Steam at 75 kPa and 8 percent quality is contained in a spri

Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston-cylinder device with an initial volume of 2 m3 Steam is now heated until the pressure and volume rise to 225 kPa and 5 m3 During this
process, the pressure increases linearly with change in the volume.
(5a). Calculate the mass of steam contained in the cylinder.
(5b). Determine the internal energy at the initial and final states.
(5c). Calculate the work done during the process, in kJ.
(5d). Determine the heat transfer during the process, in kJ.

Solution

Solution:-

Initial State,    P₁ = 75 kPa,   Volume = 2 m³,     Quality = 8% or X = 0.08,

Final State, P₁ = 225 kPa, Volume = 2 m³

                                                                                                                                                 Data From Steam Table,  

At, P₁ = 75 kPa, V_f =0.001037 m³/kg, V_g =2.217 m³/kg, h_f = 384.5 kJ/kg, h_g = 2663.0 kJ/kg

Saturation Temperature = 92 deg.C

At, P₁ = 225 kPa, V_f =0.0010645 m³/kg , V_g =0.790 m³/kg , h_f = 520.0 kJ/kg, h_g = 2711.6 kJ/kg

Saturation Temperature = 124 deg.C

(Where V_f and V_g are the specific volumes in fluid and gas phase respectively and h_f and h_g are the specific enthalpy in fluid and gas phase respectively)

(a) Average Specific Volume = V

V = V_f + X ( V_g - V_f ) = 0.001037 + 0.08 (2.217 - 0.001037) = 0.17831404 m³/kg

Density will be 1/V = 1/0.17831404 = 5.608083357 kg/m³

So, Mass of Steam = Volume x Density = 2 x 5.608083357 = 11.216 kg

(b) Since, Mass of steam will be same = 11.216 kg  

     So, specific volume = 5/11.216 = 0.44579 m³/kg

Now, V = V_f + X ( V_g - V_f )    

0.44579 = 0.0010645 + X(0.790 - 0.0010645)       or    X = 0.5637

Enthapy of Steam at Initial state = h_f + X(h_g - h_f)

h₁ = 384.5 + 0.08 (2663.0 - 384.5) = 566.78 kJ/kg

So, Enthapy of Steam at final state = h_f + X(h_g - h_f)

h₂ = 520.0 + 0.5637 (2711.6 - 520.0) = 1755.405 kJ/kg

So, Net Change in Enthalpy ( This will be same as net change in internal energy)

               = m ( h₂ - h₁ ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

(c) Work Done (Area Under PV curve) = 1/2 x (P₁ + P₂) x ( V₁ - V₂)

                                                       = 1/2 x ( 75 + 225) x (5 - 2)

                                                   W = 450 KJ

(d) From First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

                            = 13331.618 + 450

                         Q = 13781.618 kJ/kg