Respuesta :
Solution:
Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg
Temperature = 200 °C
And 25% of the volume by liquid water is steam.
State 1
[tex]$m=\frac{V}{v}$[/tex]
[tex]$m=m_f+m_g$[/tex]
[tex]$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$[/tex]
[tex]$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$[/tex] (taking the value of [tex]$v_g$[/tex] and [tex]$v_g$[/tex] at 200°C )
[tex]$V=6.304 \times 10^{-3}$[/tex]
Now quality of vapor
[tex]$x=\frac{m_g}{m}$[/tex]
[tex]$=3.377 \times 10^{-3}$[/tex]
Internal energy at state 1 can be found out by
[tex]$u_1=u_f+xu_{fg}$[/tex]
[tex]$=850.65+3.377\times10^{-3}\times 1744.65$[/tex]
= 856.54 kJ/kg
After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor [tex]$v_2=v_g \text { and }\ x=1$[/tex]
Tank is rigid, so volume of tank is constant.
[tex]$v_g=v_2=\frac{V}{m}$[/tex]
[tex]$v_g=\frac{6.304\times 10^{-3}}{1.4}$[/tex]
[tex]$v_g=4.502 \times 10^{-3} \ m^3 /kg$[/tex]
Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature
[tex]$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$[/tex]
= 369.11° C
Internal energy at state 2
[tex]$u_2=2154.9 \ kJ/kg$[/tex]
Now power rating of the resistor
[tex]$P=\frac{m(u_2-u_1)}{t}$[/tex]
[tex]$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$[/tex]
= 1.51 kW
