<u>Answer :</u>
(a) The Initial temperature of the steam is 247.557°C
(b) The enthalpy change per unit mass of the steam is -1771 kJ/kg
(c) The final pressure is 1555 kPa, and the liquid vapor mixture contains small mass of vapor.
<u>Explanation :</u>
From the given statements, we know
Initial steam pressure () = 3.5 MPa
Degree of Super heat = 5°C
Saturation Temperature () = 242.557°C [From Steam Table]
Now, the Initial temperature () of steam is given by-
=
+ Degree of Superheat
= 242.557 + 5 = 247.557°C
(b)To find: The enthalpy change per unit mass of the steam is
We know enthalpy (h) change is given by
Δh = –
From steam Table, properties of the gas at state 1 and state 2
= 2821.1 kJ/kg
= 1049.7 kJ/kg
On substituting the values, we get
or, Δh = 1049.7 - 2821.1 = -1771 kJ/kg
(c) To Find: The final pressure of the liquid vapor mixture contains small mass of vapor.
We find that the gas at the final position, i.e., at position 3
Final Temperature () = 200°C [Given] ……………………………..(1)
Specific Volume () = Specific Volume (
) = 0.00123
/kg ….....(2)
Using steam table for corresponding values of (1) and (2), we get
Final Pressure = 1555 kPa
Dryness fraction () = 0.0006.