ENGR Thermodynamics

Transcription

1 ENGR hermodynamics W #5 Problem : he Increase of Entropy Principle - 2 pts 11-May-11 Will the entropy of steam increase, decrease or remain the same as it flows through a real adiabatic turbine? Read : Answers : he usual problem solving procedure does not really apply th this "problem" because it is really a short-answer question. he entropy of steam will increase as it flows through a real adiabatic turbine. Real turbines are not reversible. hey are irreversible. Irreversibilities cause entropy generation. Because the process is adiabatic, there is no way to decrease the entropy. herefore, the entropy generation due to irreversibilities in the real turbine produce an INREASE in the entropy of the steam as it passes through the turbine. Dr. - ENGR 224 hw5-sp11.xlsm, /9/2011

2 ENGR hermodynamics W #5 Problem : he Increase of Entropy Principle - 2 pts 11-May-11 Will the entropy of the working fluid in an ideal arnot ycle increase, decrease or remain the same during the isothermal heat addition process? Read : Answer : he usual problem solving procedure does not really apply th this "problem" because it is really a short-answer question. he entropy of the working fluid in an ideal arnot ycle will increase during the isothermal heat addition process. Because the cycle is a arnot ycle, each step in the cycle is completely reversible. Nonetheless, adding heat to the working fluid INREASES its entropy. Keep in mind that the heat AME from a thermal reservoir. When the heat "left" the thermal reservoir, the entropy of the reservoir decreased by the same amount that the entropy of the working fluid in the cycle increased. he result is a zero net change in the entropy of the universe, despite the fact that the entropy of the working flid INREASED. Dr. - ENGR 224 hw5-sp11.xlsm, /9/2011

3 ENGR hermodynamics W #5 Problem : he Increase of Entropy Principle - 2 pts 11-May-11 Steam is accelerated as it flows through a real, adiabatic nozzle. Will the entropy of the steam at the nozzle exit be greater than, equal to or less than the entropy at the nozzle inlet? Read : Answer : he usual problem solving procedure does not really apply th this "problem" because it is really a short-answer question. he entropy of the steam at the nozzle exit will be greater than the entropy at the nozzle inlet. Real nozzles are not reversible. hey are irreversible. Irreversibilities cause entropy generation. Because the process is adiabatic, there is no way to decrease the entropy. herefore, the entropy generation due to irreversibilities in the real nozzle produce an INREASE in the entropy of the steam as it is accelerated through the nozzle. Dr. - ENGR 224 hw5-sp11.xlsm, /9/2011

4 ENGR hermodynamics W #5 Problem : ΔS Sys, ΔS Res, and ΔS Univ, for a.. Process - 4 pts 11-May-11 During the isothermal heat rejection process of a arnot ycle, the working fluid experiences an entropy change of -0.7 Btu/ o R. If the temperature of the heat sink is 95 o F, determine a.) b.) c.) he amount of heat transfer he entropy change of the heat sink he total entropy change of the universe for this process. Read : In part (a) the system is the EX in which heat is rejected to the cold reservoir. Because this step is isothermal and reversible, S gen = 0 and we can evaluate S fluid using the definition of entropy. Part (b) is very similar because the cold reservoir also undergoes a reversible process and remains at a constant temperature. So, we can also evaluate S sink from the definition of entropy. Finally, we can evaluate S univ because it is just the sum of S fluid and S res. Given : S fluid -0.7 Btu/ o R 95 o F o R Find : a.)??? Btu b.) S sink??? Btu/ o R c.) S univ??? Btu/ o R Diagram : = 95 o F S 1 S 2 EX S fluid = 0.7 Btu/ o R Assumptions : 1 - he heat transfer process is completely reversible because it is a step in a arnot ycle. 2 - he reservoir is internally reversible. 3 - he working fluid absorbs heat isothermally. Equations / Data / Solve : Part a.) Let's begin by applying the entropy balance equation to the EX in which the isothermal heat rejection step in the arnot ycle takes place. Sfluid S gen,int Eqn 1 Because this EX is part of a arnot ycle, it is completely reversible and so the entropy generation is ZERO. So, Eqn 1 reduces to the definition of S. S fluid Eqn 2 Now, because the heat transfer process is reversible, the temperature of the system at which the heat leaves the EX must be the same as the temperature of the reservoir,. As a result, Eqn 2 can be simplified to : S fluid Eqn 3 Next, we can solve Eqn 3 for and plug in values to complete this part of the problem. S fluid Eqn 4 Now, we can plug numbers into Eqn 4 : Btu he negative value for in association with our usual sign convention, indicates that heat is transferred out of the working fluid. In terms of cycle or tie-fighter diagrams, all and W values are positive with their direction indicated by an arrow. herefore : Btu Dr. - ENGR 224 hw5-sp11.xlsm, 7.28E 5/9/2011

5 Part b.) ere we take our system to be the reservoir that provides heat to the arnot ycle. he reservoir is internally reversible. herefore, Eqn 1 becomes : S res Eqn 4 Now, because the heat transfer process is reversible and the temperature of the reservoir,, is constant, Eqn 4 simplifies to : S res Eqn 5 Now, we can plug numbers into Eqn 5 : S res kj/-k Part c.) Now we can calculate S univ directly for S fluid and S res using : Suniv Sfluid Sres Eqn 6 Plugging values into Eqn 6 yields : S univ kj/-k S his makes sense since : univ gen,total Eqn 7 S S 0 And, for a completely reversible process like this : gen,total Eqn 8 Verify : None of the assumptions made in this problem solution can be verified. Answers : a.) 388 Btu b.) S res kj/-k c.) S univ 0 kj/-k Dr. - ENGR 224 hw5-sp11.xlsm, 7.28E 5/9/2011

6 ENGR hermodynamics W #5 Problem : Entropy hange in the Evaporator of a Refrigerator - 6 pts 11-May-11 R-134a enters the coils of the evaporator of a refrigeration system as a saturated vapor-liquid mixture at a pressure of 200 kpa. he refrigerant absorbs 120 kj of heat from the cooled space, which is maintained at -5 o, and leaves the evaporator as a saturated vapor at the same pressure. Determine... a.) he entropy change of the refrigerant. b.) he entropy change of the refrigerated space. c.) he entropy change of the universe for the process. Read : he key here is that the temperature of the R-134a does not change in the evaporator, so it behaves as a true thermal reservoir and we can determine the change in entropy from the definition of entropy. Given : P 200 kpa 120 kj x 2 1 kg vap/kg -5.0 deg K Find : a.) S R-134a??? kj/k b.) S res??? kj/k c.) S univ??? kj/k Diagram : P 1 = 200 kpa 0 < x 1 < 1 2 = sat 1 2 Evaporator old Reservoir = -5 o = 120 kj P 2 = 200 kpa x 2 < 1 2 = sat Assumptions : 1 - he pressure everywhere in the evaporator is 200 kpa so that P 2 = P he temperature of the refrigerated space remains constant throughout the process so that it behaves as a true thermal reservoir. 3 - he temperature of the R-134a in the evaporator is always equal to the saturation temperature at 200 kpa. As a result, the R-134a also behaves as a true thermal reservoir Equations / Data / Solve : Parts a & b) Because both the refrigerated space and the R-134a behave as true thermal reservoirs in this process, we can calculate the change in entropy for each from the definition of entropy. S Int Re v Eqn 1 Part c.) We can immediately apply Eqn 1 to the refrigerated space because we know its temperature. Just keep in mind that from the perspective of the refrigerated space = -120 kj because heat is being transferred out of that system (the refrigerated space). sat S res kj/k We can look up the saturation temperature of the R-134a in the saturation pressure table for R-134a and then plug values into Eqn deg K S R-134a kj/k he entropy change of the universe is just the sum of the entropy change for the refrigerated space and the R-134a. Suniv Sres SR 134a Eqn 2 S univ kj/k Verify : None of the assumptions made in this problem solution can be verified. Answers : a.) S R-134a kj/k c.) S univ kj/k b.) S res kj/k Dr. - ENGR 224 hw5-sp11.xlsm, /9/2011

7 ENGR hermodynamics W #5 Problem : S Univ Upon uenching an Iron Block - 6 pts 11-May-11 A 12 kg iron block initially at 350 o is quenched in an insulated tank that contains 100 kg of water at 22 o. Assuming the water that vaporizes during the process condenses back into the liquid phase inside the tank, determine the entropy change of the universe for this process. Data : P,2O 4.18 kj/kg-k P,Fe 0.45 kj/kg-k Read : his problem requires the application of the 1st Law to determine 2 and then one of the Gibbs equations to evaluate S. he problem is simplified by the fact that the heat capacites are constant. Given : 1,Fe 350 o m Fe 12 kg K m 2O 100 kg 1,2O 22 o K Find : S univ??? kj/k Diagram : Iron m Fe = 12 kg 1 = 350 o 2 =?? o m 2O = 100 kg 1,2O = 22 o 2 =?? o Assumptions : 1 - All of the water is in the liquid phase in the final state. 2 - he heat capacities of iron and water are constants. 3 - he process is adiabatic. All the heat leaving the iron enters and stays in the water. No heat or work is exchanged with the surroundings. 4 - Both the water and the iron are perfectly incompressible. he volume of each remains constant throughout this process. Equations / Data / Solve : he universe, in this problem, is made up of the water and the iron block. herefore : S S S univ Fe 2O We can use Gibbs 2nd Equation for incompressible substances with constant heat capacities to evaluate S Fe and S 2O. ds d Eqn 2 ˆ ˆ 2 S Ln Eqn 3 1 Because the iron and water are incompressible, P = V =. So, all we need to do in order to use Eqns 3 & 1 to complete this problem is to determine 2, the final, equilibrium temperture fo both the water and the iron. Let's begin by applying the 1st Law for clases sytems with negligible changes in kinetic and potential energies to the iron and the water separately. Eqn 1 W m Uˆ Fe b,fe Fe Fe Eqn 4 W m Uˆ O 2 b,o 2 O 2 O 2 Eqn 5 If we assume that the volume of the water does not change in the process and that the volume of the iron does not change in the process, then there is no boundary work in the process. Since all of the heat leaving the iron goes into the water: O 2 Fe Eqn 6 Dr. - ENGR 224 hw5-sp11.xlsm, /9/2011

8 ombining Eqns 4-6 yields : Uˆ Uˆ Uˆ Uˆ Uˆ Uˆ O 2 2,O 2 1,O 2 Fe 1,Fe 2,Fe Eqn 7 Because the heat capacites of both iron and water are assumed to be constant, we can express the change in internal energy in terms of the heat capacities as : 2 Û ˆ dˆ 1 V V 2 1 Because the iron and water are incompressible, P = V =. So, we can replace V with and combine Eqns 7 & 8 to obtain : ˆ m ˆ m O 2 O 2 2 1,O 2 Fe Fe 1,Fe 2 Notice that at the final state, the water and the iron are at the same temperature, 2. Now, we need to solve Eqn 9 for the unknown 2. m ˆ m ˆ 2 m ˆ m ˆ O 2 O 2 1,O 2 Fe Fe 1,Fe O 2 O 2 Fe Fe Eqn 8 Eqn 9 Eqn 10 Plugging values into Eqn 7 yields : m Fe * Fe 5.4 kj/k o m 2O * 2O 418 kj/k K Now we can use Eqn 3 for the water and then for the iron to evaluate the entropy change for each. And then plug these values into Eqn 1 to complete the problem. S Fe kj/k S 2O kj/k S univ kj/k Verify : None of the assumptions made in this problem solution can be verified. Answers : S univ kj/k Dr. - ENGR 224 hw5-sp11.xlsm, /9/2011

9 ENGR hermodynamics W #5 Problem : WB-1 - Efficiency of an Int. Rev. E with Multiple eat ransfers - 4 pts 11-May-11 A system executes a power cycle while receiving 750 kj by heat transfer at its boundary where the temperature is 1500 K and discharges 100 kj by heat transfer at another portion of its boundary where its temperature is 500 K. Another heat transfer from the system occurs at a portion of the system boundary where the temperature of the system is 1000 K. No other heat transfer crosses the boundary of the system. If no internal irreversibilities are present, determine the thermal efficiency of the power cycle. Read : he key to this problem is that the cycle is internally reversible and that the temperatures within the system at which these heat transfers occur must remain constant. his allows us to use an entropy balance and the 2nd Law to determine the amount of heat rejected to reservoir 3. he 1st Law then allows to evaluate W cycle. We can then calculate th from its definition. Given : kj kj K K K Find : th??? Diagram : res, W cycle res,2 2 E R 3 3 res,3 Assumptions : 1 - he power cycle is internally reversible. 2 - eat transfers can be irreversible, but the temperatures within the system at which the heat trasfers occur must remain constant at the given values. 3 - eat is exchanged only with the three reservoirs mentioned in the problem statement. Dr. - ENGR 224 hw5-sp11.xlsm, WB-1 5/9/2011

10 Equations / Data / Solve : Let's begin with the definition of a thermal efficiency of a power cycle : Wcycle W th cycle in 1 Eqn 1 Next, we need to apply the 1st Law to the cycle to help us determine W cycle : Wcycle Eqn 2 Now, all we need to do is determine 2 and then we can use Eqns 1 & 2 to evaluate th. he key to determining 2 is the fact that the power cycle in INERNALLY reversible. his means that the entropy generation within the cycle is ZERO. Write the entropy balance equation for the cycle : Scycle Sgen,int 0 Eqn 3 Evaluating the cyclic integral in terms of the 's and 's in this cycle gives us : Eqn 4 Notice that because 1, 2 and 3 are based on the tie-fighter diagram, they are all positive numbers. heir direction is represented by the arrows in the tie-fighter diagram, above. Because 2 and 3 represent heat transfers out of the system, they become negative when we drop the sign convention as we move from Eqn 3 to Eqn 4. Now, we can solve Eqn 4 for 3 : Eqn 5 Now, plug values intoeqns 5, 2 & 3 : kj W cycle 350 kj th Verify : None of the assumptions made in this problem solution can be verified. Answers : th 46.7% Dr. - ENGR 224 hw5-sp11.xlsm, WB-1 5/9/2011

11 ENGR hermodynamics Problem : WB-2 - Efficiency and res for Rev. and Irrev. ycles - 6 pts W #5 11-May-11 omplete the following involving reversible and irreversible cycles. a.) Reversible and irreversible power cycles each discharge to a cold reservoir at and receive energy from hot reservoirs at and ', respectively. here are no other heat transfers involved. Show that ' >. b.) Read : Reversible and irreversible refrigeration cycles each discharge to a hot reservoir at and receive energy from cold reservoirs at and ', respectively. here are no other heat transfers involved. Show that ' >. he easiest approach to this problem is to carefully define your systems so that ALL of the irreversibilities lie inside the system boundary. We accomplish this by choosing the system boundary to be bigger than the E's and Ref's. he boundaries are so big, in fact, that they OU the reservoirs. his permits us to assume that the temperature of this extended system, at which the heat enters or leaves the extended system, is EUAL O E RESERVOIR. By defining our systems in this manner, we have made ALL of the irreversibilities INERNAL. But the important goal of choosing these larger systems is that, regardless of whether the cycle is reversible or not, the temperature at which heat transfer occurs is known and constant. Given : See diagrams, below. Find : a.) Show that ' >. b.) Show that ' >. Diagram : See below. Assumptions : 1 - Both heat engines operate on thermodynamic cycles. 2 - hanges in kinetic and potential energies are negligible. 3 - he mixer is adiabatic. 4 - Flow work is the only form of work that crosses the system boundary. Equations / Data / Solve : Part a.) Diagram : W R E R ' E I W I he trick to this problem is to VERY AREFULLY choose your system! If you choose your system boundary to be snug on each cycle, you run into a problem on the irreversible cycle. ere is why. he completely reversible cycle is no problem. he heat transfers must be reversible, so the temperatures within the system and the temperatures of the reservoirs must be equal. owever, in the irreversible cycle, the temperature within the system MAY NO BE EUAL to the temperature of the reservoir that is giving or receiving the heat! he solution is to choose your systems to be LARGER, so that at the boundary of this larger system, = reservoir! See the diagram, above. Dr. - ENGR 224 hw5-sp11.xlsm, WB-2 5/9/2011

12 Let's apply the entropy balance equation to the larger, extended systems surrounding each E. SR Sgen,int,R 0 SI Sgen,int,I 0 Eqn 1 Eqn 2 he entropy generation for the completely reversible cycle (in Eqn 1) is ZERO. Now, express Eqns 1 & 2 in terms of and. 0 Eqn 3 gen,int,i S 0 Eqn 4 Notice that because and are based on the tie-fighter diagrams, they are positive numbers. heir direction is represented by the arrows in the tie-fighter diagram, above. Because represents heat transfer out of the system, it becomes negative when we drop the sign convention as we move from Eqns 1 & 2 to Eqns 3 & 4. For an irreversible process or cycle,the 2nd Law tells us that : Sgen,int,I 0 Eqn 5 S 0 Solve Eqn 4 for S gen,int,i : gen,int,i Eqn 6 Now, solve Eqn 3 for / and use the result to eliminate this term from Eqn 6. Eqn 7 Manipulate Eqn 8 to obtain a relationship between and ' : 1 1 Eqn 8 Eqn 9 Part b.) Diagram : W R W I Ref R Ref I ' his part of the problem is very similar to part (a). he key is still to choose your systems wisely. Dr. - ENGR 224 hw5-sp11.xlsm, WB-2 5/9/2011

13 Apply the entropy balance equation to the larger, extended systems surrounding each refrigerator. SR Sgen,int,R 0 SI Sgen,int,I 0 Eqn 10 Eqn 11 he entropy generation for the completely reversible cycle (in Eqn 1) is ZERO. Now, express Eqns 1 & 2 in terms of and. 0 Eqn 12 gen,int,i S 0 Eqn 13 Notice that because and are based on the tie-fighter diagrams, they are positive numbers. heir direction is represented by the arrows in the tie-fighter diagram, above. Because represents heat transfer out of the system, it becomes negative when we drop the sign convention as we move from Eqns 10 & 11 to Eqns 12 & 13. For an irreversible process or cycle,the 2nd Law tells us that : Sgen,int,I 0 Eqn 14 S 0 Solve Eqn 13 for S gen,int,i : gen,int,i Eqn 6 Now, solve Eqn 12 for / and use the result to eliminate this term from Eqn 6. Eqn 7 Manipulate Eqn 8 to obtain a relationship between and ' : 1 1 Eqn 8 Eqn 9 Verify : None of the assumptions made in this problem solution can be verified. Answers : a.) b.) Dr. - ENGR 224 hw5-sp11.xlsm, WB-2 5/9/2011

14 ENGR hermodynamics W #5 Problem : WB-3 - Specific Entropy hange Using abluar Data - 4 pts 11-May-11 Determine the change in specific entropy in kj/kg-k for : a.) Water: P 1 = 10 MPa, 1 = 400 o and P 2 = 10 MPa, 2 = 100 o. ( Use the NIS Webbook ) b.) R-134a: 1 = kj/kg, 1 = - 40 o and P 2 = 5 bar, x 2 = 1.0. (Use the NIS Webbook with the default reference state: U = 0 and S = 0 for saturated liquid at 0.01 o ) c.) Air (IG): 1 = 27 o, P 1 = 2 bar and 2 = 327 o, P 2 = 1 bar. d.) ydrogen ( 2, IG): 1 = 727 o, P 1 = 1 bar and 2 = 25 o, P 2 = 3 bar. Read : Parts (a) and (b) of this problem are exercises in the use of thermodynamic property tables to evaluate S. Parts (c) and (d) are exercises in the use of the ideal gas property tables and Gibbs' 2nd Equation to evaluate S for ideal gases. Given : a.) Water P kpa P kpa o o b.) R-134a kj/kg P kpa 1-40 o x 2 1 kg vap/kg c.) Air P kpa P kpa 1 27 o o R J/mole-K MW g/mole d.) 2 P kpa P kpa o 2 25 o R J/mole-K MW g/mole Find : S??? kj/kg-k for each part of the problem. Assumptions : For parts (c) and (d), assume that air and 2 behave as ideal gases, as instructed in the problem statement. For parts (c) and (d), assume that air and 2 have variable heat capacities and use the Ideal Gas Property ables. Diagram: None needed. Equations / Data / Solve : Part a.) ere we can use the NIS Webbook to evaluate the specific entropy of water and then directly calculate the change in specific entropy. I used the isothermal properties table with the default reference state to obtain : S kj/kg-k S kj/kg-k S kj/kg-k Part b.) he first step here is to use 1 to obtain data from the Saturation properties temperature increments option. his will allow us to compare the given value of to the values of sat liq and sat vap in order to determine the phase or phases present in the system at equilibrium. ere is the relevant data : Pressure emp. (kj/kg) S (kj/kg-k) (kpa) ( o ) Sat. Liq Sat. Vap Sat. Liq Sat. Vap Since : ˆ ˆ ˆ sat liq sat vap the system contains a saturated mixture and the temperature must be equal to the saturation temperature. Dr. - ENGR 224 hw5-sp11.xlsm, WB-3 5/9/2011

15 Last, we must use the given value of the enthalpy to determine the quality of the water in the system. he key equation is: ˆ x1 ˆ 1 sat liq sat vap ˆ ˆ satliq Eqn 1 Plugging values into Eqn 1 yields : x kg vap/kg Sˆ x Sˆ 1x Sˆ sat sat sat mix vap liq Eqn 2 S kj/kg-k Because state 2 is a saturated vapor at 500 kpa, we can get S 2 directly from the saturation pressure table. S kj/kg-k S kj/kg-k Part c.) Gibbs's 2nd equation is most useful in this case because the initial and final pressures are given. ˆ ˆo ˆ o R P 2 S S2 S1 Ln MW P1 Eqn 3 We can immediately lookup S o 2 and S o 1 in the Ideal Gas Property able for air because we know 1 and 2. S o kj/kg-k 1 S o kj/kg-k 2 Now, we can plug values into Eqn 3 to get : S kj/kg-k Part d.) he same procedure is used in part (d) as in part (c). S o kj/kg-k 1 S o 0 kj/kg-k 2 Now, we can plug values into Eqn 3 to get : S kj/kg-k Verify : a.) No assumptions. b.) No assumptions. c.) We can use the Ideal Gas EOS to verify that the molar volume of the air in both states 1 and 2 is greater than 5 L/mole (diatomic gas) using Eqn 5. PV R R Eqn 4 V Eqn 5 P V L/mole V L/mole d.) Since both V 1 and V 2 are greater than 5 L/mole, the ideal gas assumption is valid. We can use the Ideal Gas EOS to verify that the molar volume of the air in both states 1 and 2 is greater than 5 L/mole (diatomic gas) using Eqn 5. V L/mole V L/mole Since both V 1 and V 2 are greater than 5 L/mole, the ideal gas assumption is valid. Answers : a.) S kj/kg-k c.) S kj/kg-k b.) S kj/kg-k d.) S kj/kg-k Dr. - ENGR 224 hw5-sp11.xlsm, WB-3 5/9/2011

16 ENGR hermodynamics W #5 Problem : WB-4 - "Show hat" for a ycle Interacting with hree Reservoirs - 4 pts 11-May-11 he system shown in the figure undergoes a cycle while receiving energy at the rate surr from the surroundings at temperature surr, receiving from a heat source at temperature, and rejecting to a thermal reservoir at. Derive an expression for the maximum theoretical value of in terms of and 0, S and U only. here is no work produced by this system. his is an absorption heat pump system. he heat source might be a propane flame and the heat sink might be the air inside your camper. So, it is important to note that S > U > surr. int:, surr and are not important in the solution of this problem! Read : Given : Find : he key here is to recognize that, and surr do not directly effect the solution of this problem and then determine how entropy generation affects the heat output of this heat pump,. Because S gen > 0 we will be able to determine a maximum value for as a function of and the system temperatures at which heat is transferred. S U 0 fxn,,, S U 0 Diagram : See the problem statement. Assumptions : 1 - he system operates on a thermodynamic cycle Equations / Data / Solve : S U 0 he system only exchanges heat with the three reservoirs shown in the diagram. Let's begin by appplying the entropy generation form of the 2nd Law to the system shown in the diagram. Scycle Sgen,int 0 Eqn 1 he change in any property, including entropy, for a cycle is zero. We can now expand the cyclic integral in terms of the three heat transfers that actually occur during the cycle. surr gen,int S 0 U S 0 Eqn 2 We can now use the 1st Law to eliminate surr from Eqn 2. surr gen,int S 0 U S 0 Eqn 3 Eqn 4 Now, we can solve Eqn 4 for S gen,int 0 U 0 S Eqn 5 Dr. - ENGR 224 hw5-sp11.xlsm, WB-4 5/9/2011

17 Because : S U Eqn 6 Eqn 7 0 U 0 S herefore, because S gen > 0, is maximized when S gen,int = 0. hat is to say this absorption heat pump will deliver the most heat to the heated space, per Joule of heat supplied from the heat source, if it is reversible. his should come as no surprise. So, for a reversible process, Eqn 5 becomes : S S,max U U 0 Eqn 8 Verify : None of the assumptions made in this problem solution can be verified. Answers :,max 0 1 S 0 1 U Dr. - ENGR 224 hw5-sp11.xlsm, WB-4 5/9/2011

18 ENGR hermodynamics W #5 Problem : WB-5 - hree-step, Ideal Gas ycle Analysis - 8 pts 11-May-11 A quantity of air undergoes a thermodynamic cycle consisting of three internally reversible processes in series. Assume that the air behaves as an ideal gas. his may not be a good assumption, but let's work with it here anyway. Step 1-2 : Step 2-3 : Step 3-1 : a.) b.) c.) d.) Isothermal expansion at 350 K from 4.75 bar to 1.0 bar. Adiabatic compression to 4.75 bar. Isobaric cooling. Sketch the cycle on a PV diagram. Sketch the cycle on a S diagram. Determine 3 in Kelvin If the cycle is a power cycle, determine its thermal efficiency. If the cycle is a refrigeration cycle, determine its OP. Read : In part (c), you can use the 2nd Gibbs Equation and the Ideal Gas Entropy able for air to determine S o 3 and then 3. In part (d), we cannot use the reservoir temperatures alone to determine OPR because the cycle is only internally reversible. It is not completely reversible. We can determine 12 = from an entropy balance, taking advantage of the fact that there is no internal entropy generation because the cycle isinternally reversible. We can evaluate W 31 from the integra of P dv and then apply the 1st Law to determine 31 =. Finish up by plugging and into the definition of OP R. Given : 1 = K R J/mol-K P kpa MW g/mole P kpa Find : a.) PV diagram c.) 3??? K b.) S diagram. d.) OP R or??? Diagram : See the solutions to parts (a) and (b) below. Assumptions : 1 - he air is contained in a closed system. 2 - Air is an ideal gas in all three states. his assumption can be verified. 3 - All three steps in the cycle are internally reversible. Equations / Data / Solve : Part a.) Part b.) P 1 3 Isobaric I s o t h e r m a l Adiabatic Isobaric Isentropic 1 Isothermal 2 P 1 P 2 V S Dr. - ENGR 224 hw5-sp11.xlsm, WB-5 5/9/2011

19 Part c.) he key to determining 3 is to recognize that since step 2-3 is both adiabatic and internally reversible, it is also isentropic. he entropy change of an ideal gas with variable heat capacities is given by : ˆ ˆ o ˆ o R P 3 S S3 S2 Ln MW P2 S o is a function of only, so we can lookup S o 1 in the Ideal Gas Property able for air. S o 2 = S o kj/kg-k Eqn 1 Since S o is a function of only, we can solve Eqn 1 for S o 2 and then interpolate on the Ideal Gas Property able for air to determine 2. ˆo ˆo R P 3 S3 S2 Ln MW P2 Eqn 2 Plugging values into Eqn 2 yields : S o kj/kg-k Interpolation on the Ideal (K) S o Gas Property able for air : K Part d.) First we must abserve that this cycle is a refrigeration cycle. here are many ways to reach this conclusion, but the easiest is to observe that the cycle progresses in a counter-clockwise direction on the PV and S diagrams. So, our goal is to determine the value of OP R. We cannot determine OP R using : R Eqn 3 / 1 here are two reasons. 1- his cycle is only internally reversible, not completely reversible (like a arnot ycle). 2- If this cycle were to be completely reversible, what would be the temperature of the hot reservoir? It would need to change during step 3-1 from 3 to 1. Which would you use? If you simply use 3, then the heat transfer in step 3-1 is not reversible and Eqn 3 does not apply. Instead, we can determine OP R from its definition. OP 1 OP R ˆ ˆ 1 Wˆ ˆ ˆ ˆ /ˆ 1 cycle Eqn 4 Note : Because of our use of the sign convention that heat transfer is positive when heat is transferred into the system, = 12 and = Let's begin by determining 12. Because this step is internally reversible and isothermal, we can use : ˆ ˆ ˆ S ˆ 12 Sgen,int 12 1 Eqn 5 ˆ Sˆ We can solve Eqn 5 for 12 : Eqn 6 Dr. - ENGR 224 hw5-sp11.xlsm, WB-5 5/9/2011

20 Now, we can apply Eqn 1 to step 1-2 to evaluate S 12. ˆ ˆ o ˆo R P 2 S12 S2 S1 Ln MW P1 Eqn 7 Because 1 = 2, S o 2 = S o 1 and Eqn 7 simplifies to : Ŝ R P Ln 2 12 MW P1 Eqn 8 Plugging values into Eqns 8 & 6 yields : S kj/kg-k 12 = kj/kg We can determine by applying the 1st Law for closed systems. ˆ W ˆ U ˆ Eqn 9 Step 3-1 is internally reversible and also isobaric, so the boundary work can be determined as follows : V1 W PdVˆ PVˆ PVˆ 31 V3 ombining Eqns 9 & 10 and making use of the definition of enthalpy yields : ˆ U ˆ W ˆ U ˆ PV ˆ ˆ ˆ ˆ Eqn 10 Eqn 11 Because we know 1 and 3 and both states are assumed to be ideal gas states, we can lookup 1 and 3 in the Ideal Gas Property able for air. o 3 requires interpolation. (K) o 1 o kj/kg o o kj/kg Now, we can plug values into Eqn 11 to evaluate 31 and then plug 12 and 31 into Eqn 4 to evaluate OP R. 31 = kj/kg OP Verify : We can use the Ideal Gas EOS to verify that the molar volume of the air in all three states is greater than 5 L/mole (diatomic gas). PV R R Eqn 12 V Eqn 13 P V L/mole V L/mole V L/mole Since all three molar volumes are greater than 5 L/mole, the ideal gas assumption is valid to about 2 significant figures. Answers : a.) See the PV diagram, above. c.) K b.) See the S diagram, above. d.) OP 3.87 Dr. - ENGR 224 hw5-sp11.xlsm, WB-5 5/9/2011

21 ENGR hermodynamics W #5 Problem : WB-6 - Maximum Work From an Adiabatic urbine - 5 pts 11-May-11 Steam enters an adiabatic turbine at 800 psia and 900 o F and leaves at a pressure of 40 psia. Determine the maximum amount of work that can be delivered by this turbine. Read : he key to this problem is the fact that the maximum work will be developed by a reversible turbine. A turbine that is both reversible and adiabatic is isentropic. Because S 2 = S 1, we can use P 2 and S 2 to determine 2. hen, we can apply the 1st Law to evaluate the work output from the isentropic turbine. Given : P psia P 2 40 psia o F Find : W sh,max??? Btu/lb m Diagram : P 1 = 800 psia 1 = 900 o F 1 urbine W sh =??? kj/kg 2 P 2 = 40 psia 2 =??? o F Assumptions : 1 - he turbine is adiabatic. 2 - hanges in kinetic and potentail energies are negligible. Equations / Data / Solve : Let's begin by applying the 1st Law for open systems to the turbine, assuming changes in kinetic and potential energies are negligible. ˆ W ˆ ˆ sh Eqn 1 W ˆ ˆ ˆ Because the turbine is adiabatic, Eqn 1 becomes : Eqn 2 sh 1 2 Because we know both 1 and P 1, we can immediately lookup 1 in the steam tables. At P 1 = 800 psia : sat Since 1 > sat, the feed is superheated steam Btu/lb m It is a bit harder to determine 2 because we only know P 2, not 2. But we do have one more piece of information that we can use. he maximum shaft work will be produced by an internally reversible turbine. A device that is both adiabatic and internally reversible is also isentropic: S 2 = S 1. We can immediately lookup S 1 because we know P 1 and 1. S 1 = S 2 Now, knowing P 2 and S 2, we can determine 2. Let's begin by determining what phases are present. o F Btu/lb m - o R he first step here is to use P 2 to obtain data from the saturation pressure table. his will allow us to compare the given value of S 2 to the values of S sat liq and S sat vap in order to determine the phase or phases present in the system at equilibrium. ere is the relevant data : Pressure emp. (Btu/lb m ) S (Btu/lb m - o R) (psia) ( o F) Sat. Liq Sat. Vap Sat. Liq Sat. Vap Since : Sˆ Sˆ Sˆ sat 2 sat liq vap the system contains a saturated mixture and the temperature must be equal to the saturation temperature. Dr. - ENGR 224 hw5-sp11.xlsm, WB-6 5/9/2011

22 Next, we must use S 2 to determine the quality of the water in the system. he key equation is: x 2 Sˆ Sˆ 2 satliq sat vap Sˆ Sˆ sat liq Eqn 3 Plugging values into Eqn 3 yields : x lb m vap/lb m Sˆ x Sˆ 1x Sˆ sat sat sat mix vap liq Eqn 4 Finally, we can plug values back into Eqn 2 to evaluate the maximum shaft work for the turbine Btu/lb m Verify : None of the assumptions made in this problem solution can be verified. W sh,max Btu/lb m Answers : W sh 312 Btu/lb m Dr. - ENGR 224 hw5-sp11.xlsm, WB-6 5/9/2011

23 ENGR hermodynamics W #5 Problem : WB-7 - S for eat ransfer to R-134a in a Rigid ank - 6 pts 11-May-11 A 0.5 m 3 rigid tank contains R-134a initially at 200 kpa and 40% quality. eat is transferred to the refrigerant from a source at 35 o until the pressure rises to 400 kpa. Determine a.) b.) c.) he entropy change of the R-134a. he entropy change of the heat source. he entropy change of the universe for this process. Read : he keys to this problem are that the specific volume is the same in states 1 and 2 and that the heat source can be treated as a true thermal reservoir (its S can be determined from the definition of entropy). Given : V 0.5 m 3 P kpa P kpa source 35 o x kg vap/kg K Find : a.) S R-134a??? kj/k b.) S res??? kj/k c.) S univ??? kj/k Diagram : 1 2 eat Source 35 o =?? kj ank P 1 = 200 kpa 1 = sat x 1 = 0.4 kg vap/kg ank P 2 = 400 kpa 2 =?? o x 2 =?? kg vap/kg Assumptions : 1 - hanges in kinetic and potential energies are negligible. 2 - he temperature of the heat source is constant and so it acts as a true thermal reservoir. 3 - he tank is a closed system. he mass of R-134a in the tank is constant. Equations / Data / Solve : Part a.) We can determine the change in the specific entropy directly by determining the initial and final specific entropy of the R-134a in the tank. hen, we can multiply by the mass of R-134a in the tank to get S R-134a. S m Sˆ Sˆ R 134a 2 1 Let's begin by determining the mass of R-134a in the tank using : Eqn 1 m V ˆV tank 1 Eqn 2 While we are looking up V 1 in the saturation pressure table we might as well lookup all the other relevant properties. Pressure emp. V (m 3 /kg) U (kj/kg) S (kj/kg-k) (kpa) ( o ) Sat. Liq Sat. Vap Sat. Liq Sat. Vap Sat. Liq Sat. Vap We can now use the quality to determine V 1, U 1, and S 1 using : Mˆ x Mˆ 1x Mˆ 1 1 sat 1 sat vap liq Eqn 3 V m 3 /kg U kj/kg S kj/kg-k Plugging values into Eqn 2 yields : m kg Now, we need S 2. We can get it because we know P 2 and we know that the volume of the tank and the mass of R-134a in the tank are both constant in this process. So: V 2 = V m 3 /kg Dr. - ENGR 224 hw5-sp11.xlsm, WB-7 5/11/2011

24 Next, we determine which phase or phases are present in state 2 by comparing V 2 to V sat liq and V sat vap at 400 kpa. Pressure emp. V (m 3 /kg) U (kj/kg) S (kj/kg-k) (kpa) ( o ) Sat. Liq Sat. Vap Sat. Liq Sat. Vap Sat. Liq Sat. Vap Since : Vˆ Vˆ Vˆ sat 2 sat liq vap the system contains a saturated mixture and the temperature must be equal to the saturation temperature. Next, we must use V 2 to determine the quality of the water in the system. he key equation is: x 2 Vˆ Vˆ 2 sat liq sat vap Vˆ Vˆ sat liq Eqn 4 Plugging values into Eqn 4 yields : x kg vap/kg We can now use the quality to determine U 1, and S 1 using Eqn 3. U kj/kg S kj/kg-k Finally, plugging values into Eqn 1 yields : S R-134a kj/k Part b.) We can determine the entropy change of the heat source from the definition of entropy because we assume the source behaves as a true thermal reservoir. Sres Int Re v Eqn 5 I put the minus sign in front of in this equation because we will apply the 1st Law to the tank to determine. So, is the amount of heat transferred out of the source and with our sign convention, - is the heat transferred when we apply Eqn 1 to the heat source. Apply the 1st Law for closed systems with negligible changes in kinetic and potential energies to the tank. W m Uˆ b Eqn 6 he volume of the tank is constant, so there is no boundary work and Eqn 6 becomes: m Uˆ Uˆ 2 1 Eqn 7 Since we calculated U 1 and U 2 in part (a), all we need to do is plug them into Eqn 7 to determine, and put back into Eqn 5 to get S res kj S source kj/k Part c.) he entropy change of the universe is just the sum of the entropy change for the refrigerated space and the R-134a. Suniv Sres SR 134a Eqn 2 S univ kj/k Verify : None of the assumptions made in this problem solution can be verified. Answers : S R-134a kj/k S source kj/k kj/k S univ Dr. - ENGR 224 hw5-sp11.xlsm, WB-7 5/11/2011