4F-5 : Performance of an Ideal Gas Cycle 10 pts

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1 4F-5 : Perfrmance f an Cycle 0 pts An ideal gas, initially at 0 C and 00 kpa, underges an internally reversible, cyclic prcess in a clsed system. The gas is first cmpressed adiabatically t 500 kpa, then cled at a cnstant pressure f 500 kpa t 0 C, and finally expanded isthermally t its riginal state. a.) b.) c.) d.) Carefully draw this prcess in a traditinal pistn-and-cylinder schematic. Sketch the prcess path fr this cycle n a P Diagram. Put a pint n the diagram fr each state. Be sure t include and label all the imprtant features fr a cmplete P diagram fr this system. Calculate Q,, U and H, in J/mle, fr each step in the prcess and fr the entire cycle. Assume that C P = (7/) R. Is this cycle a pwer cycle r a refrigeratin cycle? Explain. Calculate the thermal efficiency r COP f the cycle, whichever is apprpriate. Read : Sketch carefully. Understanding what is ging n in the prblem is half the battle. Apply the st Law, the definitins f bundary wrk, C P and C t a cycle n an ideal gas with cnstant heat capacities. Take advantage f the fact that step - is bth adiabatic and reversible, s it is isentrpic. Pwer cycles prduce a net amunt f wrk and prceed in a clck-wise directin n a P Diagram. Given : T 0.5 K P 00 kpa T??? K P 500 kpa T 0.5 K P 500 kpa Q 0 J/mle R 8.4 J/mle-K C P J/mle-K Find : Fr each f the three steps and fr the entire cycle: U??? J/mle H??? J/mle Q??? J/mle??? J/mle Diagrams : Part a.) T > T P = 500 kpa < T = 0 C P = 00 kpa Adiabatic Cmpressin Isthermal Expansin Isbaric Cling T = 0 C P = P <

2 Part b.) Assumptins: - Step - is adaibatic, Step - is isbaric, Step - is isthermal. - The entire cycle and all f the steps in the cycle are internally reversible. - Changes in kinetic and ptential energies are negligible. 4 - Bundary wrk is the nly frm f wrk interactin during the cycle. 5 - The PT behavir f the system is accurately described by the ideal gas EOS. Equatins / Data / Slve : Part c.) Let's begin by analyzing step -, the adiabatic cmpressin. Begin by applying the st law fr clsed systems t each step in the Carnt Cycle. Assume that changes in kinetic and ptential energies are negligible. =-D U= U- U Eqn Because internal energy is nt a functin f pressure fr an ideal gas, we can determine U by integrating the equatin which defines the cnstant vlume heat capacity. The integratin is simplified by the fact that the heat capacity fr the gas in this prblem has a cnstant value. æ du C = dt T Eqn U - U = ò C Eqn dt= C T -T T Cmbining Eqns & yields: = C T -T Eqn 4 The prblem is that we d nt knw T. S, ur next task is t determine T. Since the entire cycle is reversible and this step is als adiabatic, this step is isentrpic. The fastest way t determine T is t use ne f the PT relatinships fr isentrpic prcesses. -g -g g g = T P T P Eqn 5

3 -g æp g T = T P ø Slve Eqn 5 fr T : Eqn 6 C g= C P Nw, we need t evaluate : Eqn 7 CP = C+R But fr ideal gases : Eqn 8 C = CP-R Slving Eqn 8 fr C yields : Eqn 9 Plugging values int Eqn 9 and then Eqn 7 yields : C J/mle-K.4 Nw, plug values int Eqn 5 t get T T 480. K and plug that int Eqn 4 t get : J/mle Plugging values int Eqn yields : U J/mle æ D H=D U+Dç ç P Nw, we can get H frm its definitin : Eqn 0 P= But, the gas is an ideal gas: Eqn D H=D U+ R T -T Cmbining Eqns 0 & gives us : Eqn Nw, we can plug values int Eqn : H 550. J/mle Next, let's analyze step -, isbaric cling. T K P 500 kpa T 0.5 K P 500 kpa The apprpriate frm f the st Law is: Eqn Q - =DU Because we assumed that bundary wrk is the nly frm f wrk that crsses the system bundary, we can determine wrk frm its definitin. = ò Pd æ = P - Eqn 4 Isbaric prcess: Eqn 5 = R T -T Because the system cntains and ideal gas: Eqn J/mle

4 Next we can calculate U by applying Eqn t step -: U U C T T - = - U J/mle Eqn 7 Nw, slve Eqn t determine Q : Q = +DU Eqn 8 Nw, we apply Eqn t step - t determine H : D H=D U+ R T - T =D U+ = Q Q J/mle Eqn 9 H J/mle Next, we analyze step -, isthermal expansin. Fr ideal gases, U and H are functins f T nly. Therefre : U 0.0 J/mle H 0.0 J/mle The apprpriate frm f the st Law is: Eqn 0 Q = Again, because we assumed that bundary wrk is the nly frm f wrk that crsses the system bundary, we can determine wrk frm its definitin. = ò Pd Slve Eqn fr P and substitute the result int Eqn t get : Eqn EOS : Eqn P = Eqn = ò d Eqn 4 Integrating Eqn 4 yields : Eqn 5 e can use the EOS t avid calculating and as fllws: P = Apply Eqn t bth states and : Eqn 6 P= P Cancelling terms and rearranging leaves : = Eqn 7 P Use Eqn 7 t eliminate the 's frm Eqn 5 : ép ù = Ln ê P ú ë û Eqn 8 Nw, plug values int Eqn 8 and then Eqn 0 : J/mle Q J/mle é ù = êë úû Ln P=

5 Finally, we can calculate Q cycle and cycle frm : cycle = + + Q cycle = Q + Q+ Q Eqn 9 Eqn 0 Plugging values int Eqns 9 & 0 yields : cycle J/mle Q cycle J/mle This result cnfirms what an applicatin f the st Law t the entire cycle tells us: Q cycle = cycle Part d.) The cycle is a refrigeratin cycle because bth cycle and Q cycle are negative. The cefficient f perfrmance f a refrigeratin cycle is defined as : COP Q C is the heat absrbed by the system during the cycle. In this case, Q C = Q. is the wrk input t the system during the cycle. In this case, = - cycle. R QC = Eqn Therefre : Q C J/mle 09.7 J/mle Plug values int Eqn t get : COP R.709 erify : The ideal gas assumptin needs t be verified. e need t determine the specific vlume at each state and check if : > 5L/ml Eqn 5.0 L/ml 7.98 L/ml = Eqn P 5.04 L/ml The specific vlume at each state is greater than 5 L/ml fr all states and Air can be cnsidered t be a diatmic gas, s the ideal gas assumptin is valid. Answers : a.) See diagram abve. b.) See diagram abve. c.) Step U H Q Cycle d.) Refrigeratin r Heat Pump Cycle. COP R.7 COP HP 4.7 All values in this table are in J/mle.

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