Solution Manual for Thermodynamics An Integrated Learning System by Schmidt Full file at https://TestbankDirect.eu/ Prob 1.1 8/18/04 PROBLEM 1.1 PROBLEM STATEMENT: Determine the SI values for the following quantities: ft, 200 lbm, 70o F, 25 lbf/ft2 , and atm GIVEN: ft, 200 lbm, 70o F, 25 lbf/ft2 , and atm FIND: SI values for each of the given quantities ASSUMPTIONS: None GOVERNING RELATIONS: ft = 0.3048 m lbm = 0.45359 kg T(oF)= [1.8*T(oC)] + 32 lbf/ft2 =47.88 Pa = 47.88 N/m2 atm = 1.01325 bar = 101325 N/m2 PROPERTY DATA: Not applicable QUANTITATIVE SOLUTION: ft = ft * 0.3048 m/ft = 1.8288 m 200 lbm = 200 lbm * 0.45359 kg/lbm = 90.72 kg 70oF = 1.8 * T(oC) + 32 (70 – 32)/1.8 = 21.11oC 25 lbf/ft2 = 25 lbf/ft2 * 47.88 (N/m2 ) / (lbf/ft2 ) = 1197 N/m2 atm = 1.01325 bar = 101325 N/m2 DISCUSSION OF RESULTS: Self-explanatory Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/