Solution manual for thermodynamics an integrated learning system by schmidt

FIND: SI values for each of the given quantities GOVERNING RELATIONS: 1.. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 Un

Prob 1.1 8/18/04

PROBLEM 1.1

ft, 200 lbm, 70o F, 25 lbf/ft2 , and 1 atm

GIVEN: 6 ft, 200 lbm, 70o F, 25 lbf/ft2 , and 1 atm

FIND: SI values for each of the given quantities

GOVERNING RELATIONS:

1 1 ft = 0.3048 m

2 1 lbm = 0.45359 kg

3 T(oF)= [1.8*T(oC)] + 32

4 1 lbf/ft2 =47.88 Pa = 47.88 N/m2

5 1 atm = 1.01325 bar = 101325 N/m2

QUANTITATIVE SOLUTION:

6 ft = 6 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

1 atm = 1.01325 bar = 101325 N/m2

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Solution Manual for Thermodynamics An Integrated Learning System by Schmidt

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