An insulated pistoncylinder initially contains 8 kg of water

An insulated piston-cylinder initially contains 8 kg of water, of which 3 kg is in the vapor phase. The pressure inside is such that 180 kPa in maintained throughout process. Steam at 0.5 MPa and 300 C enters the cylinder from a supply line until all liquid in cylinder is vaporized. What is the final temperature, and how much mass from the steam has entered?

Solution

Answer:

Initial condition:

P1=180 kPa=1.8 bar (assumed absolute)

m1= 8 kg

m_v1 = 3 kg (vapour)

m_l1 = 5 kg(liquid)

We use ASME steam tables for calculating steam properties

h_{l1 = 490.67 kJ/kg}

h_v1 = 2701.4 kJ/kg

Therefore enthalpy at initial condition H₁= m_l1 h_l1 +m_v1 h_v1=490.67x5=2701.4x3 = 10557.55 kJ

Let amount of steam admitted at 0.5 MPa, 300 deg C is y

Sp. enthalpy at 0.5 MPa, 300 deg C h₂= 3064.6 kJ/kg

Therefore total enthalpy at final condition H= 10557.5 +3064.6 y

Total mass at final condition = 8+y

Now specific saturated vapour enthalpy at 180 kPa= 2701.4 kJ/kg

Therefore from conservation of energy final enthalpy = intial enthalpy+ external enthalpy flow

Final sp enthalpy = final enthalpy/final mass

(10557.5+3064.6 y)/(8+y)=2701.4

therefore y= (8x2701.4-10557.5)/(3064.6-2701.4)

= 30.43 kg