Thermodynamics Questions & Answers  
Question by Student 201214353
Professor
I have a question about Assignment 6 #5-(c).
To solve this question, i find $\bar{v} $ from generalized compressibility chart.
And then I used Vander waals Eq to find $T_2$

But suddenly i thought, using $v'_R$ and generalized compressibility factor in state 2 is might be possible. (Not using Vander waals Eq)
However difference between this two values($T_2$) is so big. Why is it so different? If I'm wrong, which one has error?
06.06.14
Your question is not so clear, you should give more details. But I think the problem in this case is that you are mixing two strategies together. To find $\overline{v}_2$ and $T_2$ try to use only one strategy: you can use either Van der waals or the generalized compressibility charts, but try to avoid using both on top of each other. The generalized compressibility charts is the strategy that is the most accurate and should be preferred in this case. I'll give you 1 point bonus for this question.
Question by Student 201127101
I have a question about assign 8-6. After obtaining $\dot{m}$ and $h$, professor used the energy conservation equation. $$\dot{m}_{v4} h_{v4} + \dot{m}_{A4} h_{A4} + \dot{m}_{L2} - \dot{m}_{v4} h_{v3} - \dot{m}_{A4} h_{A3} - \dot{m}_{L1} = 0$$ After calculation, equation is $$\dot{m}_{v4} h_{v4} + \dot{m}_{A4} h_{A4} = 6417 [kJ/s] \cdots Ⅰ$$ And $h_{A4}$ = ${(C_p)}_A$ $T_4$ = 1kJ/kg $T_4$, $\dot{m}_{v4}$ = 0.63kg/s, $\dot{m}_{A4}$ = 15.25kg/s. Finally, $$h_{v4} + 24.2 T_4 [kJ/kgK] = 10186 [kJ/kg]$$ To solve this equation, you used following equation $$h_v ({T_4}^n) + 24.2 {T_4}^{n+1} [kJ/kgK] = 10186 [kJ/kg]$$ So answer is $$ ∴ T_4 = 314K $$ But I try to solve this problem by $\tilde{h}$ = $\frac{\dot{m}_{mix} h_{mix}}{\dot{m}_{A}}$. So I divided by $\dot{m}_{A4}$ = 15.25kg/s in Eq Ⅰ, $$\frac{\dot{m}_{v4} h_{v4} + \dot{m}_{A4} h_{A4}}{\dot{m}_{A4}} = 421 [kJ/kg]$$ But $$\frac{\dot{m}_{v4} h_{v4} + \dot{m}_{A4} h_{A4}}{\dot{m}_{A4}} = \frac{\dot{m}_{mix4} h_{mix4}}{\dot{m}_{A4}} = \tilde{h}_{4}$$ Therefore, $$ ∴ \tilde{h}_{4} = 421 [kJ/kg dryair] $$ But I can't obtain $T_4$ because psychro chart is small(Maximum $\tilde{h}$ = 100 [kJ/kg dryair]at psychro chart). Is my way wrong? Or Is graph small only ? $$$$ Thank you, I understood this problem. But $\omega_4$ is 0.0413... I can't use psychro chart because maximum $\omega$ is 0.03 at psychro chart.
There is a mistake in your reasoning. You define correctly the mixture enthalpy as used in the psychrometric chart: $$ \tilde{h}_{4}=\frac{\dot{m}_{v4} h_{v4} + \dot{m}_{A4} h_{A4}}{\dot{m}_{A4}} = \frac{\dot{m}_{mix4} h_{mix4}}{\dot{m}_{A4}} $$ But you must express $h_{\rm A4}=C_P T_{\rm A4}$ with $T_{\rm A4}$ in Celcius. Also, when calculating the other air enthalpy $h_{\rm A3}$ you must use Celcius degrees, not Kelvin. Celcius must be used and not Kelvin because those who made the psychrometric chart defined their enthalpies this way. Try it again ;) I'll give you 1.5 point bonus boost for this question.

After trying again, you still can't find the solution because $\omega$ is off the chart. You are doing nothing wrong this time, you have just encountered a limitation of the psychro chart when determining the enthalpies. Here, there's no choice but to use table a2 and to iterate....
Question by Student 200746306
Hi, professor If I don't late, I ask some question. We solve the problem entropy and P-V-T by assuming ideal gas. but assignment is non-ideal gas so, I can't use the these equations $$ \Delta s= C_P \cdot \ln\left(\frac{T_2}{\rm T_1}\right) - R \cdot \ln \left( \frac{P_2}{\rm P_1} \right) $$ $$ \frac{P_2}{\rm P_1}=(\frac{ρ_2}{\rm ρ_1})^γ $$ But, I see that the lecture note these equations is assumed constant mass, calorically perfect gas and thermally perfect gas. So, may be I can use this equation to solve the assignment 7 #4,#5?
06.08.14
Please typeset your mathematics correctly. Use the preview command and make sure the math expressions appear as intended, that the same notation is used as in class, and that the text is easy to read for the others. After you do that, I'll answer your question.

You can only use the two expressions you are mentioning when the gas is calorically perfect and thermally perfect... Unfortunately, when using the Van der Waal equation of state, the gas is not thermally perfect and the latter can not be used.. You need to find another way to solve the problem. I'll give you 1 point bonus boost for this question.
The course is over. Keep your questions for next year ;)
06.10.14
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