A 50 cubic foot tank contains saturated 134a at 100F and a quality of 50%

[joejoe121]

Homework Statement

A tank of 50 cubic feet contains saturated 134a at 100 F and a quality of 50%. Vapor is withdrawn from the top of the tank until the temperature is 50 F. The process is adiabatic.

Relevant Equations

u = u_f + x(u_g-u_f)
v= = v_f +x(v_g-v_f)
x = (u - u_f)/(u_g-u_f)
h_avg = ((h_g)_100F _ (h_g)_50F)/2
mass_tank = volume/specific volume_final
mass_tank_final = 50 cubic feet / specific volume_final
mass_vapor = mass_tank - mass_final

I've attached all my work and data table I used to answer the questions but there isn't an answer key so I would like a second opinion.
a) The initial specific internal energy is.......Btu/lbm
b) The initial mass is....lbm
c)The average enthalpy of the withdrawn vapor is.....Btu/lbm
d)The final quality in the tank........%
e)The mass of 134 that is withdrawn is......lbm

 

[Chestermiller]

What are you asked to determine?

 

[joejoe121]

What are you asked to determine?

I'm so sorry i forgot to post the questions. I'll edit it.

 

[Chestermiller]

I don't follow what you did in parts (d) and (e). Can you please explain?

 

[joejoe121]

I don't follow what you did in parts (d) and (e). Can you please explain?

in D, i used the formula to find quality using internal energy values and in E I'm wanting to find mass, so I use the volume of the tank divided by the specific volume of the tank at 50 F which I used the same quality formula after I found the quality.

 

[Chestermiller]

in D, i used the formula to find quality using internal energy values and in E I'm wanting to find mass, so I use the volume of the tank divided by the specific volume of the tank at 50 F which I used the same quality formula after I found the quality.

I don't follow. My approach would be to let $x_f$ be the final quality in the tank and $m_f$ be the final mass in the tank. Then $$U_f-U_i=m_fu_f-m_iu_i=-(m_i-m_f)\bar{h}$$and$$v_{Lf}+(v_{Vf}-v_{Lf})x_f=\frac{V}{m_f}$$where $\bar{h}$ was the average specific enthalpy of the vapor exiting the tank.

 

[joejoe121]

I don't follow. My approach would be to let $x_f$ be the final quality in the tank and $m_f$ be the final mass in the tank. Then $$U_f-U_i=m_fu_f-m_iu_i=-(m_i-m_f)\bar{h}$$and$$v_{Lf}+(v_{Vf}-v_{Lf})x_f=\frac{V}{m_f}$$where $\bar{h}$ was the average specific enthalpy of the vapor exiting the tank.

I'm confused about the $$v_{Lf}$$ and $$v_{Vf}$$

 

[Chestermiller]

I'm confused about the $$v_{Lf}$$ and $$v_{Vf}$$

The final specific volumes of the saturated liquid and vapor.

 

[joejoe121]

The final specific volumes of the saturated liquid and vapor.

for your first formula you posted. What do I equate -(m_i-m_f)h_{avg} to?
And did i solve for my average enthalpy correctly?

 

[joejoe121]

I don't follow. My approach would be to let $x_f$ be the final quality in the tank and $m_f$ be the final mass in the tank. Then $$U_f-U_i=m_fu_f-m_iu_i=-(m_i-m_f)\bar{h}$$and$$v_{Lf}+(v_{Vf}-v_{Lf})x_f=\frac{V}{m_f}$$where $\bar{h}$ was the average specific enthalpy of the vapor exiting the tank.

Is it correct to say that the initial internal energy can be used to find the final quality?

 

[Chestermiller]

Is it correct to say that the initial internal energy can be used to find the final quality?

No, because mass has left the tank with enthalpy $\bar{h}$.

 

[joejoe121]

No, because mass has left the tank with enthalpy $\bar{h}$.

Is what I did in part C correct or should I say the average enthalpy of the withdrawn vapor just be the saturated vapor value at 100 F?
And I'm really confused on finding quality at this point.