Respuesta :
Answer:
Qin = 448.23 kJ
Explanation:
given data
mass m1 = 0.6 kg
volume v1 = 0.1 m³
pressure P1 = P2 = 800 kPa
pressure Pi = 5 MPa
temperature Ti = 500°C
temperature T2 = 250°C
volume V2 = 2V1 = 2 (0.1) = 0.2 m³
solution
we get here some value from steam table
as we get first for initial state at pressure 800 kPA
v1 = [tex]\frac{V1}{m1}[/tex] ...........1
v1 = [tex]\frac{0.1}{0.6}[/tex]
v1 = 0.1667 m³/kg
and u1 = 2004.4 kJ/kg
and
for final state at pressure 800 kPa and temperature 250
u2 = 2715.9 kJ/kg
and
v2 = 0.29321 m³/kg
now we get for stream in supply line that is
hi = 3434.7 kJ/kg
for the preesure 5 MPa and temperature 500
and
now we get mass m2 = [tex]\frac{V2}{v2}[/tex] ............2
m2 = [tex]\frac{0.2}{0.29321}[/tex]
m2 = 0.6821 kg
so
now we consider here tank as control system
so that steam cross control surface is
as mass balance
mi = m2-m1
mi = 0.6821 - 0.6
mi = 0.0821 kg
and
now we neglect microscopic for energies
so energy balance is
Qin - W (b) + mihi = m2u2 - m1u1 ............3
Qin - P(V2-V1) + mihi = m2u2 - m1u1
Qin - 800 (0.2-0.1) + 0.082 (3434.7) = 0.6821 × 2715.9 - 0.6 × 2004.4
Qin = 448.23 kJ
