User:Lonny/Erratum Schaums Outlines Thermodynamics for Engineers

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This is unofficial corrections for answers in Schaum's Outline of Thermodynamics for Engineers. These corrections come from User:Lonny. Please feel free to add correction or critiques of these corrections.

Book Details[edit | edit source]

Schaum's Outline of Thermodynamics for Engineers, 2ed (Schaum's Outline Series)
Merle Potter (Author), Ph.D., Craig Somerton (Author)
Publication Date: May 20, 2009 | ISBN-10: 0071611673 | ISBN-13: 978-0071611671 | Edition: 2

Erratum[edit | edit source]

1.19e[edit | edit source]

Convert [math]\frac{200 ft \cdot lb_f}{sec}[/math] to watts

  • [math]\frac{1 ft \cdot lb_f}{sec}=1.356 W[/math]
Book says the answer is 2,712W
Should be: [math]\frac{200 ft \cdot lb_f}{sec}\times\frac{1.356 W \cdot sec}{ft \cdot lb_f}=271.2W[/math]
In addition, with proper significant figures, this should probably be just 271W.

2.20[edit | edit source]

A rigid vessel contains water at 400oF. Heat is added so that the water passes through the critical point. Find the quality at 440oF.

  • At critical point, [math]v=\frac{0.05053 ft^3}{lb_m}[/math]

  • At 440oF, [math]v_f=\frac{0.019260 ft^3}{lb_m}[/math]

  • At 440oF, [math]v_g=\frac{1.2192 ft^3}{lb_m}[/math]
If it is a rigid vessel the specific volume cannot change. There is only one specific volume at the critical point.
Book says: 0.01728
Should be: <math>x=\frac{v-v_f}{v_g-v_f}=\frac{0.05053 - 0.01926}{1.2192 - 0.01926}=0.02606