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 #66265, Physics / Mechanics | Relativity

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

Solution:

a) We use the table of Saturated water—Pressure. We use the next equation:

$$v = v_f + x(v_g - v_f)$$ $$0.2 \, \text{m}^3/\text{kg} = 0.001 \, \text{m}^3/\text{kg} + x(0.3993 - 0.001) \, \text{m}^3/\text{kg}$$ $$0.2 = 0.001 + 0.3983x$$ $$0.3983x = 0.2 - 0.001$$ $$0.392x = 0.199$$ $$x = 0.199 \, (\text{m}^3/\text{kg}) / 0.3983 \, (\text{m}^3/\text{kg})$$ $$x = 0.4996$$

In the quality region the temperature is given as $T = 155{}^{\circ}\text{C}$

b)

$$v = v_f + x(v_g - v_f)$$ $$0.2 \, \text{m}^3/\text{kg} = 0.0011 \, \text{m}^3/\text{kg} + x(0.3044 - 0.0011) \, \text{m}^3/\text{kg}$$ $$0.2 = 0.001 + 0.3044x$$ $$0.3044x = 0.2 - 0.0011$$ $$0.392x = 0.1989$$ $$x = 0.1989 \, (\text{m}^3/\text{kg}) / 0.3044 \, (\text{m}^3/\text{kg})$$ $$x = 0.6534$$

In the quality region the temperature is given as $T = 166{}^{\circ}\text{C}$

Answer: $155{}^{\circ}\text{C}; T = 166{}^{\circ}\text{C}$

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