Thermodynamics I - Exam 3 Solutions - Fall 2005 | ME 20000, Exams of Thermodynamics

Download Thermodynamics I - Exam 3 Solutions - Fall 2005 | 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 3 This exam is open book/open notes. Use the standard solution format for Problems 2-4, except that Given and Find are already listed. A system sketch is also provided for each problem. Show your work clearly and list all needed assumptions and basic equations. Show clearly your control volume for each problem. Put your name and Thermo Number on top of every page turned 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 /10 2 135 3 /20 4 735 Total /100 Name: Soluf TOK Ke v Thermo Number: (Last) (First) (10/100 points) Problem 1: Short answer questions. (a) A reversible heat engine operates between a high-temperature reservoir at 1000°C and a low- temperature reservoir at 0°C. The rate of heat input from the high-temperature reservoir is 100 kW. What is the rate of heat rejection to the low-temperature reservoir? (5 pts) O.=) 214 kw _ Qe . J. Te Nyy = 1-2 TH Qu = We Qe W-rer Ot = 16-6 RW Qo = Qu-W = 264 kW 272 . = |~ “37% O7E6 (b) Anirreversible refrigeration cycle operates between a high-temperature reservoir at 40°C and a low-temperature reservoir at -20°C. The manufacturer claims that the COP for the refrigeration cycle is 3.8. Is this COP possible or impossible to achieve? Justify your answer with a calculation of the maximum theoretical COP for the cycle. (5 pts) COP is | pos<:ble | to achieve. revel = “GQ, <Q.” TyTe 7” 315-283 COP ney ref = 4.2 Since CO Petua| < COP peu, the. manufacd urer's Clam 1S poss ble COP Ge Te 253 Name: _ 50 } ul cOn boy Thermo Number: (Last) (First) (20/100 points) Problem 3: Given: A copper block with a mass of 10 kg and an initial temperature of 900 K is brought into contact with an aluminum block with a mass of 50 kg and an initial temperature of 200 K in a well-insulated enclosure. Let the copper block be labeled block A and the aluminum block be labeled block B. The specific heat for the copper is given by CAT) =a + BT = 0.100 + 5.0x104 T [C.]=k/(kg-K), — T is abs temp in K The specific heat for the aluminum block is C,(7) =C, = 0.895 kJ/(kg—K) , and can be assumed constant as a function of temperature. The final temperature of the blocks is determined to be 256 K from a first law analysis. Find: a. Calculate the total entropy change for the process. For this calculation, do not use an average specific heat for block A, include the temperature dependence in the calculation. ( POR AS IE TTTT 3 i: (FY m= 10 ke A mi=10kg z FE (FY T14-900K i To = 256K > | { : see \ Lz Md State 1 State 2 bp) S,-S,= +689 ee Assumptions: G) Close system (2) Lucom press: ble golds BV Q@=O &) We Basic Equations: Tncompvess.b le Sa I ad S273) = v <O Ay — Name: Ss oO l uct SOW, Key Thermo Number: (Last) (First) Solut Sonu! U Te S,-8,= ma | Aly Mg <6 n() Tin = ma" (S+pat- Me Cp ln re) TA = ig & In (SE) + my B(-Ta) (2 + mMeCwn ly =) = (10 1g CO eg -u)In( ie) +( ‘ola os (256- 906 b) + (50 I \(6 4S.) In ) in (Be) = = $f) 257-3220 4 M047 = 46,57 +» ote « neve [x] - a Lél> spe Name: So (uch Sou key Thermo Number: (Last) (First) (35/100 points) Problem 4: Given: An adiabatic steam turbine operates at steady state with water as the working fluid. At the turbine inlet, section 1, the pressure is 20 MPa and the temperature is 700°C, At the turbine exit, section 2, the pressure is 30 kPa. The isentropic efficiency of the turbine is 0.80. The power output from the turbine is 20 MW. Neglect kinetic and potential energy changes. Find: (a) What is the state of the fluid at the turbine exit? (b) Determine the exit temperature (°C). (c) Draw the process on a T-s diagram, indicating states 1, 2, and 2s. (d) Calculate mass flow rate in kg/s. state=| SLUM T= 69,10 °° Assumptions: A) steady ctdeahe m= 1 10S kgs (2) Oy 0 @) OKES oO (4) Ape =O Peake T CC) | T, = 700°C = 20 MPa Basic Equations: | . . T=? 6 1 7 8 Consy M4SS. Mem, = M2 s [kJ/(kg-K)] Consy energy j ae O-= fo Wy mh, mir 0 Wy = ~” (hi-h) Neauvr b 2 pe Solukon: Stale]. P= 20M Fa T,= 700°C nN, =~ 309 Whey $,= 6.7993 LI /ke- K