct and mass is conserved in every unit. he 1st Law is satisfied by every process to within the precision of the data presented in the table. m dot H x tream (kg/s) (ka) ( o C) (kj/kg) (kj/kg-k) (kg vap/kg) hase 1 260.8 12000 170 725.31 2.0276 N/A ub Liq. 2 260.8 12000 520 3403.4 6.5585 N/A uper. Vap. 3 260.8 1000 179.88 2880.0 6.8025 N/A uper. Vap. 4 69.2 1000 179.88 2880.0 6.8025 N/A uper. Vap. 5 191.6 1000 179.88 2880.0 6.8025 N/A uper. Vap. 6 191.6 6 36.16 1905.9 6.1929 0.7264 VLE 7 69.2 1000 179.88 762.52 2.1381 0 at'd Liq. 8 69.2 6 36.16 762.52 2.4963 0.2530 VLE 9 260.8 6 36.16 1602.5 5.2122 0.6008 VLE 10 260.8 6 36.16 151.48 0.52080 0 at'd Liq. 11 260.8 12000 36.48 163.51 0.52080 N/A ub. Liq. a.) b.) For two points for each of the 9 processes in the flow diagram, determine whether the process is totally reversible, irreversible or impossible. For four more points for each of the 9 processes in the flow diagram, support your answer from part (a) with a logical explanation and equations. tate any assumptions you make. Calculations are only required for one of the processes, but they would be helpful for some of the other processes as well. his problem is not short and these points are not free. how your work. Dr. Baratuci - ENGR 224 3, final-p11.xlsm 6/15/2011
Reversible Read : ince the problem statement tells us that each and every process obeys the principle of conservation of mass and that the 1st Law is not violated by any of the processes. o, the only test available for us to apply is the 2nd Law. Given : H 500 o C H 25 o C 773.15 K 298.15 K For all units other than the Boiler and the Condenser : Q = 0 kj E kin = 0 kj E pot = 0 kj Find : a.) Boiler H urbine plitter L urbine Mixer Condenser ump Closed FWH Expanison Valve b.) Explain & support each answer. Diagram : ee the diagram in the problem statement. Assumptions : 1-2 - he accuracy of the data in the table is limited to 5 digits. All of the states that make up the cycle are equilibrium states. 3-4 - Changes in kinetic and potential energies are negligible in every process in this system. All of the mass flow rates are correct and mass is conserved in every unit. 5 - he 1st Law is satisfied by every process to within the precision of the data presented in the table. 6 - All of the processes are adiabatic except the boiler and the condenser, which of course exchange heat with the hot and cold thermal reservoirs, respectively. Equations / Data / olve : For most of the processes in this system, we can determine whether the process obeys the 2nd Law by evaluating the rate of entropy generation in the process, gen. If :, then the process is : 0 If :, then the process is : Reversible 0 If :, then the process is : 0 For the ingle-input, ingle Outlet (IO) processes, we can evaluate gen from its defining equation: ˆ Q m res gen,tot Eqn 1 Q Next, we can solve Eqn 1 for gen : m ˆ Eqn 2 res For the Boiler and the Condenser, Eqn 2 can be simplified because heat is transferred to one thermal reservoir at a constant temperature. he results are : m ˆ ˆ gen,tot,cond 9 10 9 Q 9 10 C Q Eqn 3 1 2,boil m1 ˆ ˆ Eqn 4 2 1 H All the other processes are adiabatic, so Eqn 3 can be simplified to : ˆ Eqn 5 gen,tot m For the Multiple-Inlet, Multiple Outlet (MIMO) processes, Eqn 5 must be modified. he result is : Now, we can evaluate gen,tot for each process. m m Outlets Inlets ˆ ˆ Eqn 6 Dr. Baratuci - ENGR 224 3, final-p11.xlsm 6/15/2011
Boiler Because 2 > H, the boiler would require heat to flow spontaneously from the colder thermal reservoir into the hotter working fluid. his is a violation of the 2nd Law. H urbine plitter Q = 0 and H 3 < H 2 therefore W > 0. o the turbine does produce work, as required. Because 3 > 2, Eqn 5 shows that gen,tot > 0. All properties of all three streams are the same and Q = 0. 2nd Law : m4 ˆ m ˆ ˆ 5 m3 gen,tot 4 5 3 But: 3 = 4 = 5. herefore becomes : ˆ 4 m4 m5 m3 Eqn 8 But because mass is conserved, m 4 + m 5 = m 3. Reversible herefore, gen,tot = 0. L urbine Mixer Q = 0 and H 3 < H 2 therefore W > 0. o the turbine does produce work, as required. Because 6 < 5, Eqn 5 shows that gen,tot < 0. 6 = 8 = 9. 6 = 8 = 9. If we assume no friction is present, then adiabatically mixing streams at the same temperature and pressure is reversible. Calculations are not required. 2nd Law : m9 ˆ m ˆ ˆ 8 m6 lugging values into yields : out m 1359.3 kw in m 1359.3 kw gen,tot 0.0 kw/k gen,tot 9 8 6 o the precision of the data in the table, gen,tot = 0. Reversible Condenser Before we can use Eqn 3 to evaluate for the condenser, we must apply the 1st Law to determine Q 9-10. 1st Law : Q W ˆ ˆ,cond m H H Eqn 8 9 10 10 9 A condenser has no moving parts and so no shaft work crosses the system boundary. o, Eqn 8 becomes : Q m Hˆ Hˆ 9 10 10 9 Eqn 9 lugging values into Eqns 9 & 3 yields : gen,tot > 0. Q 9-10 gen,tot -378426 kw 45.73 kw/k 0.1753 kj/kg-k ump Exp. Valve Q = 0 and H 11 > H 10 therefore W < 0. o the pump does require work, as it should. Because 10 < 11, Eqn 5 shows that gen,tot = 0. Reversible Q = 0 and W = 0. because a valve does not usse or produce shaft work. he 1st Law simplifies to H 8 < H 7. he data conforms to this requirement. he Expansion Valve is isenthalpic. Because 8 < 7, Eqn 5 shows that gen,tot > 0. Dr. Baratuci - ENGR 224 3, final-p11.xlsm 6/15/2011
Reversible Closed FWH Exchanging heat between two streams at different temperatures is irreversible. Calculations are not required. 2nd Law : m ˆ m ˆ m ˆ m ˆ gen,tot 4 7 1 1 4 4 1 11 lugging values into yields : out m dot 676.8 kw in m dot 606.6 kw gen,tot 70.2 kw/k gen,tot > 0. Verify : None of the assumptions can be verified based on the given information or the results. Answers : a.) b.) Explain Boiler X 2 > H H urbine X gen,tot > 0 plitter X gen,tot = 0 L urbine X gen,tot < 0 Mixer X gen,tot = 0 Condenser X gen,tot > 0 ump X gen,tot = 0 Closed FWH X gen,tot > 0 Expanison Valve X gen,tot > 0 Dr. Baratuci - ENGR 224 3, final-p11.xlsm 6/15/2011