Question

A piston-cylinder device contains 0.56 kg of steam at 300 degrees C and 1.4 MPa. Steam is cooled at constant pressure until one-half of the mass condenses.

(a) Find the final temperature. T₁= _______ degrees celcius

(b) Determine the volume change in m³

Answer 1

Given that :

masss of the steam, m = 0.56 kg

initial temperature, T₀ = 300 ⁰C

initial Pressure, p₁ = 1.4 MPa

At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure,

(a) the final temperature will be given as :

According to saturated steam table by pressure, we get

T₂ = T_{[email protected] MPa} = 195.04

(b) Steam is cooled at constant pressure until one-half of the mass condenses. it means that, we have

x₂ = 0.5

At the final stage, p₂ = 1.4 MPa , T₂ = T_{[email protected] MPa} = 195.04

On saturated steam table -

Specific Volume of Saturated Steam, v_g = 0.140768 m³/kg

Specific Volume of Saturated Water, v_f = 0.00114892 m³/kg

The specific volume at the final stage which is given as :

v₂ = v_f + x₂ (v_g - v_f)                                                                        { eq.1 }

inserting the values in eq.1,

v₂ = (0.00114892 m³/kg) + (0.5) [(0.140768 m³/kg) - (0.00114892 m³/kg)]

v₂ = (0.00114892 m³/kg) + (0.5) (0.13961908 m³/kg)

v₂ = (0.00114892 m³/kg) + (0.06980954 m³/kg)

v₂ = 0.0709 m³/kg

On saturated steam table -

Specific Volume of Superheated Steam, v₁ = 0.1823 m³/kg

Now, the volume change in m³ will be given as ::

v = m (v₂ - v₁)                                                                   { eq.2 }

inserting the values in eq.2,

v = (0.56 kg) [(0.0709 m³/kg) - (0.1823 m³/kg)]

v = (0.56 kg) (-0.1114 m³/kg)

v = - 0.062 m³