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 ${\displaystyle {\frac {200ft\cdot lb_{f}}{sec}}}$ to watts

Givens:

• ${\displaystyle {\frac {1ft\cdot lb_{f}}{sec}}=1.356W}$

Book says the answer is 2,712W

Should be: ${\displaystyle {\frac {200ft\cdot lb_{f}}{sec}}\times {\frac {1.356W\cdot sec}{ft\cdot lb_{f}}}=271.2W}$

In addition, with proper significant figures, this should probably be just 271W.

2.20[edit | edit source]

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

Givens:

• At critical point, ${\displaystyle v={\frac {0.05053ft^{3}}{lb_{m}}}}$

• At 440^oF, ${\displaystyle v_{f}={\frac {0.019260ft^{3}}{lb_{m}}}}$

• At 440^oF, ${\displaystyle v_{g}={\frac {1.2192ft^{3}}{lb_{m}}}}$

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