Question

Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. What is the fraction of the steam in its initial state?

Answer 1

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

Answer 2

Answer:

93%

Explanation:

Hello,

In this case, based on the second state, at 101.33 kPa and 105 °C a saturated vapor is formed, which has an enthalpy of:

$hg_{101.325kPa,105^0C}=2683.4kJ/kg$

Now, since such condition was attained at constant enthalpy, the previous enthalpy equals the mixture's enthalpy at 1100 kPa, thus, for such condition:

$hmix_{1100kPa}=2638.4kJ/kg$

In such a way, the quality $x$, or fraction of the steam at the initial condition turns out:

$x=\frac{hmix-hf}{hfg}=\frac{2638.4kJ/kg-781.03kJ/kg}{1999.6kJ/kg} =0.93=93\%$

Best regards.