changes.mady.by.user Arthur Aslanyan (Nafta College)
Saved on Feb 07, 2019
...
\frac{d Q^{\downarrow}_{AQ}}{dt} = q^{\downarrow}_{AQ}(t)
q^{\downarrow}_{AQ}(t)= C_a \cdot \frac{\partial p_a(t,r)}{\partial r} \bigg|_{r=r_e}
p_a(t, r)= p(0) + \int_0^t p_1 \left(\frac{r_e^2}{\chi}(t-\tau), \frac{r/}{r_e} \right) \dot p(\tau) d\tau
\frac{\partial p_1}{\partial t_D} = \frac{\partial^2 p_1}{\partial r_D^2} + \frac{1}{r_D}\cdot \frac{\partial p_1}{\partial r_D}
p_1(t = 0, r_D)= 0
p_1(t, r_D=1) = 1
\frac{\partial p_1}{\partial r_D}(t, \bigg_ {r_D=r_a/r_e)} = 0
\frac{\partial p_a}{\partial t} = \chi \cdot \left[ \frac{\partial^2 p_a}{\partial r^2} + \frac{1}{r}\cdot \frac{\partial p_a}{\partial r} \right]
