Intermediate Thermodynamics Questions & Answers  
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 ? $$$$ Thank you, I understood this problem. But $\omega_4$ is 0.0413... I can't use psychro chart because maximum $\omega$ is 0.03 at psychro chart.
06.06.14
There is a mistake in your reasoning. You define correctly the mixture enthalpy as used in the psychrometric chart: $$ \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 $$ But, I see that the lecture note these equations is assumed constant mass, calorically perfect gas and thermally perfect gas. So, may be I can use this equation to solve the assignment 7 #4,#5?
06.07.14
You can only use the two expressions you are mentioning when the gas is calorically perfect and thermally perfect... Unfortunately, when using the Van der Waal equation of state, the gas is not thermally perfect and the latter can not be used.. You need to find another way to solve the problem. I'll give you 1 point bonus boost for this question.
Question by Student 201427145
In the example problem about Ideal gas law[Micro size, focusing on 1 particule, R=$\frac{{J}}{K*kg}$], we have used $$ P=\rho RT, $$ to calculate $\xi$=$\frac{{Nq}}{4}$ [N=number density, q=speed] but there are another way using other form of Ideal gas law $$ \xi=N\int_{x=0}^{infinite} \left({v_x}f(v_x) \right) dv_x=N{(\frac{{m}}{2\pi k_BT}})^{0.5}\int_{x=0}^{infinite} \left({v_x}e^\frac{{-mv^2_x}}{2k_BT} \right) dv_x=N(\frac{{k_bT}}{2\pi m})^{0.5} $$ Considering N form of Ideal gas law, $$ P=Nk_bT, N=\frac{{P}}{k_BT} $$ $$ \xi=\frac{{P}}{{{(2\pi mk_BT)}}^{0.5}}=\frac{{101300(\frac{{N}}{m^2})}}{{{(2\pi 0.028(\frac{{kg}}{mole})1.3806*10^{-23}(\frac{{J}}{K})*300(K))}}^{0.5}}=2.91*10^{27}(\frac{{particules}}{m^2s}) $$ and there is about 9% error with Answer in class[It was 3.17*$10^{27}$] So I am wondering why there are error between the answer using another form of Ideal gas law. we regarded square root of q^2 as same for q ignoring 15%error in the class, Is that the cause? Can I seek the answer using the N form of ideal gas law ignoring some error?
03.15.19
Yes you are correct. The 9% error comes from the fact that we assumed $\bar{q}\approx\sqrt{\overline{q^2}}$. Very good observation. But for completeness you should explain what the definition of $f(v_x)$ is.
Question by Student 201427145
I'm sorry for omiting a explain about $f(v_x)$. It is a Maxwell velocity distribution function. Generally total number of molecules is so large and the velocity has a continuous distribution value. We can determine $\Delta$v very smaller(extremly to dv) to integrate the number of molecules between v and v + dv. Here, the ratio of molecules moving at the velocity between $v_x$ and $v_x+ dv_x$ along the x axis when the system consisting of N or more ideal gas molecules is in equilibrium thermally with the environment is as follows: $$ \frac{{dN}}{N}=(\frac{{m}}{2\pi k_BT})^{0.5}e^{\frac{{-mc^{2}}}{2k_BT}}dv_x=f(v_x)dv_x $$ and we can define $f(v_x)=(\frac{{m}}{2\pi k_BT})^{0.5}$, Maxwell velocity distrubution function along x axis.
Great, this makes your previous post clearer!
Question by Student 201327136
Professor, i have a question about Q5-a of Assign #1. There is a ppm in the script, but i can not definite that ppm is mole per mole or mass per mass or otherelse. Can you definite it?
03.18.19
Parts per million of species X means number of molecules/atoms of species X per millions of molecules/atoms in the mixture.
Question by Student 201427145
I would like to ask about applying polytropic process when I get the heat transferred to air in Assignment Q#6-(b). $\delta$Q, dV, dT, dP is not Zero. So I assumed it's a Polytropic Process that various processes were mixed. Given that $T_1$=294K, $V_1=0.0283m^2$, $V_2=0.038m^3$(It coule be induced through 6-(a)), $P_1=103kN/m^2, P_2=206kN/m^2, n_{air}$=1.2mole(From Ideal gas law), $C_{v.air}$=717J/kgK
Definition of Polytropic Process:
$P_1V_1^n=P_2V_2^n$---n=-2.352(It's negative)
Ideal Gas Law-$PV=n\bar{R}T$:
$T_2=\frac{P_2V_2}{n\bar{R}}$---$T_2=790K$
1st Law of Thermodynamics:
d(me)+PdV=$\delta$Q ($\delta$W=0 because it's frictionless)
Recall, $PV^n=C(Polytropic), u=c_vT$(Calorically Perfect gas) and Integrate on both side:
Q=$\int_{T_1}^{T_2}{mC_vdT}$+$\int_{V_1}^{V_2}{PdV}=mC_v(T_2-T_1)$+$\int_{V_1}^{V_2}{CV^{-n}}$
$\int_{V_1}^{V_2}{CV^{-n}}$=$\frac{P_2V_2^nV_2^{1-n}-P_1V_1^nV_1^{1-n}}{1-n}$=$\frac{P_2V_2-P_1V_1}{1-n}$
Q=$mC_v(T_2-T_1)$+$\frac{P_2V_2-P_1V_1}{1-n}$=1.2mole*0.029kg/mole*717J/kgK*(790K-294K)+$\frac{206kN/m^2*0.038m^2-103kN/m^2*0.0283m^2}{1+2.352}$=13874.7J
It's 1.5% error with answer even though Polytropic exponent is negative. In the wikipedia, negative exponents reflect a process where work and heat flow simultaneously in or out of the system. For this reason, Can not I apply the polytropic process?
03.25.19
It's a good effort but your solution and the answer are off. You can not assume $n$ is constant in this process for the air. As the heat is added, $n$ will vary. You need to rather set the pressure of the argon equal to the one of the air. If you assume a polytropic process for the air, I'll give you A0 at the most. Also, please use \$\$ instead of \$ for long mathematical expressions — this will make your post easier to read.
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