Hm, I guess you are confused about the last term in your equation. Essentially, the mass conservation of water includes only water vapor and liquid water, not air. Thus, assuming there is no liquid water coming out because all the liquid evaporates, mass conservation of water entails that: $$\underbrace{\dot{m}_{\rm V2}}_{\textrm{vapor going out}}=\underbrace{\dot{m}_{\rm W}}_{\textrm{liquid going in}}+\underbrace{\dot{m}_{\rm V1}}_{\textrm{vapor going in}}$$ Then, it can be easily shown that: $$\dot{m}_{\rm V1}=\frac{\omega_1 \dot{m}_1}{1+\omega_1}$$ Thus: $$\underbrace{\dot{m}_{\rm V2}}_{\textrm{vapor going out}}=\underbrace{\dot{m}_{\rm W}}_{\textrm{liquid going in}}+\underbrace{\frac{\omega_1 \dot{m}_1}{1+\omega_1}}_{\textrm{vapor going in}}$$ Please fix your notation again: $w$ should read $\omega$. Then I'll give you 1.5 point bonus boost for this question.
 05.28.14
 Question by Student 200927141 Professor, This is problem about enthalpy. When I solve the problem4 on Assignment #8, I tried to use Table A-2. But i can't solve the problem. But When i use the Fig A-9, It was ok. so i just substitute Celsius temperature on formula instead of Kelvin. It's working. i thought enthalpy is function of Kelvin. But this one need to solve using Celsius... Previous question (1,2) usually need difference of temperature, so i didn't need to worry. Can i know the reason why it's working only for a celsius?
I don't understand... How exactly do you find the enthalpy from Figure A-9 (psychrometrics chart)? Please explain better your question, then I can answer it.
 Question by Student 201214353 Professor, I have a question about chemical reaction rate lecture. In lecture, you say $$\bar {q_N} \approx \sqrt{\frac{3k_B T_N}{m_N}} \\ \bar {q_N} = \sqrt{\frac{8k_B T_N}{\pi m_N}}$$ Thus : $$\sqrt{\frac{3k_B T_N}{m_N}} \approx \sqrt{\frac{8k_B T_N}{\pi m_{N}}}$$ But I don't understand how are these two equations the same. And what is the meaning of $$\bar {q_N} = \sqrt{\frac{8k_B T_N}{\pi m_N}}$$
 06.02.14
The equations should read $\overline{q}_{\rm N}=...$ not $\overline{q}_{\rm n}=...$, and the particule mass should read $m_{\rm N}$ not $m_{\rm N_2}$. Please make these changes in your question..
Yes in class, I mentioned that: $$\underbrace{\sqrt{\frac{3 k_{\rm B} T_{\rm N}}{m_{\rm N}}}}_\textrm{approximate} \approx \underbrace{\sqrt{\frac{8 k_{\rm B} T_{\rm N}}{\pi m_{\rm N}}}}_\textrm{exact}$$ The approximate solution is easy to find from the definition of the temperature as outlined in the tables. The exact solution is difficult to obtain and its derivation is beyond the scope of this course. Just remember that the approximate solution is a very good approximation to the exact solution (less than 15% error). In the exam, you can use either the approximate or the exact solution. I'll give you 1 point bonus boost for this question.
 Question by Student 201127116 Professor, I have question about Steady-state. I guess, All molecules doesn't move or there isn't any other change of form. Is it right? And why we could erase $$\frac{d}{dt}\int_V \rho dV$$ in Stdeay-state? You had taught about Steady-state. But i didn't make sense about Steady-state.
 06.03.14
At steady state, the thermodynamic properties such as $\rho$, $P$, $T$, or the gas/liquid macroscopic velocity vector $\vec{v}$ do not change in time. Therefore, all time derivatives involving the macroscopic properties and thermodynamic properties should be set to zero at steady-state. However, the microscopic properties are not necessarily constant in time. For instance, the speed of one molecule is never zero and varies in time whether the problem is steady-state or not.. I'll give you 1 point bonus boost for this question.
 Question by Student 200927141 Hi, professor. I'm trying to solve the assignment#9. In the problem2,3,5, we used the enthalpy formula including enthalpy of formation and additional enthalpy when it's not 298K. But in problem6, how do i use entropy form? Do i just use the entropy change in a perfect gas ??
When the problem involves chemical reactions, you have to calculate the entropy for each species similarly to how we calculate the enthalpy. That is: $$s=\frac{\overline{s}^0}{\cal M} + \Delta s$$ where $\overline{s}^0$ is obtained from Table A-24 and $\Delta s$ is the entropy addition calculated using perfect gas relationships if the gas is not at 298 K and 1 atm. That is: $$\Delta s= C_P \cdot \ln\left(\frac{T}{\rm 298~K}\right) - R \cdot \ln \left( \frac{P}{\rm 1~atm} \right)$$ I'll give you 1 point bonus boost for this question.
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 $\pi$