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.

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

Convert ${\displaystyle \frac{200ft\cdot l{b}_f}{sec}}$ to watts

Givens:

• ${\displaystyle \frac{1ft\cdot l{b}_f}{sec}=1.356W}$

Book says the answer is 2,712W

Should be: ${\displaystyle \frac{200ft\cdot l{b}_f}{sec}\times \frac{1.356W\cdot sec}{ft\cdot l{b}_f}=271.2W}$

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

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, ${\displaystyle v=\frac{0.05053f{t}^3}{l{b}_m}}$

• At 440^oF, ${\displaystyle v_f=\frac{0.019260f{t}^3}{l{b}_m}}$

• At 440^oF, ${\displaystyle v_g=\frac{1.2192f{t}^3}{l{b}_m}}$

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