Answer:
the work done by the steam during this process = 119.575 kJ
Explanation:
From the information given:
At state 1
Let obtain the specific volume of the saturated vapor from the Saturated water- Pressure table A-5 at the pressure of 100 kPa
[tex]v_1 = v_{g \ 100 \ kPa)[/tex]
[tex]v_1 = 1.6941 \ m^3 /kg[/tex]
where [tex]v_1[/tex] is the specific volume of the saturated vapor at state 1.
At state 2:
From the tables A-6 of Superheated water at the pressure of 100 kPa or 0.1 MPa and at the temperature of 200°C, the specific volume [tex]v_2 = 2.1724 \ m^3 /kg[/tex]
where [tex]v_2[/tex] is the specific volume of the superheated water at state 2.
The workdone by the steam during the process can be expressed by the formula:
[tex]W = P(V_2 -V_1)[/tex]
[tex]W = mP(v_2-v_1)[/tex]
where;
m = mass of the saturated water vapor
P = pressure of the saturated water vapor
[tex]V_2 =[/tex] volume of the superheated water at state 2
[tex]V_1 =[/tex] volume of the saturated water at state 1
Replacing our values ;
W = 2.5 (100) ( 2.1724 -1.6941)
W = 250(0.4783 )
W = 119.575 kPa.m³ [tex]\times \dfrac{1 \ kJ}{1 \ kPa.m^3}[/tex]
W = 119.575 kJ
∴
the work done by the steam during this process = 119.575 kJ
