1. Air contained in a piston cylinder device undergoes a new power cycle in whic

Question

 1.        Air contained in a piston cylinder device undergoes a new power cycle in which
the processes are as follows:
1-2 constant temperature compression from 300 K, 100 kPa, with a compression ratio
of 10
2-3 constant volume heat addition until the temperature is 500 K 
3-4 constant pressure heat addition 
4-1 isentropic expansion
Assume constant specific heats at 300 K.  Sketch the system showing all energy
actions at the boundary, the P-v diagram, and the T-s diagram.  Show the development
of all equations from conservation of energy concepts and process relations. 
Calculate the pressures and temperatures around the cycle, the heat added, heat
rejected, net work done, and the thermal efficiency.  SELECTED ANSWERS: P3=1667 kPa,
T4=670.2 K, wnet =116.4 kJ/kg, ETA_th=37%

Explanation / Answer

A power cycle for a piston-cylinder device is described by the following four processes: 1→2 Isothermal compression from T1 = 300K, P1 = 100 kPa to P2 = 600 kPa. 2→3 Constant pressure heat addition until the temperature is T3 = 800K. 3→4 Isentropic expansion until the volume at state 4 equals the volume at state 1. 4→1 Constant volume heat rejection until the temperature is 300K. @ R О Ð ÑŠ ÑŒ 0.900 kJ/(kg-K), R = 0.300 kJ/(kg-K), k = 1.500. (a) Sketch the P-v diagram for this cycle (b) Sketch the T-s diagram for this cycle (c) Determine the heat rejected during process 1→2, in kJ/kg. (d) Determine the heat added during process 2→3, in kJ/kg. (e) Determine the total cycle expansion work done by the gas, in kJ/kg. Could you please fill in this table, this will help organize all of the processes. State P (kPa) T (K) V (m3/kg) 1 100 300 2 600 300 3 600 800 4 SOLUTION.doc SOLUTION For an ideal gas following equations are applicable : i) PV = nRT where n = quantity of gas in moles, P = Pressure, V = Volume, T = Absolute temperature, R = Universal gas constant = 8.31 J/mole/K ……….(1) Or PV = m(R/M)T where m = mass of gas, P = Pressure, V = Volume, T = Absolute temperature, M = Molecular mass of the gas, R/M = Gas constant for unit mass of the given gas (different for different gases) …………(2) ii) For adiabatic expansion or compression, PVk = Constant where k = Cp/Cv ……(3) iii) From the first law of thermodynamics we have : dQ = dU + dW where dQ = Quantity of heat given to the system, dU = Change in the internal energy of the system(gas), dW = External work done by the gas. ………….(4) iv) Work done by the gas at constant pressure : dW = PdV ………(5) v) Work done by the gas at constant volume = 0 vi) Heat transferred to/from the gas at constant volume = mcvdT where cv= specific heat of the gas at constant volume ……(6) vii) Heat transferred to/from the gas at constant pressure = mcpdT where cp= specific heat of the gas at constant pressure ……(7) viii) Cp – Cv = R where Cp = Molar specific heat of the gas at constant pressure (i.e. specific heat for one mole), Cv = Molar specific heat of the gas at constant volume ……(8) Cp = Mcp and Cv = Mcv Substituting in (8) we get : Mcp - Mcv = R or cp - cv = R/M …….(9) (a) Sketch the P-v diagram for this cycle 1→2 Isothermal compression from T1 = 300K, P1 = 100 kPa to P2 = 600 kPa. Assuming 1 kg of gas, m = 1 kg. From (9) R/M = cp - cv = (0.9 – 0.6) kJ/(kg-K) = 0.3 kJ/(kg-K) = 0.3x103 J/(kg-K) Substituting values in (2) for initial condition T1 = 300K, P1 = 100 kPa, we get : 100x103 x V1 = 0.3 x 103 x 300 V1 = 0.9 m3 For isothermal compression, T = constant. Hence (2) becomes : PV = Constant P1V1 = P2V2 Or V2 = P1V1/P2 = 100 x 0.9/600 = 0.15 m3 V2 = 0.15 m3 2→3 Constant pressure heat addition until the temperature is T3 = 800K At constant pressure, (2) becomes : P2V2 = m(R/M)T2 (T2 = T1 = 300K as process 1 – 2 is isothermal) P2V3 = m(R/M)T3 (P3 = P2 as pressure is constant) Or V3/V2 = T3/T2 Or V3 = 800 x 0.15/300 = 0.4 m3 V3 = 0.4 m3 3→4 Isentropic expansion until the volume at state 4 equals the volume at state 1. A reversible isentropic process is also adiabatic. V4 = V1 = 0.9 m3 Applying (3) we get : P3(V3)k = P4(V4)k Substituting V4 = 0.9 m3, V3 = 0.4 m3, P3 = P2 = 600 kPa, we get : P4 = P3(V3/V4)k = 600 (0.4/0.9)1.5 = 178 kPa P4 = 178 kPa From (2) P3V3/T3 = P4V4/T4 Or T4 = (P4/P3) (V4/V3) T3 Substituting, V3 = 0.4 m3, V4 = 0.9 m3, P3 = 600 kPa, P4 = 178 kPa, T3 = 800K, we get : T4 = (178/600)(0.9/0.4)800 = 534K T4 = 534K 4→1 Constant volume heat rejection until the temperature is 300K. As both volume and temperature are back to the initial values, from (2) the pressure also returns to its initial value. Following table gives the values of P,V and T at states 1,2,3 and 4. State P (kPa) T (K) V (m3/kg) 1 100 300 0.9 2 600 300 0.15 3 600 800 0.4 4 178 534 0.9 PV diagram P(kPa) 600 2 3 500 400 300 200 4 100 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 V (m3/kg) (b) Sketch the T-s diagram for this cycle Change in Entropy is defined as : dS = dQ/T 1→2 Isothermal compression from T1 = 300K, P1 = 100 kPa to P2 = 600 kPa In an isothermal process change in the internal energy is zero (as the temperature is constant). Hence, from (4), dQ = dW. Also we know that the work done by the gas in an isothermal process is given by : W = (R/M)T logeVf/Vi where Vf and Vi are the final and initial volumes respectively In the present case Vi = V1 = 0.9 m3/kg and Vf = V2 = 0.15 m3/kg, R/M = 0.3 kJ/(kg-K) Work done between states 1 and 2 = W12 = 0.3 x 300 x loge(0.15/0.9) = -160 kJ/kg As for isothermal process dQ = dW, dQ = -161 kJ/kg (negative sign implies heat is flowing out of the system) Change in entropy S2 – S1 = dQ/T = - 160/300 = - 0.54 kJ/(kg-K) 2→3 Constant pressure heat addition until the temperature is T3 = 800K For constant pressure heat addition: dQ = cpdT dS = cpdT/T Integrating we get : S3 – S2 = cp loge(T3/T2) cp = 0.900 kJ/(kg-K), T2 = 300K and T3 = 800K S3 – S2 = cp loge(T3/T2) = 0.9 x loge(800/300) = 0.88 kJ/(kg-K) 3→4 Isentropic expansion until the volume at state 4 equals the volume at state 1 As the entropy is constant S4 = S3 ; D F u w • І Ñ– Ò‘ µ Б Ð’ ÑŠ ÑŒ < > ? h E F v w Ñ‹ ÑŒ = > Z [ ¦ § С gd ? @ E F Q R W X { | Ѓ ‚ Ð Ñ’ †• Ñž Ò ¦ § С h h h h h h h h 5С Т 7stant volume heat rejection until the temperature is 300K For constant volume heat rejection : dQ = cvdT dS = cvdT/T Integrating we get : S4 – S1 = cv loge(T1/T4) Cv = 0.600 kJ/(kg-K), T4 = 534K and T1 = 300K S1 – S4 = cv loge(T1/T4) = 0.6 x loge(300/534) = - 0.34 kJ/(kg-K) T-S diagram Note that we do not know the absolute values of entropy at the four states but only the changes in the entropy between the states. Hence, the entropy axis does not give the absolute values. 800 3 T (K) 700 600 534 4 500 400 300 2 1 0.54 200 0.88 100 S kJ/(kg-K) (c) Determine the heat rejected during process 1→2, in kJ/kg As calculated above under (b) heat rejected during process 1 – 2 = 161 kJ/kg (d) Determine the heat added during process 2→3, in kJ/kg For constant pressure heat addition: dQ = cpdT cp = 0.900 kJ/(kg-K), T2 = 300K and T3 = 800K dQ = 0.9 x (800 – 300) = 450 kJ/kg Heat added during process 2 - 3 = 450 kJ/kg (e) Determine the total cycle expansion work done by the gas, in kJ/kg Work done during process 1-2 = -160 kJ/kg (as calculated above under (b)) Work done during process 2-3 = Pressure x Change in volume = 600 x (0.4 – 0.15) = 150 kJ/kg Work done during process 3-4 Work done in adiabatic process is given by : W = (R/M)(Tf – Ti)/(1 – k) Substituting R/M = 0.3 kJ/(kg-K), Ti = T3 = 800 K, Tf = T4 = 534 K, k = 1.5 5) = 160 kJ/kg Work done during process 4-1 = 0 (as change in volume = 0) Net work done = -160 + 150 + 160 = 150 kJ/kg

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