I just realized that I forgot to include the negative sign before the parentheses - my mistake! Now that I've corrected it, I have the same expression as yours.
I suppose that ##\Delta h_{vap}\frac{v_V}{(v_V-v_L)}## will be equivalent to ## \Delta h##?
Thank you! After doing that I got:
$$
Q = \frac{m_\text{out} \left[u_L v_V + u_V v_V - 2 u_V v_L + P v_V^2 - P v_L v_V\right]}{v_V - v_L}
$$
I'm not sure about this one. I think I might have missed something.
Okay so the general formula for the u will be given by:
$$u = u_L + \frac{(u_V - u_L) (V/m - v_L)}{v_V - v_L}$$
What exactly is the next approach? Should I consider the given formula for entalphy and get something similar to this?
$$h = u_L + P v_L + \frac{v - v_L}{v_V - v_L} (u_V - u_L) +...
Okay. So this is what I got:
$$x_{1} = \frac{m_{v1}}{m_{v1}+m_{l1}} = \frac{m_{v1}}{m_{1}} $$
$$x_{2} = \frac{m_{v2}}{m_{v2}+m_{l2}} = \frac{m_{v2}}{m_{2}} $$
$$u_{1} = u_{l}+\frac{(u_{v}-u_{l})*m_{v1}}{m_{1}}$$
$$u_{2} = u_{l}+\frac{(u_{v}-u_{l})*m_{v2}}{m_{2}}$$
So now I guess I should...
**Question:**
Why do these two different approaches lead to the same result in this case? Is one of them more appropriate or valid than the other? I am looking for an explanation and clarification on this matter.Thank you in advance for your insights.