Question
Specific volume of wet steam with dryness fraction $x$ is(a) $x v_{f}$(b) $v_{f}+x v_{f g}$(c) $x^{2} v_{g}$(d) $x^{3} v_{g}$
Step 1
e., not mixed with liquid water), - \(v_{f g}\) is the specific volume of the dry steam. Show more…
Show all steps
Your feedback will help us improve your experience
Yogesh R and 82 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Specific volume: If X is the dryness fraction, Vf is the specific volume of saturated liquid, and Vg is the specific volume of saturated vapor, then the specific volume of wet steam (at dry and saturated conditions) is given by the formula: (1-x)Vf + xVg.
Find the dryness fraction of a steam having a specific volume of 15 m3/kg after expansion in a turbine. The specific volumes of saturated liquid and saturated vapor corresponding to the pressure are 0.001 m3/kg and 15.25 m3/kg, respectively.
The specific volume of water (the liquid) will be written as $V_{l}$. Since $V_{v}^{\prime}>>V_{l}^{\prime}$, most of the weight is due to water. Thus if $m_{l}$ is mass of the liquid and $m_{v}$ that of the vapour then $$ \begin{gathered} m=m_{l}+m_{v} \\ V=m_{l} V_{l}^{\prime}+m_{v} V_{v}^{\prime} \quad \text { or } \quad V-m V_{l}^{\prime}=m_{v}\left(V_{v}^{\prime}-V_{l}^{\prime}\right) \end{gathered} $$ So $m_{v}=\frac{V-m V_{l}^{\prime}}{V_{v}^{\prime}-V_{1}^{\prime}}=20 \mathrm{gm}$ in the present case. Its volume is $m_{v} V_{v}=1.01$
Thermodynamics And Molecular Physics
Phase Transformations
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD