Thermodynamics I - Exam 4 Solutions Key | ME 20000, Exams of Thermodynamics

Download Thermodynamics I - Exam 4 Solutions Key | ME 20000 and more Exams Thermodynamics in PDF only on Docsity! ME 200 — Spring 2005 Last Name, First Name: Circle your lecture below: Thermo No.: Div. 1 ~~ 8:30 A.M. Div. 2 — 10:30 A.M. Div. 3 — 12:30 P.M. Div. 4 - 2:30 P.M. Prof. Chen Prof. Lucht Mr. Paek Dr. Zheng Exam 4 This exam is open book/open notes. Use the standard solution format for the problems, except that Given and Find are already listed. Asystem sketch is also provided for Problems 1 and 3. Show your work clearly and list all needed assumptions and basic equations. Put your name and Thermo Number on top of every page tumed in and make sure the pages are in the correct order. Work only on the front side of each page. If you need extra space, work on the extra yellow paper that is provided and clearly indicate which problem is being solved on each extra sheet of paper. Place your final answers in the boxes provided for each part. Erase or mark through work that you do not wish to have graded. You will receive only partial credit if multiple calculations are presented for the same problem even if one is completely correct. Problem Score 1 130 2 135 3 135 Total /100 name Solution Key _ Thermo Numbers (Last) (First) (30/100 points) Problem 1: Given: A steady flow compressor is used to compress air from an initial state of 100 kPa, 300 K to a final state of 4 MPa, 880 K. The rate of work input to the compressor is 2 MW. Heat transfer at the rate of 0.1 MW occurs from the compressor to the ambient environment at a temperature of 300 K. Consider a control volume drawn around the compressor so that the heat transfer occurs at the ambient temperature. Use the ideal gas air tables to analyze the compression process. Show your control volume. Find: a. Calculate the air mass flow rate in kg/s, b. Calculate the rate of entropy generation for the compressor in KW/K. ~ Assumptions: > ; (1) $4 eady flow Qqy = 0.1 MW (2) atv 1s ‘deal gas Woy =-2MW (3) AKE =O <— (4) APE =0 (5\ unvform {low P;=100kPa Tr = 300K P,=4 MPa Tz = 880K (a) m = Bill kgs (6) Soon = | 40.526 Ee Basic Equations: . . Consy mass, WM M,~ Me ~ e ’ . e Consv energy: tg =0= Oo ~ Wey ah, -mNe2 entropy Balence « db Sev “qt = O- Qev $MS,-MS + gen b Soldtons Stole Li T= 300K P,= 100 FPa h,= 30019 KT/kg $2- 1.702703 Nhe k Site 2! T,- "60K Pi> 4 MPa hy = 910156 KI/kg Sy2 = 2.82344 0% wane Solution Key (Last) (First) Thermo Number: Un = €10 "I/ ko ee Qin = han ha = 2908" RSE = 1249 3/b 9 Pout = Uyr~ U4 = 610 ~ 2€b = 526 b/es Wied = Gin~ Gout = T24 MKS News Unmet. 72 - Rt 2 o,5¢0 aA [24 © € = (0.2 key C724 KT /1g) pes kT Name: So | ut fron K e 4 Thermo Number: (Last) (First) (35/100 points) Problem 3: Given: A steady-flow Brayton cycle engine has a mass flow rate of 2.0 kg/s of air. The temperature and pressure at the compressor inlet are 100 kPa and 280 K. The compressor pressure ratio is 15, and the temperature at the turbine inlet is 1400 K. The isentropic efficiencies of the compressor and turbine are 1.0 and 0.80, respectively. Use the ideal gas air tables. pick closest values, do not interpolate. Indicate control volumes on the diagram below. Find: a. Sketch the cycle on the T-s diagram shown below. a. Calculate the net power for the Brayton-cycle engine in kW. b. Calculate the thermodynamic efficiency for the cycle. (6) W,,,= | G12 kw (c) m4, =| 0-33 F Assumptions: Oy aiv-standarc cycle (2\ steady stot a (BY\ACE= O fH) APR =O (\ aie 76 idead Aas (b) unilor my few Basic Equations: COnsSy mass nm = m™,= Vita = Me _ Wr a emp = pa(ha-hn) Wap = vn (h3- he) Qin ma (ls- b> ) v ev e -h Mu = Wret/ Gen a 5, tor V6 vopre POS me Pre Pa Zokelion. Stade 1) T= 790% = [00 bPa = 26013 hy P= 1, 0864 Name: S 0 ur ‘0 “ £9 (Last) Thermo Number: (Firs) Stole 2: S,=S, ova 1B pr = (A222 = |[2= 600 & h, = 607 IT /kg Stete St “Tz = HOOK a7 1515 “Yk P3= Py = 1500 - Pa Pox = 450.6 Le ‘ = Siete Ws§ Sus= Sa Pry = Pay JI = 4S5SOIS_ . Pras “OTe 50,08 => “Tas = TOK hese 724 '/e5 Stadt e “e b,-he ee | ha hes hy= ha O@(hahye) ~ psi oe (isis— 724) My hy =~ 662 "hcg Nar = O.% = one ui (g-he) = (2.0 ba) (1515-482 Weg) = |2b6b EW Weemp = reli h) = (20) ¢07r 2g0) = 654 kw Oin = vil hela) ~ 20\(1515 ~607) = ] 716 KW 4 Wret = Wy pm Wrowp = @le kW Ny = Wat 82 _~ 9 2279 Cin mi