A closed, rigid vessel has total volume 0.15m^3, and contains a mixture of liqui

Question

A closed, rigid vessel has total volume 0.15m^3, and contains a mixture of liquid and vapor water. The volume of the liquid is 0.001m^3. The volume of the vapor is 0.149m^3. The oressure in the vessel is 5 bar.

a) What is the temperature in the vessel?

b)What is the mass of the liquid water? THe mass of the vapor? What is the quality of water in the vessel?

c)Now heat is slowly added to the system until the last droplets just about disappear, and all water is now just about in the saturated vapor state. Estimate the pressure(to within 1 bar) and temp(to within 5c) in the vessel at that instant.

d)estimate how much heat is needed(to within 20kj) is needed for process c.

e) sketch p-v and T-v diagrams

Explanation / Answer

a)

m = m_f + m_g

m_f = V_f / v_f = 0.001 / v_f............1

m_g = V_g / v_g = 0.149 / v_g...............2

m = (0.001 / v_f) + (0.149 / v_g).........3

v = V/m = 0.15 / [(0.001 / v_f) + (0.149 / v_g)]

Quality x = m_g / m

x = (0.149 / v_g) / [(0.001 / v_f) + (0.149 / v_g)]

From steam properties, at P = 5 bar, we get T = 152 deg C, v_f = 0.00109 m^3/kg, u_f = 640 kJ/kg, h_f = 640 kJ/kg, v_g = 0.375 m^3/kg, u_g = 2560 J/kg, h_g = 2750 kJ/kg

a)

Therefore T = 152 deg C.

b)

From eqn1,

m_f = 0.001 / 0.00109 = 0.917 kg

From eqn2,

m_g = 0.149 / 0.375 = 0.397 kg

m = m_f + m_g = 0.917 + 0.397 = 1.314 kg

Quality x = m_g / m = 0.397 / 1.314 = 0.302

c)

Pressure and temperature will remain consant during phase change process.

Hence, P = 5 bar, T = 152 deg C

d)

Heat added Q = m*(h_g - h_f)

= 1.314*(2750 - 640)

= 2772.5 kJ

e)

p-v and T-v plots = line will be horizontal within the dome moving from left to right.

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