Topic: Thermodynamics
Calculate the volume of 3 kg of steam at a pressure of 1.3 bar. The dryness fraction of the sample of steam is 0.8. The specific volume of water at 1.3 bar is 0.001 $m^{3}/kg$ and specific volume of dry stream is $1.3~m^{3}/kg$ at 1.3 bar.
- A. $5.64~m^{3}$
- B. $3.12~m^{3}$
- C. $4.24~m^{3}$
- D. $2.64~m^{3}$
Answer: B. $3.12~m^{3}$
Explanation: Specific volume of wet steam $v = v_f + x(v_g - v_f)$. Since $v_f$ is very small compared to $v_g$, $v \approx x v_g$.
$v = 0.001 + 0.8(1.3 - 0.001) = 0.001 + 0.8(1.299) \approx 0.8 \times 1.3 = 1.04~m^3/kg$.
Total volume $V = m \times v = 3 \times 1.04 = 3.12~m^3$.