Answer:
given,
liquid water in the rigid tank = 1.4 kg
The saturated liquid properties of water at 200◦C are
[tex]v_f = 0.001157 m^3/s\ and \ u_f = 850.46 kJ/kg[/tex]
a) tank initially contains saturated liquid and water so, volume occupied by water
[tex]V_1 = m v_1 [/tex]
= 1.4 × 0.001157
= 0.001619 m³
total volume =
= [tex]\dfrac{1}{0.25}\times 0.001619[/tex]
= [tex]0.006476 m^3[/tex]
b)[tex] v_2 = \dfrac{V}{m} =\dfrac{0.006476}{1.4}[/tex]
=0.004626 kg/m³
for v₂ = 0.004626 kg/m³
T₂ = 371.3 ⁰C P₂ = 21,367 kPa u₂ = 2201.5 kJ/kg
c) total internal energy
[tex]\Delta U = m(u_2-u_1)[/tex]
= 1.4 (2201.5 - 850.46) kJ/kg
= 1892 kJ
