Question

Problem 4 (25 points) A piston cylinder contains 0.01 kg air at 290 K, 96 kPa. A reversible adiabatic compression brings it to 1/10 of the initial volume. After this process combustion adds 1000 kJ/kg of energy (assume heat transfer) in a constant P process to state 3. a) Plot the two process in a P-v and T-s diagram. b) Find P and T for state 2. c) Find T for state 3. d) Find the work in the two processes.

Answer 1

(b) pressure P2 = 2411.42 kpa.

Temperature T2 = 728.38k

(C) temperature T3=1723.405k

(d) work done on compression W1-2= 3.1455kJ

Work done at constant pressure process

W2-3 =2.855kJ

Givendalar piston Cylinder contains initially inass, of air m = 0.01 kg Temperature T = 290°k pressure Pia 96 kpa Reversible adabalic Compression brings it to Yo of The initial volume to stake a volume V₂ = 1 Volume Vi After this process combustion adds Energy in a constant pressure process to state 3 process in 2 proces 2.3 is reversible adabal'c compression Constant pressure teal addition Qin - loss wyling i T Initial Conditions of air at stale I Pa 96kpa T = 290k m=0.olhe

from Ideal gas equation, we can find Initial volume (vi) of air ided gas canalon PremRT - R = 0.287 17/18k Gas constrant of air Pivoam RTI (96) v = (0.01) (0.287)(290) V = 0.008669 m² After reversible adabahic Compression to stake a the volume is reduced to % of volume u Volume vs = love Vysto (0.008 669) Vy = 0.0008 669 m2 from reversible adabah'c comprension (77= 1.4) Pivil a po uz ****(C) Py = 96 60 000 kr INT

Py : 2411.41 Hpa ve SSUNY and Stalle les 2411.41 kapal le Howerer processing and process 9.3 is assumed to be closed olces because of Piston a cylinder assemble. "mals of the syslem is constant stakea Ps : 2411.41 kra V 0.0008 669 m m=ooling R = 0.287 listing T2=? Note's Assuming Ideal gas having constant specific proporlies through out process coga 2.3 from Ideal gas equation - P 2 V2 = MRT (2 4 14) (0-0008561) - 0:01 10:26# * , T2 = 728.38°r (6) Temperature at stack 2 Tz = 728,88k / Process 2-3 is constant pressure Heat addition in Comblestion chambur of piston cylinder assembly Heal addition at Constant pressure & in = m((T3-12)

specific Head of air at constant pressure (p=1.005! Q = 100015ling m= 0.01ig 12 = 728.38k T3=? from all the above daha Heal transfer pes up Q = 4 AT unit balanc (kling) = (kit lg k) (k) Q in - Cp (?3-[z) 1000 = 11005 (75 -728.38) (I3-728.38) = 995.024 Ty= 1723.405 OK temperature at strak 3 T3 = 1723 1405k/ chake 3 P2 = P3=2411.41 kpa Y= 7230405k m=.01 M R = 0-287 Ksliga Nizz? from Idel gas equation P2 Ug = MR T

(2411.41) Vg = 0.01 *0.287 (177235405) Vy= 0.082051 m² (d) work in two processes work in process ins (comprestan) PIN - Peve W conserned 8-1 Reversible adabalic compression work formula & Pino a PLV2 27 -1 z 96 (0.008 669) - (2011,4)) (0.000567) 1:47 e-3.14556 kJ here are sing represents that work done on the comprosser to compressor consumed 3.145567 of energy | Wiega-3, 14556 45 work in proces 2.3 prouss 2-3 is ted Addition that means volume Gripanded at constant pressure at constant pressure

work done at constrant pressure Waars = P(Vgr V₂) Were = By (Wg - VA) -2411.41 (0.002 091 - 00000 8669) (Wang= 2.855 95 ] . Wing : 3.14556 kJ in work f-2 process (consumed) Work in procen 207 Wyan = 2.85545 (produted)