User:Lonny/Erratum Schaums Outlines Thermodynamics for Engineers

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

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

1.19e

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

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^o.

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