(F)
 
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===2.20===
===2.20===
A rigid vessel contains water at 400<sup>o</sup>F. Heat is added so that the water passes through the critical point. Find the quality at 440<sup>o</sup>.
A rigid vessel contains water at 400<sup>o</sup>F. Heat is added so that the water passes through the critical point. Find the quality at 440<sup>o</sup>F.
:Givens:
:Givens:
:*At critical point, <math>v=\frac{0.05053 ft^3}{lb_m}</math>
:*At critical point, <math>v=\frac{0.05053 ft^3}{lb_m}</math>

Latest revision as of 23:29, 22 August 2017

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 to watts

Givens:
Book says the answer is 2,712W
Should be:
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.

Givens:
  • At critical point,


  • At 440oF,


  • At 440oF,
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
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