Explanation Given data: Quantity Common Units Dimensions M L T θ N J I (a) calories per kilogram kelvin cal kg ⋅ K (b) pound -mass per square foot hour lb m ft 2 ⋅ h (c) pound-force per square inch lb f in 2 Discussion: The first given quantity is calories per kilogram kelvin. The dimensional equivalent of cal is { = } ML 2 T − 2 . The dimensional equivalent of kg is { = } M . The dimensional equivalent of K is { = } Θ . Substitute ML 2 T − 2 for cal, M for kg and Θ for K in cal kg ⋅ K . cal kg ⋅ K = { ML 2 T − 2 M Θ } = L 2 T − 2 Θ = L 2 T − 2 Θ − 1 The second given quantity is pound-mass per square foot hour. Consider the conversion factor for lb m to kg. 1 lb m = 0.4536 kg Consider the conversion factor to convert ft to m. 1 ft = 0.3048 m 1 ft 2 = 0.0929 m 2 Consider the conversion factor to convert h to s. 1 h = ( 60 × 60 ) s = 3 , 6 00 s Convert the values using the above conversion factor. 1 lb m ft 2 ⋅ h = 0.4536 kg 0.0929 m 2 × 3600 s 1 lb m ft 2 ⋅ h = 4 .133 × 1 0 − 4 kg m 2 ⋅ s (1) From equation (1), the constant number 4.133 × 10 − 4 is unitless and dimensionless. The dimensional equivalent of kg is { = } M . The dimensional equivalent of m 2 is { = } L 2

Thinking Like an Engineer: An Active Learning Approach (3rd Edition)

3rd Edition

ISBN: 9780133593211

Author: Elizabeth A. Stephan, David R. Bowman, William J. Park, Benjamin L. Sill, Matthew W. Ohland

Publisher: PEARSON

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Chapter 7, Problem 21ICA

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Find out the fundamental dimension of the given quantities.

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Chapter 7, Problem 20ICA

Chapter 7, Problem 22ICA

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Thinking Like an Engineer: An Active Learning Approach (3rd Edition)

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