1/2kg of a liquid-vapor water mixture in a rigid tank initially at 2 bar is heat

Question

1/2kg of a liquid-vapor water mixture in a rigid tank initially at 2 bar is heated by 120V heater for 5 minutes until it becomes a saturated vapor at 4 bar. changes in kinetic and potential energies can be ignored. Determine:
a) the volume of the tank.
b) the initial quality of the mixture.  
c) the change in internal energy.
d) the change in internal energy.
e) the current of the heater assuming all the power consumed by the heater is transferred as heat to the mixture.              
              

Explanation / Answer

FINAL

P2 = 4 bar

saturated vapor properties at P2 are

v2 = vg = 0.4625 m^2/kg

u2 = ug = 2554 kJ/kg

Volume of the tank, V = m*v2 = 0.5*0.4625 = 0.232 m^3

INITIAL

P1 = 2 bar

Saturation properties at P1 are

vf = 0.001061 and vg = 0.8857 m^3/kg

uf = 504.5 and ug = 2529 kJ/kg

Also for rigid tank

v1 = v2 = 0.4625 m^3/kg

v1 = 0.4625 = (1-x)*vf + x*vg

x = 0.5216

u1 = (1-x)*uf + x*ug

u1 = 1561 kJ/kg

c)

dU = m*(u2-u1) = 0.5*(2554-1561)

dU = 496.5 kJ

d)

dU = V*I*t

496.5*10^3 = 120*I*5*60

I = 13.8 A

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.