Question

For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and ( b ) 630 kPa. What is the temperature of each case?

Answer 1

Answer on Question #66213, Physics / Solid State Physics

For a specific volume of $0.2\mathrm{m}^3/\mathrm{kg}$, find the quality of steam if the absolute pressure is (a) $40\mathrm{kPa}$ and (b) $630\mathrm{kPa}$. What is the temperature of each case?

Find: $x - ?T - ?$

Given:

$$\begin{array}{l} v = 0.2 \, \mathrm{m}^3/\mathrm{kg} \\ p_1 = 40 \, \mathrm{kPa} \\ p_2 = 630 \, \mathrm{kPa} \\ \end{array}$$

Solution:

(a) Using information from Table (Properties of Saturated H2O-Pressure Table), we calculate the quality as follows:

$$\begin{array}{l} v = v_f + x(v_g - v_f) \quad (1), \\ \text{Of } (1) \Rightarrow 0.2 = 0.001 + x(3.993 - 0.001) \quad (2) \\ \text{Of } (2) \Rightarrow x = 0.04985 \\ \end{array}$$

The temperature is found in Table next to the pressure entry: $T = 75.9{}^{\circ}\mathrm{C}$.

(b) We must interpolate to find the correct values in Table. Using the values at 0.6 and 0.8 MPa we have:

$$\begin{array}{l} v_g = \left(\frac{0.03}{0.2}\right) \times (0.2404 - 0.3157) + 0.3157 = 0.3044 \quad (3) \\ v_f = 0.0011 \quad (4) \\ \end{array}$$

(3) and (4) in (1): $0.2 = 0.0011 + x(0.3044 - 0.0011)$ \quad (5)

$$\text{Of } (5) \Rightarrow x = 0.6558$$

The temperature is interpolated to be:

$$T = \left(\frac{0.03}{0.2}\right) \times (170.4 - 158.9) + 158.9 = 160.6{}^{\circ}\mathrm{C}$$

Answer:

(a) $x = 0.04985$, $T = 75.9{}^{\circ}\mathrm{C}$;

(b) $x = 0.6558$, $T = 160.6{}^{\circ}\mathrm{C}$

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