Question by Student 200746306 Hello professor, I have a question about assignment 9 question #3. Products on a dry air have 86.85% nitrogen particulars. So, I think chemical formula that a CH4 + b O2 + c N2 ----> 9.7 CO2 + 0.5 CO + 2.95 O2 + d H2O + 86.85 N2 And I found that c is 92.4. But nitrogen doesn't reaction other particles So, I think that c is same value 86.85. I was confused to this thing. Isn't my calculate wrong?
 06.04.14
Yes, your calculation is wrong.. Since $\rm N_2$ doesn't react with other species, it should be conserved through the reaction. I'll give you 0.5 point bonus for this question. I would have given more if you would have typeset the chemical reaction better with the correct subscripts..
 Question by Student 200746306 Professor, I'm sorry to bother you by a question I have some question about assignment 9 question number 6. When water enter the chamber, Water, butane and air are all mixture same place?
 06.05.14
I guess what you mean is whether the water mixes with the butane. The answer is no, the water remains in the cooling jacket and does not mix with the fuel or the combustion products (the combustion takes place in one duct, while the water flows in another duct surrounding the first one). I already explained this in class... I'll give you 0.5 point bonus boost for this question.
Question by Student 201127144
Professor I have some question about Our latest assingment #8, #9 There is many 'ejected rate(m dot)' is exist
 Focus on air, I think
 "Total ' air's ' volume" = " ' vapor's ' volume in air " + " ' dry air's ' volume"
 But you say
 "Total air's volume" = " vapor's volume in air " = " dry air's volume"
'Vapor' and 'dry air' are exist in total air, however why they have same volume?
Hm, the question doesn't make sense. Explain better what is the problem — I didn't mention about the volume but the mass flow rates. Also, typeset your mathematics correctly.
 Question by Student 201127134 Solving assignment 8, no.5, i have to figure out mass flow rate of each duct, to obtain this, i have to divide volume flow rate[cubic meter per second] by specific volume[cubic meter per kg]. Therefore, i could find specific volume in the table, Figure A9, i could find specific volume. What i am confused is that, in the Fig A9, whether specific volume is (air+vapor)'s volume per mass of dry air or consists of only dry air's volume per mass of dry air. I think you have not mentioned about this in the class. Plus, expression of this in the Figure A9 is quite ambiguous for me.
 06.06.14
I strongly recommend not to use theories that I have not covered in class... All problems in the assignments and exams can be solved using the theory shown in class — do that and you'll score very well on the test! I will not give you a bonus boost for this question because it is out of the scope of this course.
 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?
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....
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