Recent content by DiogoSousa9630

Ah, I finally understand it now. All makes sense. Thank you so much for your guidance.

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 after making the substitutions and simplifying I got the following: $$m_2u_2-m_1u_1 = -m_{out} \frac{(u_Vv_L - u_Lv_V)}{v_V - v_L}$$ With: $$m_1u_1 = m_1(u_L + \frac{(u_V - u_L) (V/m_1 - v_L)}{v_V - v_L})$$ $$m_2u_2 = m_2(u_L + \frac{(u_V - u_L) (V/m_2 - v_L)}{v_V - v_L})$$

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...

Yes, thank you! So far so good!

**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.