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ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 3.26 - Steam NIST / TFT Fundamentals - 2 pts Complete the following table for water using either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in. Use the default reference state for both the NIST and TFT. T (oC) P (kPa) 200 H (kJ/kg) 140 Phase Description 1800 950 500 800 80 Read : x (kg vap/kg) 0.7 0 3162.2 This problem is designed to test how well you understand how to use the NIST Webbook tables of thermodynamic properties and/or the Thermal/Fluid Properties (TFP) plug-in for Excel. It also tests your understanding and ability to use quality, x. Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent. Given : Two pieces of data for each part, (a) through (e). Find : Complete the table. Assumption: The system is in an equilibrium state. Solution : Part a.) Given : P 200 kPa x 0.7 kg vap/kg NIST Because we are given a quality we know that vapor and liquid both exist in the system in equilibrium. Therefore, we use the Saturation properties — pressure increments option. Results : o 120.21 C T Hsat liq 504.70 kJ/kg 2706.2 kJ/kg Hsat vap ˆ ˆ ˆ H sat x H sat 1 x H sat mix Eqn 1 liq vap H TFT 2045.8 kJ/kg In order to determine Tsat at 200 kPa, use the following cell formula : =TFProp("Water","SI_C","P",200000,"X",0.7,"T") Eqn 2 T 120.23 o C In order to determine H at 200 kPa and x = 0.7 use the following cell formula : =TFProp("Water","SI_C","P",200000,"X",0.7,"H") Eqn 3 H Part b.) Given : o 140 C T 2045.7 H kJ/kg 1800 kJ/kg NIST Because we are given a temperature and we need to determine which phases are present, the first thing we need to do is generate a saturation temperature table using the NIST Webbook and the Saturation properties — temperature increments option. The key results are : Since : ˆ H ˆ H ˆ H sat sat vap liq Temp. (oC) Pressure (kPa) 140 361.54 H (kJ/kg) Sat. Liq 589.16 Sat. Vap 2733.4 the system contains a saturated mixture and the pressure must be equal to the saturation pressure. P= Psat 361.54 kPa Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is: x ˆ H ˆ H sat liq ˆ ˆ H H sat vap Eqn 4 sat liq x Plugging values into Eqn 4 yields : Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.26 0.5647 kg vap/kg 4/12/2011 TFT Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present. Hsat liq Hsat vap =TFProp("Water","SI_C","T",140,"X",0,"H") =TFProp("Water","SI_C","T",140,"X",1,"H") Hsat liq Hsat vap 588.72 kJ/kg 2733.40 kJ/kg ˆ H ˆ H ˆ H sat sat Since : Eqn 5 Eqn 6 vap liq P the system contains a saturated mixture and the pressure must be equal to the saturation pressure. =TFProp("Water","SI_C","T",140,"X",1,"P") Eqn 7 361.29 kPa P= Psat Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is: x =TFProp("Water","SI_C","T",140,"H",1800,"X") Eqn 7 x Part c.) Given : P 950 kPa 0.5648 kg vap/kg x 0 kg vap/kg NIST Because the quality is zero, we immediately know the system contains a saturated liquid. This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given). So, we can use 950 kPa in the Saturation properties — pressure increments option. Pressure (kPa) 950 Here is the relevant data : Temp. (oC) H (kJ/kg) Sat. Liq Sat. Vap 752.74 2775.1 177.66 o T = Tsat 177.66 C H = Hsat liq 752.74 kJ/kg TFT Because the quality is zero, we immediately know the system contains a saturated liquid. This tells us that T = Tsat , H = Hsat liq and Psat = P = 950 kPa (given). Part d.) x =TFProp("Water","SI_C","P",950000,"X",0,"T") Eqn 8 T 177.69 oC H =TFProp("Water","SI_C","P",950000,"X",0,"H") Eqn 9 H 752.68 kJ/kg Given : T o 80 C P 500 kPa NIST Here we are given both the T and P. Let's use the given P, and the Saturation properties- pressure increments option, to determine the Tsat associated with P. Then, we can determine the phases present by comparing the given T to Tsat(P). Tsat(500 kPa) = 151.83 oC Since T = 80oC is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined ! So, we need to look at either the Isothermal properties option or the Isobaric properties option in the NIST Webbook. Either way, no interpolation is required because we know and enter botht he T and P values ! Here is the relevant data : Temp. (oC) 80 Pressure (kPa) 500 H (kJ/kg) 335.37 H 335.37 kJ/kg Here we still need to determine Tsat associated with P. TFT Tsat(500 kPa) =TFProp("Water","SI_C","P",500000,"X",0,"T") Eqn 10 Tsat(500 kPa) = o 151.86 C o Since T = 80 C is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined ! But it is easy to use the TFT to determine H. H =TFProp("Water","SI_C","P",500000,"T",80,"H") H 334.78 kJ/kg Eqn 11 Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.26 4/12/2011 Part e.) Given : P 800 kPa H 3162.2 kJ/kg NIST The first step here is to use the given P to obtain data using the Saturation properties — pressure increments option. This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases present in the system at equilibrium. Since : Temp. (oC) Pressure (kPa) 800 Here is the relevant data : 170.41 H (kJ/kg) Sat. Liq Sat. Vap 720.86 2768.3 the system contains a superheated vapor, quality is undefined and we must use the Isobaric properties option to determine T. ˆ H ˆ H sat vap This time we must enter a range of T values so that we bracket the given H value. I chose to go from 100oC to 500oC by 10oC steps. Here are the two rows in the resulting table that bracket the H value of 3161.7 kJ/kg. Temp. (oC) 340 ??? 350 Pressure (kPa) 800 800 800 H (kJ/kg) 3141.1 3162.2 3162.2 Now, we can determine H from this table by inspection ! T 350.00 oC TFT Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present. Hsat liq Hsat vap =TFProp("Water","SI_C","P",800000,"X",0,"H") =TFProp("Water","SI_C","P",800000,"X",1,"H") Hsat liq Hsat vap Since : 720.68 kJ/kg 2768.67 kJ/kg Eqn 13 the system contains a superheated vapor, quality is undefined and we can directly calculate T using the following formula. ˆ H ˆ H sat vap T Eqn 12 =TFProp("Water","SI_C","P",800000,"H",3161.7,"T") Eqn 14 T Verify : None of the assumptions made in the solution of this problem can be verified based on the given information. Answers : NIST Webbook T (oC) 120.21 140 177.66 80 350.2 P (kPa) 200 361.54 950 500 800 H (kJ/kg) 2045.75 1800 752.74 335.37 3162.2 x (kg vap/kg) 0.7 0.565 0 N/A N/A Phase Description Sat'd Mixture (VLE) Sat'd Mixture (VLE) Sat'd Liquid Subcooled Liquid Superheated Vapor P (kPa) 200 361.29 950 500 800 H (kJ/kg) 2045.72 1800 752.68 334.78 3162.2 x (kg vap/kg) 0.7 0.565 0 N/A N/A Phase Description Sat'd Mixture (VLE) Sat'd Mixture (VLE) Sat'd Liquid Subcooled Liquid Superheated Vapor 350.23 oC TFT T (oC) 120.23 140 177.69 80 350.2 Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.26 4/12/2011 ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 3.29E - R-134a NIST/TFT Fundamentals - 2 pts Complete the following table for R-134a using either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in. Use the default reference state for both the NIST and TFT. T (oF) P (psia) H (Btu/lbm) 80 78 x (lbm vap/lbm) 15 0.6 10 70 180 129.46 110 Read : Phase Description 1.0 This problem is designed to test how well you understand how to use the NIST Webbook tables of thermodynamic properties and/or the Thermal/Fluid Properties (TFP) plug-in for Excel. It also tests your understanding and ability to use quality, x. Finally, it makes you think about the relationship between the data tables and the physical system (phases) that they represent. Given : Two pieces of data for each part, (a) through (e). Find : Complete the table. Assumptions: The system is in an equilibrium state. Solution : Part a.) P 80 psia H NIST : The first step here is to use the given P to obtain data from the Saturation properties — pressure increments option. This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases present in the system at equilibrium. Pressure (psia) 80 Here is the relevant data : Since : 78 Btu/lbm Temp. (oF) H (Btu/lbm) Sat. Liq Sat. Vap 97.162 176.02 65.922 the system contains a subcooled liquid and temperature must be less than the saturation temperature. We must interpolate on Isobaric Properties . ˆ H ˆ H sat liq This time we must enter a range of T values so that we bracket the given H value. I chose to go from 0oF to 100oF by 10oF steps. Here are the two rows in the resulting table that bracket the H value of 78 Btu/lbm . Temp. (oF) 0 ??? 10 Pressure (psia) 80 80 80 H (Btu/lbm) 75.994 78 79.107 Now, we can determine T from this table by interpolation. o 6.44 F T The quality of the water in the system is not defined because it is a subcooled liquid. TFT : Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present. Hsat liq Hsat vap =TFProp("R-134a","EE_F","P",80,"X",0,"H") =TFProp("R-134a","EE_F","P",80,"X",1,"H") Hsat liq Hsat vap Since : Eqn 2 Eqn 3 97.11 Btu/lbm 175.90 Btu/lbm the system contains a subcooled liquid and temperature must be less than the saturation temperature. We can now use the TFT to directly evaluate T. ˆ H ˆ H sat liq T =TFProp("R-134a","EE_F","P",80,"H",78,"T") o 15 F Eqn 4 T o 6.62 F x 0.60 lbm vap/lbm Part b.) T NIST : Because the quality lies between 0 and 1, we immediately know the system contains a saturated mixture. Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.29E 4/12/2011 This tells us that T = Tsat = 15oF (given) and P = Psat . So, we can look up 15oF in the Saturation properties — temperature increments option. Temp. (oF) Here is the relevant data : Pressure (psia) 29.739 15.00 H (Btu/lbm) Sat. Liq Sat. Vap 80.634 169.07 29.739 psia P = Psat Because both saturated liquid and saturated vapor are present in the system at equilibrium, we must use the quality to evaluate the overall average specific internal energy, as follows. The key equation that relates the specific internal energy of the system to the specific internal energy of the saturated liquid and of the saturated vapor is: ˆ ˆ ˆ H sat x H sat 1 x H sat mix Eqn 6 liq vap Plugging values into Eqn 2 yields : TFT : H 133.70 kJ/kg Because the quality lies between 0 and 1, we immediately know the system contains a saturated mixture. This tells us that T = Tsat = 15oF (given) and P = Psat . In order to determine Tsat at 15oF, use the following cell formula : =TFProp("R-134a","EE_F","T",15,"X",0.6,"P") Eqn 7 P 29.739 psia 133.60 Btu/lbm In order to determine H at 15oF and x = 0.6 use the following cell formula : =TFProp("R-134a","EE_F","T",15,"X",0.6,"H") Eqn 8 H o 10 F Part c.) T NIST : Here we are given both the T and P. Let's use the given P, and the Saturation properties — pressure increments option, to determine the Tsat associated with P. Then, we can determine the phases present by comparing the given T to Tsat(P). P 70 psia Tsat(70 psia) = o 58.338 F Since T = 10oF is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined ! So, we need to look at either the Isothermal properties option or the Isobaric properties option. Temp. (oF) 10 Here is the relevant data : Pressure (psia) 70 H (Btu/lbm) 79.099 H TFT : 79.099 Btu/lbm Here we still need to determine Tsat associated with P. Tsat(70 psia) =TFProp("R-134a","EE_F","P",70,"X",0,"T") Eqn 9 Tsat(70 psia) o 58.339 F o Since T = 10 F is less than Tsat, we can conclude that the sytem contains a subcooled or compressed liquid. The quality is undefined ! But it is easy to use the TFT to determine H. H =TFProp("R-134a","EE_F","P",70,"T",10,"H") Eqn 10 H 180 psia H 79.045 Btu/lbm Part d.) P 129.46 Btu/lbm NIST : The first step here is to use the given P to generate data using the Saturation properties — pressure increments option. This will allow us to compare the given value of U to the values of Usat liq and Usat vap in order to determine the phase or phases present in the system at equilibrium. Pressure (psia) 180 Here is the relevant data : Since : ˆ H ˆ H ˆ H sat sat liq vap Temp. (oF) 117.73 H (Btu/lbm) Sat. Liq Sat. Vap 115.28 181.79 the system contains a saturated mixture and the temperature must be equal to the saturation temperature. T = Tsat 117.73 oF Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is: Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 3.29E 4/12/2011 x ˆ H ˆ H sat liq ˆ ˆ H H sat vap Eqn 11 sat liq x Plugging values into Eqn 11 yields : TFT : 0.2132 kg vap/kg Here, we must also determine the Hsat vap and Hsat liq values to determine what phase or phases are present. Hsat liq Hsat vap =TFProp("R-134a","EE_F","P",180,"X",0,"H") =TFProp("R-134a","EE_F","P",180,"X",1,"H") Hsat liq Hsat vap Eqn 12 Eqn 13 115.20 Btu/lbm 181.67 Btu/lbm the system contains a saturated mixture and the temperature must be equal to the saturation temperature. Since : ˆ H ˆ H ˆ H sat sat T =TFProp("R-134a","EE_F","P",180,"X",0,"T") liq vap Eqn 13 o 117.74 F T Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is: x =TFProp("R-134a","EE_F","P",180,"H",129.46,"X") Eqn 13 o 0.2145 F x o 110 F Part e.) T 1.0 lbm vap/lbm NIST : Because the quality is 1.0, we immediately know the system contains a saturated vapor. This tells us that T = Tsat = 100oF (given), H = Hsat vap and P = Psat . x So, we can look up 110oF using the Saturation properties — temperature increments option. Temp. H (Btu/lbm) Here is the relevant data : Pressure (oF) (psia) Sat. Liq Sat. Vap 161.07 110 112.46 181.05 161.07 psia P = Psat TFT : H = Hsat vap 181.05 Btu/lbm In order to determine Psat at 110oF, use the following cell formula : P =TFProp("R-134a","EE_F","T",110,"X",1,"P")' Eqn 14 P 161.05 psia H 180.93 Btu/lbm In order to determine H at 110oF and x = 1.0 use the following cell formula : H =TFProp("R-134a","EE_F","T",110,"X",1,"H")' Eqn 15 Verify : None of the assumptions made in the solution of this problem can be verified based on the given information. Answers : NIST T (oF) 6.44 15 10 117.73 110 P (psia) 80 29.74 70 180 161.07 H (Btu/lbm) 78 133.70 79.10 129.46 181.05 x (lbm vap/lbm) N/A 0.6 N/A 0.2132 1.0 Subcooled Liquid Sat'd Mixture (VLE) Subcooled Liquid Sat'd Mixture (VLE) Saturated Vapor P (psia) 80 29.74 70 180 161.05 H (Btu/lbm) 78 133.60 79.05 129.46 180.93 x (lbm vap/lbm) N/A 0.6 N/A 0.2145 1.0 Subcooled Liquid Sat'd Mixture (VLE) Subcooled Liquid Sat'd Mixture (VLE) Saturated Vapor Phase Description TFT T (oF) 6.62 15 10 117.74 110 Dr. Baratuci - ChemE 260 Phase Description hw2-sp11.xlsm, 3.29E 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : 4.8 - Isobaric Expansion of R-134a - 6 pts A piston-and-cylinder device with a set of stops contains 0.3 kg of steam at 1.0 MPa and 400oC. The location of the stops corresponds to 60% of the initial volume. Now, the steam is cooled. Determine the compression work if the final state is… P2a = 1000 kPa T2a = 250 oC V2a > 0.6 V1 m = 0.3 kg P1 = 1000 kPa T1 = 400 oC P2b = 500 kPa V2b = 0.6 V1 a.) b.) c.) 1.0 MPa and 250oC 500 kPa Determine the temperature at the final state in part (b) Read : The key to determining properties for state 1 is that we know both the pressure and temperature. So, we can determine every other property, including the specific volume and internal energy. The key for states 2a and 2b is whether the piston has fallen all the way down to rest on the stops or not. Since P2a = P1, the piston cannot be resting on the pins. If the piston were resting on the pins, P2a < P1. Consequently, we expect to find V2a > 0.6 * V1 and V2b = 0.6 * V1. Boundary work can be determined from its definition if we assume the process is quasi-equilibrium. Then, we can apply the 1st Law to the process to determine Q. Given : m T1 P1 TFT Parameters : Find : a.) b.) c.) Diagram : See above. T2a P2a P2b V2b/V1 0.3 kg o 400 C 1000 kPa W ba W bb T2b Dr. Baratuci - ChemE 260 Water ??? ??? ??? o 250 C 1000 kPa 500 kPa 0.6 SI_C kJ kJ o C hw2-sp11.xlsm, 4.8 Qa Qb ??? ??? kJ kJ 4/17/2011 Assumptions: Solution : - The steam in the cylinder is a closed system. - Boundary work is the only form of work interaction. - Changes in kinetic and potential energies are negligible. - The expansion is a quasi-equilibrium process. - The initial and final states are both equilibrium states. As problems become more complex, I think it helps to organize the information you collect about each state in a table like the one shown below. The bold green values represent the information obtained from the problem statement. Not every value needs to be filled in, but the table gives you a concise way to keep track of what you know and what you do not know. T P Tsat Phase X v V U H 1 400 1000 179.9 Super 2a 250 1000 179.9 Super 0.3066 0.091976 2956.8 3263.4 0.2327 0.069805 2709.4 2942.1 2b 151.9 500 151.9 Sat Mix 0.4892 0.1840 0.055186 1579.2 1671.2 o C kPa o C kg vap/kg m3/kg m3 kJ/kg kJ/kg The overall objective of this problem is to determine Q and W. This almost always requires the application of the 1st Law. Apply the integral form of the 1st Law to process 1-2a and 1-2b. Q12 W12 U Ekin Epot Eqn 1 If we assume that changes in kinetic and potential energies are negligible, then Eqn 1 simplifies to : ˆ U ˆ Q12 W12 U U 2 U1 m U 2 1 Eqn 2 Consider the steam inside the cylinder to be our system. If we assume that the process is a quasi-equilibrium process and boundary work is the only type of work that crosses the system boundary, then we can use the definition of boundary work to evaluate W12. 2 W12 P dV Eqn 3 1 Part a.) The process in part (a) is isobaric. This allows us to simplify Eqn 3, as follows. ˆ V ˆ W12a P dV P dV P V2a V1 m P V 2a 1 2a 2a 1 1 Eqn 4 Combining Eqns 2 & 4 yields : ˆ U ˆ mP V ˆ V ˆ m U ˆ U ˆ m H ˆ H ˆ Q12a W12a m U 2a 1 2a 1 2a 1 2a 1 Eqn 5 Since we know both T1 and P1 and T2a and P2a, we can lookup any other intensive property for states 1 and 2a in the steam tables. I used the TFT add-in to determine specific V, U and H for states 1 and 2a. V U H State 1 0.30659 2956.8 3263.4 State 2a 0.23268 2709.4 2942.1 m3/kg kJ/kg kJ/kg We can now plug values into Eqns 4 & 5 to evaluate W12a and Q12a. Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 4.8 W12a -22.17 kJ Q12a -96.38 kJ 4/17/2011 Part b.) This part is a bit trickier because the process is not isobaric and we need to determine what phase or phases are present in state 2b. This is worrisome because we cannot apply Eqn 3 to evaluate boundary work unless we know the process path. The process 1-2b consists of two parts. In the first part of the process, before the piston comes to rest on the pins, pressure is constant. This part is an isobaric process and we can use Eqn 4 to evaluate the boundary work. In the second part of the process in part (b), the volume does not change because the piston is resting on the pins. This is an isochoric process and Wb = 0 kJ ! W12b -36.79 kJ Now, we can rearrange Eqn 2 algebraically to help us evaluate Q12b. ˆ U ˆ Q12b W12b m U 2b 1 Eqn 6 So, all we need to do before we use Eqn 6 to finish this part of the problem is determine U2. The problem is that we only know the value of one intensive variable for state 2b, P2b. But we can determine V2b based on the given relationship: V2b = 0.6 * V1 . Since the mass inside the system is constant, the specific volume in state 2b is also 0.6 times the sprecific volume in state 1. 0.18395 m3/kg V2b This gives us the 2nd intensive property value that we need to evalaute any other intensive property. In this case, we need U2b so we can use Eqn 6 to evaluate Q12b. 1579.2 kJ/kg U2b Part c.) Q12b -450.1 kJ We can determine the temperature in state 2b by using the TFT add-in to determine the specific volume of saturated liquid and saturated vapor at P2b and comparing the actual specific volume, V2b. At P2b = 500 kPa Vsat liq = Since : ˆ ˆ ˆ V sat liq < V2b < Vsat vap 0.00109279 m3/kg Verify : None of the assumptions made in this problem solution can be verified. Answers : a.) Dr. Baratuci - ChemE 260 0.3748911 m3/kg T2b = Tsat at P2b = 500 kPa The TFT add-in yields the saturation temperature at 500 kPa : c.) Vsat vap = W12a -22.17 kJ Q12a -96.38 kJ T2b 151.9 o b.) T2b 151.9 o W12b -36.79 kJ Q12b -450.1 kJ C C hw2-sp11.xlsm, 4.8 4/17/2011 ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 4.42 - Compression of Cooling Air by a Linear Spring - 5 pts Steam at 75 kPa and 8% quality is contained in a spring-loaded piston-and-cylinder device as shown below, with an initial volume of 2 m3. Steam is now heated until its volume is 5 m3 and its pressure is 225 kPa. Determine the heat transferred to and the work produced by the steam during this process. Steam Linear Spring Read : The key to solving this problem is the linear nature of the relationship between P and V due to the linear nature of the spring. This makes it relatively easy to evaluate the boundary work for the process. Because the process path is linear on a PV Diagram, the boundary work (area under the path) is just the average pressure times the change in the volume. I demonstrate why this is so in the solution, below. Since initial and final P and V values are given, Wb can be calculated without any other Given the two intensive variables for the initial state, P1 and x1, we can evaluate all of the other properties, including V1 and U1. This will allow us to determine the mass of water in the cylinder from the given initial volume and the specific volume we determined from P1 and x1. This mass and the final volume lets us calculate the final specific volume. As a result, we know the values of two intensive variables in the final state, P2 and specific volume, so we can evaluate U2. Once we know U2, U1 and Wb, we can apply the 1st Law to evaluate Q for the process. Given : P1 = V1 = x1 = 75 2 0.08 kPa m3 kg vap/kg P2 = V2 = 225 5 kPa m3 Find : Wb = ??? kJ Q ??? kJ Assumptions: - The steam in the cylinder is a closed system. - The process occurs slowly enough that it is a quasi-equilibrium process. - There is no friction between the piston and the cylinder wall. - The spring force varies linearly with position. - Changes in kinetic and potential energies are negligible. - Boundary work is the only form of work that crosses the system boundary. Solution : Let's begin with the 1st Law. Ultimately, this is the equation that will give us the relationship between Q and W. The 1st Law for a closed system with negligible change in kinetic and potential energies is : Q - W = DU Eqn 1 If we assume that boundary work is the only form of work that crosses the system boundary, then W becomes Wb in Eqn 1. Q - Wb = DU Eqn 2 2 We can evaluate Wb from the relationship : Wb P dV Eqn 3 1 Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 4.42 4/12/2011 Because the spring is linear, P is a linear function of V. This means : æ P - P1 ö ÷÷ ⋅ ( V - V ) P = P1 + ççç 2 1 èç V2 - V1 ÷ø÷ Eqn 4 2 V2 P P1 P2 P1 V22 V1 V2 1 V1 V1 Wb P1 2 V V1 dV P1 V2 V1 V2 V1 2 1 V2 V1 2 Eqn 5 Now, for some algebraic magic… Wb P1 V2 V1 1 P2 P1 V22 2 V1 V2 V12 2 V2 V1 Wb P1 V2 V1 1 P2 P1 2 V2 V1 2 V2 V1 Eqn 7 Wb P1 V2 V1 1 P2 P1 V2 V1 2 Eqn 8 Wb Eqn 6 P2 P1 V2 V1 2 Eqn 9 OK. That was a lot of work, but now we can plug in numbers and evaluate Wb. Wb 450 kJ So, all we need to do is evaluate U2 and U1 and we can use Eqn 10 to complete the solution of this problem. We can evaluate U1 because we know both the pressure and the quality, P1 and x1. Here is the relevant data : TFT : Chemical : Units : Temp. Pressure (kPa) 75 Water SI_C (oC) 91.78 U (kJ/kg) Sat. Liq Sat. Vap 384.92 2496.2 The key equation that relates the specific internal energy of the system to the specific internal energy of the saturated liquid and of the saturated vapor is: ˆ ˆ ˆ U sat x U sat 1 x U sat mix vap Eqn 10 liq U1 Plugging values into Eqn 10 yields : 553.82 kJ/kg The key to evaluating U2 is to recognize that the mass of H2O inside the cylinder is the same throughout the process. This allows us to modify Eqn 2, as follows. ( ˆ -U ˆ Q = Wb + DU = Wb + m ⋅ U 2 1 ) Eqn 11 Since the initial state is completely determined because we know P1 and x1, we can evaluate V1 and use it to determine the mass of H2O in the cylinder. Here is the relevant data : I used the TFT add-in. Temp. Pressure (kPa) 75 o ( C) 91.78 V (m3/kg) Sat. Liq Sat. Vap 0.001037 2.2171 ˆ ˆ ˆ V sat x Vsat 1 x Vsat mix vap Eqn 12 liq Plugging values into Eqn 12 yields : V1 Now, we can determine the mass of H2O in the cylinder using : m= Plugging values into Eqn 13 yields : m Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, 4.42 0.1783 m3/kg V1 Eqn 13 V̂1 11.22 kg 4/12/2011 Now we know both the volume and the mass in the final state, so we can evaluate the specific volume from : V̂2 = Plugging values into Eqn 15 yields : V2 V2 m Eqn 14 0.4458 m3/kg Now we know the values of two intensive properties in the final state, P2 and specific volume, so we can evaluate any other intensive property. In this case, we need to determine U2 so we can apply Eqn 12 to complete the solution of this problem. Pressure (kPa) 225 Here is the relevant data : I used the TFT add-in. Since : ˆ V ˆ V ˆ V sat 2 sat liq vap Temp. o ( C) 124.00 V (m3/kg) Sat. Liq Sat. Vap 0.001064 0.79325 U (kJ/kg) Sat. Liq Sat. Vap 520.34 2533.1 the system contains a saturated mixture and the temperature must be equal to the saturation temperature. In order to determine U2, we must first evaluate the quality, x2. x2 ˆ V ˆ V 2 sat liq ˆ ˆ V V sat vap Eqn 15 sat liq 0.5614 kg vap/kg x2 Plugging values into Eqn 15 yields : Next, we can use the following equation to evaluate U2. ˆ x U ˆ ˆ U 2 2 sat 1 x 2 U sat vap U2 Plugging values into Eqn 16 yields : Eqn 16 liq 1650.3 kJ/kg Finally, we can plug values into Eqn 11 to evaluate Q and complete this solution. Verify : None of the assumptions made in this problem solution can be verified. Answers : Wb Dr. Baratuci - ChemE 260 450 kJ hw2-sp11.xlsm, 4.42 Q 12747.9 kJ Q 12750 kJ 4/12/2011 ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 4.59E - Determining H Using Heat Capacity Polynomials - 4 pts Determine the enthalpy change of oxygen (O2), in Btu/lbm, as it is heated from 800 to 1500oR, using: a.) b.) c.) The empirical specific heat equation as a function of temperature (Table A-2Ec). The CoP value at the average temperature (Table A-2Eb). The CoP value at room temperature, 80oF, (TableA-2Ea). Read : The empirical specific heat equation will yield the most accurate estimate of the enthalpy change. Assuming a constant value of Cp determined at the average temperature should yield a reasonable estimate of H as well. Using the Cp value at room temperature should not be very accurate. We can compare this result to the value we get in part (a). Given : T1 Find : H1-2 = 800 oR Assumptions: Solution : Part a.) ??? 1500 oR T2 Btu/lbm a.) b.) c.) Oxygen behaves as an ideal gas Oxygen behaves as an ideal gas with a constant heat capacity evaluated at Tavg. Oxygen behaves as an ideal gas with a constant heat capacity evaluated at 80oF. Let's begin by collecting the data we need from Table A-2Ec : oP a b T c T 2 d T 3 C Eqn 1 Where: CP [=] Btu/lbmole-oR 32.00 lbm/lbmole MW T [=] oR 491-3240 oR 6.085 2.017E-03 -5.275E-07 5.372E-11 Temp (K) a b c d The enthalpy change associated with a temperature change for an ideal gas can be determined from : H 1 2 T2 C o P dT Eqn 2 T1 Combining Eqns 1 & 2 and integrating yields : a (T T ) H 2 1 b 2 c d (T2 T12 ) (T23 T13 ) (T24 T14 ) 2 3 4 Eqn 3 H Plug in values for the temperatures and the constants to get : Ĥ Btu / lbm Btu/lbmole Btu / lbmole H Eqn 4 MW lb m / lbmole H1-2 Part b.) 5442.3 We can use Table A-2Eb to evaluate the heat capacity at Tavg : Tavg 170.1 Btu/lbm 1150 oR 690.33 oF CP T (oF) Btu/lbm-oR 600 690.33 700 0.239 Cp(1150oR) 0.242 Cp(1150oR) 0.2417 Btu/lbm-oR Now we can use Eqn 2 to determine H1-2. It is easier to integrate this time because CP is assumed to be constant. ˆo Ĥ1 2 C P,avg T2 T1 H1-2 Plugging values into Eqns 5 yields : Part c.) Eqn 5 We can use Table A-2Ea to evaluate the heat capacity at 80oF: Cp(80oF) Now, we can apply Eqn 5 again because CP is assumed to be a constant. H1-2 169.2 (not bad !) Btu/lbm 0.2190 Btu/lbm-oR 153.3 Btu/lbm This is more than a 10% error. This is unacceptable. Verify : None of the assumptions made in the solution of this problem can be verified based on the given information. Answers : a.) H1-2 170.1 Btu/lbm b.) H1-2 169.2 Btu/lbm ( Error ~ 0.5% ) c.) H1-2 153.3 Btu/lbm ( Error > 10% ) Dr. Baratuci - ChemE 260 hw2-sp11.xlsm , 4.59E 4/12/2011 ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 4.60 - Determining U Using Heat Capacity Polynomials - 4 pts Determine the internal energy change of hydrogen (H2), in kJ/kg, as it is heated from 200 to 800 K, using: a.) b.) c.) The empirical specific heat equation as a function of temperature (Table A-2c). The CoV value at the average temperature (Table A-2b). The CoV value at room temperature, 300 K, (TableA-2a). Read : The empirical specific heat equation will yield the most accurate estimate of the enthalpy change. Assuming a constant value of CV determined at the average temperature should yield a reasonable estimate of U as well. Using the CV value at room temperature should not be very accurate. We can compare this result to the value we get in part (a). Given : T1 Find : U1-2 = 200 K Assumption: Solution : Part a.) ??? T2 800 K kJ/kg a.) b.) c.) Hydogen behaves as an ideal gas Hydogen behaves as an ideal gas with a constant heat capacity evaluated at Tavg. Hydogen behaves as an ideal gas with a constant heat capacity evaluated at 80oF. Let's begin by collecting the data we need from Table A-2c : a b T c T2 d T3 C o P Eqn 1 Where: MW Temp (K) a b c d T [=] K CP [=] J/mol-K 2.016 g/mol 273-1800 K 29.11 -1.916E-03 4.003E-06 -8.704E-10 The enthalpy change associated with a temperature change for an ideal gas can be determined from : H 1 2 T2 C o P dT Eqn 2 T1 Combining Eqns 1 & 2 and integrating yields : a (T T ) H 2 1 b 2 c d (T2 T12 ) (T23 T13 ) (T24 T14 ) 2 3 4 Eqn 3 H Plug in values for the temperatures and the constants to get : 17474.9 J/mol Next, we have to determine U from H. Consider the differential form of the definition of enthalpy : U PV H But for ideal gases : Eqn 4 Eqn 5 Combining Eqns 4 & 5 yields : IG U IG R TIG H Eqn 6 Solving Eqn 6 for U yields : IG H IG R TIG U Eqn 7 Plugging values into Eqn 7 yields : Û kJ / kg Part b.) R T PV J / mol U 1000 g / kg Eqn 8 MW g / mol 1000 J / kJ R U 8.314 12486.5 J/mol-K J/mol U1-2 6193.7 kJ/kg We can use Table A-2b to evaluate the heat capacity at Tavg : Tavg Fortunately, there is an entry in Table A-2b at 500 K, so we can read CV : CV 500 K 10.389 (kJ/kg-K) The internal energy change associated with a temperature change for an ideal gas can be determined from : Û1 2 T2 Cˆ o V dT Eqn 9 T1 Because CV is assumed to be constant, Eqn 9 is easy to integrate and obtain : Dr. Baratuci - ChemE 260 hw2-sp11.xlsm , 4.60 4/12/2011 ˆo Û1 2 C V ,avg T2 T1 Part c.) Eqn 10 Plugging values into Eqn 10 yields : U1-2 We can use Table A-2a to evaluate CV at 300 K: CV(300 K) Now, we can apply Eqn 6 again because CP is assumed to be a constant. H1-2 6233.4 (not bad !) kJ/kg 10.1830 Btu/lbm-oR 6109.8 Btu/lbm This is less than a 2% error. Not so bad. Why ? Because CV for H2 doesn't change so much with respect to T. Verify : None of the assumptions made in the solution of this problem can be verified based on the given information. Answers : a.) U1-2 6194 Btu/lbm b.) U1-2 6233 Btu/lbm ( Error ~ 0.64% ) c.) U1-2 6110 Btu/lbm ( Error ~ 1.4% ) Dr. Baratuci - ChemE 260 hw2-sp11.xlsm , 4.60 4/12/2011 ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 WB-1 - Clapeyron & Clausius-Clapeyron Equations - 5 pts Estimate the latent heat of vaporization, in Btu/lbm, of ammonia at -10oF using: a.) b.) c.) The Clapeyron equation The Clausius-Clapeyron equation The ammonia tables Read : The key to this problem is using the saturated NH3 tables to obtain P*(T) data for use in the Clapeyron and Clausius-Clapeyron Equations. This data is used to estimate dP*/dT in the Clapeyron Equation and d[Ln(P*)] / d(1/T) in the Clausius-Clapeyron Equation. The Clausius-Clapyron Equation is less accurate because it requires two more simplfying assumptions than the Clapeyron Equation. But, the C-C Eqn does not require that you know the Vvap. So, it is a trade off. Given : T o -10 F Assumptions: P*(T) is approximately linear beteen 0 and -20 oF. 1- Solution : Part a.) H d P* vap dT T V vap The Clapeyron Equation is : Eqn 1 If P*(T) is approximately linear over a narrow temperature interval, then : d P * P* T dT Eqn 2 Now, we can substitute Eqn 2 into Eqn 1 and solve for Hvap, as follows. H P * vap T T Vvap Eqn 3 H vap T Vvap P * T Eqn 4 We can look up the vapor pressure of NH3 at 0oF and -20oF in the saturation temperature table and use this data to evaluate P* / T. T (oF) T (oR) P* (psia) 0 -20 459.67 439.67 30.397 18.279 P* / T 1 psia = P* / T o 0.6059 psia/ R 144 lbf / ft2 87.25 lbf / ft2-oR Next, we must use the saturated NH3 Tables again to evaluate Vvap. At -10oF : Vsat liq Vsat vap 0.023929 ft3 / lbm 11.502 ft3 / lbm Now, we can plug values into Eqn 4 : Vvap 11.47807 ft3 / lbm T Hvap o 449.67 R 4.503E+05 ft-lbf / lbm Hvap 578.70 Btu / lbm Unit conversions: 1 Btu = Dr. Baratuci - ChemE 260 778.17 ft-lbf hw2-sp11.xlsm, WB-1 4/12/2011 Part b.) The Clausius-Clapeyron Equation is : H vap Ln P* R 1 C T Eqn 5 This equation tells us that Ln[P*] vs. 1/T is linear and the slope is : Slope Ln P2* Ln P1* 1 / T2 1 / T1 H vap Eqn 6 R We can use the data in the table from part (a) to evaluate the slope. Slope o -5139.39 R Next we can solve Eqn 6 for Hvap and plug in values to evaluate it. H vap R Slope Eqn 7 R Hvap Hvap 17.03 lbm/lbmole MW Part c.) The saturated NH3 tables tell us that : Verify : None of the assumptions made in the solution of this problem can be verified based on the given information. Answers : a.) Hvap 578.7 Btu / lbm b.) Hvap 599.6 Btu / lbm c.) Hvap 576.3 Btu / lbm Hsat liq Hsat vap Hvap o 1.987 Btu/lbmole- R 10211.97 Btu/lbmole 599.65 Btu / lbm 31.982 Btu / lbm 608.26 Btu / lbm 576.28 Btu / lbm The Clausius is quite accurate, but the Clausius-Clapeyron Equation is off by about 4%. The additional assumptions for the Clausius-Clapeyron Equation cause this increase in the error. The molar volume of the saturated liquid is only about 2% of the molar volume of the saturated vapor, Vvap ~ Vsat vap is not a very bad assumption. The Clausius-Clapeyron Equation also requires that the saturated vapor behave as an ideal gas. V The molar volume of the saturated vapor at -10oF is : 3 195.88 ft / lbmole 12.07 L/mole Because the molar volume is much less than 20 L/mole, it is not accurate to treat the NH3 as an ideal gas. This is the principle source of error in using the Clausius-Clapeyron Equation in this problem. Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, WB-1 4/12/2011 ENGR 224 - Thermodynamics Problem : WB-2 - Hypothetical Process Paths and the Latent Heat of Vaporization - 8 pts Use the hypothetical process path shown here to help you determine the change in enthalpy in Joules for 20.0 g of heptane (C7H16) as it changes from a saturated liquid at 300 K to a temperature of 370 K and a pressure of 58.7 kPa. Calculate the H for each step in the path. Do not use tables of thermodynamic properties, except to check your answers. Instead, use the Antoine Equation to estimate the heat of vaporization of heptane at 300 K. Obtain the Antoine constants from the NIST Webbook. Use the average heat capacity of heptane gas (also from from the NIST Webbook) over the temperature range of interest. Assume heptane gas is an ideal gas at the relevant temperatures and pressures. Read : Given: Baratuci HW #2 14-Apr-11 4 3 1 Saturated Liquid P1, T1 Superheated Vapor P2, T2 Superheated Vapor P1, T2 2 Saturated Vapor P1, T1 Step 1-2 is a bit tricky. We can use the Antoine Eqn with the Clausius-Clapeyron Eqn to estimate Hvap. Step 2-3 is straightforward because the problem instructs us to use an average Cp value. The only difficulty will be that Cp values may not be avail m= 20 g H1-2 ??? J T1 = 300 K H2-3 ??? J x1 = T2 = P2 = 0.0 (sat'd liq) 370 K 58.7 kPa H3-4 H1-4 ??? ??? J J Assumptions : Find: - Clausius-Clapeyron applies: - The saturated vapor is an ideal gas - The molar volume of the saturated vapor is much much greater than the molar volume of the saturated liquid. - The heat of vaporizatioon is constant over the temperature range of interest. - The superheated vapor also behaves as an ideal gas. - The heat capacity of the vapor is nearly constant over the temperature range of interest so that using the average value is a reasonable approximation. Solution : First we can observe that H1-2 = Latent heat of vaporization at 300 K We can estimate the heat of vaporization using the Clausius -Clapeyron Equation. vap H Ln P* R 1 C T Eqn 1 If we plot Ln P* vs. 1/T(K), the slope is - Hvap/R. We can calculate the vapor pressures at two different temperatures using the Antoine equation. Use temperatures near the temperature of interest, 300 K. Use the two points to estimate the slope over this small range of temperatures. Ln Pb Ln Pa 1 / Tb 1 / Ta Eqn 2 Log10(P*) = A - (B / (T + C)) Eqn 3 Slope Antoine Equation: P is in bar T is in Kelvin The Antoine constants from the NIST WebBook are: A 4.02832 B 1268.636 C -56.199 From the Antoine Eqn: Ta = Tb = 299.5 K 300.5 K Dr. Baratuci - ChemE 260 Pa = Pb = 6.52 kPa 6.85 kPa hw2-sp11.xlsm , WB-2 Slope = -4423.1 K 4/12/2011 Next we use this slope with Eqn 1 to determine the latent heat of vaporization at 300 K : R= n 8.314 J/mol K m MW Eqn 4 Hvap = 36773 J/mol MW = 100.20 g/mol n 0.1996 H(1-2) = mol 7,340 J Next, let's consider the enthalpy change from states 2 to 3, saturated vapor to superheated vapor: Because we assumed a constant heat capacity, we can evaluate H using: T C H p Eqn 5 The heat capacities are tabulated in the NIST WebBook, under the Name Search option. Interpolate to estimate Cp at both T1 and T2. Then, average these two values of Cp to obtain the average heat capacity. This is equivalent to determining a linear equation between T1 and T2 and integrating. Gas phase heat capacity data from the NIST WebBook: Solve: Cp(T1) = Cp(T2) = Cp, avg = Temperature (K) 300 400 500 Cp,gas (J/mol*K) 165.98 210.66 252.09 There are many different ways to estimate Cp(T1) and Cp(T2). 166.0 J/mole-K 197.3 J/mole-K 181.6 J/mole-K H(2-3) = Now, just multiply by the number of moles, n, to get H2-3 : 12,713 J/mol H(2-3) = 2,538 J Last, we need to determine the enthalpy change from states 3 to 4, in which the pressure of the superheated vapor is reduced. Recall the assumption that the vapor behaves as an ideal gas. Because enthalpy is only a function of T for ideal gases, and since T3 = T4 : H(3-4) = Finally: H1-4 = H1-2 + H2-3 + H3-4 = 0 J 9,878 J Verify : None of the assumptions made in the solution of this problem can be verified based on the given information. Answers : H1-2 7340 J H3-4 0 J H2-3 2538 J H1-4 9,878 J Dr. Baratuci - ChemE 260 hw2-sp11.xlsm , WB-2 4/12/2011 ENGR 224 - Thermodynamics Problem : Baratuci HW #2 14-Apr-11 WB-3 - Work and Heat Transfer for a Closed, 3-Step Cycle - 6 pts A closed system undergoes a thermodynamic cycle consisting of the following processes: Process 1-2: Adiabatic compression from P1 = 50 psia, V1 = 3.0 ft3 to V2 = 1 ft3 along a path described by : ˆ 1.4 constant PV Process 2-3: Process 3-1: Constant volume. Constant pressure with U1 - U3 = 46.7 Btu. There are no significant changes in kinetic or potential energies in any of these processes. a.) b.) c.) Sketch this cycle on a PV Diagram. Calculate the net work for the cycle in Btu. Calculate the heat transfer for process 2-3. Read : The sketch requires a lot of thought about the different types of processes involved. Determine the Wb value for each step in the cycle and then add them up to get Wcycle. Apply the 1st law to each step in the process and make use of the known value of U1 - U3 to evaluate Q23. Given: Step 1-2: P1 50 psia 1.4 Step 2-3 V3 = V2 = 3 1.0 ft Step 3-1: P3 = P2 U1 - U3 = 46.7 Btu Find: Assumptions: Solution : Part a.) a.) b.) c.) 3 3.0 ft 3 1.0 ft V1 V2 Sketch this cycle on a PV Diagram. W cycle ??? Btu Q23 ??? Btu - The gas is a closed system - Boundary work is the only form of work interaction - Changes in kinetic and potential energies are negligible. It might be easier to construct the PV Diagram after we complete the rest of the problem. But we might learn something by constructing the diagram first. Let's try. We know V1 > V2. We know step 1-2 is an adiabatic compression, so we know that P2 > P1 and W12 < 0. Because adiabatic compression causes the fluid to heat up because of the addition of energy in the form of work, T2 > T1. That is all we need to sketch path 1-2 on our PV Diagram. See below. Because step 2-3 is isochoric, we know that the path for step 2-3 is a vertical line and W23 = 0. Because we also know that step 3-1 is isobaric, we know that path 3-1 is a horizontal line at P3 = P1. This shows us that step 2-3 results in a decrease in pressure at constant volume. The only way to accomplish this is to cool the system, so we expect T3 < T2 and Q23 < 0. Now, we know everything we need to sketch path 2-3 on the PV Diagram. Step 3-1 is an isobaric expansion in which U increases by 46.7 Btu. Since the volume increases, we know that W31 > 0. The addition of heat must be the cause of the increase in U and the expansion of the working fluid. So, we expect Q 31 > 0. P T1 We are able to determine what the path for this cycle looks like without calculating anything ! T3 I 2 s o c h o r i c 3 T2 PV1.4 = Constant Isobaric 1 V Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, WB-3 4/12/2011 Part b.) Since Wcycle = W12 + W23 + W31, we will work our way around the cycle and calculate each work term along the way. In step 1-2: P V = C. We can use V or specific V here because the system is closed and the mass inside is constant. The boundary work for a polytropic process is : dV P2 V2 P1 V1 1 1 V 2 2 W12 P dV C 1 Eqn 1 All we need to do is determine P2 and then we can use Eqn 1 to evaluate W12. The way to go here is to evaluate the constant, C, from P1 and V1 : P1 V1 C Eqn 2 232.777 psia-ft C 3 Now, apply the polytropic path equation to state 2, as follows. P2 V2 C P2 Eqn 3 C V2 P2 Eqn 4 232.8 psia Now, we can plug values back into Eqn 1 to determine W12 : 1 ft2 1 Btu 2 144 in 778.17 ft-lbf W 12 Step 2-3 is isochoric, so there is no baoundary work: W 23 3 -206.94 psia-ft -38.295 Btu 0 Btu Step 3-1 is isobaric, so we can evaluate W23 from : P dV P dV P V V P V W 31 1 3 Part c.) 1 1 3 3 Eqn 5 Here, P is either P1 or P3, because they are the same. Plugging in values yields : W 31 Now, we add up the work terms to get : Wcycle = 3 100.00 psia-ft 18.505 Btu -19.79 Btu In order to determine Q23, we need to apply the 1st Law to the step. Q23 - W23 = U3 - U2 = 0 Eqn 6 But, W23 = 0, so Eqn 6 becomes : Q23 = U3 - U2 Eqn 7 Next, let's apply the 1st Law to step 1-2. Q12 - W12 = U2 - U1 Eqn 8 Plugging in the value of W12 yields : U2 - U1 = 38.295 Btu We were given that : U1 - U3 = 46.7 Btu U3 - U2 = - ( U2 - U1 ) - ( U1 - U3 ) Therefore : Eqn 9 U3 - U2 = Q23 Therefore : Verify : None of the assumptions made in this problem solution can be verified. Answers : a.) See the PV Diagram, above. b.) Wcycle -19.79 -84.995 Btu -84.995 Btu Btu What kind of cycle is this ? This is a refrigeration or heat pump cycle because Wcycle < 0. For a cycle, Qcycle = Wcycle, so Qcycle < 0 as well. The cycle rejects all the heat that is absorbs plus it converts a net amount of input work into heat and also rejects that energy as heat. We cannot tell whether it is a refrigeration or heat pump cycle because the only difference is the goal or desired quantity. If this cycle were a refrigerator, the goal would be to absorb Q23 from the refrigerated space. If the cycle were a heat pump, the goal would be to reject Q31 to the heated space. Notice that more energy would be removed from the refrigerated space than the net amount of work put into the cycle. If this cycle were a heat pump, it would deliver Q31 to the heated space and Q31 is MUCH larger than Wcycle. This advantage is why heat pumps are ofter preferred to electrical resistance heaters. Q31 65.205 Btu Applying the 1st Law to step 3-1 yields : COPHP Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, WB-3 3.29 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : WB-4 - Heat Conduction Through a Composite Wall - 3 pts A composite plane wall consists of a 9 inch thick layer of brick and a 4 inch thick layer of insulation. The outer surface temperatures of the brick and insulation are 1260oR and 560oR, respectively, and there is perfect contact at the interface between the brick and the insulation. At steady-state, determine the heat conduction flux through the wall in Btu/h-ft2 and the temperature in oR at the interface between the brick and the insulation. Data : Brick 1.4 k Insulation 0.05 Btu/h-ft-oR Read : This problem is an application of Fourier's Law of Conduction. The key is to assume that the process operates at steady-state. In this case, all of the heat that arrives at the interface between the brick and the insulation by conduction through the brick must leave the interface as heat conduction through the insulation. Otherwise, the temperature at the interface would rise or fall as energy accumulated or was depleted at the interface. Given : Lbrick 9 0.75 1.4 1260 kbrick Tbrick Find : q ??? Diagram : in ft Lins o Btu/h-ft-oR R kins Tins Btu/h-ft2 Tint Lbrick = 9 in Tins = 1260 oR 4 0.3333 0.05 560 ??? in ft Btu/h-ft-oR R o o R Lins = 4 in k = 0.05 Btu/h-ft2-oR Tint = ? oR k = 1.4 Btu/h-ft2-oR Assumptions : Solution : Tins = 560 oR 1- The heat transfer is a steady-state process. No variable changes with respect to time. 2- The thermal conductivity of the brick is uniform and constant, not a function of temperature. 3- The thermal conductivity of the insulation is uniform and constant, not a function of temperature. The key to this problem is that, at steady-state, all of the heat that arrives at the interface between the brick and the insulation by conduction through the brick must leave the interface as heat conduction through the insulation. Otherwise, the temperature at the interface would rise or fall as energy accumulated or was depleted at the interface. In terms or Fourier's Law of Conduction, this means : dT dT q brick k brick q ins k ins dx brick dx ins Eqn 1 If we assume that the thermal conductivity of brick is constant and the thermal conductivity of the insulation is also constant, then the temperature profiles in the brick and in the insulation are linear. This allows us to replace the derivatives in Eqn 1 with the slope of the temperature profile in the brick and in the insulation. T T k brick k ins x brick x ins Eqn 2 Now, we can express the slopes of the temperature profiles in terms of the known temperatures on the outer faces of the wall and the unknown temperature at the interface between the brick and insulation layers. T Tint Tins Tint k brick brick k ins Lbrick L ins Dr. Baratuci - ChemE 260 hw2-sp11.xlsm, WB-4 Eqn 3 4/12/2011 The only unknown in Eqn 3 is Tint, so we can solve for it. k k k k Tint ins brick Tins ins Tbrick brick L L L L brick ins brick ins Tins Tint Eqn 4 k ins k Tbrick brick L ins L brick k ins k brick L ins L brick Plugging values into Eqn 5 yields : Eqn 5 kbrick/Lbrick kins/Lins Tint 1.867 Btu/h-ft2-oR 0.150 Btu/h-ft2-oR o 1207.9 R Now that we know Tint, we can evaluate the temperature gradient or slope in each material and then use either part of Eqn 1 to evaluate q. (T/x)brick qbrick -69.42 o R/ft (T/x)brick 2 -97.19 Btu/h-ft qins Verify : None of the assumptions made in this problem solution can be verified. Answers : q Dr. Baratuci - ChemE 260 2 -97.2 Btu/h-ft Tint hw2-sp11.xlsm, WB-4 o -1943.80 R/ft 2 -97.19 Btu/h-ft o 1208 R 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : WB-5 - Combined Convection and Radiation Heat Loss - 3 pts A 3.0 m2 hot black surface at 80oC is losing heat to the surrounding air at 25oC by convection with a convection heat transfer coefficient of 12 W/m2-oC, and by radiation to the surrounding surfaces at 15oC. Determine the total rate of heat loss from the surface in W. Read : The total rate of heat loss from the black surface is the sum of the heat loss rates by convection and radiation. The key assumption here is that the emissivity of the surface is 1.0 because it is described as "black". Given : AHT h Find : Qtotal 3.0 m2 2 12 W/m -K 1 ??? Diagram : Ts Tair Tsurr W Tair = 25 oC Ts = 80 oC Assumptions : 1234- Solution : o 80 C o 25 C o 15 C Tsurr = 15 oC The heat transfer is a steady-state process. No variable changes with respect to time. The surface only exchanges heat by radiation with the surroundings and only exchanges heat by convection to the air. The emissivity of the surface and of the surroundings are assumed to be 1.0. The absorptivity, , of the surface and the surroundings are 1.0. The key here is that the total rate of heat loss from the surface is the sum of the heat loss rate by radiation and convection. Q total Qconv Qrad Eqn 1 Heat transfer by convection can be determined using Newton's Law of Cooling. Q conv h A HT Ts Tair Eqn 2 If we assume that the emissivity of the black surface and of the surroundings are both 1.0, then we can evaluate the radiation heat transfer rate using : 4 Qrad A HT Ts4 Tsurr Eqn 3 Where is the Stefan-Boltzmann Constant : Plugging values into Eqns 3, 2 & 1 yields : Qconv Qrad 1980 W 1473.0 W Qtotal 3453 W Ts Tsurr 353.15 K 288.15 K Verify : None of the assumptions made in this problem solution can be verified. Answers : Qtotal Dr. Baratuci - ChemE 260 2 4 5.67E-08 W/m -K 3450 W hw2-sp11.xlsm, WB-5 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : WB-6 - Minimum Insulation Thickness for a Hot Surface - 3 pts A flat surface is covered with insulation with a thermal conductivity of 0.08 W/m-K. The temperature at the interface between the surface and the insulation is 300oC. The outside of the insulation is exposed to air at 30oC and the convection heat transfer coefficient between the insulation and the air is 10 W/m2-K. Ignoring radiation heat transfer, determine the minimum thickness of the insulation, in m, such that the outside surface of the insulation is not hotter than 60oC at steady-state Read : This is a straight-forward application of Fourier's Law of Conduction and Newton's Law of Cooling. The key is that, at steady-state, energy cannot accumulate at the surface where the air touches the insulation. As a result, the rate at which heat is conducted out through the insulation must be equal to the rate at which heat is removed by convection from the surface of the insulation. Given : kins h Qrad Find : Lins Tint Tair Ts 0.08 W/m-K 2 10 W/m -K 0 W ??? o 300 C o 30 C o 60 C cm Diagram : Tair = 30oC Ts = 60oC Tint = 300oC Lins Assumptions : Solution : 123- The heat transfer is a steady-state process. No variable changes with respect to time. The thermal conductivity of the brick is uniform and constant, not a function of temperature. The thermal conductivity of the insulation is uniform and constant, not a function of temperature. The key to this problem is that, at steady-state, all of the heat that arrives at the surface of the insulation by conduction through the insulation must leave the surface as heat transfer by convection. Otherwise, the temperature at the surface would rise or fall as energy accumulated or was depleted at the surface. In terms or Fourier's Law of Conduction and Newton's Law of Cooling, this means : dT q ins k ins q conv h Ts Tair dx ins Eqn 1 If we assume that the thermal conductivity of the insulation is constant, then the temperature profile in the insulation is linear. This allows us to replace the derivative in Eqn 1 with the slope of the temperature profile in the insulation. T k ins h Ts Tair x ins Eqn 2 Now, we can express the slopes of the temperature profiles in terms of the known temperature at the interface between the surface and the insulation and the temperature at the outer surface of the insulation. T Ts k ins int h Ts Tair L ins Eqn 3 The only unknown in Eqn 3 is Lins, so we can solve for it. L ins k ins Tint Ts h Ts Tair Eqn 4 Plugging values into Eqn 4 yields : Lins Verify : None of the assumptions made in this problem solution can be verified. Answers : Lins Dr. Baratuci - ChemE 260 0.064 m 0.064 m hw2-sp11.xlsm, WB-6 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : WB-7 - 1st Law Analysis of Steam in a Closed System - 3 pts As shown in the figure below, 5.0 kg of steam contained within a piston-and-cylinder device undergoes an expansion from state 1 where the specific internal energy is 2709.9 kJ/kg to state 2 where the specific internal energy is 2659.6 kJ/kg. During the process, there is heat transfer to the steam with a magnitude of 80 kJ. Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5 kJ. There is no significant change in the kinetic or gravitational potential energies of the steam. Determine the work done by the steam on the piston during the process in kJ. Read : This problem is an application of the 1st Law. Making and applying the proper assumptions are the keys to solving this problem correctly. Given : Q W prop m Find : Wb Diagram : See above. Assumptions: Solution : U1 U2 80 kJ -18.5 kJ 5 kg ??? 2709.9 kJ/kg 2659.6 kJ/kg kJ - The steam is a closed system - Boundary work and the work done by the paddle wheel are the only forms of work in this process. - Changes in kinetic and potential energies are negligible. - The process is a quasi-equilibrium process. - The initial and final states are both equilibrium states. Let's begin by writing the 1st Law, closed system, for this process. ˆ U ˆ Q Wtotal m U 2 1 Eqn 1 Propeller work and boundary work are the only forms of work that cross the system boundary in this process. Therefore : Wtotal Wprop Wb Combining Eqns 1 & 2 yields : Eqn 2 ˆ U ˆ Q Wprop Wb m U 2 1 Eqn 3 We can solve Eqn 3 for the unknown Wb, as follows. ˆ U ˆ Wb Q Wprop m U 2 1 Plugging values into Eqn 4 yields : Eqn 4 Wb Verify : None of the assumptions made in this problem solution can be verified. Answers : Wb Dr. Baratuci - ChemE 260 350 kJ 350 kJ hw2-sp11.xlsm, WB-7 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : 6.23 - Power Plant Efficiency - 2 pts In 2001, the United States produced 51% of its electricity in the amount of 1.878 x 1012 kW-h from coal-fired plants. Taking the average thermal efficiency to be 34%, determine the amount of thermal energy rejected by the coal-fired power plants in the United States. Read : Assume that the shaft work produced by turbines in electric power plants is 1.878 x 1012 kW-h. Use the definition of the thermal efficiency of a heat engine to determine QH and then apply the 1st law to all of the power generation facilities in the country, collectively, to determine QC. Given : W HE Find : QC 1.878E+12 kW-h ??? 0.34 kW-h Diagram : Hot QH = ??? kW-h = 0.34 HE WHE = 1.878 x 1012 kW-h QC = ??? kW‐h Cold 3.646E+12 kW-h - Assume that the shaft work produced by turbines in electric power plants is 1.878 x 1012 kW-h. Assumptions: Solution : QC Let's begin by writing the equation that defines the efficiency of a power cycle. Desired Output WHE Re quired Input QH Eqn 1 QH We can solve Eqn 1 for QH : WHE Plugging values into Eqn 2 yields : Eqn 2 QH 5.524E+12 kW-h We can determine the annual energy rejected, QC, by applying the 1st Law to all the heat engines in the US. Solving Eqn 3 for QC yields : QH = WHE + Q C Eqn 3 Q C = QH - WHE Eqn 4 Plugging values into Eqn 4 yields : Verify : Answers : QC The big assumption on which this solution cannot be readily verified. QC 3.65E+12 Dr. Baratuci - ChemE 260 3.646E+12 kW-h This is an enormous amount of heat to dump into the environment ! kW-h hw2-sp11.xlsm, 6.23 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : 6.41 - Work, Heat and COP in a Refrigeration Cycle - 2 pts A refrigerator used to cool a computer requires 3 kW of electrical power and has a COP of 1.4. Calculate the cooling effect of this refrigerator, in kW. Hot QH Ref Wcycle QC Cold Read : This is an application of the the definiton of COP for a refrigeration cycle. Given : COPR 1.4 Find : QC ??? Assumptions : 1- W 3 kW kJ When the refrigerator is operating, it operates at steady-state. Equations / Data / Solve : QC QC Wref Q H QC The definition of COPR is : COPR Solve Eqn 1 for QC : QC COPR Wref Eqn 1 Eqn 2 QC Plugging values into Eqn 2 yields : 4.2 kW We pay for just 3 kW of electrical power to operate our refrigerator and the device removes heat from the refrigerated space at a rate of 4.2 kW. Pretty cool, eh ? Verify : Answers : None of the assumptions made in this problem solution can be verified. QC Dr. Baratuci - ChemE 260 4.2 kW hw2-sp11.xlsm, 6.41 4/12/2011 ENGR 224 - Thermodynamics Baratuci HW #2 14-Apr-11 Problem : 6.55 - Heat Pump COP and Monthly Operating Cost - 2 pts A house that was heated by electric resistance heaters consumed 1200 kW-h of electric energy in a winter month. If this house were heated instead by a heat pump that has an average COP of 2.4, determine how much money the home owner would have saved that month. Assume a price of $0.085/kW-h for electricity. Read : This problem is an application of the definition of the coefficient of performance of a heat pump. Notice that money "saved" is the difference between the monthly operating cost of the electrical resistance heater and the monthly operating cost of the heat pump. Given : COPHP Price 2.4 0.085 $/kW-h Savings ??? $/month Find : W 1200 kW-h Diagram : Hot QH COPHP = 2.4 HP WHP = 1200 kW-h QC Cold Assumptions: Solution : 1- When the heat pump is operating, it operates at steady-state. Let's begin by determining the cost to heat the house with the electrical resistance heaters. In an electrical resistance heater, each kW-h of electrical energy is converted directly into a kW-h of heat that is transferred into the house. The cost of this heat is just the product of the electrical energy used and the price of the electricity. Cost Wmonth kW h / month * Pr ice[$ / kW h] Eqn 1 costresist 102 $/month Now, let's consider a heat pump that will also deliver the 1200 kW-h of heat into the house. In terms of our traditional tiefighter diagram, this means QH = 1200 kW-h for our heat pump. Now, we have to determine how much electrical work we must supply to the heat pump in order to get QH = 1200 kW-h. QH 1200 kW-h Let's write the equation that defines the COP for a heat pump cycle. COPHP WHP We can solve Eqn 2 for WHP, as follows : QH Desired Output Re quired Input WHP Eqn 2 QH COPHP Plugging values into Eqn 3 yields : Eqn 3 W HP 500 kW costHP 42.5 $/month We can determine the monthly energy consumption of the heat pump using Eqn 1. We can now determine the savings that could be achieved with this heat pump from : Savings = costresist - costHP Verify : Answers : Eqn 4 Savings 59.5 $/month None of the assumptions made in this problem solution can be verified. Savings Dr. Baratuci - ChemE 260 59.5 $/month hw2-sp11.xlsm, 6.55 4/12/2011