Introduction to Thermodynamics - Problems with Solutions | ME 205, Assignments of Thermodynamics

Download Introduction to Thermodynamics - Problems with Solutions | ME 205 and more Assignments Thermodynamics in PDF only on Docsity! PROBLEM 2.56 enowl! A gar conimined in & pirton~cplvader assembly undergoes & Conptant -peeiture expansion while bang slowly heated. State data are provided: Fred? Por the $45, evaluate work aad heat transfer, fer the praten, Cratvete work and choige rh patenttat eaters y, Schematic 7 Gen DATA: ENGINETEING MODEL! 1. AS shoum In He Schematic, tue Closed systems are conrivercd: the Gar and ths Piston. 2. The gar vadergeer a conctand~ preckurc Paocets. 3. farthe gas there Go no change tn 4 120.1 V_= 0.02 oP 09-025 potentiod energy (tec Example 43) Gnd ne overall Chuage ne Erne h'e * energy, @ 4. For tha pshm,ythere oe he heat trancfer, BIS; there fs ne change in internal Energy, ne everal! Range in Kinetic eneegy, and ae ferchen. Fig, Pz.56 AMALTTS! (4) Tating Phe sas as the sytiein , the work é optained fran, By 2th We ("pd¥= ply] = Gano, Your -ate (3% |= 440 <—— Redvcing an energy balance, AU + > +t QPpe = Ow Qrw+-sw 3 ay J BPt ~*~ a Qt aed +oae eT 5 42507 ~—_- Qe) Teterng the pitien at the tyatim, an energy dans ter bg wore occurs on tha bottom taefucn Gren the gas, AP Th Hep surface phe pists dees cverk on +hG atmectums) f Be A Woaten® [Fda = (Patel — PA) AR = CRytm-p) (AAR) : = Ondtm. PAY (hp gerchon} 2 C1 kW Kono) | iim) aeBRI <_— An energy balance for the piSten reducer a5 feifousie (A + oc res] paten = Seaton <r = APE pixton = ~Wevten e+ aed — © d, Overall eneraq “Yabanca Sheet” in terms of maguitudes! Tapati Q2 22567 ‘Dispesthon of +he energy input! @ Stored ae AVE in The ga O.2SeT @ Bored aa OPE in tht, Qirfen! | 2-60 eT © Trancier by work te HHS aoe eT aqme phere pared ;  PROBLEM 2-63 KNOWN: 4 gas of known mass is compressed in a piston-cylinder assembly from a specified initial pressure to a known tinal state. The pressure-volume relationship for the process is given, and the specific internal energy change is known. FIND: Determine the heat transfer for the process. SCHEMATIC AND GIVEN DATA: State 1 Process 1 2 pF3 = constant Au = 50 kidkg Py = 8 bar ¥,= 0.02 mi yu, + SO KIL RG ENGINEERING MODEL: 1. ‘The gas is a closed system. 2. The system undergoes a polytropic process in which pl" = constant. 3. Kinclic and potential energy effects are negligible i ANALYSIS: The heat transfer can be determined from the energy balance AKE + APE+AU=Q~ W Neglecting changes in kinetic energy (AKE = 0) and potential energy (APE = 0) and solving for heat transfer give Q=aU+W internal energy change is determined trom the given mass and change in specific internal energy : AU =m Au = 0.2 kg)(50 Kg) = 10k) PRORLEM 2.75" KNOWN: A closed system undergoes a cycle consisting of twee processes. FIND: Determine the net work of the cycle and the heat transfer for process 2-3, identify whether the cycle is a power cycle or a refrigeration cycle and explain. SCHEMATIC AND GIVEN DATA: State 2 2m Process }-+ 2 100K] prowess 2m 3 p¥—constuat | oman yolane ty = 100 10 ] State } ——_— Process 3— 1 Gas - constant pressure = tar | adtabatic pc 1bar =02m' =m P, 2 {Ve constant 3 1 ey ENGINEERING MODEL: . The gas is a closed system. . Kinetic and potential energy effects are negligible . Process 1-2 is polytropic in which p¥= constant. . Process 2-3 is constant volume. . Process 3-1 is constant pressure and adiabalic. Weepe ANALYSIS: Cycle work is the sum of work associated with each process in the cycle Weycte = Wir Was + Way Process 1-2 is a polywopic process with pV = constant. Therefore, p = constantiV. vr = (constant) af 2 =p (4) 1 dV = ee = (constant) f ¥ ¥ oS Ik 02m Wy = (i barj2 m3 }—_™ =I) -= 460.5 KI iam bar Yar 107 N-m (#22 PROBLEM 2.75 (Continued ) Process 2-3 is constant volume; thus Wis~ ("pd = 0k Process 3-1 is coustant pressure; thus Ws, = [' nd = pl, -¥%) hos Wai ~ (tbar\2 m? -0.2m} |__| 6 309 Tbar [20° N-m Substituting the work associated with cach process yields the cycle work Woycie = 460.5 SJ) = 04S + 180 KF - 280.5 kb Since the net cycle work is into the cycle (negative), the cycle is a refrigeration cycle. For process 2-3, an energy balance is AKEn; + APEas + (Ua U2) = Gas - Was Neglecting changes in kinetic energy (AKEps = 0) and potential energy (APEzs = 0), substituting #5; = 0 (determined above), and solving for Ox yield Ooi = U3 — Ub For the cycle, (a Uys (Us ~ U2) + (U1 - Oa) = 0 Solving for (1/3 ~ U4) yields (5 ~ Up) = (U2 — Oh) — (U4 - 04) From the problem statement, (U2 - U1) = 100 kT For process 3-1, an energy balance is AKEs + APEs, + (U1 — 03) = Ost ~ 31 Neglecting changes in kinetic energy (AK Es, = 0} and potential energy (APEa1 = @), substituting Qa) = 0 since process 3-1 is adiabatic, and solving for (U; — U3) give (U,- U3) = Ws, =-280 KI Substituting for changes in internal energy gives (Us ~ U2) = (100 kJ) — (C180 kJ) = 80 bd © Solving for G2; @3 = Us— Ur 80 kd ‘The heat transfer is positive during process 2-3, denoting cncrgy transfer by beat into the gas during this process. As an alternative solution, for the overall cycle, Qo = O12 + Qos + Os: = Woyete Thus, Q2s = Woycie - G2 - Os Hor process 1-2, an energy balance is AKEn + APE + AU: = Gr ~ Wiz Neglecting changes in kinetic energy (AKE,2 = 0) and potential energy (APEi2 = 0) and solving for heat transfer give Qi2 = AU 2 + Wrz = 100 kJ + (460.5 kJ) =-360.5 bY The heat transter is negative during process 1-2, denoting energy transfer by heat from the gas during this process. Since process 3-1 is adiabatic, Ga) = 0 KJ, Substituting values for W,,. and heat transter associated with ewch process yields Qi; (280.5 kJ) - (-360.5 k]}—0 kJ = 80.) ‘The beat transfer is pasitive during process 2-3, denating energy transfer by heat into the gas during this process.