Assignment 9 with Solutions for Introduction to Thermodynamics | ME 205, Assignments of Thermodynamics

Download Assignment 9 with Solutions for Introduction to Thermodynamics | ME 205 and more Assignments Thermodynamics in PDF only on Docsity! PROBLEM 5.43 5.43 A refrigeration cycle operating between two reservoirs receives energy Qe trom a cold reservoir at Te = 274 K and rejects energy Gy to a hot reservoir at Tj; = 315 K. For each of the following cases determine whether the cycle operates reversibly, operates irreversibly, or is impossible: KNOWN: A rettigeration cycle operates between two reservoirs with specified temperatures. FIND: Whether cach of four cycles operates reversibly, operates irreversibly, or is impossible. SCHEMATIC AND GIVEN DATA: \ | x. {| Retrigeration [| Mace 1] Cycle ' | t ! oo dg = 205K Cold Reservoir ENGINEERING MODEL: 1. The contrel volume detined by the dashed line on the accompanying diagram undergoes a refrigeration cyele. ANALYSIS: The meximuro coefficient of performance for a refrigeration cycle operating between two i reservoirs is 279K _~ 6,875 . SISK-275K PROBLEM 5.43 (Continued) Coefficient of performance for any refrigeration cycle is ede (a) Given Qe = 1000 KI, Meyae = 80 kJ, the coefficient of performance determined using these energy data is | { i i tb) Given Qc = 1200 kJ, Gu = 2000 1, cycle work can be determined trom i Woyte = Qu— Qc = 2000 kJ - 1200 kd = 800 kT i \ | The coefficient of performance determincd using these energy data is 1200K _ | 5 : 800k) cle operates irreversibly. (c) Given Qu = 1575 XJ, Myete = 200 KJ, heat transfer associated with the cold reservoir can be determined from Weyee = Qi - Oc + Qe = Qu— Weycte = 1575 kJ — 200 KI = 1375 kd é ‘The coefficient of performance determined using these energy data is Since B= = 6.875, the evele operates reversibly. an (d) Given f= 6, the cycie is irreversible. Since 8=6 < Bou = 6.875, the cycle vperates irreversibly. —§ <———-—~ PRoGLEM 5.54 AL stendy state, « retrigeration cycle uperating between hot and cold reservoirs at 300 K and 274 K, respectively, qmbves citeryy by heat transfer {rom the wold seservoie c 4 rate of 600 kW, (2) If the eyele's eoctficient of performance és 4, determine the power input required, in kW. ©) Determine the mioimuni thcoreticul power required, in KW, for any such eyete, dvaursts: 5 - = Qe. Gooxw = 1SDKWW ESS ap se = Se 2 Ga0KW - mo PES Bo + &) S fais Geof Te Soften yy BS Rudy => Waue? Here 27 PEE ]< ycle wf 25K ] & Weyete bone (28) Ca) ce) By Weycle BSG. SRN PROBLEM 5.67 5.67 By supplying energy at an average rate of 24,000 kJ/h, a heat pump maintains the temperature of a dwelling at 20°C, If electricity costs 8.5 cents per kW-h, determine the minimum theoretical operating cost for each day of operation if the heat pump receives energy by heat transfer from. (a) the outdoor air, at -7°C. (6) the ground, at 5°C. KNOWN: ‘A heat pump maintains the temperature within 2 dwelling under known conditions. FIND: Determine the minimum theoretical operating cost for each day of operation if the heat pump receives energy by heat transfer from the outdoor air, at -7°C; and the ground ar sec, SCHEMATIC AND GIVEN DATA: 000 ki/h W, oyele ! Heat pump , cycle CoStacenicity = 8.5 cents/k Wh ENGENEERING MODEL: (1) ‘The system shown on the accompanying diagram undergoes a heat pump cycle in steady state. (2) The dwelling’s interior acts as the hot reservoir. (3) The outdoor air and ground act as the cold reservoirs in parts a and b, respectively. PROBLEM 5.67 (Continued) ANALYSIS: The coefficient of performance for the heat pump is less than, or equal to, the coefficient of performance for a reversible heat pump operating between reservoirs at Te and Th. i Then with Eq. 5.6 on a time rate basis and Eq. 5.11 we get. i Vows 2 KI gare (Fy 7) _ 28000 O95K—T)) ay 4 oe OT T 293K 3600s! is “7, By a) ‘The corresponding minimum theoretical operating cost for each day of operation is as follows. , 24h |$0.085| 2: 527 | Ae saoes ° Use Eqs. (1) and (2) to determine the minimum theoretical operating cust for each day of operation if the heat pump receives energy by heat mansfer from different cold reservoirs. (a) For a cold reservoir af To-7°C=266 K, re) 24,000— (293K — 266K) Woe RW cae 293K 3600sf 5 KI ‘oute = 8 30. a 1.254 — KW: 5 = $1.25iday s=(ocrsuw| #2 00) (b) For a cold reservoir at 5°C=278 K, kJ 34,000-— (293K -278K) th law Woe 2- | 1 = 0.341 kW ee 293K. 3600s; KS] bos $2 (034kw)—| 24 h [50.085] =$0.70/day _ I day KW 1. Owing to effects of irreversibilities, the daily operating cost of an actual heat purp providing the specified heating will be significantly greater than the calculated minimum values.