# HW3s - Spring 2006 Thermodynamics Homework #3, Solutions 1....

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Spring 2006 1 ME 201 Thermodynamics Homework #3, Solutions 1. Convert the following temperatures to ° F, ° C, K, R (each temperature ¼ pt) a. 98.6 ° F T(°F)=98.6 ° F, T( ° C)=(98.6-32)/1.8=37 ° C, T(K)=(98.6+460)/1.8=310.3 K, T(R)=98.6+460=558.6 R b. 298 K T( ° F)=(298)1.8-460=76.4 ° F, T( ° C)=298-273=25 ° C, T(K)=298 K, T(R)=(298)1.8=536.4 R c. 5715 ° F T( ° F)=5715 ° F, T( ° C)=(5715-32)/1.8=3157.6 ° C, T(K)=(5715+460)/1.8=3430.6 K, T(R)=5715+460=6175 R d. 460 R T( ° F)=460-460=0 ° F, T( ° C)=(460)/1.8-273= -17.4 ° C, T(K)=460/1.8=255.6 K, T(R)=460R e. 100 ° C T( ° F)==(100)1.8+32=212 ° F, T( ° C)=100 ° C, T(K)=100+273=373 K, T(R)=(100+273)1.8=671.4 R 2. Convert the following pressures to psia and kPa. (each pressure ½ pt) a. 760 mm of Hg P(psia) = 760 (mm of Hg) x 0.019 (psia/mm of Hg) = 14.4 psia P(kPa) = 760 (mm of Hg) x 0.133 (kPa/mm of Hg) = 101.1 kPa b. 101 bar P(psia) = 101 (bar) x 14.50 (psia/bar) = 1464.5 psia

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## This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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HW3s - Spring 2006 Thermodynamics Homework #3, Solutions 1....

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