Question

A rigid tank of volume 0.3 m³ contains water initially at its critical point. The water is now cooled to 150°C

a) What mass of water is in the tank?

b) Has the average specific volume of this water changed during the cooling process?

c) Find the mass of liquid water and the mass of water vapor in the tankat the end of the process.

d) Find the volume of liquid water and the volume of water vapor in the tank at the end of the process.

e)Find the quality of the final mixture. Is it meaningful to talk about the quality of the water at the initial state of this process? Why or why not?

f) Find the enthalpy change of the water in the tank.

Answer 1

At critical point the temperature of water is 373.9 ^oC. Density of water at critical point is 0.322 g/cc.

So the mass of this water is 322 x 0.3 = 96.6 Kg.

a) Since we are not adding any water and just coolong the mass of the water will remain same which is 96.6 Kg

b) When this is cooled to 150 ^oC then density increases so the average specific volume will change specifically it will decrease.

Density at 150 oC is 918 kg/m³

average specific volume will be 1/918 kg/m³ = 0.00108 m³/Kg

c) We can calculate the vapour pressure of water at 150 ^oC using the formula, this is used by engineers.

p = (t^oC/100)⁴ in bar

which states theat at 150 ^oC the vapour pressure of water is aroung 5. The exact value is 4.76.

The average specific volume, , is the ratio of the total volume to the total mass of the system

From property table we find that Vf at 150 C is 0.00109 m3/Kg and Vg is 0.3748 m³/Kg

V = m_fV_f + m_gV_g

m_f = 96.6-m_g

0.3=(96.6-m_g) x 0.00109 + m_g x 0.3748

0.3 = 0.1052 - 0.00109m_g + 0.3748m_g

0.3 - 0.105 = 0.3737 m_g

m_g = 0.195/0.3737

m_g = 0.5218 Kg

SO the mass of liquid water is 0.5218 Kg and the mass of water vapour is 96.6-0.5218 = 96.07 Kg

d) Volume of liquid water is 0.3748 x 0.5218 = 0.1955 m³

volume of water vapour = 96.07 x 0.00109 = 0.104 m³