Question by Student 201327103 Professor, Question is in the jpg file

 03.11.14
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. $$v_x = \frac {1}{4} v_{total}$$
 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.
 $\pi$