Intermediate Thermodynamics Questions & Answers  
Question by Student 201327103
Professor, Question is in the jpg file
question.jpg    
03.12.14
No questions in attached files are allowed. Please edit your post and type your question directly.
Question by Student 201027111
professor, I'd like to ask about today(3.17) problem when solve B part( find heat needed to go from b2 to b3) prof said 1st law thermo can't use because unknown veriable is two with pdv and Q but in process V is constant so isn't that dv=0 ? set dv=0 than d(me)=Q than intergral and we can get (me)_b3 - (me)_b2 = Q and assume calorically perfect gas than e=CvT so Q=mCv(T(b3)-T(b2)) we can get solve this way is wrong?
03.17.14
This is a good question. But before I answer, please edit your post and write the mathematics properly (with correct subscripts and same notation as in class).
Question by Student 200927141
Professor. When we derive the h=CpT, we assume that P is constant. But, today when we solve the problem about 3) heat needed to obtain state B3 from B2, Why can we apply h=CpT despite of Pressure is changed from 2.49MPa to 3.29MPa?
Please use the right notation when asking a question. Cp is not the same as $C_P$, h=CpT is not the same as $h=C_P T$, etc. After you edit your question and change the notation, I'll answer it. Also, don't ask me two questions at the same time. Please wait before I answer the question, and then ask a new one. Otherwise I'll delete the second question.
Question by Student 201127101
I don't know proof is right, but I'll try it. $v$ is magnitude of any velocity. $\theta$ is an angle that is between the x-axis and the straight line of the plane xy. $\alpha$ is an angle that is between the straight line of v and plane xy. When considering only half of the x-axis, the total velocity is shown below. $$v_{total} = \int_{- \pi / 2}^{\pi / 2} \int_{- \pi / 2}^{\pi / 2} v cos \alpha cos \theta d \theta d \alpha $$ $$v_{total} = 4v $$ $v$ is magnitude of any velocity, so v can be seen as the x-axis velocity. $$v_x = \frac {1}{4} v_{total}$$
It's getting there, but not close enough yet. Think about it more ;)
Question by Student 200927141
Professor. When we derive the $$ h=C_PT$$ we assume that P is constant. But, today when we solve the problem about 3) heat needed to obtain state B3 from B2, Why can we apply $$ h=C_PT$$ despite of Pressure is changed from 2.49MPa to 3.29MPa?
No, we did not assume that $P$ is constant when deriving the calorically perfect gas relationship $h=C_P T$. Rather we assumed that $C_P$ is a constant. What may be confusing you here is that $C_P$ is the so-called specific heat at constant pressure, which corresponds to: $$ C_P =\left(\frac{\partial h}{\partial T}\right)_P $$ The latter is true in the general case when the enthalpy depends on both pressure and temperature. But $h$ doesn't depend on pressure for a thermally perfect gas. Thus, for a thermally perfect gas $C_P$ becomes: $$ C_P =\frac{d h}{d T} $$ Thus, assuming constant $C_P$, we can then use $h=C_P T$ and this does not entail that $P$ should be constant. I'll give you one point for this question. I would have given more if you would have asked it with no typesetting mistake the first time around.
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