ME 201 Thermodynamics

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Domenic Holmes
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1 Fall 204 ME 20 Thermodynamics Homework #8 Solutions. A rigid wall, insulated container is divided into two regions by a removable wall. One region contains 0. lb m of CO at 00 F, while the other region contains 0.2 lb m of CO at 50 F. A stirrer is inserted into the container and when the wall between the two regions is removed the stirrer provides a shaft work input of 5 Btu. Determine the final temperature of the CO. Solution: Working Fluid: CO (Ideal Gas) Process: Constant Volume Initial State Final State State State 2 State 3 T 00 F K T 2 50 F K T K 42. F m 0. lb m kg m lb m kg m lb m kg u 9.62 kj/kg u kj/kg u kj/kg Bold values are calculated. Q 0 W sh -5 Btu (into system) kj W bnd 0 Final State: UNKNOWN Conservation of Mass: m + m 2 m 3 st Law: m 3 u 3 (m u + m 2 u 2 ) -W sh Approach: To fix the final state we will use our process description along with the conservation of mass to determine the final mass. Using the conservation of energy, we can determine the final internal energy. This will then fix the final state and we can go to the ideal gas tables to determine the final temperature Using conservation of mass m3 m + m kg Applying our conservation of energy we can write m3u3 - m u m2 u2 - Wsh From the ideal gas tables for CO we find u 9.62 kj/kg and u kj/kg Using this in our conservation of energy equation and solving for u 3 (m u m2 u2) - Wsh u3 m3 Substituting our values and calculating we have u kj/kg

2 ME 20 Thermodynamics Fall 204 Going to the CO table, we find that T K 42. F 2. A balloon containing 0.5 kg of Refrigerant-2 is removed from the freezer at 00 kpa and - 40 C and placed on the kitchen counter. Determine the heat transfer required to (a) raise the R-2 temperature to its boiling point (b) boil the liquid (c) raise the resulting vapor to the room temperature of 23 C You may assume that the kitchen pressure is 00 kpa. Solution: For this problem we will consider the kitchen as our system. Working Fluid: R-2 (Phase Change substance) Process: Constant pressure State State 2 State 3 State 4 T 0 C T C T C T 4 23 C P 00 kpa P 2 00 kpa P 3 00 kpa P 4 00 kpa v m 3 /kg v m 3 /kg v m 3 /kg v m 3 /kg u kj/kg u kj/kg u kj/kg u kj/kg phase: sub.liq phase: sat.liq phase: sat.vap. phase: sup.vap. Process -2 Process 2-3 Process 3 Q -2 UNKNOWN Q 2-3 UNKNOWN Q 3 UNKNOWN W sh 0 W sh 0 W sh 0 W bnd,-2 UNKNOWN W bnd,2-3 UNKNOWN W bnd,3 UNKNOWN Final State: Fixed Final State: Fixed Final State: Fixed Italicized values are from steam tables. Conservation of Mass m m 2 m 2 m 3 m 3 m 4 st Law u 2 - u q -2 - w bnd,-2 u 3 -u 2 q w bnd,2-3 u 4 -u 3 q 3 - w bnd,3 W bnd w bnd,-2 P (v 2 -v ) w bnd,2-3 P 2 (v 3 -v 2 ) w bnd,3 P 3 (v 4 -v 3 ) Approach: The three heat transfers will be calculated from the first law, but to do this we must first evaluate the boundary work using the appropriate expression for Pdv. Since all of our states are fixed it will simply be a matter of evaluating our needed properties. We begin by noting that all four states can be treated as phase change substances, so that if we desire to use the R-2 tables we can write v v f (at 0 C) m 3 /kg u u f (at 0 C) kj/kg v 2 v f (at 00 kpa) m 3 /kg u 2 u f (at 00 kpa) 8.88 kj/kg v 3 v g (at 00 kpa) m 3 /kg u 3 u g (at 00 kpa) kj/kg v 4 v(at 23 C, 00 kpa) m 3 /kg u 4 u(at 23 C, 00 kpa) kj/kg 2

3 ME 20 Thermodynamics Fall 204 Now calculating our boundary work w bnd,-2 P (v 2 -v ) (00)( ).3000 x 0-3 kj/kg w bnd,2-3 P 2 (v 3 -v 2 ) (00)( ).5794 x 0 kj/kg w bnd,3 P 3 (v 4 -v 3 ) (00)( ) kj/kg and our heat transfer is q -2 u 2 - u + w bnd, (-0.07) + (.3000 x 0-3 ) 8.95 kj/kg q 2-3 u 3 - u 2 + w bnd, (.5794 x 0 ) kj/kg q 3 u 4 - u 3 + w bnd, (4.0793) 3.44 kj/kg or on a total basis Q -2 (0.5)(8.95) 4.48 kj Q 2-3 (0.5)(65.30) kj Q 3 (0.5)(3.44) 5.72 kj 3. Three of the processes that occur in the piston cylinder device of an internal combustion engine are: Process : Constant pressure heat addition during which the volume doubles Process 2: Isentropic expansion during which the volume increases 9 times Process 3: Constant volume heat removal to a final pressure of 00 kpa At the start of process the engine contains liters of air at 950 K and 5.7 MPa. For each of the three processes the air undergoes find the work and heat transfer in kj. What is the total work and heat transfer for all three processes together? Solution: We will work this problem process by process. Working Fluid: Air (ideal gas) Process: Isobaric State State 2 T 950 K T K P 5700 kpa P kpa V liters V x 0-5 m x 0-5 m 3 m 7.37 x 0 kg m x 0 kg u kj/kg u kj/kg v x 0-2 m 3 /kg v x 0-2 m 3 /kg Italicized values are from ideal gas relations or air tables. Bold values are calculated. Q UNKNOWN W sh 0 W bnd P(V 2 -V ) (isobaric) Final State: UNKNOWN Conservation of Mass: m m 2 st Law: U 2 - U Q-W bnd 3

4 ME 20 Thermodynamics Fall 204 Approach: To fix the final state we will use our process description along with the conservation of mass to determine the final specific volume. The boundary work is then calculated. Finally we use the conservation of energy to determine the heat transfer. We start by determining our mass. Using the ideal gas law we have RT (0.287)(950) v x 0 m /kg P 5700 V (3.5 x 0 ) m 7.37 x 0 kg v -2 (4.783 x 0 ) From the air tables we find u kj/kg By conservation of mass we have m 2 m 7.37 x 0 kg Since our process is isobaric P 2 P 5700 kpa We are told that the volume doubles, so that V 2 2V 7.0 x 0-5 m 3 v 2 2v x 0-2 m 3 /kg The temperature at state 2 is then calculated from -5 P v (5700)(9.566 x 0 ) T K R Using the air tables: u kj/kg Our boundary work is calculated from W bnd P(V 2 -V ) (5700)(7-3.5)(0-5 ) kj The first law is then used to calculate the heat transfer or Q m(u kj - u ) + W bnd -2 (7.37 x 0 4 )( ) For our second process we have Working Fluid: Air (ideal gas) Process: Isentropic State 2 State 3 T K T K P kpa P kpa V 2 7 x 0-5 m 3 V x 0 m 3 m x 0 kg m x 0 kg u kj/kg u kj/kg v x 0-2 m 3 /kg v m 3 /kg v r v r Italicized values are from ideal gas relations. Bold values are calculated. Q UNKNOWN W sh 0

5 ME 20 Thermodynamics Fall 204 P V - P V W bnd (isentropic) - k Final State: UKNOWN Conservation of Mass: m 2 m 3 st Law: U 3 - U 2 Q - W bnd Approach: For the final state we will apply our process description to get the final volumes, then the isentropic condition to determine the temperature and then calculate our pressure from the ideal gas law. Then our boundary work will be calculated from PdV. Finally, we use the conservation of energy to determine the heat transfer. We begin by applying conservation of mass to write m 3 m x 0 kg We are told that our volume increases nine fold, so that V 3 9V x 0 m 3 v 3 9v x 0 - m 3 /kg To determine the temperature at 3 we find that we can use the v r in the air tables as follows. For an isentropic process we have vr3 v3 vr2 v2 Then v3 vr3 vr2 ( )(9) v2 where v r2 is obtained from the air tables at 900 K. Now going to the air tables for a v r of we find T K and u kj/kg We then calculate the pressure by RT3 (0.287)(948.) P3 36.kPa v3 (0.8609) The boundary work is calculated from P3V3 - P2V2 (36.)(6.3 x 0 ) - (5700)(7 x 0 ) Wbnd - k kj where k has been taken at 424 K. Now using our conservation of energy to determine the heat transfer Q m(u - u ) + W 3 2 (7.37 x kJ bnd )( )

6 ME 20 Thermodynamics Fall 204 Finally, we have our third process Working Fluid: Air (ideal gas) Process: Isotropic State 3 State 4 T K T K P kpa P 4 00 kpa V x 0 m 3 V x 0 m 3 m x 0 kg m x 0-5 kg u kj/kg u 4.44 kj/kg v m 3 /kg v m 3 /kg Italicized values are from ideal gas relations or tables. Bold values are calculated. Q UNKNOWN W sh 0 W bnd 0 (constant volume process) Final State: UNKNOWN Conservation of Mass: m 4 m 3 st Law: U 4 - U 3 Q Approach: To fix the final state we will use our process description. The ideal gas law can then be used to determine the temperature and the internal energy is found from the air tables. Finally, we use the conservation of energy to determine the heat transfer. Our conservation of mass yields m 4 m x 0 kg Our process is isometric so that V 4 V x 0 m 3 v 4 v m 3 /kg The temperature is determined from P v (00)(0.8609) T K R (0.287) Then from the air tables at 300 K u 4.44 kj/kg The first law is then used to calculate the heat transfer or Q m(u 4 - u 3 ) (7.37 x 0 )( ) kj Then the total heat transfer and boundary work are kj and W kj Q bnd 6

ME 201 Thermodynamics

ME 201 Thermodynamics ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and